Chem Notebook COmplete

enmanuelg.

By: Enmanuel Garrido

Enmanuel’s Chemistry Notebook


Honors Chemistry

Class Policies and Grading

The students will receive a Unit Outline at the beginning of each Unit. It will

have information about the assignments that they will do, what it’s grade

classification will be, what action they will need to do to complete the

assignment and when it is due.

The students will receive a Weekly Memo of the activities they will be

responsible for that week. It will serve to inform the students of the learning

goal for the week. It will also give the students any special information

about that week.

The students will also receive daily lectures and assignments that are

designed to teach and re-enforce information related to the learning goal.

This will be time in which new material will be taught and reviewed and will

give the students the opportunity to ask questions regarding the concepts

being taught.

The students will work with a Lab partner and also be in a Lab group, but it

will be up to the individual student to do his or her part of all assignments

and the individual student will ultimately be responsible for all information

presented in the class.

The students will be required to follow all District and School Policies and to

follow all Lab Safety Procedures, which they will be given and will sign,

while performing labs. Students should come to class on time and with the

supplies needed for that class.

The following grading policy will be used.

Percent of Final Grade

Notebook 40%

Test/Projects 30%

Labs/Quizzes 20%

Work 10%

The students will be given a teacher generated Mid Term and a District

Final.


Unit 1

Measurement Lab

Separation of Mixtures Lab with Lab Write Up

Unit 2

Flame Test Lab

Nuclear Decay Lab

Element Marketing Project

Unit 3

Golden Penny Lab with Lab Write Up

Molecular Geometry

Research Presentation on a Chemical

Mid Term

Unit 4

Double Displacement Lab

Stoichiometry Lab with Lab Write Up

Mole Educational Demonstration Project

Unit 5

Gas Laws Lab with Lab Write Up

States of Matter Lab

Teach a Gas Law Project

Unit 6

Dilutions Lab

Titration Lab

District Final


Unit 1 (22 days)

Chapter 1 Introduction to Chemistry

Honors Chemistry

2016/2017 Syllabus

3 days

1.1 The Scope of Chemistry 1.3 Thinking Like a Scientist

1.2 Chemistry and You 1.4 Problem Solving in Chemistry

Chapter 2 Matter and Change

2.1 Properties of Matter 2.3 Elements and Compounds

2.2 Mixtures 2.4 Chemical Reactions

Chapter 3 Scientific Measurement

9 days

10 days

3.1 Using and Expressing Measurements 3.3 Solving Conversion Problems

3.2 Units of Measurement

Unit 2 (15 days)

Chapter 4 Atomic Structure

5 days

4.1 Defining the Atom 4.3 Distinguishing Among Atoms

4.2 Structure of the Nuclear Atom

Chapter 5 Electrons in Atoms

5 days

5.1 Revising the Atomic Model 5.2 Electron Arrangement in Atoms

5.3 Atomic Emission Spectrum and the Quantum Mechanical Model

Chapter 6 The Periodic Table

6.1 Organizing the Elements 6.3 Periodic Trends

6.2 Classifying Elements

Unit 3 (22 days)

Chapter 25 Nuclear Chemistry

25.1 Nuclear Radiation 25.3 Fission and Fusion

25.2 Nuclear Transformations 25.4 Radiation in Your Life

Chapter 7 Ionic and Metallic Bonding

7.1 Ions 7.3 Bonding in Metals

7.2 Ionic Bonds and Ionic Compounds

Chapter 8 Covalent Bonding

5 days

6 days

8 days

8 days

8.1 Molecular Compounds 8.3 Bonding Theories

8.2 The Nature of Covalent Bonding 8.4 Polar Bonds and Molecules

Unit 4 (14 days)

Chapter 9 Chemical Names and Formulas

6 days

9.1 Naming Ions 9.3 Naming & Writing Formulas Molecular Compounds

9.2 Naming and Writing Formulas for Ionic Compounds 9.4 Names for Acids and Bases

Chapter 22 Hydrocarbons Compounds

22.1 Hydrocarbons 22.4 Hydrocarbon Rings

Chapter 23 Functional Groups

4 days

4 days

23.1 Introduction to Functional Groups 23.4 Alcohols, Ethers, and Amines


Unit 5 (28 days)

Chapter 10 Chemical Quantities 8 days

10.1 The Mole: A Measurement of Matter 10.3 % Composition & Chem. Formulas

10.2 Mole-Mass and Mole-Volume Relationships

Chapter 11 Chemical Reactions 8 days

11.1 Describing Chemical Reactions 11.3 Reactions in Aqueous Solutions

11.2 Types of Chemical Reactions

Chapter 12 Stoichiometry 12 days

12.1 The Arithmetic of Equations 12.3 Limiting Reagent and % Yield

12.2 Chemical Calculations

Unit 6 (22 days)

Chapter 13 States of Matter 6 days

13.1 The Nature of Gases 13.3 The Nature of Solids

13.2 The Nature of Liquids 13.4 Changes in State

Chapter 14 The Behavior of Gases 10 days

14.1 Properties of Gases 14.3 Ideal Gases

14.2 The Gas Laws 14.4 Gases: Mixtures and Movement

Chapter 15 Water and Aqueous Systems 6 days

15.1 Water and its Properties 15.3 Heterogeneous Aqueous Systems

15.2 Homogeneous Aqueous Systems

Unit 7 (18 days)

Chapter 16 Solutions 8 days

16.1 Properties of Solutions 16.3 Colligative Properties of Solutions

16.2 Concentrations of Solutions 16.4 Calc. Involving Colligative Property

Chapter 17 Thermochemistry 5 days

17.1 The Flow of Energy 17.3 Heat in Changes of State

17.2 Measuring and Expressing Enthalpy Change 17.4 Calculating Heats in Reactions

Chapter 18 Reaction Rates and Equilibrium 5 days

18.1 Rates of Reactions 18.3 Reversible Reaction & Equilibrium

18.2 The Progress of Chemical Reactions 18.5 Free Energy and Entropy

Unit 8 (14 days)

Chapter 19 Acid and Bases 10 days

19.1 Acid-Base Theories 19.4 Neutralization Reactions

19.2 Hydrogen Ions and Acidity 19.5 Salts in Solutions

19.3 Strengths of Acids and Bases

Chapter 20 Oxidation-Reduction Reactions 4 days

20.1 The Meaning of Oxidation and Reduction 20.3 Describing Redox Equations

20.2 Oxidation Numbers


Lorenzo Walker Technical High School

MUSTANG LABORATORIES

Chemistry Safety

Safety in the MUSTANG LABORATORIES - Chemistry Laboratory

Working in the chemistry laboratory is an interesting and rewarding experience. During your labs, you will be actively

involved from beginning to end—from setting some change in motion to drawing some conclusion. In the laboratory, you

will be working with equipment and materials that can cause injury if they are not handled properly.

However, the laboratory is a safe place to work if you are careful. Accidents do not just happen; they are caused—by

carelessness, haste, and disregard of safety rules and practices. Safety rules to be followed in the laboratory are listed

below. Before beginning any lab work, read these rules, learn them, and follow them carefully.

General

1. Be prepared to work when you arrive at the lab. Familiarize yourself with the lab procedures before beginning the lab.

2. Perform only those lab activities assigned by your teacher. Never do anything in the laboratory that is not called for in

the laboratory procedure or by your teacher. Never work alone in the lab. Do not engage in any horseplay.

3. Work areas should be kept clean and tidy at all times. Only lab manuals and notebooks should be brought to the work

area. Other books, purses, brief cases, etc. should be left at your desk or placed in a designated storage area.

4. Clothing should be appropriate for working in the lab. Jackets, ties, and other loose garments should be removed. Open

shoes should not be worn.

5. Long hair should be tied back or covered, especially in the vicinity of open flame.

6. Jewelry that might present a safety hazard, such as dangling necklaces, chains, medallions, or bracelets should not be

worn in the lab.

7. Follow all instructions, both written and oral, carefully.

8. Safety goggles and lab aprons should be worn at all times.

9. Set up apparatus as described in the lab manual or by your teacher. Never use makeshift arrangements.

10. Always use the prescribed instrument (tongs, test tube holder, forceps, etc.) for handling apparatus or equipment.

11. Keep all combustible materials away from open flames.

12. Never touch any substance in the lab unless specifically instructed to do so by your teacher.

13. Never put your face near the mouth of a container that is holding chemicals.

14. Never smell any chemicals unless instructed to do so by your teacher. When testing for odors, use a wafting motion to

direct the odors to your nose.

15. Any activity involving poisonous vapors should be conducted in the fume hood.

16. Dispose of waste materials as instructed by your teacher.

17. Clean up all spills immediately.

18. Clean and wipe dry all work surfaces at the end of class. Wash your hands thoroughly.

19. Know the location of emergency equipment (first aid kit, fire extinguisher, fire shower, fire blanket, etc.) and how to use them.

20. Report all accidents to the teacher immediately.

Handling Chemicals

21. Read and double check labels on reagent bottles before removing any reagent. Take only as much reagent as you

need.

22. Do not return unused reagent to stock bottles.

23. When transferring chemical reagents from one container to another, hold the containers out away from your body.

24. When mixing an acid and water, always add the acid to the water.

25. Avoid touching chemicals with your hands. If chemicals do come in contact with your hands, wash them immediately.

26. Notify your teacher if you have any medical problems that might relate to lab work, such as allergies or asthma.

27. If you will be working with chemicals in the lab, avoid wearing contact lenses. Change to glasses, if possible, or notify

the teacher.

Handling Glassware

28. Glass tubing, especially long pieces, should be carried in a vertical position to minimize the likelihood of breakage and

to avoid stabbing anyone.

29. Never handle broken glass with your bare hands. Use a brush and dustpan to clean up broken glass. Dispose of the

glass as directed by your teacher.


30. Always lubricate glassware (tubing, thistle tubes, thermometers, etc.) with water or glycerin before attempting to insert

it into a rubber stopper.

31. Never apply force when inserting or removing glassware from a stopper. Use a twisting motion. If a piece of glassware

becomes "frozen" in a stopper, take it to your teacher.

32. Do not place hot glassware directly on the lab table. Always use an insulating pad of some sort.

33. Allow plenty of time for hot glass to cool before touching it. Hot glass can cause painful burns. (Hot glass looks cool.)

Heating Substances

34. Exercise extreme caution when using a gas burner. Keep your head and clothing away from the flame.

35. Always turn the burner off when it is not in use.

36. Do not bring any substance into contact with a flame unless instructed to do so.

37. Never heat anything without being instructed to do so.

38. Never look into a container that is being heated.

39. When heating a substance in a test tube, make sure that the mouth of the tube is not pointed at yourself or anyone

else.

40. Never leave unattended anything that is being heated or is visibly reacting.

First Aid in the MUSTANG LABORATORIES - Chemistry Laboratory

Accidents do not often happen in well-equipped chemistry laboratories if students understand safe laboratory procedures

and are careful in following them. When an occasional accident does occur, it is likely to be a minor one.

The instructor will assist in treating injuries such as minor cuts and burns. However, for some types of injuries, you must

take action immediately. The following information will be helpful to you if an accident occurs.

1. Shock. People who are suffering from any severe injury (for example, a bad burn or major loss of blood) may be in a

state of shock. A person in shock is usually pale and faint. The person may be sweating, with cold, moist skin and a weak,

rapid pulse. Shock is a serious medical condition. Do not allow a person in shock to walk anywhere—even to the campus

security office. While emergency help is being summoned, place the victim face up in a horizontal position, with the feet

raised about 30 centimeters. Loosen any tightly fitting clothing and keep him or her warm.

2. Chemicals in the Eyes. Getting any kind of a chemical into the eyes is undesirable, but certain chemicals are

especially harmful. They can destroy eyesight in a matter of seconds. Because you will be wearing safety goggles at all

times in the lab, the likelihood of this kind of accident is remote. However, if it does happen, flush your eyes with water

immediately. Do NOT attempt to go to the campus office before flushing your eyes. It is important that flushing with water

be continued for a prolonged time—about 15 minutes.

3. Clothing or Hair on Fire. A person whose clothing or hair catches on fire will often run around hysterically in an

unsuccessful effort to get away from the fire. This only provides the fire with more oxygen and makes it burn faster. For

clothing fires, throw yourself to the ground and roll around to extinguish the flames. For hair fires, use a fire blanket to

smother the flames. Notify campus security immediately.

4. Bleeding from a Cut. Most cuts that occur in the chemistry laboratory are minor. For minor cuts, apply pressure to the

wound with a sterile gauze. Notify campus security of all injuries in the lab. If the victim is bleeding badly, raise the

bleeding part, if possible, and apply pressure to the wound with a piece of sterile gauze. While first aid is being given,

someone else should notify the campus security officer.

5. Chemicals in the Mouth. Many chemicals are poisonous to varying degrees. Any chemical taken into the mouth

should be spat out and the mouth rinsed thoroughly with water. Note the name of the chemical and notify the campus

office immediately. If the victim swallows a chemical, note the name of the chemical and notify campus security

immediately.

If necessary, the campus security officer or administrator will contact the Poison Control Center, a hospital emergency

room, or a physician for instructions.

6. Acid or Base Spilled on the Skin.

Flush the skin with water for about 15 minutes. Take the victim to the campus office to report the injury.

7. Breathing Smoke or Chemical Fumes.

All experiments that give off smoke or noxious gases should be conducted in a well-ventilated fume hood. This will make

an accident of this kind unlikely. If smoke or chemical fumes are present in the laboratory, all persons—even those who

do not feel ill—should leave the laboratory immediately. Make certain that all doors to the laboratory are closed after the

last person has left. Since smoke rises, stay low while evacuating a smoke-filled room. Notify campus security

immediately.


MUSTANG LABORATORIES

COMMITMENT TO SAFETY IN THE LABORATORY

As a student enrolled in Chemistry at Lorenzo Walker Technical High

School, I agree to use good laboratory safety practices at all times. I

also agree that I will:

1. Conduct myself in a professional manner, respecting both my personal safety and the safety of

others in the laboratory.

2. Wear proper and approved safety glasses or goggles in the laboratory at all times.

3. Wear sensible clothing and tie back long hair in the laboratory. Understand that open-toed shoes

pose a hazard during laboratory classes and that contact lenses are an added safety risk.

4. Keep my lab area free of clutter during an experiment.

5. Never bring food or drink into the laboratory, nor apply makeup within the laboratory.

6. Be aware of the location of safety equipment such as the fire extinguisher, eye wash station, fire

blanket, first aid kit. Know the location of the nearest telephone and exits.

7. Read the assigned lab prior to coming to the laboratory.

8. Carefully read all labels on all chemical containers before using their contents, remove a small

amount of reagent properly if needed, do not pour back the unused chemicals into the original

container.

9. Dispose of chemicals as directed by the instructor only. At no time will I pour anything down the

sink without prior instruction.

10. Never inhale fumes emitted during an experiment. Use the fume hood when instructed to do so.

11. Report any accident immediately to the instructor, including chemical spills.

12. Dispose of broken glass and sharps only in the designated containers.

13. Clean my work area and all glassware before leaving the laboratory.

14. Wash my hands before leaving the laboratory.

NAME _________________________

Enmanuel Garrido

PERIOD ________________________

1

PARENT NAME ____________________________

Anna Ester Garrido

PARENT NUMBER _________________________

239-200-0466

SIGNATURE ____________________________

DATE ____________________________________

08/25/16


Chapter 1

Unit 1

Introduction to Chemistry

The students will learn why and how to solve problems using

chemistry.

Identify what is science, what clearly is not science, and what superficially

resembles science (but fails to meet the criteria for science).

Students will identify a phenomenon as science or not science.

Science

Observation

Inference

Hypothesis

Identify which questions can be answered through science and which

questions are outside the boundaries of scientific investigation, such as

questions addressed by other ways of knowing, such as art, philosophy, and

religion.

Students will differentiate between problems and/or phenomenon that can and

those that cannot be explained or answered by science.

Students will differentiate between problems and/or phenomenon that can and

those that cannot be explained or answered by science.

Observation

Inference

Hypothesis

Theory

Controlled experiment

Describe how scientific inferences are drawn from scientific observations

and provide examples from the content being studied.

Students will conduct and record observations.

Students will make inferences.

Students will identify a statement as being either an observation or inference.

Students will pose scientific questions and make predictions based on

inferences.

Inference

Observation

Hypothesis

Controlled experiment

Identify sources of information and assess their reliability according to the

strict standards of scientific investigation.

Students will compare and assess the validity of known scientific information

from a variety of sources:


Print vs. print

Online vs. online

Print vs. online

Students will conduct an experiment using the scientific method and compare

with other groups.

Controlled experiment

Investigation

Peer Review

Accuracy

Precision

Percentage Error

Chapter 2

Matter and Change

The students will learn what properties are used to describe

matter and how matter can change its form.

Differentiate between physical and chemical properties and physical and

chemical changes of matter.

Students will be able to identify physical and chemical properties of various

substances.

Students will be able to identify indicators of physical and chemical changes.

Students will be able to calculate density.

mass

physical property

volume

chemical property

vapor

extensive property

Chapter 3

mixture

intensive property

solution

element

compound

Scientific Measurements

The students will be able to solve conversion problems using

measurements.

Determine appropriate and consistent standards of measurement for the

data to be collected in a survey or experiment.

Students will participate in activities to collect data using standardized

measurement.

Students will be able to manipulate/convert data collected and apply the data

to scientific situations.

Scientific notation

International System of Units (SI)

Significant figures

Accepted value

Experimental value

Percent error

Dimensional analysis


Determine appropriate and consistent standards of measurement for the data to be collected in a survey or experiment.

King Henry Died By Drinking Chocolate Milk


To use the Stair-Step method, find the prefix the original measurement starts with. (ex. milligram)

If there is no prefix, then you are starting with a base unit.

Find the step which you wish to make the conversion to. (ex. decigram)

Count the number of steps you moved, and determine in which direction you moved (left or right).

The decimal in your original measurement moves the same number of places as steps you moved and in the

same direction. (ex. milligram to decigram is 2 steps to the left, so 40 milligrams = .40 decigrams)

If the number of steps you move is larger than the number you have, you will have to add zeros to hold the

places. (ex. kilometers to meters is three steps to the right, so 10 kilometers would be equal to 10,000 m)

That’s all there is to it! You need to be able to count to 6, and know your left from your right!

1) Write the equivalent

a) 5 dm =_______m .5 b) 4 mL = ______L .004 c) 8 g = _______mg 8000

d) 9 mg =_______g .009

e) 2 mL = ______L .002 f) 6 kg = _____g 6000

g) 4 cm =_______m .04 h) 12 mg = ______ .0012 g i) 6.5 cm 3 = _______L .0065

j) 7.02 mL =_____cm 7.02

3 k) .03 hg = _______ 3 dg l) 6035 mm _____cm 60.35

m) .32 m = _______cm 32

n) 38.2 g = _____kg 0.382


2. One cereal bar has a mass of 37 g. What is the mass of 6 cereal bars? Is that more than or less

than 1 kg? Explain your answer.

37*6=222 1kg=1000g

222g


Using SI Units

Match the terms in Column II with the descriptions in Column I. Write the letters of the correct term in

the blank on the left.

Column I Column II

_____ k. 1. distance between two points

a. time

_____ e. 2. SI unit of length

_____ m. 3. tool used to measure length

_____ g. 4. units obtained by combining other units

_____ b. 5. amount of space occupied by an object

_____ h. 6. unit used to express volume

_____ f. 7. SI unit of mass

_____ c 8. amount of matter in an object

_____ d. 9. mass per unit of volume

_____ o 10. temperature scale of most laboratory thermometers

_____ l. 11. instrument used to measure mass

_____ a 12. interval between two events

_____ j. 13. SI unit of temperature

_____ i. 14. SI unit of time

_____ n. 15. instrument used to measure temperature

b. volume

c. mass

d. density

e. meter

f. kilogram

g. derived

h. liter

i. second

j. Kelvin

k. length

1. balance

m. meterstick

n. thermometer

o. Celsius

Circle the two terms in each group that are related. Explain how the terms are related.

16. Celsius degree, mass, Kelvin _____________________________________________________

________________________________________________________________________________

Both of these terms are used to measure temperaature, or the amount of energy.

17. balance, second, mass __________________________________________________________

Both of these terms are used to measure the amount of mass an object has.

________________________________________________________________________________

18. kilogram, liter, cubic centimeter __________________________________________________

Both of these terms are used to measure volume.

________________________________________________________________________________

19. time, second, distance __________________________________________________________

Both of these terms measure of time.

________________________________________________________________________________

20. decimeter, kilometer, Kelvin _____________________________________________________

Both terms are used to measure distance.

________________________________________________________________________________


1. How many meters are in one kilometer? __________

1000m

2. What part of a liter is one milliliter? __________

thousandth

3. How many grams are in two dekagrams? __________

20g

4. If one gram of water has a volume of one milliliter, what would the mass of one liter of water be in

kilograms?__________

1

5. What part of a meter is a decimeter? __________

tenth

1000ml=1l

1000ml=1000g

In the blank at the left, write the term that correctly completes each statement. Choose from the terms

listed below.

Metric SI standard ten

prefixes ten tenth

6. An exact quantity that people agree to use for comparison is a ______________ .

7. The system of measurement used worldwide in science is _______________ .

8. SI is based on units of _______________ .

9. The system of measurement that was based on units of ten was the _______________ system.

10. In SI, _______________ are used with the names of the base unit to indicate the multiple of ten

that is being used with the base unit.

11. The prefix deci- means _______________ .


Standards of Measurement

Fill in the missing information in the table below.

S.I prefixes and their meanings

Prefix

Meaning

0.001

0.01

deci- 0.1

10

hecto- 100

1000

Circle the larger unit in each pair of units.

1. millimeter, kilometer 4. centimeter, millimeter

2. decimeter, dekameter 5. hectogram, kilogram

3. hectogram, decigram

6. In SI, the base unit of length is the meter. Use this information to arrange the following units of

measurement in the correct order from smallest to largest.

Write the number 1 (smallest) through 7 - (largest) in the spaces provided.

_____ a. kilometer

_____ b. centimeter

_____ c. meter

_____ e. hectometer

_____ f. millimeter

_____ g. decimeter

_____ d. dekameter

Use your knowledge of the prefixes used in SI to answer the following questions in the spaces

provided.

7. One part of the Olympic games involves an activity called the decathlon. How many events do you

think make up the decathlon?_____________________________________________________

8. How many years make up a decade? _______________________________________________

9. How many years make up a century? ______________________________________________

10. What part of a second do you think a millisecond is? __________________________________


The Learning Goal for this assignment is: Determine appropriate and consistent

standards of measurement for the collected

in a survey or experiment.

Notes Section

10 0 =1 Larger numbers have a positive exponent,

while

numbers that are very small in value have a

negative exponent.

1. 7,485 6. 1.683

2. 884.2 7. 3.622

3. 0.00002887 8. 0.00001735

4. 0.05893 9. 0.9736

5. 0.006162 10. 0.08558

11. 6.633 X 10−⁴ 16. 1.937 X 10⁴

12. 4.445 X 10−⁴ 17. 3.457 X 10⁴

13. 2.182 X 10−³ 18. 3.948 X 10−⁵

14. 4.695 X 10² 19. 8.945 X 10⁵

15. 7.274 X 10⁵ 20. 6.783 X 10²


SCIENTIFIC NOTATION RULES

How to Write Numbers in Scientific Notation

Scientific notation is a standard way of writing very large and very small numbers so that they're

easier to both compare and use in computations. To write in scientific notation, follow the form

N X 10 ᴬ

where N is a number between 1 and 10, but not 10 itself, and A is an integer (positive or negative

number).

RULE #1: Standard Scientific Notation is a number from 1 to 9 followed by a decimal and the

remaining significant figures and an exponent of 10 to hold place value.

Example:

5.43 x 10 2 = 5.43 x 100 = 543

8.65 x 10 – 3 = 8.65 x .001 = 0.00865

****54.3 x 10 1 is not Standard Scientific Notation!!!

RULE #2: When the decimal is moved to the Left the exponent gets Larger, but the value of the

number stays the same. Each place the decimal moves Changes the exponent by one (1). If you

move the decimal to the Right it makes the exponent smaller by one (1) for each place it is moved.

Example:

6000. x 10 0 = 600.0 x 10 1 = 60.00 x 10 2 = 6.000 x 10 3 = 6000

(Note: 10 0 = 1)

All the previous numbers are equal, but only 6.000 x 10 3 is in proper Scientific Notation.


RULE #3: To add/subtract in scientific notation, the exponents must first be the same.

Example:

(3.0 x 10 2 ) + (6.4 x 10 3 ); since 6.4 x 10 3 is equal to 64. x 10 2 . Now add.

(3.0 x 10 2 )

+ (64. x 10 2 )

67.0 x 10 2 = 6.70 x 10 3 = 6.7 x 10 3

67.0 x 10 2 is mathematically correct, but a number in standard scientific notation can only

have one number to the left of the decimal, so the decimal is moved to the left one place and

one is added to the exponent.

Following the rules for significant figures, the answer becomes 6.7 x 10 3 .

RULE #4: To multiply, find the product of the numbers, then add the exponents.

Example:

(2.4 x 10 2 ) (5.5 x 10 –4 ) = ? [2.4 x 5.5 = 13.2]; [2 + -4 = -2], so

(2.4 x 10 2 ) (5.5 x 10 –4 ) = 13.2 x 10 –2 = 1.3 x 10 – 1

RULE #5: To divide, find the quotient of the number and subtract the exponents.

Example:

(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = ? [3.3 / 9.1 = .36]; [-6 – (-8) = 2], so

(3.3 x 10 – 6 ) / (9.1 x 10 – 8 ) = .36 x 10 2 = 3.6 x 10 1


Convert each number from Scientific Notation to real numbers:

1. 7.485 X 10³ =7485

6. 1.683 X 10⁰

=1.683

2. 8.842 X 10² =884.2

7. 3.622 10⁰

=3.622

3. 2.887 X 10−⁵ =.00002887

8. 1.735 X 10−⁵

=.00001735

4. 5.893 X 10−² =.05893

9. 9.736 X 10−¹

=.9736

5. 6.162 X 10−³ =.006162

10. 8.558 X 10−²

.08558

Convert each number from a real number to Scientific Notation:

11. 0.0006633 16. 1,937,000

1.937X10 6

6.633X10 -4

12. 0.0004445 17. 34,570

4.445X10 -4 3.4570X10 4

13. 0.002182 18. 0.00003948

2.182X10 -3 3.948X10 -5

14. 469.5 19. 894,500

4.695X10 2 8.945X10 5

15. 727,400 20. 678.3

7.274X10 5 6.783X10 2


The Learning Goal for this assignment is:

Notes Section:

Question Sig Figs Question Add & Subtract Question Multiple & Divide

1 4 1 55.36 1 20,000

2 4 2 84.2 2 94

3 3 3 115.4 3 300

4 3 4 0.8 4 7

5 4 5 245.53 5 62

6 3 6 34.5 6 0.005

7 3 7 74.0 7 4,000

8 2 8 53.287 8 3,900,000

9 2 9 54.876 9 2

10 2 10 40.19 10 30,000,000

11 3 11 7.7 11 1,200

12 2 12 67.170 12 0.2

13 3 13 81.0 13 0.87

14 4 14 73.290 14 0.049

15 4 15 29.789 15 2,000

16 3 16 39.53 16 0.5

17 4 17 70.58 17 1.9

18 2 18 86.6 18 0.05

19 2 19 64.990 19 230

20 1 20 36.0 20 460,000


Significant Figures Rules

There are three rules on determining how many significant figures are in a

number:

1. Non-zero digits are always significant.

2. Any zeros between two significant digits are significant.

3. A final zero or trailing zeros in the DECIMAL PORTION ONLY are

significant.

Please remember that, in science, all numbers are based upon measurements (except for a very few

that are defined). Since all measurements are uncertain, we must only use those numbers that are

meaningful.

Not all of the digits have meaning (significance) and, therefore, should not be written down. In

science, only the numbers that have significance (derived from measurement) are written.

Rule 1: Non-zero digits are always significant.

If you measure something and the device you use (ruler, thermometer, triple-beam, balance, etc.)

returns a number to you, then you have made a measurement decision and that ACT of measuring

gives significance to that particular numeral (or digit) in the overall value you obtain.

Hence a number like 46.78 would have four significant figures and 3.94 would have three.

Rule 2: Any zeros between two significant digits are significant.

Suppose you had a number like 409. By the first rule, the 4 and the 9 are significant. However, to

make a measurement decision on the 4 (in the hundred's place) and the 9 (in the one's place), you

HAD to have made a decision on the ten's place. The measurement scale for this number would have

hundreds, tens, and ones marked.

Like the following example:

These are sometimes called "captured zeros."

If a number has a decimal at the end (after the one’s place) then all digits (numbers) are significant

and will be counted.

In the following example the zeros are significant digits and highlighted in blue.

960.

70050.


Rule 3: A final zero or trailing zeros in the decimal portion ONLY are

significant.

This rule causes the most confusion among students.

In the following example the zeros are significant digits and highlighted in blue.

0.07030

0.00800

Here are two more examples where the significant zeros are highlighted in blue.

When Zeros are Not Significant Digits

4.7 0 x 10−³

6.5 0 0 x 10⁴

Zero Type # 1 : Space holding zeros in numbers less than one.

In the following example the zeros are NOT significant digits and highlighted in red.

0.09060

0.00400

These zeros serve only as space holders. They are there to put the decimal point in its correct

location.

They DO NOT involve measurement decisions.

Zero Type # 2 : Trailing zeros in a whole number.

In the following example the zeros are NOT significant digits and highlighted in red.

200

25000

For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point)

of the numbers ONLY. Here is what to do:

1) Count the number of significant figures in the decimal portion of each number in the problem. (The

digits to the left of the decimal place are not used to determine the number of decimal places in the

final answer.)

2) Add or subtract in the normal fashion.

3) Round the answer to the LEAST number of places in the decimal portion of any number in the

problem

The following rule applies for multiplication and division:

The LEAST number of significant figures in any number of the problem determines the number of

significant figures in the answer.

This means you MUST know how to recognize significant figures in order to use this rule.


How Many Significant Digits for Each Number?

1) 2359 = ______

2) 2.445 x 10−⁵= ______

3) 2.93 x 10⁴= ______

4) 1.30 x 10−⁷= ______

5) 2604 = ______

6) 9160 = ______

7) 0.0800 = ______

8) 0.84 = ______

9) 0.0080 = ______

10) 0.00040 = ______

11) 0.0520 = ______

12) 0.060 = ______

13) 6.90 x 10−¹= ______

14) 7.200 x 10⁵= ______

15) 5.566 x 10−²= ______

16) 3.88 x 10⁸= ______

17) 3004 = ______

18) 0.021 = ______

19) 240 = ______

20) 500 = ______


For addition and subtraction, look at the decimal portion (i.e., to the right of the decimal point) of the

numbers ONLY. Here is what to do:

1) Count the number of significant figures in the decimal portion of each number in the problem. (The

digits to the left of the decimal place are not used to determine the number of decimal places in the

final answer.)

2) Add or subtract in the normal fashion.

3) Round the answer to the LEAST number of places in the decimal portion of any number in the

problem.

Solve the Problems and Round Accordingly...

1) 43.287 + 5.79 + 6.284 = _______

2) 87.54 - 3.3 = _______

3) 99.1498 + 6.5397 + 9.7 = _______

4) 5.868 - 5.1 = _______

5) 59.9233 + 86.21 + 99.396 = _______

6) 7.7 + 26.756 = _______

7) 66.8 + 2.3 + 4.8516 = _______

8) 9.7419 + 43.545 = _______

9) 4.8976 + 48.4644 + 1.514 = _______

10) 4.335 + 35.85 = _______

11) 9.448 - 1.7 = _______

12) 75.826 - 8.6555 = _______

13) 57.2 + 23.814 = _______

14) 77.684 - 4.394 = _______

15) 26.4496 + 3.339 = _______

16) 9.6848 + 29.85 = _______

17) 63.11 + 2.5412 + 4.93 = _______

18) 11.2471 + 75.4 = _______

19) 73.745 - 8.755 = _______

20) 6.5238 + 1.7 + 27.79 = _______


The following rule applies for multiplication and division:

The LEAST number of significant figures in any number of the problem determines the number of

significant figures in the answer.

This means you MUST know how to recognize significant figures in order to use this rule.

Solve the Problems and Round Accordingly...

1) 0.6 x 65.0 x 602 = __________

2) 720 ÷ 7.7 = __________

3) 929 x 0.3 = __________

4) 300 ÷ 44.31 = __________

5) 608 ÷ 9.8 = __________

6) 0.06 x 0.079 = __________

7) 0.008 x 72.91 x 7000 = __________

8) 73.94 x 67 x 780 = __________

9) 0.62 x 0.097 x 40 = __________

10) 600 x 10 x 5030 = __________

11) 5200 ÷ 4.46 = __________

12) 0.0052 x 0.4 x 107 = __________

13) 0.099 x 8.8 = __________

14) 0.0095 x 5.2 = __________

15) 8000 ÷ 4.62 = __________

16) 0.6 x 0.8 = __________

17) 2.84 x 0.66 = __________

18) 0.5 x 0.09 = __________

19) 8100 ÷ 34.84 = __________

20) 8.24 x 6.9 x 8100 = __________


Dimensional Analysis

This is a way to convert from one unit of a given substance to

another unit using ratios or conversion units. What this video

www.youtube.com/watch?v=aZ3J60GYo6U

Let’ look at a couple of examples:

1. Convert 2.6 qt to mL.

First we need a ratio or conversion unit so that we can go from quarts to milliliters. 1.00 qt = 946 mL

Next write down what you are starting with

2.6 qt

Then make you conversion tree

2.6 qt

Then fill in the units in your ratio so that you can cancel out the original unit and will be left with the

unit you need for the answer. Cross out units, one at a time that are paired, and one on top one on

the bottom.

2.6 qt mL

qt

Now fill in the values from the ratio.

2.6 qt 946 mL

1.00 qt

Now multiply all numbers on the top and multiply all numbers on the bottom and write them as a

fraction.

2.6 qt 946 mL = 2,459.6 mL

1.00 qt 1.00

Now divide the top number by the bottom number and write that number with the unit that was not

crossed out.


1qt=32 oz 1gal = 4qts 1.00 qt = 946 mL 1L = 1000mL

2. Convert 8135.6 mL to quarts

8135.6 ml 1qt

1

946ml

=

8135.6

946

8.6000ml

3. Convert 115.2 oz to mL

115.2 oz

1qt

32oz

=

946ml 108979.2

1qt

32 3405.6ml

4. Convert 2.3 g to Liters

2.3gal

4qt 946

1gal

1qt

=

1L 8703.2

1000ml

1000

8.7L

5. Convert 8.42 L to oz

8.42L

1000mL

1.00qt

=

32oz 269440

284.82oz

1L

946mL

1qt

946

Go to http://science.widener.edu/svb/tutorial/ chose #7 “Converting Volume” and do 5 more in the

space provided.

1. Convert _________ to _________

=

2. Convert _________ to _________

=

3. Convert _________ to _________

=

4. Convert _________ to _________

=

5. Convert _________ to _________

=


Chapter 4

Unit 2

Atomic Structure

The students will learn what makes up atoms and how are

atoms of one element different from atoms of another element.

Explore the scientific theory of atoms (also known as atomic theory) by

describing changes in the atomic model over time and why those changes

were necessitated by experimental evidence.

Students will be able to draw/identify each atomic model.

Students will be able to compare/contrast the different atomic models.

Students will be able to describe how results of experimental evidence caused

the atomic model to change.

proton

electron

neutron

nucleus

electron cloud

Explore the scientific theory of atoms (also known as atomic theory) by

describing the structure of atoms in terms of protons, neutrons and

electrons, and differentiate among these particles in terms of their mass,

electrical charges and locations within the atom.

Students will compare/contrast the characteristics of subatomic particles.

atomic number

mass number

isotope

atomic mass unit (amu)

atomic mass


Chapter 5

Electrons in Atoms

The students will be able to describe the arrangement of

electrons in atoms and predict what will happen when

electrons in atoms absorb or release energy.

Describe the quantization of energy at the atomic level.

Students will participate in activities to view emission spectrums using a

diffraction grating or a spectroscope.

Students will be able to explain how the spectrum lines relate to electron motion.

energy level

atomic orbital

quantum mechanical model

Chapter 6

The Periodic Table

The student will learn what information the periodic table

provides and how periodic trends can be explained.

Relate properties of atoms and their position in the periodic table to the

arrangement of their electrons.

Students will be able to compare and contrast metals, nonmetals, and metalloids.

Students will be able to describe the traits of various families on the periodic

table.

Students will be able to explain periodicity.

Students will write/represent electron configuration of various elements.

Students will be able to use a periodic table to calculate the number of p + , e - , and

n 0 .

Students will be able to calculate the average weight of mass.

periodic law

halogen

metals

noble gas

nonmetals

transition metal

metalloid

atomic radius

alkali metal

ionization energy

alkaline earth metal

electronegativity


The Learning Goal for this assignment is:

Notes Section

http://www.learner.org/interactives/periodic/basics_interactive.html


Atoms Are Building Blocks

Atoms are the basis of chemistry. They are the basis for everything in the Universe. You

should start by remembering that matter is composed of atoms. Atoms and the study of

atoms are a world unto themselves. We're going to cover basics like atomic structure

and bonding between atoms.

Smaller Than Atoms?

Are there pieces of matter that are smaller than atoms?

Sure there are. You'll soon be learning that atoms are

composed of pieces like electrons, protons, and neutrons.

But guess what? There are even smaller particles moving

around in atoms. These super-small particles can be found

inside the protons and neutrons. Scientists have many

names for those pieces, but you may have heard of

nucleons and quarks. Nuclear chemists and physicists

work together at particle accelerators to discover the

presence of these tiny, tiny, tiny pieces of matter.

Even though super-tiny atomic particles exist, you only

need to remember the three basic parts of an atom: electrons, protons, and neutrons.

What are electrons, protons, and neutrons? A picture works best to show off the idea.

You have a basic atom. There are three types of pieces in that atom: electrons, protons,

and neutrons. That's all you have to remember. Three things! As you know, there are

almost 120 known elements in the periodic table. Chemists and physicists haven't

stopped there. They are trying to make new ones in labs every day. The thing that

makes each of those elements different is the number of electrons, protons, and

neutrons. The protons and neutrons are always in the center of the atom. Scientists call

the center region of the atom the nucleus. The nucleus in

a cell is a thing. The nucleus in an atom is a place where

you find protons and neutrons. The electrons are always

found whizzing around the center in areas called shells or

orbitals.

You can also see that each piece has either a "+", "-", or a

"0." That symbol refers to the charge of the particle. Have

you ever heard about getting a shock from a socket, static

electricity, or lightning? Those are all different types of

electric charges. Those charges are also found in tiny particles of matter. The electron

always has a "-", or negative, charge. The proton always has a "+", or positive, charge. If

the charge of an entire atom is "0", or neutral, there are equal numbers of positive and

negative pieces. Neutral means there are equal numbers of electrons and protons. The

third particle is the neutron. It has a neutral charge, also known as a charge of zero. All

atoms have equal numbers of protons and electrons so that they are neutral. If there are

more positive protons or negative electrons in an atom, you have a special atom called

an ion.


Looking at Ions

We haven’t talked about ions before, so let’s get down to basics. The

atomic number of an element, also called a proton number, tells you the

number of protons or positive particles in an atom. A normal atom has a

neutral charge with equal numbers of positive and negative particles.

That means an atom with a neutral charge is one where the number of

electrons is equal to the atomic number. Ions are atoms with extra

electrons or missing electrons. When you are missing an electron or

two, you have a positive charge. When you have an extra electron

or two, you have a negative charge.

What do you do if you are a sodium (Na) atom? You have eleven

electrons — one too many to have an entire shell filled. You need to

find another element that will take that electron away from you. When you lose that

electron, you will you’ll have full shells. Whenever an atom has full shells, we say it is

"happy." Let's look at chlorine (Cl). Chlorine has seventeen electrons and only needs

one more to fill its third shell and be "happy." Chlorine will take your extra sodium

electron and leave you with 10 electrons inside of two filled shells. You are now a happy

atom too. You are also an ion and missing one electron. That missing electron gives you

a positive charge. You are still the element sodium, but you are now a sodium ion (Na + ).

You have one less electron than your atomic number.

Ion Characteristics

So now you've become a sodium ion. You have ten electrons.

That's the same number of electrons as neon (Ne). But you

aren't neon. Since you're missing an electron, you aren't really

a complete sodium atom either. As an ion you are now

something completely new. Your whole goal as an atom was

to become a "happy atom" with completely filled electron

shells. Now you have those filled shells. You have a lower

energy. You lost an electron and you are "happy." So what

makes you interesting to other atoms? Now that you have

given up the electron, you are quite electrically attractive.

Other electrically charged atoms (ions) of the opposite charge

(negative) are now looking at you and seeing a good partner to

bond with. That's where the chlorine comes in. It's not only chlorine. Almost any ion with

a negative charge will be interested in bonding with you.


Electrovalence

Don't get worried about the big word. Electrovalence is just another word for something

that has given up or taken electrons and become an ion. If you look at the periodic table,

you might notice that elements on the left side usually become positively charged ions

(cations) and elements on the right side get a negative charge (anions). That trend

means that the left side has a positive valence and the right side has a negative

valence. Valence is a measure of how much an atom wants to bond with other atoms. It

is also a measure of how many electrons are excited about bonding with other atoms.

There are two main types of bonding, covalent and electrovalent. You may have heard

of the term "ionic bonds." Ionic bonds are electrovalent bonds. They are just groups of

charged ions held together by electric forces. When in the presence of other ions, the

electrovalent bonds are weaker because of outside electrical forces and attractions.

Sodium and chlorine ions alone have a very strong bond, but as soon as you put those

ions in a solution with H + (Hydrogen ion), OH - (Hydroxide), F - (Fluorine ion) or Mg ++

(Magnesium ion), there are charged distractions that break the Na-Cl bond.

Look at sodium chloride (NaCl) one more time. Salt is a very strong bond when it is

sitting on your table. It would be nearly impossible to break those ionic/electrovalent

bonds. However, if you put that salt into some water (H2O), the bonds break very

quickly. It happens easily because of the electrical attraction of the water. Now you have

sodium (Na + ) and chlorine (Cl - ) ions floating around the solution. You should remember

that ionic bonds are normally strong, but they are very weak in water.


Neutron Madness

We have already learned that ions are atoms that are

either missing or have extra electrons. Let's say an atom

is missing a neutron or has an extra neutron. That type of

atom is called an isotope. An atom is still the same

element if it is missing an electron. The same goes for

isotopes. They are still the same element. They are just a

little different from every other atom of the same element.

For example, there are a lot of carbon (C) atoms in the

Universe. The normal ones are carbon-12. Those atoms have 6 neutrons. There are a

few straggler atoms that don't have 6. Those odd ones may have 7 or even 8 neutrons.

As you learn more about chemistry, you will probably hear about carbon-14. Carbon-14

actually has 8 neutrons (2 extra). C-14 is considered an isotope of the element carbon.

Messing with the Mass

If you have looked at a periodic table, you may have noticed that the atomic mass of

an element is rarely an even number. That happens because of the isotopes. If you are

an atom with an extra electron, it's no big deal. Electrons don't have much of a mass

when compared to a neutron or proton.

Atomic masses are calculated by figuring out the

amounts of each type of atom and isotope there are in

the Universe. For carbon, there are a lot of C-12, a

couple of C-13, and a few C-14 atoms. When you

average out all of the masses, you get a number that is a

little bit higher than 12 (the weight of a C-12 atom). The

average atomic mass for the element is actually 12.011.

Since you never really know which carbon atom you are

using in calculations, you should use the average mass

of an atom.

Bromine (Br), at atomic number 35, has a greater variety of isotopes. The atomic mass

of bromine (Br) is 79.90. There are two main isotopes at 79 and 81, which average out

to the 79.90amu value. The 79 has 44 neutrons and the 81 has 46 neutrons. While it

won't change the average atomic mass, scientists have made bromine isotopes with

masses from 68 to 97. It's all about the number of neutrons. As you move to higher

atomic numbers in the periodic table, you will probably find even more isotopes for

each element.


Summary


Electron Configuration

Color the sublevel:

s = Red

d = Green

p = Blue

f = Orange

s

P

D

F

Write in sublevels

Write period, sublevel and super scripts.

Ctrl Shift =

gives you super scripts


The Learning Goal for this assignment is:

www.youtube.com/watch?v=jtYzEzykFdg

www.youtube.com/watch?

annotation_id=annotation_2076&feature=iv&src_vid=jtYzEzykFdg&v=cOlac8ruD_0

www.youtube.com/watch?

annotation_id=annotation_570977&feature=iv&src_vid=cOlac8ruD_0&v=lR2vqHZWb5A

Notes Section


Electron Configuration

In order to write the electron configuration for an atom you must know the 3 rules of

electron configurations.

1. Aufbau

Notation

nO e

where

n is the energy level

O is the orbital type (s, p, d, or f)

e is the number of electrons in that orbital shell

Principle

electrons will first occupy orbitals of the lowest energy level

2. Hund rule

when electrons occupy orbitals of equal energy, one electron enters each orbital until

all the orbitals contain one electron with the same spin.

3. Pauli exclusion principle

an orbital contains a maximum of 2 electrons and

paired electrons will have opposite spin


In the space below, write the unabbreviated electron configurations of the following elements:

1) sodium ________________________________________________

2) iron ________________________________________________

3) bromine ________________________________________________

4) barium ________________________________________________

5) neptunium ________________________________________________

In the space below, write the abbreviated electron configurations of the following elements:

6) cobalt ________________________________________________

7) silver ________________________________________________

8) tellurium ________________________________________________

9) radium ________________________________________________

10) lawrencium ________________________________________________

Determine what elements are denoted by the following electron configurations:

11) 1s²s²2p⁶3s²3p⁴ ____________________

12) 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹ ____________________

13) [Kr] 5s²4d¹⁰5p³ ____________________

14) [Xe] 6s²4f¹⁴5d⁶ ____________________

15) [Rn] 7s²5f¹¹ ____________________

Identify the element or determine that it is not a valid electron configuration:

16) 1s²2s²2p⁶3s²3p⁶4s²4d¹⁰4p⁵ ____________________

17) 1s²2s²2p⁶3s³3d⁵ ____________________

18) [Ra] 7s²5f⁸ ____________________

19) [Kr] 5s²4d¹⁰5p⁵ ____________________

20) [Xe] ____________________

1)sodium 1s 2 2s 2 2p 6 3s 1 2)iron 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 6

3)bromine 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 5 4)barium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2

5)neptunium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 6 7s 2 5f 5 6)cobalt [Ar] 4s 2 3d 7

7)silver [Kr] 5s 2 4d 9 8)tellurium[Kr] 5s 2 4d 10 5p 4

9)radium [Rn] 7s 2 10)lawrencium[Rn] 7s 2 5f 14 6d 1

1s 2 2s 2 2p 6 3s 2 3p 4 sulfur 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 rubidium

[Kr] 5s 2 4d 10 5p 3 antimony [Xe] 6s 2 4f 14 5d 6 osmium

[Rn] 7s 2 5f 11 einsteinium 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 4d 10 4p 5 not valid (take a look at “4d”)

1s 2 2s 2 2p 6 3s 3 3d 5 not valid (3p comes after 3s) [Ra] 7s 2 5f 8 not valid (radium isn’t a noble gas)

[Kr] 5s 2 4d 10 5p 5 valid iodine

20)[Xe] not valid (an element can’t be its own electron configuration)


Create groups for these Scientist and explain your groupings

(use the information you got from your research)


Research the Scientist and summarize their contributions to the Atomic Theory

Antoine Henri Becquerel

Niels Bohr

Louis de Barogilie

Glenn Seaborg

Hantaro Nagaoka

Democritus

Marie and Pierre Curie

Eugene Goldstein

Dmitri Mendeleev

J.J. Thomson

James Chadwick

Erwin Shrodinger

John Dalton

Lothar Meyer

Robert Millikan

J.W. Dobereiner

Ernest Rutherford


The Learning Goal for this Assignment is

Alkali Metals

any of the elements lithium, sodium, potassium, rubidium, cesium, and francium,

occupying Group IA (1) of the periodic table. They are very reactive, electropositive,

monovalent metals forming strongly alkaline hydroxides.

Alkali Earth Metals

Transitional Metals

The alkaline earth elements are metallic elements found in the second

group of the periodic table. All alkaline earth elements have an oxidation

number of +2, making them very reactive. Because of their reactivity, the

alkaline metals are not found free in nature.

The 38 elements in groups 3 through 12 of the periodic table are called

"transition metals". As with all metals, the transition elements are both

ductile and malleable, and conduct electricity and heat.

Inter Transitional Metals

They include elements 57-71 (lanthanides) and 89-103 (actinides).

The lanthanides are very similar, and the actinides are all radioactive.

They have three incomplete outermost electron shells and are all

metals.

Metals

is typically hard, shiny, malleable, fusible, and ductile, with good electrical and thermal conductivity

(e.g., iron, gold, silver, copper, and aluminum, and alloys such as brass and steel).

Metalloids

Metalloids have properties of both metals and non-metals. Some of the metalloids, such as silicon

and germanium, are semi-conductors. This means that they can carry an electrical charge under

special conditions. This property makes metalloids useful in computers and calculators.

Non Metals

Non-metals are the elements in groups 14-16 of the periodic table. Non-metals are not able to

conduct electricity or heat very well.The non-metals exist in two of the three states of matter at

room temperature: gases (such as oxygen) and solids (such as carbon).

Noble Gases

any of the gaseous elements helium, neon, argon, krypton, xenon, and radon, occupying

Group 0 (18) of the periodic table. They were long believed to be totally unreactive but

compounds of xenon, krypton, and radon are now known.


Using Wikipedia, define the 8 categories of elements on the

left page.

Color your periodic table similar to the one on

pages 168—169 of your book.

alkali metals

alkaline metals

other metals

transitional metals

lanthanoids

metalloids

non metals

halogens

noble gases

unknown elements

actinoids


Define Atomic Size: The size of the atom

Atomic Size

Explanation:

Groups increase going from top to bottom because more energy levels are added.

Periods increase going left to right because the bigger the nucleus the more compact and smaller

the atom is. This occurs because the greater the nucleus the more gravitaitional pull and attraction

between the electrons and protons.


Ionization Energy

Define Ionization Energy: The energy required to remove an electron from an atom.

Explanation:

First ionization energy tends to decrease from top to bottom withen a group and increase from

left to right.


Electronegativity

Define Electronegativity: Is the ability of an atom of an element to attract electrons when the atom is in a

compound.

Explanation:

Electronegativity values decrease from top to bottom withen a group. For representative elements,

the values ttend to increase from left to right across a period. For example, Bromine has a higher

electonegativity than potassium because although it is in the same period it has a larger atomic

mass.


Ion Size

Define Ion Size: The charge of an atom

Explanation:

When atoms gain or lose electrons, the atom becomes an ion. When an atom gains an

electron, it becomes a negatively charged ion that we call an anion. When an atom

loses an electron it is called a cation, positively charged ion.


Unit 3

Chapter 25 Nuclear Chemistry

The students will learn what happens when an unstable

nucleus decays and how nuclear chemistry affects their lives.

Explore the theory of electromagnetism by comparing and contrasting the

different parts of the electromagnetic spectrum in terms of wavelength,

frequency, and energy, and relate them to phenomena and applications.

Students will be able to compare and contrast the different parts of the

electromagnetic spectrum.

Students will be able to apply knowledge of the EMS to real world phenomena.

Students will be able to quantitatively compare the relationship between energy,

wavelength, and frequency of the EMS.

amplitude

wavelength

frequency

hertz

electromagnetic radiation

photon

Planck’s constant

Explain and compare nuclear reactions (radioactive decay, fission and

fusion), the energy changes associated with them and their associated

safety issues.

Students will be able to compare and contrast fission and fusion reactions.

Students will be able to complete nuclear decay equations to identify the type of

decay.

Students will participate in activities to calculate half-life.

radioactivity

nuclear radiation

alpha particle

beta particle

gamma ray

positron

½ life

transmutation

fission

fusion


Chapter 7

Ionic and Metallic Bonding

The students will learn how ionic compounds form and how

metallic bounding affects the properties of metals.

Compare the magnitude and range of the four fundamental forces

(gravitational, electromagnetic, weak nuclear, strong nuclear).

Students will compare/contrast the characteristics of each fundamental force.

gravity

electromagnetic

strong

weak

Distinguish between bonding forces holding compounds together and other

attractive forces, including hydrogen bonding and van der Waals forces.

Students will be able to compare/contrast traits of ionic and covalent bonds.

Students will be able to compare/contrast basic attractive forces between

molecules.

Students will be able to predict the type of bond or attractive force between

atoms or molecules.

ionic bond

covalent bond

metallic bond

polar covalent bond

hydrogen bond

van der Waals forces

London dispersion forces

Chapter 8

Covalent Bonding

The students will learn how molecular bonding is different

than ionic bonding and electrons affect the shape of a

molecule and its properties.

Interpret formula representations of molecules and compounds in terms of

composition and structure.

Students will be able to interpret chemical formulas in terms of # of atoms.

Students will be able to differentiate between ionic and molecular compounds.

Students will be able to list various VSEPR shapes and identify examples of

each.

Students will be able to predict shapes of various compounds.

Molecule

empirical formula

Atom

Electron

Element

Compound


Enmanuel Garrido

Name ____________________

Go to the web site www.darvill.clara.net/emag

1. Click on “How the waves fit into the spectrum” and fill in this table:

>: look out for the

RED words on the web site!

Low __________, frequency Long wavelength

High frequency, Short ______________

wavelength

Radio Waves

Microwaves Infra-red Visible Light Ultra-violet X-rays

Gamma rays

2. Click on “Radio waves”. They are used for _______________________

communications

3. Click on “Microwaves”. They are used for cooking, mobile _________, Wifi speed _______ cameras cameras and _________. radar

4. Click on “Infra-red”. These waves are given off by _____ hot _________. objects They are used for remote controls,

cameras in police ____________ helicopters , and alarm systems.

5. Click on “Visible Light”. This is used in DVD ___ players and _______ laser printers, and for seeing where we’re going.

6. “UV” stands for “ ________ Ultra ___________”. Violet This can damage the _________ retina in your eyes, and cause

sunburn and even _______ skin cancer. Its uses include detecting forged ______ bank _______. notes

7. X-rays are used to see inside people, and for _________ airport security.

8. Gamma rays are given off by some ________________ radioactive substances. We can use them to kill ________ cancer cells,

which is called R_______________ adiotherapy .

9. My Quiz score is ____%. 100


10. Name ________________________________

Go to the web site www.darvill.clara.net/emag

Name How they’re made Uses Dangers

Gamma rays

X-rays

Ultra-violet

Visible Light

Infra-red

Microwaves

Radio Waves

Stars radioactive substances

Stars

X-ray machines

Nebula

Sun

Special lamps, sun beds

anything hot enough to glow

light bulbs

Hot objects body

Stars, lamps, flames

Magnetrons "chips"

Extremely high frequency radio waves

Stars, Sparks, and lightning

Transmiters

_____ Frequency _____ frequency,

Short wavelength ______ Wavelength


Learning Goal for this section:

Explain and compare nuclea reactions (radioactive decay, fission and fusion), the energy changes associated

with them and their associated saftey issues.

Notes Section:

6 protons 8 neutrons 14 mass Carbon -> Nitrogen 14 mass 7 protons

Beta Particle

Negative Electron

Positive Electron Positron

Gamma Energy

very high frequency so it can penetrate deeply.

Half-Life 100g of a substance, if it has a 20min half-life, after 20min it has

50g, after another 20min, it has 25g later. The initial substance is conver

into a different element. So that after the first half-life the substance has 5

intitail substance and the other 50g is changed into a different substance


The Nucleus

A typical model of the atom is called the Bohr Model, in

honor of Niels Bohr who proposed the structure in 1913. The Bohr atom consists of a central nucleus

composed of neutrons and protons, which is surrounded by electrons which “orbit” around the nucleus.

Protons carry a positive charge of one and have a mass of about 1 atomic mass unit or amu (1 amu =1.7x10-

27 kg, a very, very small number). Neutrons are electrically “neutral” and also have a mass of about 1 amu. In

contrast electron carry a negative charge and have mass of only 0.00055 amu. The number of protons in a

nucleus determines the element of the atom. For example, the number of protons in uranium is 92 and the

number in neon is 10. The proton number is often referred to as Z.

Atoms with different numbers of protons are called elements, and are arranged in the periodic table with

increasing Z.

Atoms in nature are electrically neutral so the number of electrons orbiting the nucleus equals the number of

protons in the nucleus.

Neutrons make up the remaining mass of the nucleus and provide a means to “glue” the protons in place.

Without neutrons, the nucleus would split apart because the positive protons would repel each other. Elements

can have nucleii with different numbers of neutrons in them. For example hydrogen, which normally only has

one proton in the nucleus, can have a neutron added to its nucleus to from deuterium, ir have two neutrons

added to create tritium, which is radioactive. Atoms of the same element which vary in neutron number are

called isotopes. Some elements have many stable isotopes (tin has 10) while others have only one or two. We

express isotopes with the nomenclature Neon-20 or 20 Ne 10, with twenty representing the total number of

neutrons and protons in the atom, often referred to as A, and 10 representing the number of protons (Z).

Alpha Particle

Decay

Alpha decay is a radioactive process in which a

particle with two neutrons and two protons is

ejected from the nucleus of a radioactive atom. The particle is identical to the nucleus of a helium atom.

Alpha decay only occurs in very heavy elements such as uranium, thorium and radium. The nuclei of these

atoms are very “neutron rich” (i.e. have a lot more neutrons in their nucleus than they do protons) which makes

emission of the alpha particle possible.

After an atom ejects an alpha particle, a new parent atom is formed which has two less neutrons and two less

protons. Thus, when uranium-238 (which has a Z of 92) decays by alpha emission, thorium-234 is created

(which has a Z of 90).

Because alpha particles contain two protons, they have a positive charge of two. Further, alpha particles are

very heavy and very energetic compared to other common types of radiation. These characteristics allow alpha

particles to interact readily with materials they encounter, including air, causing many ionizations in a very short

distance. Typical alpha particles will travel no more than a few centimeters in air and are stopped by a sheet of

paper.


Beta Particle Decay

Beta decay is a radioactive process in which an electron is emitted from the nucleus of a radioactive

atom Because this electron is from the nucleus of the atom, it is called a beta particle to distinguish it

from the electrons which orbit the atom.

Like alpha decay, beta decay occurs in isotopes which are “neutron rich” (i.e. have a lot more

neutrons in their nucleus than they do protons). Atoms which undergo beta decay are located below

the line of stable elements on the chart of the nuclides, and are typically produced in nuclear reactors.

When a nucleus ejects a beta particle, one of the neutrons in the nucleus is transformed into a proton.

Since the number of protons in the nucleus has changed, a new daughter atom is formed which has

one less neutron but one more proton than the parent. For example, when rhenium-187 decays

(which has a Z of 75) by beta decay, osmium-187 is created (which has a Z of 76). Beta particles

have a single negative charge and weigh only a small fraction of a neutron or proton. As a result, beta

particles interact less readily with material than alpha particles. Depending on the beta particles

energy (which depends on the radioactive atom), beta particles will travel up to several meters in air,

and are stopped by thin layers of metal or plastic.

Positron emission or beta plus decay (β+ decay) is a subtype of radioactive decay called beta decay,

in which a proton inside a radionuclide nucleus is converted into a neutron while releasing a positron

and an electron neutrino (νe). Positron emission is mediated by the weak force.

An example of positron emission (β+ decay) is shown with magnesium-23 decaying into sodium-23:

23 Mg12 → 23 Na11 + e +

Because positron emission decreases proton number relative to neutron number, positron decay

happens typically in large "proton-rich" radionuclides. Positron decay results in nuclear transmutation,

changing an atom of one chemical element into an atom of an element with an atomic number that is

less by one unit.

Positron emission should not be confused with electron emission or beta minus decay (β− decay),

which occurs when a neutron turns into a proton and the nucleus emits an electron and an

antineutrino.


Gamma

Radiation

After a decay reaction, the nucleus is often in an

“excited” state. This means that the decay has

resulted in producing a nucleus which still has

excess energy to get rid of. Rather than emitting another beta or alpha particle, this energy is lost by

emitting a pulse of electromagnetic radiation called a gamma ray. The gamma ray is identical in

nature to light or microwaves, but of very high energy.

Like all forms of electromagnetic radiation, the gamma ray has no mass and no charge. Gamma rays

interact with material by colliding with the electrons in the shells of atoms. They lose their energy

slowly in material, being able to travel significant distances before stopping. Depending on their initial

energy, gamma rays can travel from 1 to hundreds of meters in air and can easily go right through

people.

It is important to note that most alpha and beta emitters also emit gamma rays as part of their decay

process. However, their is no such thing as a “pure” gamma emitter. Important gamma emitters

including technetium-99m which is used in nuclear medicine, and cesium-137 which is used for

calibration of nuclear instruments.

Half Life

Half-life is the time required for the quantity of a

radioactive material to be reduced to one-half its

original value.

All radionuclides have a particular half-life, some

of which a very long, while other are extremely

short. For example, uranium-238 has such a

long half life, 4.5x109 years, that only a small fraction has decayed since the earth was formed. In

contrast, carbon-11 has a half-life of only 20 minutes. Since this nuclide has medical applications, it

has to be created where it is being used so that enough will be present to conduct medical studies.


The Learning Goal for this assignment is:

Distinguish between bonding forces holding compound

together and other attractive forces, including hydrogen bonding

and van der waald forces.

Introduction to Ionic Compounds

Those molecules that consist of charged ions with opposite charges are called IONIC. These ionic

compounds are generally solids with high melting points and conduct electrical current. Ionic

compounds are generally formed from metal and a non-metal elements. See Ionic Bonding below.

Ionic Compound Example

For example, you are familiar with the fairly benign unspectacular behavior of common white

crystalline table salt (NaCl). Salt consists of positive sodium ions (Na + ) & negative chloride ions (Cl - ).

On the other hand the element sodium is a silvery gray metal composed of neutral atoms which react

vigorously with water or air. Chlorine as an element is a neutral greenish-yellow, poisonous, diatomic

gas (Cl2).

The main principle to remember is that ions are completely different in physical and chemical

properties from the neutral atoms of the elements.

The notation of the + and - charges on ions is very important as it conveys a definite meaning.

Whereas elements are neutral in charge, IONS have either a positive or negative charge depending

upon whether there is an excess of protons (positive ion) or excess of electrons (negative ion).

Formation of Positive Ions

Metals usually have 1-4 electrons in the outer energy level. The electron arrangement of a rare gas is

most easily achieved by losing the few electrons in the newly started energy level. The number of

electrons lost must bring the electron number "down to" that of a prior rare gas.

How will sodium complete its octet?

First examine the electron arrangement of the atom. The atomic number is eleven, therefore, there

are eleven electrons and eleven protons on the neutral sodium atom. Here is the Bohr diagram and

Lewis symbol for sodium:


This analysis shows that sodium has only one electron in its outer level. The nearest rare gas is neon

with 8 electron in the outer energy level. Therefore, this electron is lost so that there are now eight

electrons in the outer energy level, and the Bohr diagrams and Lewis symbols for sodium ion and

neon are identical. The octet rule is satisfied.

Ion Charge?

What is the charge on sodium ion as a result of losing one electron? A comparison of the atom and

the ion will yield this answer.

Sodium Atom

Sodium Ion

11 p+ to revert to 11 p + Protons are identical in

12 n an octet 12 n

the atom and ion.

Positive charge is

11 e- lose 1 electron 10 e-

caused by lack of

0 charge + 1 charge

electrons.

Formation of Negative Ions

How will fluorine complete its octet?

First examine the electron arrangement of the atom. The atomic number is nine, therefore, there are

nine electrons and nine protons on the neutral fluorine atom. Here is the Bohr diagram and Lewis

symbol for fluorine:

This analysis shows that fluorine already has seven electrons in its outer level. The nearest rare gas

is neon with 8 electron in the outer energy level. Therefore only one additional electron is needed to

complete the octet in the fluorine atom to make the fluoride ion. If the one electron is added, the Bohr

diagrams and Lewis symbols for fluorine and neon are identical. The octet rule is satisfied.


Ion Charge?

What is the charge on fluorine as a result of adding one electron? A comparison of the atom and the

ion will yield this answer.

Fluorine Atom Fluoride Ion *

9 p+ to complete 9 p + Protons are identical in

10 n octet 10 n

9 e- add 1 electron 10 e-

0 charge - 1 charge

the atom and ion.

Negative charge is

caused by excess

electrons

* The "ide" ending in the name signifies a simple negative ion.

Summary Principle of Ionic Compounds

An ionic compound is formed by the complete transfer of electrons from a metal to a nonmetal and

the resulting ions have achieved an octet. The protons do not change. Metal atoms in Groups 1-3

lose electrons to non-metal atoms with 5-7 electrons missing in the outer level. Non-metals gain 1-4

electrons to complete an octet.

Octet Rule

Elemental atoms generally lose, gain, or share electrons with other atoms in order to achieve the

same electron structure as the nearest rare gas with eight electrons in the outer level.

The proper application of the Octet Rule provides valuable assistance in predicting and explaining

various aspects of chemical formulas.

Introduction to Ionic Bonding

Ionic bonding is best treated using a simple

electrostatic model. The electrostatic model

is simply an application of the charge

principles that opposite charges attract and

similar charges repel. An ionic compound

results from the interaction of a positive and

negative ion, such as sodium and chloride in

common salt.

The IONIC BOND results as a balance

between the force of attraction between

opposite plus and minus charges of the ions

and the force of repulsion between similar

negative charges in the electron clouds. In

crystalline compounds this net balance of

forces is called the LATTICE ENERGY.

Lattice energy is the energy released in the

formation of an ionic compound.

DEFINITION: The formation of an IONIC

BOND is the result of the transfer of one or

more electrons from a metal onto a nonmetal.


Metals, with only a few electrons in the outer energy level, tend to lose electrons most readily. The

energy required to remove an electron from a neutral atom is called the IONIZATION POTENTIAL.

Energy + Metal Atom ---> Metal (+) ion + e-

Non-metals, which lack only one or two electrons in the outer energy level have little tendency to lose

electrons - the ionization potential would be very high. Instead non-metals have a tendency to gain

electrons. The ELECTRON AFFINITY is the energy given off by an atom when it gains electrons.

Non-metal Atom + e- --- Non-metal (-) ion + energy

The energy required to produce positive ions (ionization potential) is roughly balanced by the energy

given off to produce negative ions (electron affinity). The energy released by the net force of

attraction by the ions provides the overall stabilizing energy of the compound.

Notes Section:


The Learning Goal for this assignment is:

Introduction to Covalent Bonding:

Bonding between non-metals consists of two electrons shared between two atoms. Using the Wave

Theory, the covalent bond involves an overlap of the electron clouds from each atom. The electrons

are concentrated in the region between the two atoms. In covalent bonding, the two electrons shared

by the atoms are attracted to the nucleus of both atoms. Neither atom completely loses or gains

electrons as in ionic bonding.

There are two types of covalent bonding:

1. Non-polar bonding with an equal sharing of electrons.

2. Polar bonding with an unequal sharing of electrons. The number of shared electrons depends on

the number of electrons needed to complete the octet.

NON-POLAR BONDING results when two identical non-metals equally share electrons between

them. One well known exception to the identical atom rule is the combination of carbon and hydrogen

in all organic compounds.

Hydrogen

The simplest non-polar covalent molecule is hydrogen. Each hydrogen

atom has one electron and needs two to complete its first energy level.

Since both hydrogen atoms are identical, neither atom will be able to

dominate in the control of the electrons. The electrons are therefore

shared equally. The hydrogen covalent bond can be represented in a

variety of ways as shown here:

The "octet" for hydrogen is only 2 electrons since the nearest rare gas is

He. The diatomic molecule is formed because individual hydrogen atoms

containing only a single electron are unstable. Since both atoms are

identical a complete transfer of electrons as in ionic bonding is

impossible.

Instead the two hydrogen atoms SHARE both electrons equally.

Oxygen

Molecules of oxygen, present in about 20% concentration in air are

also covalent molecules. See the graphic on the left of the Lewis Dot

Structure.

There are 6 electrons in the outer shell, therefore, 2 electrons are

needed to complete the octet. The two oxygen atoms share a total of

four electrons in two separate bonds, called double bonds.

The two oxygen atoms equally share the four electrons.


POLAR BONDING results when two different non-metals unequally share electrons between them.

One well known exception to the identical atom rule is the combination of carbon and hydrogen in all

organic compounds.

The non-metal closer to fluorine in the Periodic Table has a greater tendency to keep its own electron

and also draw away the other atom's electron. It is NOT completely successful. As a result, only

partial charges are established. One atom becomes partially positive since it has lost control of its

electron some of the time. The other atom becomes partially negative since it gains electron some of

the time.

Hydrogen Chloride

Hydrogen Chloride forms a polar covalent molecule. The graphic

on the left shows that chlorine has 7 electrons in the outer shell.

Hydrogen has one electron in its outer energy shell. Since 8

electrons are needed for an octet, they share the electrons.

However, chlorine gets an unequal share of the two electrons,

although the electrons are still shared (not transferred as in ionic

bonding), the sharing is unequal. The electrons spends more of the

time closer to chlorine. As a result, the chlorine acquires a "partial"

negative charge. At the same time, since hydrogen loses the

electron most - but not all of the time, it acquires a "partial" charge.

The partial charge is denoted with a small Greek symbol for delta.

Water

Water, the most universal compound on all of the earth, has the property of

being a polar molecule. As a result of this property, the physical and

chemical properties of the compound are fairly unique.

Dihydrogen Oxide or water forms a polar covalent molecule. The graphic on

the left shows that oxygen has 6 electrons in the outer shell. Hydrogen has

one electron in its outer energy shell. Since 8 electrons are needed for an

octet, they share the electrons.

Notes Section:

1. Count the Valence e-

2.Find central atom and bond other atom to it. Submit bonds from total and put lone pairs and double bonds and

triple bonds as needed.

3. Find formal charges. Try tp get as close to zero as possible.


C 2 H 6 O Ethanol CH 3 CH 2 O

Step 1

Find valence e- for all atoms. Add them together.

C: 4 x 2 = 8

H: 1 x 6 = 6

O: 6

Total = 20

Step 2

Find octet e- for each atom and add them together.

C: 8 x 2 = 16

H: 2 x 6 = 12

O: 8

Total = 36

Step 3

Subtract Step 1 total from Step 2.

Gives you bonding e-.

36 – 20 = 16e-

Step 4

Find number of bonds by diving the number in step 3 by 2

(because each bond is made of 2 e-)

16e- / 2 = 8 bond pairs

These can be single, double or triple bonds.

Step 5

Determine which is the central atom

Find the one that is the least electronegative.

Use the periodic table and find the one farthest

away from Fluorine or

The one that only has 1 atom.


Step 6

Put the atoms in the structure that you think it will

have and bond them together.

Put Single bonds between atoms.

Step 7

Find the number of nonbonding (lone pairs) e-.

Subtract step 3 number from step 1.

20 – 16 = 4e- = 2 lone pairs

Step 8

Complete the Octet Rule by adding the lone

pairs.

Then, if needed, use any lone pairs to make

double and triple bonds so that all atoms meet

the Octet Rule.

See Step 4 for total number of bonds.


Linear

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp AX 2 None 180

BeCl 2

Beryllium Dichloride

CI

Be

CI

element bond lone pair

C


Trigonal Planar

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp 2 AX 3 None 120

BF 3

Boron Trifluoride

F

B

F

F

element bond lone pair

C


Bent

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp 3 AX 2 E 2 2 104.5

OF 2

Oxygen Difluoride

O

F

F

element bond lone pair

C


Tetrahedral

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp 3 AX 2 None 109.5

Phosphate

PO 4

3-

O

O

P

O

O

element bond lone pair

C


Bent

Molecular Geometry

Orbital Equation Lone Pairs Angle

Sp 3 AX 2 E 2 1 107

Phosphorus

Trihydride

H

PH 3

P H

C

element bond lone pair

C


Bent

Molecular Geometry

Orbital Equation Lone Pairs Angle

Sp 2 AX 2 E 1 116

Trioxide

O 3

O O O

element bond lone pair

C


Trigonal Bi Pyramidal

Molecular Geometry

Orbital Equation Lone Pairs Angle

Sp 3 d AX 5 None 120/90

Phosphorus

pentachloride

PCl 5

Cl

Cl

Cl

P

Cl

Cl

element bond lone pair

C


T-Shaped

Molecular Geometry

Orbital Equation Lone Pairs Angle

Sp 3 d Ax 3 E 2 2 90

Chlorine

Trifluoride

ClF 3

F

Cl

F

F

element bond lone pair

C


Octahedral

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp 3 d 2 AX 6 None 90

Sulfur hexafluoride

SF 6

F

F

F

S

F

F

F

element bond lone pair

C


Square Planar

Molecular Geometry

Orbital Equation Lone Pairs Angle

sp 3 d 2 AX 4 E 2 2 90

Iodine Tetrachloride ion

ICl 4

-

C

C

I

C

C

element bond lone pair

C


Orbitals Equation Lone Pairs Angle

Name

sp AX2 None 180

Linear

sp 2 AX3 None 120

Trigonal Planar

sp 2 AX2E 1 116

Bent

sp 3 AX4 None 109.5

Tetrahedral

sp 3 AX3E 1 107

Trig. Pyramidal

sp 3 AX2E2 2 104.5

Bent

sp 3 d AX5 None 120/90

Trig. Bipyramidal

sp 3 d AX3E2 2 90

T-Shaped

sp 3 d 2 AX6 None 90

Octahedral

sp 3 d AX4E2 1 90

Square Planar


Name Formula Charge

Dichromate Cr₂O₇ 2-

Sulfate SO₄ 2-

Hydrogen Carbonate HCO₃ 1-

Hypochlorite ClO 1-

Phosphate PO₄ 3-

Nitrite NO₂ 1-

Chlorite ClO₂ 1-

Dihydrogen phosphate H₂PO₄ 1-

Chromate CrO₄ 2-

Carbonate CO₃ 2-

Hydroxide OH 1-

Hydrogen phosphate HPO₄ 2-

Ammonium NH₄ 1+

Acetate C₂H₃O₂ 1-

Perchlorate ClO₄ 1-

Permanganate MnO₄ 1-

Chlorate ClO₃ 1-

Hydrogen Sulfate HSO₄ 1-

Phosphite PO₃ 3-

Sulfite SO₃ 2-

Silicate SiO₃ 2-

Nitrate NO₃ 1-

Hydrogen Sulfite HSO₃ 1-

Oxalate C₂O₄ 2-

Cyanide CN 1-

Hydronium H₃O 1+

Thiosulfate S₂O₃ 2-


Chapter 9

Unit 4

Chemical Names and Formulas

The students will learn how the periodic table helps them

determine the names and formulas of ions and compounds.

Chapter 22 Hydrocarbon Compounds

The student will learn how Hydrocarbons are named and the

general properties of Hydrocarbons.

Describe how different natural resources are produced and how their rates

of use and renewal limit availability.

Students will explore local, national, and global renewable and nonrenewable

resources.

Students will explain the environmental costs of the use of renewable and

nonrenewable resources.

Students will explain the benefits of renewable and nonrenewable resources.

Nuclear reactors

Natural gas

Petroleum

Refining

Coal


Chapter 23 Functional Groups

The student will learn what effects functional groups have on

organic compounds and how chemical reactions are used in

organic compounds.

Describe the properties of the carbon atom that make the diversity of carbon

compounds possible.

Identify selected functional groups and relate how they contribute to

properties of carbon compounds.

Students will identify examples of important carbon based molecules.

Students will create 2D or 3D models of carbon molecules and explain why this

molecule is important to life.

covalent bond

single bond

double bond

triple bond

monomer

polymer


http://www.bbc.co.uk/education/guides/zm9hvcw/revision

methane- is a natural gas used for cooking and heating

propane- is gas used in gas cylinders for BBQ etc

octane- used in petrol for cars

Parent Molecule- the longest unbranched

chain containing the functional group.

The position of the functional group is labled

with a number

Methane - Monsters

Ethane - Eat The rule is a comma between numbers, and a

Propane - Pupils dash between # and letters.

Butane - But

Pentane - Prefer

Hexane - Hairy

Alkenes- all end in -ene

Heptane - Haggis

all contain a carbon to carbon

Octane - Occasionally

double bond-unsaturated

only contain single bonds

saturated.

uses are fuels, solvents,

plastics, and vinegar


Unit 5

Chapter 10 Chemical Quantities

The student will learn why the mole is important and how the

molecular formula of a compound can be determined

experimentally.

Chapter 11 Chemical Reactions

The students will learn how chemical reactions obey the law of

conservation of mass and how they can predict the products

of a chemical reaction.

Characterize types of chemical reactions, for example: redox, acid-base,

synthesis, and single and double replacement reactions.

Students will be able to identify the type of chemical reaction that occurs.

Students will be able to compare/contrast reactants and products of various

types of chemical reactions.

Students will be able to predict the product of various reactants.

Students will be able to write balanced chemical equations for each type of

reaction.

Decomposition

Combustion

Redox

Acid-Base

Synthesis

single-replacement

double-replacement

Differentiate between chemical and nuclear reactions.

Students will compare/contrast chemical and nuclear reactions.

fission

fusion


Chapter 12 Stoichiometry

The students will learn how balanced chemical equations are

used in stoichiometric calculations and how to calculate

amounts of reactants and products in a chemical equation.

Apply the mole concept and the law of conservation of mass to calculate

quantities of chemicals participating in reactions.

Students will be able to use a balanced equation to determine mole ratios.

Students will be able to apply law of conservation of mass to chemical equations.

Students will be able to calculate empirical and molecular formulas.

Students will be able to calculate the % composition of a compound.

Students will be able to calculate theoretical yield.

Students will be able to calculate % error.

Students will be able to calculate molar mass.

Students will be able to perform stoichiometric calculations, including limiting

reagents.

mole

Avogadro’ s number

molar mass

gram formula mass


The Mole

www.youtube.com/watch?v=AsqEkF7hcII

www.youtube.com/watch?v=tEn0N4R2dqA

www.youtube.com/watch?v=Pft2CASl0M0

www.youtube.com/watch?v=rwhJklbK8R0

atomic mass unit (amu) = molar mass (g)


www.youtube.com/watch?v=BTRm8PwcZ3U

www.youtube.com/watch?v=F9NkYSKJifs

www.youtube.com/watch?v=xPdqEX_WMjo

Molar Mass


Categories of Reactions

All chemical reactions can be placed into one of six categories. Here they are, in no

particular order:

1) Synthesis: A synthesis reaction is when two or more simple compounds combine to form a

more complicated one. These reactions come in the general form of: A+B ---> AB

One example of a synthesis reaction is the combination of iron and sulfur to form iron (II) sulfide:

8 Fe + S8 ---> 8 FeS

If two elements or very simple molecules combine with each other, it’s probably a synthesis reaction.

The products will probably be predictable using the octet rule to find charges.

2) Decomposition: A decomposition reaction is the opposite of a synthesis reaction - a

complex molecule breaks down to make simpler ones. These reactions come in the general form:

AB ---> A+B

One example of a decomposition reaction is the electrolysis of water to make oxygen and hydrogen

gas:

2 H2O ---> 2 H2 + O2

If one compound has an arrow coming off of it, it’s probably a decomposition reaction. The products

will either be a couple of very simple molecules, or some elements, or both.

3) Single displacement: This is when one element trades places with another element in a

compound. These reactions come in the general form of:

One example of a single displacement reaction is when magnesium replaces hydrogen in water to

make magnesium hydroxide and hydrogen gas:

Mg + 2 H2O ---> Mg(OH)2 + H2

A+BC ---> B+ AC

If a pure element reacts with another compound (usually, but not always, ionic), it’s probably a single

displacement reaction. The products will be the compounds formed when the pure element switches

places with another element in the other compound.

Important note: these reactions will only occur if the pure element on the reactant side of the equation

is higher on the activity series than the element it replaces.


4) Double displacement: This is when the anions and cations of two different molecules

switch places, forming two entirely different compounds. These reactions are in the general form:

One example of a double displacement reaction is the reaction of lead (II) nitrate with potassium

iodide to form lead (II) iodide and potassium nitrate:

Pb(NO3)2 + 2 KI ---> PbI2 + 2 KNO3

AB + CD ---> AD + CB

If two ionic compounds combine, it’s probably a double displacement reaction. Switch the cations

and balance out the charges to figure out what will be made.

Important note: These reactions will only occur if both reactants are soluble in water and only one

product is soluble in water.

5) Acid-base: This is a special kind of double displacement reaction that takes place when an

acid and base react with each other. The H + ion in the acid reacts with the OH - ion in the base,

causing the formation of water. Generally, the product of this reaction is some ionic salt and water:

HA + BOH ---> AB + H20

One example of an acid-base reaction is the reaction of hydrobromic acid (HBr) with sodium

hydroxide:

HBr + NaOH ---> NaBr + H2O

If an acid and a base combine, it’s an acid-base reaction. The products will be an ionic compound

and water.

6) Combustion: A combustion reaction is when oxygen combines with another compound to

form water and carbon dioxide. These reactions are exothermic, meaning they produce heat. An

example of this kind of reaction is the burning of napthalene: Organic CBase + O2 ----> H20 + CO

C10H8 + 12 O2 ---> 10 CO2 + 4 H2O

If something that has carbon and hydrogen reacts with oxygen, it’s probably a combustion reaction.

The products will be CO2 and H2O.

Follow this series of questions. When you can answer "yes" to a question, then

stop!

1) Does your reaction have two (or more) chemicals combining to form one chemical? If yes, then it's

a synthesis reaction

2) Does your reaction have one large molecule falling apart to make several small ones? If yes, then

it's a decomposition reaction

3) Does your reaction have any molecules that contain only one element? If yes, then it's a single

displacement reaction

4) Does your reaction have water as one of the products? If yes, then it's an acid-base reaction

5) Does your reaction have oxygen as one of it's reactants and carbon dioxide and water as

products? If yes, then it's a combustion reaction

6) If you haven't answered "yes" to any of the questions above, then you've got a double

displacement reaction.


List what type the following reactions are:

1) NaOH + KNO3 --> NaNO3 + KOH

double displacement

2) CH4 + 2 O2 --> CO2 + 2 H2O

combustion

3) 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na

single displacement

4) CaSO4 + Mg(OH)2 --> Ca(OH)2 + MgSO4

double displacement

5) NH4OH + HBr --> H2O + NH4Br

Acid Base

6) Pb + O2 --> PbO2

Synthesis

7) Na2CO3 --> Na2O + CO2

Decompisition


Determine the Type of Reaction for each equation.

Then predict the products of each of the following chemical reactions. If a reaction will not occur,

explain why not.

Then Balance the equation.

1+ 2- 1+ 1- 1+ 1- 1+ 2-

1) __Ag2SO4 + __NaNO3 → 2AgNO3+__ Na2SO4

Ag 2

S 1

O7 (10)

Na1 (2)

N1 (2)

1+ 1- 2+ 2-

(2) 1 Ag

1 S

(10) 7 O

2 Na

(2)1 N

2) __NaI + __CaSO4 →

Na 1 (2)

I 1 (2)

Ca 1

S1

O4

2 Na

2 I

1 Ca

1 S

4 O

3) __HNO3 + __Ca(OH)2 → H2O+__ Ca(NO3)2

H 3 (4)

(4) 2 H

O 5 (8)

(8) 7 O

N 1 (2)

2 N

Ca 1

1 Ca

4) __CaCO3 → __ Ca + __ CO2

5) __AlCl3 + __(NH4)PO4 →

6) __Pb + __Fe(NO3)3 →

1+ 2- 2+ 1-

Na2SO4+__ CaI2

1+ 1- 2+ 1- 2+ 1-

Ca 1

C 1

O3

1 Ca

1 C

3 O

3+ 1- 1+ 3- 3+ 3- 1+ 1-

AlPO4 + 3NHACl

O 4

Al 1

Cl 3

N 3

H 12

P 1

4 O

1 Al

(3) Cl

3 1N

3 1H

1P

7) __C3H6 + __O2 →

8) __Na + __CaSO4 →


How to Balance Chemical Equations

A chemical equation is a theoretical or written representation of what happens during a chemical

reaction. The law of conservation of mass states that no atoms can be created or destroyed in a

chemical reaction, so the number of atoms that are present in the reactants has to balance the

number of atoms that are present in the products. Follow this guide to learn how to balance chemical

equations.

Step 1

Write down your given equation. For this example, we will use:

C3H8 + O2 --> H2O + CO2

Step 2

Write down the number of atoms that you have on each side of the equation. Look at the subscripts

next to each atom to find the number of atoms in the equation.

Left side: 3 carbon, 8 hydrogen and 2 oxygen

Right side: 1 carbon, 2 hydrogen and 3 oxygen


Step 3

Always leave hydrogen and oxygen for last. This means that you will need to balance the carbon

atoms first.

Step 4

Add a coefficient to the single carbon atom on the right of the equation to balance it with the 3 carbon

atoms on the left of the equation.

C3H8 + O2 --> H2O + 3CO2

The coefficient 3 in front of carbon on the right side indicates 3 carbon atoms just as the subscript 3

on the left side indicates 3 carbon atoms.

In a chemical equation, you can change coefficients, but you should never alter the subscripts.


Step 5

Balance the hydrogen atoms next. You have 8 on the left side, so you'll need 8 on the right side.

C3H8 + O2 --> 4H2O + 3CO2

On the right side, we added a 4 as the coefficient because the subscript showed that we already

had 2 hydrogen atoms.

When you multiply the coefficient 4 times the subscript 2, you end up with 8.

Step 6

Finish by balancing the oxygen atoms.

Because we've added coefficients to the molecules on the right side of the equation, the number of

oxygen atoms has changed. We now have 4 oxygen atoms in the water molecule and 6 oxygen

atoms in the carbon dioxide molecule. That makes a total of 10 oxygen atoms.

Add a coefficient of 5 to the oxygen molecule on the left side of the equation. You now have 10

oxygen molecules on each side.

C3H8 + 5O2 --> 4H2O + 3CO2.

The carbon, hydrogen and oxygen atoms are balanced. Your equation is complete.


1) ___ NaNO3 + ___ PbO ___ Pb(NO3)2 + ___ Na2O

Na 1(2)

N 1(2)

O 4(7)

Pb 1(1)

2 Na

2 N

7 O

1 Pb

2) ___ AgI + ___ Fe2(CO3)3 ___ FeI3 + ___ Ag2CO3

Ag 1

I 1

Fe 2

C 3

O 9

2 Ag

(6)3 I

2)1 Fe

1 C

3 O

3) ___ C2H4O2 + ___ O2 ___ CO2 + ___ H2O

C 2

O 6

H 4

(2) 1C

(6) 4 2 O

4 H

4) ___ ZnSO4 + ___ Li2CO3 ___ ZnCO3 + ___ Li2SO4

C 1

O 7

S 1

Li 2

Zn 1

1 C

7 O

1 S

2 Li

1 Zn

5) ___ V2O5 + ___ CaS ___ CaO + ___ V2S5

Ca 1 (5)

O 5

V 2

S 1 (5)

(5) 1 Ca

(5) 1 O

2 V

5 S


6) ___ Mn(NO2)2 + ___ BeCl2 ___ Be(NO2)2 + ___ MnCl2

7) ___ AgBr + ___ GaPO4 ___ Ag3PO4 + ___ GaBr3

8) ___ H2SO4 + ___ B(OH)3 __ B2(SO4)3 + ___ H2O

9) ___ S8 + ___ O2 ___ SO2

10) ___ Fe + ___ AgNO3 ___ Fe(NO3)2 + ___ Ag


1) 2 NaNO3 + PbO Pb(NO3)2 + Na2O

2) 6 AgI + Fe2(CO3)3 2 FeI3 + 3 Ag2CO3

3) C2H4O2 + 2 O2 2 CO2 + 2 H2O

4) ZnSO4 + Li2CO3 ZnCO3 + Li2SO4

5) V2O5 + 5 CaS 5 CaO + V2S5

6) Mn(NO2)2 + BeCl2 Be(NO2)2 + MnCl2

7) 3 AgBr + GaPO4 Ag3PO4 + GaBr3

8) 3 H2SO4 + 2 B(OH)3 B2(SO4)3 + 6 H2O

9) S8 + 8 O2 8 SO2

10) Fe + 2 AgNO3 Fe(NO3)2 + 2 Ag

Additional Notes:

Always start with carbon.


The Learning Goal for this assignment is:

Stoichiometry and Balancing Reactions

Stoichiometry is a section of chemistry that involves using relationships between reactants and/or

products in a chemical reaction to determine desired quantitative data. In Greek, stoikhein means

element and metron means measure, so stoichiometry literally translated means the measure of

elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to

first understand the relationships that exist between products and reactants and why they exist, which

require understanding how to balanced reactions.

Balancing

In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The

reactants are displayed on the left side of the equation and the products are shown on the right, with

the separation of either a single or double arrow that signifies the direction of the reaction. The

significance of single and double arrow is important when discussing solubility constants, but we will

not go into detail about it in this module. To balance an equation, it is necessary that there are the

same number of atoms on the left side of the equation as the right. One can do this by raising the

coefficients.

Reactants to Products

A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a

chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as

well as their states, and the proportion for how much of each particle is create relative to one another,

through the stoichiometric coefficient. The following equation demonstrates the typical format of a

chemical equation:

2Na(s) + 2HCl(aq) → 2NaCl(aq) + H2(g)

In the above equation, the elements present in the reaction are represented by their chemical

symbols. Based on the Law of Conservation of Mass, which states that matter is neither created nor

destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and

products, though the elements they are paired up with often change in a reaction. In this reaction,

sodium (Na), hydrogen (H), and chloride (Cl) are the elements present in both reactants, so based on

the law of conservation of mass, they are also present on the product side of the equations.

Displaying each element is important when using the chemical equation to convert between

elements.

Stoichiometric Coefficients

In a balanced reaction, both sides of the equation have the same number of elements. The

stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical

reaction to balance the number of each element on both the reactant and product sides of the

equation. These stoichiometric coefficients are useful since they establish the mole ratio between

reactants and products. In the balanced equation:

2Na(s)+2HCl(aq)→2NaCl(aq)+H2(g)


we can determine that 2 moles of HCl will react with 2 moles of Na(s) to form 2 moles of NaCl(aq) and 1

mole of H2(g). If we know how many moles of Na we start out with, we can use the ratio of 2 moles

of NaCl to 2 moles of Na to determine how many moles of NaCl were produced or we can use the

ration of 1 mole of H2 to 2 moles of Na to convert to NaCl. This is known as the coefficient factor. The

balanced equation makes it possible to convert information about one reactant or product to

quantitative data about another element. Understanding this is essential to solving stoichiometric

problems.

Example 1

Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction.

Solution

___Pb(OH)4 +___H2SO4→___Pb(SO4)2 +___H2O

Start by counting the number of atoms of each element.

Unbalanced

Pb 1 1 Pb

O 8 9 O

H 6 2 H

S 1 2 S

The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does

not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the

equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a

coefficient of 2 should be added in front of H2SO4to have an equal number of sulfur on both sides of

the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4

coefficient should be added in front of H2O where there is a deficiency of oxygen. Count the number

of elements now present on either side of the equation. Since the numbers are the same, the

equation is now balanced.

Pb(OH)4 + 2H2SO4→ Pb(SO4)2 + 4H2O

Balanced

Pb 1 1 Pb

O 8 12 12 9 O

H 6 8 8 2 H

S 1 2 2 2 S

Balancing reactions involves finding least common multiples between numbers of elements present

on both sides of the equation. In general, when applying coefficients, add coefficients to the

molecules or unpaired elements last.

A balanced equation ultimately has to satisfy two conditions.

1. The numbers of each element on the left and right side of the equation must be equal.

2. The charge on both sides of the equation must be equal. It is especially important to pay

attention to charge when balancing redox reactions.


Stoichiometry and Balanced Equations

In stoichiometry, balanced equations make it possible to compare different elements through the

stoichiometric factor discussed earlier. This is the mole ratio between two factors in a chemical

reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show

how stoichiometric factors are useful.

Example 2

There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How

many party invitations can be sent?

Solution

The equation for this can be written as

I+2S→IS2

where

I represent invitations,

S represents stamps, and

IS 2 represents the sent party invitations consisting of one invitation and two stamps.


Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation.

Invitations Stamps Party Invitations Sent

In this example are all the reactants (stamps and invitations) used up? No, and this is normally the

case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the

one that runs out first, prevents the reaction from continuing and determines the maximum amount of

product that can be formed.

Example 3

What is the limiting reagent in this example?

Solution

Stamps, because there was only enough to send out invitations, whereas there were enough

invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can

be solved using stoichiometric factors.

12 I x 1IS2 = 12 IS2 possible

1I

20 S x 1IS2 = 10 IS2 possible

2S


When there is no limiting reagent because the ratio of all the reactants caused them to run out at the

same time, it is known as stoichiometric proportions.

Types of Reactions

There are 6 basic types of reactions.

Combustion: Combustion is the formation of CO2 and H2O from the reaction of a chemical

and O2

Combination (synthesis): Combination is the addition of 2 or more simple reactants to form a

complex product.

Decomposition: Decomposition is when complex reactants are broken down into simpler

products.

Single Displacement: Single displacement is when an element from on reactant switches with

an element of the other to form two new reactants.

Double Displacement: Double displacement is when two elements from on reactants

switched with two elements of the other to form two new reactants.

Acid-Base: Acid- base reactions are when two reactants form salts and water.

Molar Mass

Before applying stoichiometric factors to chemical equations, you need to understand molar mass.

Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual

element as listed in the periodic table established this relationship for atoms or ions. For compounds

or molecules, you have to take the sum of the atomic mass times the number of each atom in order to

determine the molar mass.

Example 4

What is the molar mass of H2O?

Solution

Molar mass = 2 × (1.00g/mol) + 1×(16.0g/mol) = 18.0g/mol

Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of

products or vice versa.

Example 5: Combustion of Propane

Propane (C3H8) burns in this reaction:

C3H8 + 5O2 → 4H2O + 3CO2

If 200 g of propane is burned, how many g of H2Ois produced?

Solution

Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products

you must convert from grams of C3H8 to moles of C3H8 then from moles of C3H8 to moles of H2O.

Then convert from moles of H2O to grams of H2O.


Step 1: 200g C3H8 is equal to 4.54 mol C3H8.

Step 2: Since there is a ratio of 4:1 H2O to C3H8, for every 4.54 mol C3H8 there are 18.18 mol

H2O.

Step 3: Convert 18.18 mol H2O to g H2O 18.18 mol H2O is equal to 327.27 g H2O.

Variation in Stoichiometric Equations

Almost every quantitative relationship can be converted into a ratio that can be useful in data

analysis.

Density

Density (ρ) is calculated as mass/volume. This ratio can be useful in determining the volume of a

solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse

relationship would be used.

Volume x (Mass/Volume) = Mass

Mass x (Volume/Mass) = Volume

Percent Mass

Percent establishes a relationship as well. A percent mass states how many grams of a mixture are of

a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams

are of the stated element or compound. This is useful in determining mass of a desired substance in

a molecule.

Example 6

A substance is 5% carbon by mass. If the total mass of the substance is 10.0 grams, what is the

mass of carbon in the sample? How many moles of carbon are there?

Solution

10 g sample x 5 g carbon = 0.5 g carbon

100 g sample

0.5g carbon x 1 mol carbon = 0.0416 mol carbon

12.0g carbon

Molarity

Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it

is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical

equations and dilutions.


Example 7

How much 5M stock solution is needed to prepare 100 mL of 2M solution?

Solution

100 mL of dilute solution (1 L/1000 mL) (2 mol/1L solution) (1 L stock solution/5 mol solution) (1000

ml stock solution/1L stock solution) = 40 mL stock solution.

These ratios of molarity, density, and mass percent are useful in complex examples ahead.

Determining Empirical Formulas

An empirical formula can be determined through chemical stoichiometry by determining which

elements are present in the molecule and in what ratio. The ratio of elements is determined by

comparing the number of moles of each element present.

Example 8

1. Find the molar mass of the empirical formula CH2O.

12.0g C + (1.00g H) * (2H) + 16.0g O = 30.0 g/mol CH2O

2. Determine the molecular mass experimentally. For our compound, it is 120.0 g/mol.

3. Divide the experimentally determined molecular mass by the mass of the empirical formula.

(120.0 g/mol) / (30.0 g/mol) = 3.9984

4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a

slight error in the experimentally determined molecular mass. If the answer is not close to a whole

number, there was either an error in the calculation of the empirical formula or a large error in the

determination of the molecular mass.

5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular

formula.

CH2O * 4 =?

C: 1 * 4 = 4

H: 2 * 4 = 8

O 1 * 4 = 4

CH2O * 4 = C4H8O4

6. Check your result by calculating the molar mass of the molecular formula and comparing it to the

experimentally determined mass.

molar mass of C4H8O4= 120.104 g/mol

experimentally determined mass = 120.056 g/mol

% error = | theoretical - experimental | / theoretical * 100%

% error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100%

% error = 0.040 %


Stoichiometry and balanced equations make it possible to use one piece of information to calculate

another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and

see if you can use what you learned to solve the following problems.

Problem 1

Why are the following equations not considered balanced?

a. H2O(l)→H2(g)+O2(g)

b. Zn(s)+Au + (aq) →Zn 2+ (aq) +Ag(s)

Problem 2

Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions.

Write the balanced chemical equation for this reaction.

Problem 3

Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M

solution?

Problem 4

If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the

limiting reagent and how many moles of water are produced? The unbalanced equation is provided

below.

CH4(g)+O2(g)→CO2(g)+H2O(l)


Theoretical and Actual Yields

Key Terms




(Excess reagent, limiting reagent)

Theoretical and actual yields

Percentage or actual yield

Skills to Develop

Use stoichiometric calculation to determine excess and limiting reagents in a chemical reaction



and explain why.

Calculate theoretical yields of products formed in reactions that involve limiting reagents.

Evaluate percentage or actual yields from known amounts of reactants

Theoretical and Actual Yields

Reactants not completely used up are called excess reagents, and the reactant that completely

reacts is called the limiting reagent. This concept has been illustrated for the reaction:

2Na+Cl2 →2NaCl

Amounts of products calculated from the complete reaction of the limiting reagent are called

theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of

actual yield to theoretical yield expressed in percentage is called the percentage yield.

percent yield = actual yield / theoretical yield ×100

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products.

Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical

reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the

process or inefficiency of the chemical reaction.

Example 1

Methyl alcohol can be produced in a high-pressure reaction

CO(g) + 2H2(g) →CH3OH(l)

If 6.1 metric tons of methyl alcohol is obtained from 1.2 metric tons of hydrogen reacting with excess

amount of CO, estimate the theoretical and the percentage yield?

Hint:

To calculate the theoretical yield, consider the reaction

CO(g) + 2H2(g) → CH3OH(l)

28.0 + 4.0 = 32.0 (stoichiometric masses in, g, kg, or tons)

1.2 tons H2 × 32.0 CH3OH = 9.6 tons CH3OH

4.0 H2

Thus, the theoretical yield from 1.2 metric tons (1.2x10 6 g) of hydrogen gas is 9.6 tons. The actual

yield is stated in the problem, 6.1 metric tons. Thus, the percentage yield is


%yield = 6.1 tons × 100 = 64%

9.6tons

Discussion

Due to chemical equilibrium or the mass action law, the limiting reagent may not be completely

consumed. Thus, a lower yield is expected in some cases. Losses during the recovery process of the

product will cause an even lower actual yield.

Example 2

A solution containing silver ion, Ag + , has been treated with excess of chloride ions Cl − . When dried,

0.1234 g of AgCl was recovered. Assuming the percentage yield to be 98.7%, how many grams of

silver ions were present in the solution?

Hint:

The reaction and relative masses of reagents and product are:

The calculation,

Ag + (aq) + Cl − (aq) → AgCl(s)

107.868 + 35.453 = 143.321

0.1234 g AgCl ×107.868 g Ag + =0.09287 g Ag +

143.321g AgCl

shows that 0.1234 g dry AgCl comes from 0.09287g Ag + ions. Since the actual yield is only 98.7%,

the actual amount of Ag + ions present is therefore

0.09287 g Ag + = 0.09409 g Ag +

0.987

Discussion

One can also calculate the theoretical yield of AgCl from the percentage yield of 98.7% to be

0.1234 g AgCl =0.1250 g AgCl

0.987

From 0.1250 g AgCl, the amount of Ag + present is also 0.09409 g.

Stoichiometry - A Review

Skills Taught








evaluate molecular weight for a given formula

evaluate weight (mass) percentages of elements for a given formula

evaluate amounts (in mass and mole units) produced in a chemical reaction from given

conditions

classify reactions by types: combination, combustion, displacement, formation, etc

determine the chemical formula when weight percentages are given and then evaluate the

mole percentages of elements in the formula

determine the chemical formula when weight percentages are given and molecular weight is

known

determine the amount produced, the actual yield, and other stoichiometry quantities for a given

reaction


Review Purposes






To get an overall view of stoichiometry.

Apply skills learned to perform quantitative chemical analysis.

Apply theories and rules of chemistry to solve problems.

Assess areas of strength and weakness for review purposes.

Improve problem solving strategy and learning efficiency.

Stoichiometry

STOICHIOMETRY is the quantitative relationship of reactants and products. This unit has been

divided into the following objects. A brief review is given here so that you can get a birds'-eye or

overall view of stoichiometry.

1.Amounts of substances

Express amounts of substance in mass units of g, kg, tons, and convert them to moles,

kilomoles, or millimoles.

2.Chemical formulas

Represent a substance with a formula that reflects its chemical composition, structure, and

bonding; evaluate weight and mole percentages of elements in a substance; and determine

chemical formula by elemental analysis.

3.Reaction features

Define some common features of chemical reactions; classify chemical reactions by common

features such as combination, combustion, decomposition, displacement, and redox reactions.

4.Reaction equations

Express quantitative relationships using chemical reaction equations; evaluate quantities of

reactants and products in a chemical reaction; and solve reaction stoichiometry problems.

5.Excess and limiting reagents

Define excess and limiting reagents; determine excess and limiting reagents in a reaction

mixture; and determine quantities produced in a chemical reaction.

6.Yields

Define theoretical and actual yields due to limiting reagent; apply the concept of limiting

reagent to evaluate theoretical yield; convert actual yield to percentage yields.


Use the space provided to write out the steps you take to solve different types of problems.

Use any additional space for notes. These 2 pages should be full when you turn in your notebook.

Steps to solve problems

Mole to Mole:


Unit 6

Chapter 13 States of Matter

The students will learn what are the factors that determine and

characteristics that distinguish gases liquids and solids and

how substances change from one state to another.

Differentiate among the four states of matter.

Students will measure the physical characteristics of matter such as temperature

and density.

Students will compare and contrast the physical characteristics of the 4 states of

matter.

solid

liquid

gas

plasma

Relate temperature to the average molecular kinetic energy.

Students will be able to compare and contrast the motion of particles of a sample

at various temperatures.

Kinetic energy

Kinetic theory

Temperature

Describe phase transitions in terms of kinetic molecular theory.

Students will be able to identify and describe phase changes.

Students will be able to compare and contrast the change in particle motion

for phase changes.

Students will be able to interpret heating/cooling curves and phase diagrams.

melting point

freezing point

boiling point

condensation

sublimation

phase diagram

kinetic molecular theory


Chapter 14 The Behavior of Gases

The students will learn how gases respond to changes in

pressure, volume, and temperature and why the ideal gas law

is useful even though ideal gases do not exist.

Interpret the behavior of ideal gases in terms of kinetic molecular theory.

Students will be able to describe the behavior of an ideal gas.

Students will participate in activities to apply the Ideal Gas Law and its

component laws to predict gas behavior.

Students will be able to perform temperature/pressure conversions.

Compressibility

Boyle’s Law

Charles’s Law

Gay-Lussac’s Law

Combine Gas Law

Ideal Gas Law

Partial pressure

Dalton’s Law of partial pressure

Diffusion

Effusion

Graham’s Law of effusion

Chapter 15 Water and Aqueous Systems

The students will learn how the interactions between water

molecules account for the unique properties of water and how

aqueous solutions form.

Discuss the special properties of water that contribute to Earth's suitability

as an environment for life: cohesive behavior, ability to moderate

temperature, expansion upon freezing, and versatility as a solvent.

Students will be able to prepare a solution of known molarity

Students will participate in activities to calculate molarity

Surface tension

Surfactant

Aqueous solutionSolvent

Solute


The Learning Goal for this assignment is:

Take note over the following chapter. Use the Headings provided to organize your notes. Define and number all highlighted vocabulary (total 23 ) as well

as summarize the sections. You may add pictures where needed. The pictures should be an appropriate size. Use Arial 12 for all text. This document

should be 3 pages and should be saved as a pdf before you submit it into Angel.

13.1 The Nature of Gases

Chapter 13 States of Matter

Pages 420 - 439

Kinetic Theory and a Model for Gases

a. What are the three assumptions of the kinetic theory as it applies to gases?

The particles in a gas are considered to be small, hard spheres with an insignificant volume.

The motion of the particles in a gas is rapid, constant, and random.

All collisions between particles in a gas are perfectly elastic

In elastic collision kinetic energy is transferred without loss from one particle to another

The aimless paths of the molecules in a gas take is called a “random walk”.

Within a gas there are no attractive or repulsive forces existing between the particles.

Kinetic energy- the energy an object has because of its motion

Kinetic theory- all matter consists of tiny particles that are in constant motion.

Gas Pressure

a. How does kinetic theory explain gas pressure?

Gas pressure is the result of billions of rapidly moving particles in a gas simultaneously colliding with

an object.

Although one single particle exerts a tiny amount of force it is the gas in the moving body that creates

the measurable force, just like how cells contribute to the whole which is the organism.

If there aren’t collisions between the particles present, then there is no pressure involved.

Atmospheric pressure decreases as a person climbs a mountain because the density of Earths’

atmosphere decreases as the elevation increases, and therefore there are less particles colliding

together.

Gas pressure- results from the force exerted by a gas per unit surface area of an object

Vacuum- is an empty space with no particles, and no pressure.

Atmospheric pressure- is the collisions of atoms and molecules in air with objects.

Barometer- is a device that is used to measure atmospheric pressure.

Pascal (Pa)- is the SI unit of pressure.

Standard atmosphere (atm)- is the pressure required to support 760 mm of mercury in a mercury

barometer at 25°C.

1 atm= 760 mm Hg = 101.3 kPa

Kinetic Energy and Temperature

a. What is the relationship between the temperature in kelvins and the average kinetic energy of

particles?

The Kelvin temperature of a substance is directly proportional to the average kinetic energy of the

particles of the substance.


Average kinetic energy is related to the state in which a substance is in because it has to do with

temperature and movement of particles.

The effects of temperature on a particle motion in liquids and solids are more complex than in gases.

Summary: In this chapter we learn about the fundamental and basic information and rules that

accompany the different states are matter. Gases have a high kinetic energy that is why the particles

that make them up are always in constant motion colliding with each other and creating gas pressure.

Also, kinetic energy is something directly tied to temperature of substances, as average kinetic

energy increases so does temperature, and vice versa. There are also unique characteristics of this

measurement because even though the same element that makes up three substances can have the

same amount of average kinetic energy; it is possible for each one to be in a different physical state.

13.2 The Nature of Liquids

A Model of Liquids

a. What factors determine the physical properties of a liquid?

The interplay between the disruptive motions of particles in a liquid and the attractions among the

particles determines the physical properties of liquids.

Fluids are substance that flow, and can conform, or fill in the shape of their containers.

In contrary to gases, in liquids there is a molecular attraction between the particles which keeps the

particles together, and they are much denser than gases.

Liquids have a definite volume.

Evaporation

a. What is the relationship between evaporation and kinetic energy?

During evaporation, only those molecules with a certain minimum kinetic energy can escape from the

surface of the liquid.

The reason that this process occurs is because most molecules in a liquid don’t have enough kinetic

energy to overcome the attractive forces and escape transforming into gases.

Heat and temperature also play a role in evaporation since the amount of energy gained by a particle

increases as temperature increases, making it move faster, much like how in fission the particles start

to bolt into different directions outside their specified routes.

Vaporization- is the conversion of a liquid to a gas or vapor.

Evaporation- is vaporization that occurs at the surface of a liquid that is not boiling.

Vapor Pressure

a. When can a dynamic equilibrium exist between a liquid and its vapor?

In a system at constant vapor pressure, a dynamic equilibrium exists between the vapor and the

liquid. The system is in equilibrium because the rate of evaporation of liquid equals the rate of

condensation of vapor.

When the number of particles entering the vapor increases and some of the particles condense they

return in a liquid state. Increase in temperature = increase in vapor pressure.

Vapor pressure- is a measure of the force exerted by a gas above a liquid.

This could possibly be what occurs and takes place inside a pressure cooker in order to soften and

cook the food inside.

Boiling Point

a. Under what conditions does boiling occur?

When a liquid is heated to a temperature at which particles throughout the liquid have enough kinetic

energy to vaporize, the liquid begins to boil.

Boiling point (bp)- is the temperature at which the vapor pressure of the liquid is just equal to the

external pressure on the liquid.

Normal boiling point- is defined as the boiling point of a liquid at a pressure of 101.3 kPa.

Liquids boil when its vapor pressure is equal to the external pressure.


Summary: In this chapter we learned about the specific states of matter and how they relate to

liquids, and what particular characteristics they have. I also learned how kinetic energy plays a role in

determining the nature of a substance. Fluids are not just limited to liquids, but also gases because

they can flow if they are denser than air. Something unique was that vapor pressure requires a

specialized balance to be at equilibrium. Boiling is not just about the increase in temperature of a

liquid but also vapor pressure and external pressure.

13.3 The Nature of Solids

A Model of Solids

a. How are the structure and properties of solids related?

The general properties of solids reflect the orderly arrangement of their particles and the fixed

locations of their particles.

Melting point (mp)- is the temperature at which a solid changes into a liquid.

Freezing point (fp)- is the temperature at which a liquid changes into a solid.

Crystal Structure and Unit Cell

a. What determines the shape of a crystal?

The shape of a crystal reflects the arrangement of the particles within the solid.

Crystal- a solid in which the particles are arranged in an orderly, repeating, three-dimensional pattern

called a crystal lattice.

Unit cell- is the smallest group of particles within a crystal that retains the geometric shape of a

crystal.

Allotropes- are two or more different molecular forms of the same element in the same physical state.

Amorphous solid- is a solid that lacks an ordered internal structure.

Glass- is a transparent fusion product of inorganic substances that have cooled to a rigid state

without crystalizing.

Summary: In this chapter I learned about the changes a solid take on due to increase or decrease of

temperature, melting, and how a liquid can change to a solid, freezing. I also learned about the

different types of solids which are amorphous and crystalline. Each having different rules that alter

their function and characteristics.

13.4 Changes of State

Sublimation

a. When can sublimation occur?

Sublimation occurs in solids with vapor pressures that exceed atmospheric pressure at or near room

temperature.

Sublimation- is the change of a substance from a solid to a vapor without passing through the liquid

state.

Dry ice.

Phase Diagrams

a. How are the conditions at which phases are equilibrium represented on a phase diagram?

The conditions of pressure and temperature at which two phases exist in equilibrium are indicated on

a phase diagram by a line separating the two regions representing the phases.

Phase diagram- this gives the conditions of temperature and pressure at which a substance exists as

solid, liquid, or gas (vapor).

Triple point- this describes the only set of conditions at which all three phases can exist in equilibrium

with one another.

Summary: This section taught me how there are exceptions to how matter can change its state. For

example, Dry Ice changes from a solid straight into a gas through sublimation, instead of going

through a liquid phase. I also learned about different phases which can determine a substances state

of matter based on different conditions. Triple point is when all three phases exist in equilibrium.


Name: Enmanuel G.

Name: Robens T.

Grade: 10

States of Matter Project

You and you lab partner are going to create a study aid in the form of a game for the

information in Chapter 13 States of Matter.

First, each of you, independently from each other, will summarize the chapter on 3

pages of a pdf which will be submitted in Angel by the end of class on Wednesday Feb

22.

Second, you and you lab partner will be given a game platform which you will use for

your questions and answers, either Jeopardy or Kahoot.

Third, you will fill in the information at the bottom of this page with your username,

passwords and/or websites so that you do not forget this and I have a copy in case

anything gets misplaced. This page will be submitted into Angel as a Word Document

on Wednesday February 22 during class.

Fourth, you will use your notes to generate the questions and answers.

Finally you will give me access to your game by putting the website or Game Number

on this page adding this page to your 3 pages of notes and resubmitting it in Angel as

a pdf by the end of the class on Friday Feb 24.

This page is due by the end of class on Wednesday February 22.

This Project is due by the end of class on Friday February 24.

Jeopardy (https://jeopardylabs.com)

Password:

Edit Link:

Play Link:

Kahoot (https://getkahoot.com)

Username: EnmanuelG

Email: Enmanuelg239@gmail.com

Password: chemistry239

Game PIN: 847811


P 1 : 1980 torr

V 1 : 7730 ml

P 2 : 2.95atmà 2242 torr

V 2: Unknown (in ml)

1980 torr x 7730 ml = 2242 torr x V2

1980 torr x 7730ml

=V 2

2242 torr

15305400ml

2242

V 2

= 6826

= V 2


V 1 : Unknown (in L)

T 1 : 85.4 °C à 358.4 K

V 1 : 3550mL à 3.55L

T 2: 294.0 K

V 1

358.4K = 3.55L

294K

V 1

= ( 358.4K ) 3.55L

294 K

V 1

= 1272.32L

294

V 1

= 4.33L


P 1 : 107 kPA

T 1 : 41 C à 314 K

P 2 : Unknown

T 2 : 22 C à 295 K

107kPa

314K = P 2

295K

314K x P 2

= 295 K x 107kPa

P 2

=

295 K x 107kPa

314 K

P 2

= 31565kPa

314


Combined Gas Law- The law that describes the relationship among the pressure, temperature,

and volume of an enclosed gas, inside a container.

Formula:

Example: A 5.00-­‐L air sample has a pressure of 107 kPa at a temperature of -­‐50.0°C. If the

temperature is raised to 102°C and the volume expands to 7.00 L, what will the new pressure

be?

P 1 : 107 kPa

V 1 : 5.00 L

T 1 : -­‐50.0°C. à 223 K

P 2 : Unknown

V 2 : 7.00 L

T 2 : 102° C à 375 K

107 kPa x 5.00 L

223° K

= P 2 x 7.00 L

375° K

107 kPa x 5.00 L x 375° K = P 2

x 7.00 L x 223° K

107 kPa x 5.00 L x 375° K

7.00 L x 223° K

200625 kPa

1561

= P 2

= P 2

P 2

=129 kPa


Ideal Gas Law: The gas law that includes all four variables-P, V, T, and n, (Pressure, Volume,

Temperature, and number of moles) and their relationships inside a container.

Formula:

PV = nRT

Example: What pressure will be exerted by 0.450 mol of a gas at 25°C if it is contained in a

0.650-­‐L vessel?

Pressure: unknown

Volume: 0.650-­‐L

Number of moles (n): 0.450 mol

R: 8.31

Temperature: 25°C à 298 K

P = n × R ×T

V

/L •kPa

0.450 mol ×8.31

P =

K • mol × 298 K

0.650 L

P = 1,114.371kPa

.650

P =1, 714.41692kPa

P =1.71×10 3 kPa


The Learning Goal for this assignment is: The students will learn how the interactions between water

molecules account for the unique properties of water and how aqueous solutions form.

Take note over the following chapter. Use the Headings provided to organize your notes. Define and number all highlighted vocabulary (total 22 ) as well

as summarize and take notes over the sections. You may add pictures where needed. The pictures should be an appropriate size. Use Arial 12 for all

text. This document should be 2 pages and should be saved as a pdf before you submit it into Angel.

Chapter 15 Water and Aqueous Systems

Pages 488 - 507

15.1 Water and Its Properties

Water in the Liquid State

EQ: What factor causes the high surface tension, low vapor pressure, and high boiling point of water?

a. Many unique and important properties of water including its high surface tension, low vapor

pressure, and high boiling point result from hydrogen bonding.

Many of the characteristics of water are due to the structure in which water is organized, having one

oxygen and two hydrogen atoms makes the bonds highly polar.

Water is one of the liquids with the highest surface tensions.

1.Surface tension: The inward force, or pull, that tends to minimize the surface area of a liquid.

2.Surfactant: Is any substance that interferes with the hydrogen bonding between water molecules and thereby

reduces surface tension.

Water in the Solid State

EQ: How can you describe the structure of ice?

a. The structure of ice is a regular open framework of water molecules in a hexagonal arrangement.

Extensive hydrogen bonding in ice holds the water molecules further apart in a more ordered arrangement

This is what causes water in its solid form, ice, to float when dumped in a cup of water because it has a lower

density.

Summary: In this section it is now known that waters many of water’s unique characteristics is due to its

structure, especially how the hydrogen atoms are bonded together. In a liquid, these characteristics are high

surface tension, low vapor pressure, and high boiling point. As a solid, the characteristics caused by the

hydrogen bonding are density, arrangement of molecules, and the ability to float.

15.2 Homogeneous Aqueous Systems

Solutions

EQ: What types of substances dissolve most readily in water?

a. Substances that dissolve most readily in water include ionic compounds and polar covalent

compounds.

Water dissolves so many of the substances that it comes in contact with that you won’t find chemically

pure water in nature.

Solvents and solutes may be gases, liquids, or solids.

Nonpolar covalent compounds such as methane, and compounds found in oil, grease, and gasoline do not

Dissolve in water, however, they will dissolve in gasoline, this is do to the structure of the solvent and the

solute and the attractions and bonds that exist between them. There are nearly insoluble compounds

because the attractions among the ions are stronger than the attractions exerted by water.

3.Aqueous solution: is water that contains dissolved substances.

4.Solvent: The dissolving medium in a solution.

5.Solute: The dissolved particles in a solution.

6.Solvation: The process by which the positive and negative ions of an ionic solid become surrounded by

solvent molecules.

Electrolytes and Nonelectrolytes

EQ: Why are all ionic compounds electrolytes?

a. All ionic compounds are electrolytes because they dissociate into ions.

Conduction of an electric current requires ions that are mobile and, thus, able

to carry charges throughout a liquid, this is because they ionize in a solution.

Electrolytes are essential to all metabolic processes, including the ability for cells to

carry electrical impulses internally to communicate to other cells, also electrolytes are crucial to nerve and

muscle function, this is why it is always recommended to drink a liquid with significant amounts of electrolytes

when working out, playing a sport, or doing any other physical activity that uses up electrolytes.

Some polar molecular compounds are nonelectrolytes in the pure state but become electrolytes when they

dissolve in water.


7.Electrolyte: Is a compound that conducts an electric current when it is in an aqueous solution or in a molten

state.

8.Nonelectrolyte: Is a compound that does not conduct an electric current in either an aqueous solution or the

molten state.

9.Strong electrolyte: Is a solution that contains all, or nearly all, of the solute exists as ions.

10.Weak electrolyte: An electrolyte that conducts and electric current poorly because only a fraction of the

solute in the solution exists as ions.

Hydrates

EQ: Why do hydrates easily lose and regain water?

a. The forces holding the water molecules in hydrates are not very strong, so the water is easily lost and

regained.

When writing the formula for a hydrate, use a dot to connect the formula of the compound and the number of

water molecules per formula unit.

The water molecules in hydrates are held by weak forces, so hydrates often have an appreciable vapor

pressure.

When a desiccant has absorbed all the water it can hold, the compound can be returned to its anhydrous state

using heat.

11.Water of hydration: The water contained in a crystal; water of crystallization.

12.Hydrate: A compound that contains water of hydration.

13.Anhydrous: A substance that does not contain water.

14.Effloresce: If a hydrate has a vapor pressure higher than the pressure of water vapor in the air, the hydrate

will lose its water of hydration.

15.Hygroscopic: Hydrates and other compounds that remove moisture from the air.

16.Desiccant: Is a substance that is used to absorb moisture from the air and create a dry atmosphere.

17.Deliquescent: Refers to compounds that remove sufficient water from the air to dissolve completely and

form solutions.

Summary: Water’s hydrogen bonds are able to break apart compounds, because since the water molecule is

polar, it contains kinetic energy, and it attracts molecules making up the targeted substance. Even though

water is effective as a solvent, it is unable to dissolve everything, even so, since both solvents and solutes

come in the form of gasses, solids, and liquids, gasoline for example works in place of water. Electrolytes are

important for a number of metabolic activities, because much like circuits, they transport signals and energy

across the body. Hydrates are important because they are the means by which many activities in nature take

place, due to the simple reason that they contain water.

15.3 Heterogeneous Aqueous Systems

Suspensions

EQ: What is the difference between a suspension and a solution?

a. A suspension differs from a solution because the particles of a suspension are much larger and do not

stay suspended indefinitely.

18.Suspension: is a mixture from which particles settle out upon standing.

Colloids

EQ: What distinguishes a colloid from a suspension and a solution?

a. Colloids have particles smaller than those in suspensions and larger than those in solutions.

19.Colloid: Is a heterogeneous mixture containing particles that range in size from 1 nm to 1000nm.

20.Tyndall effect: The scattering of visible light by colloidal particles.

21.Brownian motion: The chaotic movement of colloidal particles, which was first observed by the Scottish

botanist Robert Brown.

22.Emulsion: Is a colloidal dispersion of a liquid in a liquid. Mayo

Summary: The particles in a suspension can be removed by filtration. Milk is one of the most referred to as

when colloids are talked about. Both the dispersed phase and the dispersion medium in a colloid can be any

state of matter. Colloids can demonstrate both the Tyndall effect and the Brownian motion.


Unit 7

Chapter 16 Solutions

The students will learn what properties are used to describe

the nature of solutions and how to quantify the concentration

of a solution.

Chapter 17 Thermochemistry

The student will learn how energy is converted in a chemical

or physical process and how to determine the amount of

energy is absorbed or released in that process.

Differentiate among the various forms of energy and recognize that they can

be transformed from one form to others.

Students will participate in activities to investigate and describe the

transformation of energy from one form to another (i.e. batteries, food, fuels,

etc.)

Explore the Law of Conservation of Energy by differentiating among open,

closed, and isolated systems and explain that the total energy in an isolated

system is a conserved quantity.

Students will be able to calculate various energy changes:

o q = mc∆t

o ∆Hfus

o ∆Hmelt

Thermochemistry

Heat

System

Surrounding

Law of conservation of energy

Bond Making is exothermic

Bond Breaking is endothermic

Heat capacity

Specific heat

Calorimetry

Enthalpy

Thermochemical equation

Molar heat of (fusion, solidification,

vaporization, condensation, solution)

Distinguish between endothermic and exothermic chemical processes.

Students will be able to recognize exothermic and endothermic reactions through

experimentation.

Students will participate in activities (Pasco) to create exothermic and

endothermic graphs.

Endothermic

Exothermic


Create and interpret potential energy diagrams, for example: chemical

reactions, orbits around a central body, motion of a pendulum

Students will participate in activities (Pasco) to create exothermic and

endothermic graphs.

Students will be able to interpret exothermic and endothermic reaction graphs.

Potential energy diagram

Thermochemical equations

Chapter 18 Reaction Rates and Equilibrium

The student will learn how the rate of a chemical reaction can

be controlled, what the role of energy is and why some

reactions occur naturally and others do not.

Explain how various factors, such as concentration, temperature, and

presence of a catalyst affect the rate of a chemical reaction.

Students will be able to describe how each factor may affect the rate of a

chemical reaction.

Students will be able to compare the relative effect of each factor on the rate of a

chemical reaction.

Rate

Collision theory

Activation energy

Catalyst

Activated complex

Inhibitor

Explain the concept of dynamic equilibrium in terms of reversible processes

occurring at the same rates.

Students will be able to describe a system in dynamic equilibrium.

Students will be able to describe how factors may affect the equilibrium of a

reaction.

Reversible reaction

Chemical equilibrium

Le Chatelier principle

Explain entropy’s role in determining the efficiency of processes that convert

energy to work.

Students will be able to describe the change in entropy of a reaction.

Students will be able to determine if a reaction is spontaneous

Entropy

Law of disorder

Spontaneous/nonspontaneous reaction


The Learning Goal for this assignment is:

Defining Concentration

Measures of Concentration

Concentration is defined as the amount of dissolved solute in a given amount of solvent or

solution. There are several terms that describe concentration. Some of these terms are relative;

that is, they can be used only to compare the concentration of one solution to another. Dilute and

concentrated are two such terms. A dilute solution contains less dissolved solute than a

concentrated solution (in equal volumes of solution).

The terms saturated, unsaturated, and supersaturated are terms that describe concentration more

precisely.







Saturated: The maximum amount of solute is dissolved in a given amount of solvent at a

particular temperature. Such solutions are stable.

Unsaturated: Less than the maximum amount of solute is dissolved in a given amount of

solvent at a particular temperature. Such solutions are stable.

Supersaturated: More than the maximum amount of solute is dissolved in a given amount of

solvent at a particular temperature. Such solutions are unstable.

Look at the solubility curve shown below:

The solubility of NaNO3 is 86.0 g/100 mL H2O (at 20 °C). If you

prepare a solution of 50.0 g NaNO3 dissolved in 100 mL H2O, an unsaturated solution results

(Point A on the graph).

continue adding NaNO to the solution until 86.0 g are dissolved, a saturated solution results

(Point B).

heat the solution to 50 °C, 113 grams NaNO3 can be dissolved (Point C). When the solution

cools back down to 20 °C, it will be supersaturated (Point D).

Quantifying Concentration

To describe the concentration of a solution even more precisely, various measures of concentration

can be used. Some of the ways concentration can be quantified include calculating the

Mass of solute per solution mass (expressed as a percent or parts per million)

Moles of solute per kilogram solvent (molality)

Mass of solute per liter of solution (grams/liter)

Moles of solute per liter of solution (molarity)


Part 1: Mass Percent

Mass percent (also called percent by mass, weight percent, or percent by weight) compares the

mass of the solute to the entire mass of the solution.

Notes:

Part 2: Parts per Million

Parts per million (ppm) is another measure of concentration. It is similar to mass percent. But

mass percent indicates the number of grams of solute per 100 g solution. Parts per million

indicates the number of grams of solute per 1,000,000 g solution. This measure of concentration is

often used to express the concentrations of very dilute solutions.

Notes:

Part 3: Molality

Molality (m) is the ratio of the moles of solute to the kilograms of solvent. Note: this is the first

measure of concentration that is concerned with the mass of the solvent, not the mass of the

solution as a whole.

Notes:

Part 4: Grams per Liter

To express the concentration of a solution in grams per liter, you must know the mass of the solute

and the volume of the solution, not just the volume of the solvent.

Notes:

Suppose you wanted to know what the concentration would be before making the solution. Could that

be done? In order to relate the volume or mass of solvent to the volume of solution, you would have

to know the density of the solution. You will see how solution density can be used to calculate

molarity in the next section.


Part 5: Molarity

Molarity (M) is the most common measure of concentration. The concentration of most solutions

you see in the lab are expressed in terms of molarity. Just like g/L, molarity calculations require

that you know either the volume or density of the resulting solution.

Notes:

Using Density to Calculate Molarity

If the volume of the resulting solution is not known, molarity is calculated as follows:

Convert the volume of solvent to grams. (The simulation does this step for you.)

Determine the total mass of the solution (mass of solute + mass of solvent).

Convert the solution mass to volume in milliliters, using its density (volume = mass / density).

Convert the solution volume to liters (divide by 1000).

Convert solute grams to moles.

Calculate the molarity (moles solute / L solution).

Reinforcing What You've Learned

1. 35.0 g potassium dichromate are dissolved in 354.0 g distilled water. What is the concentration

of the resulting solution, expressed as percent by mass? 9.00%

2. A chemist discovers that 2587.0 g distilled water are contaminated with 13.0 g NaNO3. What is

the concentration of NaNO3, expressed in ppm? 5.00 X 10 3

3. 175.00 g H2SO4 are added to 61.49 g distilled H2O. What is the molality of the resulting solution?

29.04m

4. 225 g NaBr are dissolved in 525 g distilled water. If the volume of the resulting solution is

584 mL, what is its concentration when expressed in g/L? 385g/L

5. 470.0 g Na2CO3 are dissolved in 4230.0 g distilled water. What is the molarity of the resulting

solution (D = 1.1029 g/mL)?

1.1561 M

Applying What You've Learned

6. Think about the equations for mass percent and parts/million. Consider how you might convert

between these two measures of concentration.

Mass percent = ppm ÷ 10,000 ppm = mass percent X .0001

7. Think about the equations for grams/liter and molarity. Consider how you might convert

between these two measures of concentration.

grams / L = molarity X molar mass

molarity = grams / L ÷ molar mass


Glossary

Concentration: Amount of dissolved solute in a given amount of solvent or solution.

Density: Amount of matter per unit volume. Density is calculated by dividing an object's mass by

its volume.

Mass: Amount of matter an object contains or, more scientifically, the measure of an object's

resistance to changes in motion. The SI (Systéme International) unit for mass is the kilogram. In

the lab, mass is often measured in grams.

Mass percent: Also referred to as percent by mass and, occasionally, weight percent or percent

by weight. Mathematically, mass % = (mass of solute / mass of entire solution) x 100.

Molar mass: Mass of a compound, calculated by adding up the individual masses for its component

atoms, which are obtained from the periodic table of the elements. Molar mass is expressed in

grams/mole and is sometimes referred to as molecular mass (for molecular compounds) or formula

mass (for ionic compounds).

Molality (m): Moles of solute per kilogram of solvent. Mathematically, m = moles of solute / kilogram of

solvent.

Molarity (M): Moles of solute per liter of solution. Mathematically, M = moles of solute / liter of

solution.

Mole (mol): Counting unit used to express the large numbers of particles, such as atoms or

molecules, that are involved in chemical processes. One mole of particles contains 6.02 x 10

particles. The mass of one mole of an element, in grams, is equivalent to the atomic mass for that

element, as indicated on the periodic table.

Parts per million (ppm): Measure of concentration often used for dilute solutions. Mathematically,

ppm = (mass of solute / mass of solution) x 10.

Solubility: Measure of the maximum amount of solute that can be dissolved in a given amount of

solvent at a given temperature, forming a stable solution.

Solute: Dissolved substance in a solution. The solute is generally the solution component present

in the lesser amount.

Solution: Homogeneous mixture in which one substance has been dissolved in another.

Solvent: Substance in which a solute is dissolved to form a solution. The solvent is generally the

solution component present in the greater amount.

Volume: Amount of space an object occupies. The SI (Systéme International) unit for volume is

the cubic meter. In the lab, volume is often measured in cubic centimeters, milliliters, or liters.

SAS Curriculum Pathways VLab #866


The Learning Goal for this assignment is:

The System and the Surroundings in Chemistry

Thermochemistry

The system is the part of the universe we wish to focus our attention on. In the world of chemistry, the

system is the chemical reaction. For example:

2H2 + O2 ---> 2H2O

The system consists of those molecules which are reacting.

The surroundings are everything else; the rest of the universe. For example, say the above reaction is

happening in gas phase; then the walls of the container are part of the surroundings.

There are two important issues:

1. a great majority of our studies will focus on the change in the amount of energy, not the

absolute amount of energy in the system or the surroundings.

2. regarding the direction of energy flow, we have a "sign convention."

Two possibilities exist concerning the flow of energy between system and surroundings:

1. The system can have energy added to it, which increases its amount and lessens the energy

amount in the surroundings.

2. The system can have energy removed from it, thereby lowering its amount and increasing the

amount in the surroundings.

We will signify an increase in energy with a positive sign and a loss of energy with a negative sign.

Also, we will take the point-of-view from the system. Consequently:

1. When energy (heat or work) flow out of the system, the system decreases in its amount. This

is assigned a negative sign and is called exothermic.

2. When energy (heat or work) flows into the system, the system increases its energy amount.

This is assigned a positive sign and is called endothermic.

We do not discuss chemical reactions from the surrounding's point-of-view. Only from the system's.

Notes:


Specific Heat

Here is the definition of specific heat:

the amount of heat necessary for 1.00 gram of a substance to change 1.00 °C

Note the two important factors:

1. It's 1.00 gram of a substance

2. and it changes 1.00 °C

Keep in mind the fact that this is a very specific value. It is only for one gram going one degree. The

specific heat is an important part of energy calculations since it tells you how much energy is needed

to move each gram of the substance one degree.

Every substance has its own specific heat and each phase has its own distinct value. In fact, the

specific heat value of a substance changes from degree to degree, but we will ignore that.

The units are often Joules per gram-degree Celsius (J/g*°C). Sometimes the unit J/kg K is also used.

This last unit is technically the most correct unit to use, but since the first one is quite common, you

will need to know both.

I will ignore calorie-based units almost entirely.

Here are the specific heat values for water:

Phase J g¯1 °C¯1 J kg¯1

K¯1

Gas 2.02 2.02 x 10 3

Liquid 4.184 4.184 x 10 3

Solid 2.06 2.06 x 10 3

Notice that one set of values is simply 1000 times bigger than the other. That's to offset the influence

of going from grams to kilograms in the denominator of the unit.

Notice that the change from Celsius to Kelvin does not affect the value. That is because the specific

heat is measured on the basis of one degree. In both scales (Celsius and Kelvin) the jump from one

degree to the next are the same "distance." Sometimes a student will think that 273 must be involved

somewhere. Not in this case.

Specific heat values can be looked up in reference books. Typically, in the classroom, you will not be

asked to memorize any specific heat values. However, you may be asked to memorize the values for

the three phases of water.

As you go about the Internet, you will find different values cited for specific heats of a given

substance. For example, I have seen 4.186 and 4.187 used in place of 4.184 for liquid water. None of

the values are wrong, it's just that specific heat values literally change from degree to degree. What

happens is that an author will settle on one particular value and use it. Often, the one particular value

used is what the author used as a student.

Hence, 4.184.


The Time-Temperature Graph

We are going to heat a container that has 72.0 grams of ice (no liquid water yet!) in it. To make the

illustration simple, please consider that 100% of the heat applied goes into the water. There is no loss

of heat into heating the container and no heat is lost to the air.

Let us suppose the ice starts at -10.0 °C and that the pressure is always one atmosphere. We will

end the example with steam at 120.0 °C.

There are five major steps to discuss in turn before this problem is completely solved. Here they are:

1. the ice rises in temperature from -10.0 to 0.00 °C.

2. the ice melts at 0.00 °C.

3. the liquid water then rises in temperature from zero to 100.0 °C.

4. the liquid water then boils at 100.0 °C.

5. the steam then rises in temperature from 100.0 to 120.0 °C

Each one of these steps will have a calculation associated with it. WARNING: many homework and

test questions can be written which use less than the five steps. For example, suppose the water in

the problem above started at 10.0 °C. Then, only steps 3, 4, and 5 would be required for solution.

To the right is the type of graph which is typically used to

show this process over time.

You can figure out that the five numbered sections on the

graph relate to the five numbered parts of the list just above

the graph.

Also, note that numbers 2 and 4 are phases changes: solid

to liquid in #2 and liquid to gas in #4.

Q=mcΔT

where ΔT is (Tf – Ti)

Here are some symbols that will be used, A LOT!!

Δt = the change in temperature from start to finish in degrees Celsius (°C)

m = mass of substance in grams

c = the specific heat. Its unit is Joules per gram X degree Celsius (J / g °C is one way to write

the unit; J g¯1 °C¯1 is another)

q = the amount of heat involved, measured in Joules or kilojoules (symbols = J and kJ)

mol = moles of substance.

ΔH is the symbol for the molar heat of fusion and ΔH is the symbol for the molar heat of

vaporization.

We will also require the molar mass of the substance. In this example it is water, so the molar mass is

18.0 g/mol.

Notes:


Step One: solid ice rises in temperature

As we apply heat, the ice will rise in temperature until it

arrives at its normal melting point of zero Celsius.

Once it arrives at zero, the Δt equals 10.0 °C.

Here is an important point: THE ICE HAS NOT MELTED

YET.

At the end of this step we have SOLID ice at zero

degrees. It has not melted yet. That's an important point.

Each gram of water requires a constant amount of energy

to go up each degree Celsius. This amount of energy is

called specific heat and has the symbol c.

72.0 grams of ice (no liquid water yet!) has changed 10.0 °C. We need to calculate the energy

needed to do this.

This summarizes the information needed:

Δt = 10 °C

The mass = 72.0 g

c = 2.06 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:

q = (mass) (Δt) (c)

Why is this equation the way it is?

Think about one gram going one degree. The ice needs 2.06 J for that. Now go the second degree.

Another 2.06 J. Go the third degree and use another 2.06 J. So one gram going 10 degrees needs

2.06 x 10 = 20.6 J. Now we have 72 grams, so gram #2 also needs 20.6, gram #3 needs 20.6 and so

on until 72 grams.

With the numbers in place, we have:

q = (72.0 g) (10 °C) (2.06 J/g °C)

So we calculate and get 1483.2 J. We won't bother to round off right now since there are four more

calculations to go. Maybe you can see that we will have to do five calculations and then sum them all

up.

One warning before going on: three of the calculations will yield J as the unit on the answer and two

will give kJ. When you add the five values together, you MUST have them all be the same unit.

In the context of this problem, kJ is the preferred unit. You might want to think about what 1483.2 J is

in kJ.

Notes:


Step Two: solid ice melts

Now, we continue to add energy and the ice begins to

melt.

However, the temperature DOES NOT CHANGE. It

remains at zero during the time the ice melts.

Each mole of water will require a constant amount of

energy to melt. That amount is named the molar heat of

fusion and its symbol is ΔHf. The molar heat of fusion is

the energy required to melt one mole of a substance at its

normal melting point. One mole of solid water, one mole

of solid benzene, one mole of solid lead. It does not

matter. Each substance has its own value.

During this time, the energy is being used to overcome water molecules' attraction for each other,

destroying the three-dimensional structure of the ice.

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion

between calories and Joules is 4.184 J = 1.000 cal.

Sometimes you also see this number expressed "per gram" rather than "per mole." For example,

water's molar heat of fusion is 6.02 kJ/mol. Expressed per gram, it is 334.16 J/g.

Typically, the term "heat of fusion" is used with the "per gram" value.

72.0 grams of solid water is 0.0 °C. It is going to melt AND stay at zero degrees. This is an important

point. While the ice melts, its temperature will remain the same. We need to calculate the energy

needed to do this.

This summarizes the information needed:

ΔHf = 6.02 kJ/mol

The mass = 72.0 g

The molar mass of H2O = 18.0 gram/mol

The calculation needed, using words & symbols is:

q = (moles of water) (ΔHf)

We can rewrite the moles of water portion and make the equation like this:

q = (grams water / molar mass of water) (ΔHf)

Why is this equation the way it is?

Think about one mole of ice. That amount of ice (one mole or 18.0 grams) needs 6.02 kilojoules of

energy to melt. Each mole of ice needs 6.02 kilojoules. So the (grams water / molar mass of water) in

the above equation calculates the amount of moles.

With the numbers in place, we have:

q = (72.0 g / 18.0 g mol¯1 ) (6.02 kJ / mol)

So we calculate and get 24.08 kJ. We won't bother to round off right now since there are three more

calculations to go. We're doing the second step now. When all five are done, we'll sum them all up.


Step Three: liquid water rises in temperature

Once the ice is totally melted, the temperature can now

begin to rise again.

It continues to go up until it reaches its normal boiling

point of 100.0 °C.

Since the temperature went from zero to 100, the Δt is

100.

Here is an important point: THE LIQUID HAS NOT

BOILED YET.

At the end of this step we have liquid water at 100 degrees. It has not turned to steam yet.

Each gram of water requires a constant amount of energy to go up each degree Celsius. This amount

of energy is called specific heat and has the symbol c. There will be a different value needed,

depending on the substance being in the solid, liquid or gas phase.

72.0 grams of liquid water is 0.0 °C. It is going to warm up to 100.0 °C, but at that temperature, the

water WILL NOT BOIL. We need to calculate the energy needed to do this.

This summarizes the information needed:

Δt = 100.0 °C (100.0 °C – 0.0 °C)

The mass = 72.0 g

c = 4.184 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:

q = (mass) (Δt) (c)

Why is this equation the way it is?

Think about one gram going one degree. The liquid water needs 4.184 J for that. Now go the second

degree. Another 4.184 J. Go the third degree and use another 4.184 J. So one gram going 100

degrees needs 4.184 x 100 = 418.4 J. Now we have 72 grams, so gram #2 also needs 418.4, gram

#3 needs 418.4 and so on until 72 grams.

With the numbers in place, we have:

q = (72.0 g) (100.0 °C) (4.184 J/g °C)

So we calculate and get 30124.8 J. We won't bother to round off right now since there are two more

calculations to go. We will have to do five calculations and then sum them all up.

Notes:


Step Four: liquid water boils

Now, we continue to add energy and the water begins to

boil.

However, the temperature DOES NOT CHANGE. It

remains at 100 during the time the water boils.

Each mole of water will require a constant amount of

energy to boil. That amount is named the molar heat of

vaporization and its symbol is ΔH. The molar heat of

vaporization is the energy required to boil one mole of a

substance at its normal boiling point. One mole of liquid water, one mole of liquid benzene, one mole

of liquid lead. It does not matter. Each substance has its own value.

During this time, the energy is being used to overcome water molecules' attraction for each other,

allowing them to move from close together (liquid) to quite far apart (the gas state).

The unit for this is kJ/mol. Sometimes you see older references that use kcal/mol. The conversion

between calories and Joules is 4.184 J = 1.000 cal.

Typically, the term "heat of vaporization" is used with the "per gram" value.

72.0 grams of liquid water is at 100.0 °C. It is going to boil AND stay at 100 degrees. This is an

important point. While the water boils, its temperature will remain the same. We need to calculate the

energy needed to do this.

This summarizes the information needed:

ΔH = 40.7 kJ/mol

The mass = 72.0 g

The molar mass of H2O = 18.0 gram/mol

The calculation needed, using words & symbols is:

q = (moles of water) (ΔH)

We can rewrite the moles of water portion and make the equation like this:

q = (grams water / molar mass of water) (ΔH)

Why is this equation the way it is?

Think about one mole of liquid water. That amount of water (one mole or 18.0 grams) needs 40.7

kilojoules of energy to boil. Each mole of liquid water needs 40.7 kilojoules to boil. So the (grams

water / molar mass of water) in the above equation calculates the amount of moles.

With the numbers in place, we have:

q = (72.0 g / 18.0 g mol¯1 ) (40.7 kJ / mol)

So we calculate and get 162.8 kJ. We won't bother to round off right now since there is one more

calculation to go. We're doing the fourth step now. When all five are done, we'll sum them all up.


Step Five: steam rises in temperature

Once the water is completely changed to steam, the

temperature can now begin to rise again.

It continues to go up until we stop adding energy. In this

case, let the temperature rise to 120 °C.

Since the temperature went from 100 °C to 120°C, the Δt

is 20°C.

Each gram of water requires a constant amount of energy

to go up each degree Celsius. This amount of energy is called specific heat and has the symbol c.

There will be a different value needed, depending on the substance being in the solid, liquid or gas

phase.

72.0 grams of steam is 100.0 °C. It is going to warm up to 120.0 °C. We need to calculate the energy

needed to do this.

This summarizes the information needed:

Δt = 20 °C

The mass = 72.0 g

c = 2.02 Joules per gram-degree Celsius

The calculation needed, using words & symbols is:

q = (mass) (Δt) (c)

Why is this equation the way it is?

Think about one gram going one degree. The liquid water needs 2.02 J for that. Now go the second

degree. Another 2.02 J. Go the third degree and use another 2.02 J. So one gram going 20 degress

needs 2.02 x 20 = 44 J. Now we have 72 grams, so gram #2 also needs 44, gram #3 needs 44 and

so on until 72 grams.

I hope that helped.

With the numbers in place, we have:

q = (72.0 g) (20 °C) (2.02 J/g °C)

So we calculate and get 2908.8 J. We won't bother to round off right now since we still need to sum

up all five values.

Notes:


The following table summarizes the five steps and their results. Each step number is a link back to

the explanation of the calculation.

Converting to kJ gives us this:

1.4832 kJ

24.08 kJ

30.1248 kJ

162.8 kJ

2.9088 kJ

Step q 72.0 g of H2O

1 1483.2 J Δt = 10 (solid)

2 24.08 kJ melting

3 30124.8 J Δt = 100 (liquid)

4 162.8 kJ boiling

5 2908.8 J Δt = 20 (gas)

Summing up gives 221.3968 kJ and proper significant digits gives us 221.4 kJ for the answer.

Notice how all units were converted to kJ before continuing on. Joules is a perfectly fine unit; it's just

that 221,396.8 J is an awkward number to work with. Usually Joules is used for values under 1000,

otherwise kJ is used.

By the way, on other sites you may see kj used for kilojoules. I've also seen Kj used. Both of these

are wrong symbols. kJ is the only correct symbol.

Enthalpy

When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal

to the change in enthalpy. Enthalpy (H) is the sum of the internal energy (U) and the product of

pressure and volume (PV) given by the equation:

H=U+PV

When a process occurs at constant pressure, the heat evolved (either released or absorbed) is equal

to the change in enthalpy. Enthalpy is a state function which depends entirely on the state

functions T, P and U. Enthalpy is usually expressed as the change in enthalpy (ΔH) for a process

between initial and final states:

ΔH=ΔU+ΔPVΔ

If temperature and pressure remain constant through the process and the work is limited to pressurevolume

work, then the enthalpy change is given by the equation:

ΔH=ΔU+PΔV

Also at constant pressure the heat flow (q) for the process is equal to the change in enthalpy defined

by the equation:

ΔH=q

By looking at whether q is exothermic or endothermic we can determine a relationship between ΔH

and q. If the reaction absorbs heat it is endothermic meaning the reaction consumes heat from the

surroundings so q>0 (positive). Therefore, at constant temperature and pressure, by the equation

above, if q is positive then ΔH is also positive. And the same goes for if the reaction releases heat,


then it is exothermic, meaning the system gives off heat to its surroundings, so q


The Learning Goal for this section is: Explain how various factors, such as concentration,

temperature, and presence of a catalyst affects the rate of a chemical reaction.

You are going to answer these 15 questions first in the order they are given to you. This will be a quiz grade. You are then going to explain the

chemical concepts that are used in each question. You can do this in the order given or group them by concepts. You can use your book or information

found on the internet but all information must be written in your own word. The font needs to be Arial 12. This is due on Monday April 24 by midnight in

the drop box. This document should be a 6-page pdf.

1. B. 6. B. 11.A.

2. C. 7. B. 12.D.

3. C. 8. D. 13.C.

4. D. 9. A. 14.D.

5. A. 10.B. 15.D.

1 This graph represents the change in energy for two laboratory trials of the same reaction.

Which factor could explain the energy difference between the trials?

A Heat was added to trial #2.

B A catalyst was added to trial #2.

C Trial #1 was stirred.

D Trial #1 was cooled.

Chemical Concepts:

There are two main chemical concepts involved in this question and they are the “catalyst” and

“activation energy”. To start off, activation energy is known as the least amount of energy possibly

needed in order for a specific chemical reaction to take place. As it is evident in the graph trial #1

required a larger amount of energy in order to reach activation, thus, it can be said that it has a higher

activation rate than that of trial #2. This is where a catalyst comes into play, a catalyst is known as an

agent added into a reaction that causes the required activation energy to be lowered; therefore, the

reaction occurs at a faster rate and requires less energy.

2 Consider this balanced chemical equation:

Which will increase the rate of the reaction?

A increasing pressure on the reaction

B decreasing concentration of the reactants

C adding a catalyst to the reaction

D decreasing the temperature of the reaction

2H 2 O 2 (aq) → 2H 2 O (l) + O 2 (g)


Chemical Concepts:

The chemical concepts involved in this question are “reaction rate” and the “catalyst”. When it comes

to reaction rate, it is described as the measurement and relationship of time and the amount of

products formed by the reactants in a reaction. There are many factors that contribute to the

fluctuation of this type of measurement and those are temperature, concentration, the presence of a

catalyst, and particle size. Increasing the temperature of the reaction also increases average kinetic

energy making the possibility of colliding particles higher which in turns increases the rate of which

the products are formed. The higher the concentration the more particles are able to collide also

changing the reaction rate, much like in a crowded room you are more likely to bump into someone

than if you were in a room with only a couple of other people. The catalyst which has already been

mentioned, decreases the activation energy of the reaction, speeding it up, and causing the reaction

rate to increase. Finally, particle size which isn’t mentioned in this question but is connected with the

subject and just means that the bigger the particle the more surface area it has, meaning that there is

more space for moving particles to collide and form new products.

3 For the reaction

A + (aq) + B — (aq) → AB (s)

increasing the temperature increases the rate of the reaction. Which is the best explanation for this

happening?

A The pressure increases, which in turn increases the production of products.

B The concentration of reactants increases with an increase in temperature.

C The average kinetic energy increases, so the likelihood of more effective collisions between ions

increases.

D Systems are more stable at high temperatures.

Chemical Concepts:

The chemical concepts involved in this question are “collision theory” and “kinetic energy”. The

collision theory is the concept that units of matter such as ions, molecules, and atoms can form new

products if they collide with other units of matter with a specific force and position caused by their

speed and direction of travel. The concept of kinetic energy dwells in the understanding of the

movement of particles based on the amount of energy they contain, it is closely involved with

temperature. The more kinetic energy a particle has the faster it moves and the force of impact it

applies while colliding with another particle increases the chances of new products to be formed.

4 Which statement explains why the speed of some reactions is increased when the surface area of

one or all the reactants is increased?

A increasing surface area changes the electronegativity of the reactant particles

B increasing surface area changes the concentration of the reactant particles

C increasing surface area changes the conductivity of reactant particles

D increasing surface area enables more reactant particles to collide

Chemical Concepts:

The chemical concept associated with this question is “particle size” and how it affects reaction rate.

Particle size quite blatantly is referred to as the amount of surface area a particle, mainly the reactant,

has. This has a direct correlation with reaction rate because the greater the surface area a particle

has the more likely it is to collide with another reactant to form a new product. The reason behind this

is that collisions occur at the surface of particles.


catalyst

C 6 H 6 + Br 2 → C 6 H 5 Br + HBr

5 Which of the following changes will cause an increase in the rate of the above reaction?

A increasing the concentration of Br 2

B decreasing the concentration of C 6 H 6

C increasing the concentration of HBr

D decreasing the temperature

Chemical Concepts:

The chemical concept tied in with this question is “concentration” and how it increases the reaction

rate of a reaction. Concentration is recognized as the amount of different units, in this case, particles,

molecules, or atoms, in relation the specific space or volume which is being analyzed. The greater the

amount of particles in a fixed volume, the more likely they’re are to collide with other "reactant”

particles and in turn, the more likely they are to form new products, which ultimately increases the

reaction rate of the reaction.

2CO + O 2 → 2CO 2

6 If the above reaction takes place inside a sealed reaction chamber, then which of these procedures

will cause a decrease in the rate of reaction?

A raising the temperature of the reaction chamber

B increasing the volume inside the reaction chamber

C removing the CO 2 as it is formed

D adding more CO to the reaction chamber

Chemical Concepts:

The chemical concepts related to this question are “volume” and “laws of gases” and how they are

related to rate of reaction. Volume in science is defined by the amount of space a substance takes

up, or occupies, which when it involves gases, has a relationship with reaction rate. Gases don’t have

a definite shape and are recognized as the types of matter which contain the most active particles,

this means that they are constantly trying to escape their container and expand. If the volume inside a

container were to increase, this means that the gas inside would spread apart even more, and it

would have more space to move around freely, consequently decreasing the chances of colliding and

in turn, decreasing the reaction rate.

7 A catalyst can speed up the rate of a given chemical reaction by


A increasing the equilibrium constant in favor of products.

B lowering the activation energy required for the reaction to occur.

C raising the temperature at which the reaction occurs.

D increasing the pressure of reactants, thus favoring products.

Chemical Concepts:

The chemical concepts associated with this question are the “catalyst” and “activation energy”.

Activation energy is defined as the minimum amount of energy a reactant needs to obtain in order for

the reaction to take place. This factor is directly affected by a catalyst which is a substance that is

used to lower the activation rate, but is not consumed or used up, in the process. When the energy

required lessens the reaction takes place at a quicker rate, because the time needed to obtain the

necessary amount of energy, is less as the quota is met at a much lower standard.

8 Which reaction diagram shows the effect of using the appropriate catalyst in a chemical reaction?

A

C

B

D

Chemical Concepts:

The chemical concepts associated with this question are the “catalyst” and “activation energy”.

Activation energy is defined as the minimum amount of energy a reactant needs to obtain in order for

the reaction to take place. This factor is directly affected by a catalyst which is a substance that is

used to lower the activation rate, but is not consumed or used up, in the process. When the energy

required lessens the reaction takes place at a quicker rate, because the time needed to obtain the

necessary amount of energy, is less as the quota is met at a much lower standard. This is slightly

comparable to a hiker who intends to scale up a mountain, the energy and time necessary to

accomplish such a feat is much more than that which is needed to climb up a hill. In a sense, what a

catalyst does is convert the mountainous, by comparison, amount of energy and time required into

reaching activation, into a lower, less time and energy draining, requirement.

9 H 2 O 2, hydrogen peroxide, naturally breaks down into H 2 O and O 2 over time. MnO 2 , manganese

dioxide, can be used to lower the energy of activation needed for this reaction to take place and, thus,

increase the rate of reaction. What type of substance is MnO 2 ?

A a catalyst

B an enhancer

C an inhibitor

D a reactant


Chemical Concepts:

The chemical concepts which are tied with this question are “catalyst”, “activation energy”, and “rate

of reaction”. Activation energy is defined as the minimum amount of energy a reactant needs to

obtain in order for the reaction to take place. A catalyst is known as an agent added into a reaction

that causes the required activation energy to be lowered; therefore, the reaction occurs at a faster

rate and requires less energy. Finally, reaction rate is described as the measurement and relationship

of time and the amount of products formed by the reactants in a reaction.

10 When a reaction is at equilibrium and more reactant is added, which of the following changes is

the immediate result?

A The reverse reaction rate remains the same.

B The forward reaction rate increases.

C The reverse reaction rate decreases.

D The forward reaction rate remains the same.

Chemical Concepts:

This question contains the chemical concepts of “equilibrium”, the “reactant”, and “forward/reverse

reaction”. Equilibrium is the condition of a reaction where the changing of reactant particles and the

product particles are no longer being exhibited at separate rates, the inclination to change from one

another as time continues passing, this standstill and levelness in concentration of both sides is

known as “EQUILIBRIUM”. A reactant is the substance that changes into something else, the

product, during the manifestation of the reaction. The forward reaction describes the change of a

substance from reactant to product; reverse reaction, on the other hand, describes the change of a

substance from a product into a reactant.

11 In which of the following reactions involving gases would the forward reaction be favored by an

increase in pressure?

A A + B ⇄AB

B A + B ⇄ C + D

C 2A + B ⇄ C + 2D

D AC ⇄ A + C

Chemical Concepts:

The chemical concepts involved with this question are “pressure” and “Le Châtelier's principle”. Le

Chatelier’s principle declares that a system that has reached equilibrium can be disrupted and

derived from equilibrium through the use of a stressing agent, which can be a force that suddenly

changes. One of these changes in conditions that tilts the fine balance found in dynamic equilibrium is

pressure. Pressure is the application of a constant force on an object or substance. Changes in the

pressure of a reaction during dynamic equilibrium, can again resume the difference in rates of

reactants to products, and products to reactants. This is the case because the proximity between

particles can either extend or decrease making it more likely or less likely for collision to occur,

changing the forward and reverse reactions.

4HCl (g) + O 2(g) ⇄ 2H 2 O (l) + 2Cl 2(g) + 113 kJ

12 Which action will drive the reaction to the right?

A heating the equilibrium mixture

B adding water to the system

C decreasing the oxygen concentration

D increasing the system’s pressure


Chemical Concepts:

The chemical concepts that are found in this question are “forward reaction” and “pressure” and its

affects on equilibrium. Forward reaction describes the change of a substance from reactant to

product( from left to right). Changes in the pressure of a reaction during dynamic equilibrium, can

again resume the difference in rates of reactants to products, and products to reactants. This is the

case because the proximity between particles can either extend or decrease making it more likely or

less likely for collision to occur, changing the forward and reverse reactions.

NO 2(g) + CO (g) ⇄ NO (g) + CO2 (g)

13 The reaction shown above occurs inside a closed flask. What action will shift the reaction to the

left?

A pumping CO gas into the closed flask

B raising the total pressure inside the flask

C increasing the NO concentration in the flask

D venting some CO 2 gas from the flask

Chemical Concepts:

The chemical concept related to this question is “reverse reaction”. Reverse reaction is the term used

to describe the change of a substance from a product into a reactant(from right to left).

NH 4 Cl (s) + heat ⇄ NH 3(g) + HCl (g)

14 What kind of change will shift the reaction above to the right to form more products?

A a decrease in total pressure

B an increase in the concentration of HCl

C an increase in the pressure of NH 3

D a decrease in temperature

Chemical Concepts:

The chemical concepts related to this question are “forward reaction” and “temperature change “and

how it disrupts equilibrium. Forward reaction describes the change of a substance from reactant to

product( from left to right). Temperature is the amount of heat or energy an object has, and it is

directly related to average kinetic energy. If heat is decreased the amount of product will increase

because they will begin to stay intact more often because the reactant particles aren’t as active.

15 In a sealed bottle that is half full of water, equilibrium will be attained when water molecules

A cease to evaporate.

B begin to condense.

C are equal in number for both the liquid and the gas phase.

D evaporate and condense at equal rates.

Chemical Concepts:

The chemical concept related to this question is “equilibrium”. Equilibrium refers to the condition of a

reaction where the changing of reactant particles to and from product particles are no longer being

exhibited at separate rates, the inclination to change from one another as time continues passing, this

standstill and levelness in concentration and rate of both sides is known as dynamic equilibrium.


Unit 8

Chapter 19 Acid and Bases

The student will learn what are the different ways chemists

define aids and bases, what the pH of a solution means and

how chemist use acid-base reactions.

Relate acidity and basicity to hydronium and hydroxyl ion concentration and

pH.

Students will be able to use a pH scale to identify substances as acids or bases.

Students will be able to use various equipment (probeware, universal pH, etc.) to

identify the pH of substances.

Students will be able to calculate H3O+ and OH- concentration of various

substances.

pH scale

Hydronium ion

Arrhenius acid/base

Lewis acid/base

Bronsted-Lowry acid/base

Strong acid/base

Weak acid/base

Neutralization reaction

Titration

Chapter 20 Oxidation-Reduction Reactions

The student will learn what happens during oxidation and

reduction and how to balance redox equations.

Describe oxidation-reduction reactions in living and non-living systems.

Students will be able to compare and contrast redox reactions.

Students will be able to assign oxidation numbers to redox reactions.

Students will be able to write half reactions

Oxidation


Reduction

Oxidation reduction reaction

Oxidation number

Half reaction

Electrochemical process

Battery

Cathode

Anode

Electrolysis


The Learning Goal for this section is:

Acids and Bases

The Observable Properties of Acids and Bases

The words acid and alkaline (an older word for base) are derived from direct sensory experience.

Acid Property #1:

The word acid comes from the Latin word acere, which means "sour." All acids taste sour. Well

known from ancient times were vinegar, sour milk and lemon juice. Aspirin (scientific name:

acetylsalicylic acid) tastes sour if you don't swallow it fast enough. Other languages derive their word

for acid from the meaning of sour. So, in France, we have acide. In Germany, we have säure from

saure and in Russia, kislota from kisly.

Base Property #1:

The word "base" has a more complex history (see below) and its name is not related to taste. All

bases taste bitter. For example, mustard is a base. It tastes bitter. Many medicines, because they are

bases, taste bitter. This is the reason cough syrups are advertised as having a "great grape taste."

The taste is added in order to cover the bitterness of the active ingredient in cough syrup.

Acid Property #2:

Acids make a blue vegetable dye called litmus turn red.

Base Property #2:

Bases are substances which will restore the original blue color of litmus after having been reddened

by an acid.

Acid Property #3:

Acids destroy the chemical properties of bases.

Base Property #3:

Bases destroy the chemical properties of acids.

Neutralization is the name for this type of reaction.

Acid Property #4:

Acids conduct an electric current.

Base Property #4:

Bases conduct an electric current.

This is a common property shared with salts. Acids, bases and salts are grouped together into a

category called electrolytes, meaning that a water solution of the given substance will conduct an

electric current.

Non-electrolyte solutions cannot conduct a current. The most common example of this is sugar

dissolved in water.


So far, the properties have an obvious relationship: taste, color change, mutual destruction, and

response to electric current. This last property is related, but in a less obvious way. The property

below identifies a unique chemical reaction that acids and bases engage in.

Acid Property #5:

Upon chemically reacting with an active metal, acids will evolve hydrogen gas (H2). The key word, of

course, is active. Some metals, like gold, silver or platinum, are rather unreactive and it takes rather

extreme conditions to get these "unreactive" metals to react. Not so with the metals in this property.

They include the alkali metals (Group I, Li to Rb), the alkaline earth metals (Group II, Be to Ra), as

well as zinc and aluminum. Just bring the acid and the metal together at anything close to room

temperature and you get a reaction. Here's a sample reaction:

Zn + 2 HCl(aq) ---> ZnCl2 + H2

Another common acid reaction some sources mention is that acids react with carbonates (and

bicarbonates) to give carbon dioxide gas:

HCl + NaCO3 ---> CO2 + H2O + NaCl

Base Property #5:

Bases feel slippery, sometimes people say soapy. This is because they dissolve the fatty acids and

oils from your skin and this cuts down on the friction between your fingers as you rub them together.

In essence, the base is making soap out of you. Yes, bases are involved in the production of soap! In

the early years of soap making, the soaps were very harsh on the skin and clothes due to the high

base content. Even today, people with very sensitive skin must sometimes use a non-soap-based

product for bathing.

It was not until more modern times that the chemical nature (as opposed to observable properties) of

acids and bases began to be explored. That leads to this property that is not directly observable by

the senses.

Acid Property #6:

Acids produce hydrogen ion (H + ) in solution. A more correct formula for what is produced is that of the

hydronium ion, H3O + . Both formulas are used interchangeably.

Acid base theories: Svante Arrhenius

I. Introduction

The basic idea is that certain substances remain ionized in solution all the time. Today, everyone

accepts this without question, but it was the subject of much dissention and disagreement in 1884,

when a twenty-five-year-old Arrhenius presented and defended his dissertation.

II. The Acid Base Theory

Acid - any substance which delivers hydrogen ion (H + ) to the solution.

Base - any substance which delivers hydroxide ion (OH¯) to the solution.

Here is a generic acid dissociating, according to Arrhenius:

HA ---> H + + A¯


This would be a generic base:

XOH ---> X + + OH¯

When acids and bases react according to this theory, they neutralize each other, forming water and a

salt:

HA + XOH ---> H2O + XA

Keeping in mind that the acid, the base and the salt all ionize, we can write this:

Finally, we can drop all spectator ions, to get this:

H + + A¯ + X + + OH¯ ---> H2O + X + + A¯

H + + OH¯ ---> H2O

These ideas covered all of the known acids at the time (the usual suspects like hydrochloric acid,

acetic acid, and so on) and most of the bases (sodium hydroxide, potassium hydroxide, calcium

hydroxide and so on). HOWEVER, and it is a big however, the theory did not explain why ammonia

(NH3) was a base. There are other problems with the theory also.

III. Problems with Arrhenius' Theory

1. The solvent has no role to play in Arrhenius' theory. An acid is expected to be an acid in any

solvent. This was found to not be the case. For example, HCl is an acid in water, behaving in

the manner Arrhenius expected. However, if HCl is dissolved in benzene, there is no

dissociation, the HCl remaining as un-dissociated molecules. The nature of the solvent plays a

critical role in acid-base properties of substances.

2. All salts in Arrhenius' theory should produce solutions that are neither acidic or basic. This is

not the case. If equal amounts of HCl and ammonia react, the solution is slightly acidic. If equal

amounts of acetic acid and sodium hydroxide are reacted, the resulting solution is basic.

Arrhenius had no explanation for this.

3. The need for hydroxide as the base led Arrhenius to propose the formula NH4OH as the

formula for ammonia in water. This led to the misconception that NH4OH is the actual base,

not NH3.

In fact, by 1896, several years before Arrhenius announced his theory, it had been recognized that

characteristic base properties where just as evident in such solvents as aniline, where no hydroxide

ions were possible.

4. H + , a bare proton, does not exist for very long in water. The proton affinity of H2O is about 799

kJ/mol. Consequently, this reaction:

H2O + H + ---> H3O +

happens to a very great degree. The "concentration" of free protons in water has been estimated to

be 10¯130 M. A rather preposterous value, indeed.

The Arrhenius theory of acids and bases will be fully supplanted by the theory proposed

independently by Johannes and Thomas Lowry in 1923.


The acid base theory of Brønsted and Lowry

I. Introduction

In 1923, within several months of each other, Johannes Nicolaus Brønsted (Denmark) and Thomas

Martin Lowry (England) published essentially the same theory about how acids and bases behave.

Since they came to their conclusions independently of each other, both names have been used for

the theory name.

II. The Acid Base Theory

Using the words of Brønsted:

". . . acids and bases are substances that are capable of splitting off or taking up hydrogen ions,

respectively."

Or an acid-base reaction consists of the transfer of a proton from an acid to a base. KEEP THIS

THOUGHT IN MIND!!

Here is a more recent way to say the same thing:

An acid is a substance from which a proton can be removed.

A base is a substance that can remove a proton from an acid.

Remember: proton, hydrogen ion and H + all mean the same thing

Very common in the chemistry world is this definition set:

An acid is a "proton donor."

A base is a "proton acceptor."

In an acid, the hydrogen ion is bonded to the rest of the molecule. It takes energy (sometimes a little,

sometimes a lot) to break that bond. So the acid molecule does not "give" or "donate" the proton, it

has it taken away. In the same sense, you do not donate your wallet to the pickpocket, you have it

removed from you.

The base is a molecule with a built-in "drive" to collect protons. As soon as the base approaches the

acid, it will (if it is strong enough) rip the proton off the acid molecule and add it to itself.

Now this is where all the fun stuff comes in that you get to learn. You see, some bases are stronger

than others, meaning some have a large "desire" for protons, while other bases have a weaker drive.

It's the same way with acids, some have very weak bonds and the proton is easy to pick off, while

other acids have stronger bonds, making it harder to "get the proton."

One important contribution coming from Lowry has to do with the state of the hydrogen ion in solution.

In Brønsted's announcement of the theory, he used H + . Lowry, in his paper (actually a long letter to

the editor) used the H3O + that is commonly used today.

III. Sample Equations written in the Brønsted-Lowry Style

A. Reactions that proceed to a large extent:


HCl + H2O ⇌ H3O + + Cl¯

HCl - this is an acid, because it has a proton available to be transferred.

H2O - this is a base, since it gets the proton that the acid lost.

Now, here comes an interesting idea:

H3O + - this is an acid, because it can give a proton.

Cl¯ - this is a base, since it has the capacity to receive a proton.

Notice that each pair (HCl and Cl¯ as well as H2O and H3O + differ by one proton (symbol = H + ). These

pairs are called conjugate pairs.

HNO3 + H2O ⇌ H3O + + NO3¯

The acids are HNO3 and H3O + and the bases are H2O and NO3¯.

Remember that an acid-base reaction is a competition between two bases (think about it!) for a

proton. If the stronger of the two acids and the stronger of the two bases are reactants (appear on the

left side of the equation), the reaction is said to proceed to a large extent.

Here are some more conjugate acid-base pairs to look for:

H2O and OH¯

HCO3¯ and CO3 2¯

H2PO4¯ and HPO4 2¯

HSO4¯ and SO4 2¯

NH4 + and NH3

CH3NH3 + and CH3NH2

HC2H3O2 and C2H3O2¯

B. Reactions that proceed to a small extent:

If the weaker of the two acids and the weaker of the two bases are reactants (appear on the left side

of the equation), the reaction is said to proceed to only a small extent:

HC2H3O2 + H2O ⇌ H3O + + C2H3O2¯

NH3 + H2O ⇌ NH4 + + OH¯

Identify the conjugate acid base pairs in each reaction.

HC 2H 3O 2 and C 2H 3O 2¯

is one conjugate pair.

H 2O and H 3O + is the other.

NH 3 and NH 4

+

is one pair.

H 2O and OH¯ is the other.

Notice that H 2O in the first equation is acting as a base and in the second equation is acting as an acid.


IV. Problems with the Theory

This theory works very nicely in all protic solvents (water, ammonia, acetic acid, etc.), but fails to

explain acid base behavior in aprotic solvents such as benzene and dioxane. That job will be left for a

more general theory, such as the Lewis Theory of Acids and Bases.

The Lewis theory of acids and bases

I. Introduction

Lewis gives his definition of an acid and a base:

"We are inclined to think of substances as possessing acid or basic properties, without having a

particular solvent in mind. It seems to me that with complete generality we may say that a basic

substance is one which has a lone pair of electrons which may be used to complete the stable group

of another atom, and that an acid is one which can employ a lone pair from another molecule in

completing the stable group of one of its own atoms."

"In other words, the basic substance furnishes a pair of electrons for a chemical bond, the acid

substance accepts such a pair."

It is important to make two points here:

1. NO hydrogen ion need be involved.

2. NO solvent need be involved.

The Lewis theory of acids and bases is more general than the "one sided" nature of the Bronsted-

Lowry theory. Keep in mind that Bronsted-Lowry, which defines an acid as a proton donor and a base

as a proton acceptor, REQUIRES the presence of a solvent, specifically a protic solvent, of which

water is the usual example. Since almost all chemistry is done in water, the fact that this limits the

Bronsted-Lowry definition is of little practical consequence.

The Lewis definitions of acid and base do not have the constraints that the Bronsted-Lowry theory

does and, as we shall see, many more reactions were seen to be acid base in nature using the Lewis

definition than when using the Bronsted-Lowry definitions.

II. The Acid Base Theory

The modern way to define a Lewis acid and base is a bit more concise than above:

Acid: an electron acceptor.

Base: an electron donor.

A "Lewis acid" is any atom, ion, or molecule which can accept electrons and a "Lewis base" is any

atom, ion, or molecule capable of donating electrons. However, a warning: many textbooks will say

"electron pair" where I have only written "electron." The truth is that it sometimes is an electron pair

and sometimes it is not.

It turns out that it may be more accurate to say that "Lewis acids" are substances which are electrondeficient

(or low electron density) and "Lewis bases" are substances which are electron-rich (or high

electron density).


Several categories of substances can be considered Lewis acids:

1. positive ions

2. having less than a full octet in the valence shell

3. polar double bonds (one end)

4. expandable valence shells

Several categories of substances can be considered Lewis bases:

1. negative ions

2. one of more unshared pairs in the valence shell

3. polar double bonds (the other end)

4. the presence of a double bond

Sören Sörenson and the pH scale

I. Short Historical Introduction

In the late 1880's, Svante Arrhenius proposed that acids were substances that delivered hydrogen ion

to the solution. He has also pointed out that the law of mass action could be applied to ionic

reactions, such as an acid dissociating into hydrogen ion and a negatively charged anion.

This idea was followed up by Wilhelm Ostwald, who calculated the dissociation constants (the

modern symbol is Ka) of many weak acids. Ostwald also showed that the size of the constant is a

measure of an acid's strength.

By 1894, the dissociation constant of water (today called Kw) was measured to the modern value of

1 x 10¯14 .

In 1904, H. Friedenthal recommended that the hydrogen ion concentration be used to characterize

solutions. He also pointed out that alkaline (modern word = basic) solutions could also be

characterized this way since the hydroxyl concentration was always 1 x 10¯14 ÷ the hydrogen ion

concentration. Many consider this to be the real introduction of the pH scale.

III. The Introduction of pH

Sörenson defined pH as the negative logarithm of the hydrogen ion concentration.

pH = - log [H + ]

Remember that sometimes H3O + is written, so

pH = - log [H3O + ]

means the same thing.

So let's try a simple problem: The [H + ] in a solution is measured to be 0.010 M. What is the pH?

The solution is pretty straightforward. Plug the [H + ] into the pH definition:

pH = - log 0.010

An alternate way to write this is:

pH = - log 10¯2

Since the log of 10¯2 is -2, we have:

pH = - (- 2)

Which, of course, is 2.


Let's discuss significant figures and pH.

Another sample problem: Calculate the pH of a solution in which the [H3O + ] is 1.20 x 10¯3 M.

For the solution, we have:

pH = - log 1.20 x 10¯3

This problem can be done very easily using your calculator. However, be warned about putting

numbers into the calculator.

So you enter (-), log, 1.20, X10 n , (-), 3, enter.

The answer, to the proper number of significant digits is: 2.921.

III. Significant Figures in pH

Here is the example problem: Calculate the pH of a solution where the [H + ] is 0.00100 M. (This could

also be a pOH problem. The point being made is the same.)

OK, you say, that's pretty easy, the answer is 3. After all, 0.00100 is 10¯3 and the negative log of 10¯3

is 3.

You would be graded wrong!! Why? Because the pH is not written to reflect the number of significant

figures in the concentration.

Notice that there are three sig figs in 0.00100. (Hopefully you remember significant figures, since you

probably studied them months ago before getting to acid base stuff. THEY ARE STILL IMPORTANT!)

So, our pH value should also reflect three significant figures.

However, there is a special rule to remember with pH (and pOH) values. The whole number portion

DOES NOT COUNT when figuring out how many digits to write down.

Let's phrase that another way: in a pH (and a pOH), the only place where significant figures are

contained is in the decimal portion.

So, the correct answer to the above problem is 3.000. Three sig figs and they are all in the decimal

portion, NOT (I repeat NOT) in the whole number portion.

Practice Problems

Convert each hydrogen ion concentration into a pH. Identify each as an acidic pH or a basic pH.

1. 0.0015

2. 5.0 x 10¯9

3. 1.0

4. 3.27 x 10¯4

5. 1.00 x 10¯12

6. 0.00010


1. 2.82

2. 8.30

3. 0.00

4. 3.485

5. 12.000

6. 4.00

Sörenson also just mentions the reverse direction. That is, suppose you know the pH and you want to

get to the hydrogen ion concentration ([H + ])?

Here is the equation for that:

[H + ] = 10¯pH

That's right, ten to the minus pH gets you back to the [H + ] (called the hydrogen ion concentration).

This is actually pretty easy to do with the calculator. Here's the sample problem: calculate the [H + ]

from a pH of 2.45.

This problem can be done very easily using your calculator. However, be warned about putting

numbers into the calculator.

So you enter 2nd, 10 x , (-), 2.45, enter.

The answer, to the proper number of significant digits is: .00355.

The pH of an acidic pond is 5. What is the hydrogen ion concentration (moles per liter)?

The answer is:

pH = -log (hydrogen ion concentration)

The answer was .00001. Thus, 5 = -log (.00001).

We'll take the formula that you started with (pH = -log([H+])) and work to the answer (solve for [H+]).

pH = - log ([H+]) Given.

pH = log ([H+] (-1) ) Since logarithms are like exponents, when you multiply a log by

something, you can just move it to the inside of log as an exponent.

10 pH = 10 log ([H+] (-1)) Take each side to tenth power.

10 pH = [H+] (-1) Since "log" is just another notation for "log base 10", when you

raise a log to the tenth power, the log cancels out.

[H+] = 10 (-pH)

Take the reciprocal of both sides.

That is the general form. To answer the specific question,

5 = - log ([H+])

5 = log ([H+] (-1) )

10 5 = [H+] (-1)

10 (-5) = [H+]

[H+]

= .00001 mol/L


On your calculator you would input 10, ^, (-), 5 and you would get 0.00001.

This is also the way to find the amount of OH + that are present in a base.

To find the pH: -log(concentration)

To find the concentration: 10 -pH

Define these terms:

pH scale

Hydronium ion

Arrhenius acid/base

Lewis acid/base

Bronsted-Lowry acid/base

Strong acid/base

Weak acid/base

Neutralization reaction

Titration


The Learning Goal for this assignment is:

1. Oxidation Numbers

Redox Reactions

Oxidation and reduction

Every atom, ion or polyatomic ion has a formal oxidation number associated with it. This value

compares the number of protons in an atom (positive charge) and the number of electrons assigned

to that atom (negative charge).

In many cases, the oxidation number reflects the actual charge on the atom, but there are many

cases where it does not. Think of oxidation numbers as a bookkeeping exercise simply to keep track

of where electrons go.

2. Reduction

Reduction means what it says: the oxidation number is reduced in reduction.

This is accomplished by adding electrons. The electrons, being negative, reduce the overall oxidation

number of the atom receiving the electrons.

3. Oxidation

Oxidation is the reverse process: the oxidation number of an atom is increased during oxidation.

This is done by removing electrons. The electrons, being negative, make the atom that lost them

more positive.

I use this mnemonic to help me remember which is which: LEO the lion says GER.

LEO = Loss of Electrons is Oxidation

GER = Gain of Electrons is Reduction

Another well-known mnemonic is this: OIL RIG

OIL = Oxidation Is Loss (of Electrons)

RIG = Reduction Is Gain (of Electrons)

Another way is to simply remember that reduction is to reduce the oxidation number. Therefore,

oxidation must increase the value.

4. Reduction-Oxidation Reactions

There are many chemical reactions in which one substance gets reduced in oxidation number

(reduction) while another participating substance gets increased in oxidation number (oxidation).

Such a reaction is called called a REDOX reaction. The RED, of course, comes from REDuction and

OX from OXidation. However, it is pronounced re-dox and not red-ox.

Here is a simple example of a redox reaction:


Ag+ + Cu ---> Ag + Cu2+

I have deliberately not balanced it and I have also written it in net ionic form. I have found that kids

studying redox get confused by net ionic form and how to change a full equation into net ionic form.

Redox equations need to be balanced but, except for the simplest ones, it cannot be done by

inspection (also called trial and error). I take that back, complex ones can be done by trial and error. It

typically takes quite a bit of work, especially when compared to how long it takes when the proper

technique is used.

There is a technique used to balance redox reactions. It is called "balancing by half-reactions." The

basic plan will be to split the full equation into two simpler parts (called half-reactions), balance them

following several standard steps, then recombine the balanced half-reactions into the final answer.

This is another technique called the "ion-electron method." I plan to ignore it.

Notes:

5. Some Definitions

Oxidizing Agent - that substance which oxidizes somebody else. It is reduced in the process.

Reducing Agent - that substance which reduces somebody else. It is oxidized in the process.

It helps me to remember these definitions by the opposite nature of what happens. By that, I mean

the oxidizing agent gets reduced and the reducing agent gets oxidized.

6. Rules for Assigning Oxidation Numbers

The Oxidation Number of an element corresponds to the number of electrons, e - , that an atom loses,

gains, or appears to use when joining with other atoms in compounds. In determining the Oxidation

Number of an atom, there are seven guidelines to follow:

1. The Oxidation Number of an individual atom is 0. This includes diatomic elements such as

O2 or others like P4 and S8

2. The total Oxidation Number of all atoms in: a neutral species is 0 and in an ion is equal to the

ion charge.

3. Group 1 metals have an Oxidation Number of +1 and Group 2 an Oxidation Number of +2

4. The Oxidation Number of fluorine is -1 in compounds

5. Hydrogen generally has an Oxidation Number of +1 in compounds, except hydrides

6. Oxygen generally has an Oxidation Number of -2 in compounds, except peroxides

7. In binary metal compounds, Group 17 elements have an Oxidation Number of -1, Group 16 of -

2, and Group 15 of -3.

Note: The sum of the Oxidation Number s is equal to zero for neutral compounds and equal to the

charge for polyatomic ion species.

Now, some examples:

1. What is the oxidation number of Cl in HCl?

Since H = +1, the Cl must be -1 (minus one).


2. What is the oxidation number of Na in Na2O?

Since O = -2, the two Na must each be +1.

3. What is the oxidation number of Cl in ClO¯?

The O is -2, but since a -1 must be left over, then the Cl is +1.

4. What is the oxidation number for each element in KMnO4?

K = +1 because KCl exists. We know the Cl = -1 because HCl exists.

O = -2 by definition

Mn = +7. There are 4 oxygens for a total of -8, K is +1, so Mn must be the rest.

5. What is the oxidation number of S in SO4 2¯

O = -2. There are four oxygens for -8 total. Since -2 must be left over, the S must = +6.

Please note that, if there is no charge indicated on a formula, the total charge is taken to be zero.

Practice Problems

Find oxidation numbers

1. N in NO3¯

2. C in CO3 2¯

3. Cr in CrO4 2¯

4. Cr in Cr2O7 2¯

5. Fe in Fe2O3

6. Pb in PbOH +

7. V in VO2 +

8. V in VO 2+

9. Mn in MnO4¯

10. Mn in MnO4 2¯

Notes:


7. Half Reactions

A half-reaction is simply one which shows either reduction OR oxidation, but not both. Here is the

example redox reaction used in a different file:

Ag + + Cu → Ag + Cu 2+

It has BOTH a reduction and an oxidation in it. That is why we call it a redox reaction, from REDuction

and OXidation.

What you must be able to do is look at a redox reaction and separate out the two half-reactions in it.

To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions

from the above example:

Ag+ → Ag

Cu → Cu 2+

The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation

number went from zero to +2, so it was oxidized in the reaction. In order to figure out the halfreactions,

you MUST be able to calculate the oxidation number of an atom.

Keep in mind that a half-reaction shows only one of the two behaviors we are studying. A single halfreaction

will show ONLY reduction or ONLY oxidation, never both in the same equation.

Also, notice that the reaction is read from left to right to determine if it is reduction or oxidation. If you

read the reaction in the opposite direction (from right to left) it then becomes the other of our two

choices (reduction or oxidation). For example, the silver half-reaction above is a reduction, but in the

reverse direction it is an oxidation, going from zero on the right to +1 on the left.

There will be times when you want to switch a half-reaction from one of the two types to the other. In

that case, rewrite the entire equation and swap sides for everything involved. If I needed the silver

half-reaction to be oxidation, I'd write Ag → Ag+ rather than just doing it mentally.

The next step is that both half-reactions must be balanced. However, there is a twist. When you

learned about balancing equation, you made equal the number of atoms of each element on each

side of the arrow. That still applies, but there is one more thing: the total amount of charge on each

side of the half-reaction MUST be the same.

When you look at the two half-reactions above, you will see they are already balanced for atoms with

one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. To do

this you add electrons to the more positive side. You add enough to make the total charge on each

side become EQUAL.

To the silver half-reaction, we add one electron:

To the copper half-reaction, we add two electrons:

Ag+ + e¯ ---> Ag

Cu ---> Cu 2+ + 2e¯


One point of concern: notice that each half-reaction wound up with a total charge of zero on each

side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not

zero.

One more point to make before wrapping this up. A half-reaction is a "fake" chemical reaction. It's just

a bookkeeping exercise. Half-reactions NEVER occur alone. If a reduction half-reaction is actually

happening (say in a beaker in front of you), then an oxidation reaction is also occurring. The two halfreactions

can be in separate containers, but they do have to have some type of "chemical

connection" between them.

Half-Reaction Practice Problems

Balance each half-reaction for atoms and charge:

1) Cl2 → Cl¯

2) Sn → Sn 2+

3) Fe 2+ → Fe 3+

4) I3¯ → I¯

5) ICl2¯ → I¯ (I'm being mean on this one. Hint: the iodine is the only thing reduced or oxidized.)

Separate each of these redox reactions into their two half-reactions (but do not balance):

6) Sn + NO3¯ →SnO2 + NO2

7) HClO + Co →Cl2 + Co 2+

8) NO2 →NO3¯ + NO

Here are the two half-reactions to be combined:

Here is the rule to follow:

Ag+ + e¯ → Ag

Cu → Cu 2+ + 2e¯

the total electrons MUST cancel when the two half-reactions are added.

Another way to say it:

the number of electrons in each half-reaction MUST be equal when the two half-reactions are

added.

What that means is that one (or both) equations must be multiplied through by a factor. The value of

the factor is selected so as to make the number of electrons equal.

In our example problem, the top reaction (the one with silver) must be multiplied by two, like this:

2Ag+ + 2e¯ → 2Ag


Notice that each separate substance is increased by the factor amount. Occasionally, a student will

multiply ONLY the electrons by the factor. That is incorrect.

When the two half-reactions are added, we get:

2Ag+ + 2e¯ + Cu → 2Ag + Cu 2+ + 2e¯

With two electrons on each side, they may be canceled, resulting in:

2Ag+ + Cu → 2Ag + Cu 2+

This is the correct answer. Notice that there are two silvers on each side and one copper. Notice also

that the total charge on each side is +2. It is balanced for both atoms and charge. Sometimes, I am

asked if the order matters, if the Cu could be first on the left-hand side. The answer is that the order

does not matter. There happen to be certain styles about where particular substances are put in the

final answer, but these are only styles. They do not affect the chemical correctness of the answer.

Notes:

Practice Problems

Separate into half-reactions, balance them and then recombine.

1) Sn + Cl2 ---> Sn 2+ + Cl¯

2) Fe 2+ + I3¯ ---> Fe 3+ + I¯

1. N in NO 3¯

The O is -2 and three of them makes -6. Since -1 is left over, the N must be +5

2. C in CO 3


The O is -2 and three of them makes -6. Since -2 is left over, the C must be +4

3. Cr in CrO 4


The O is -2 and four of them makes -8. Since -2 is left over, the Cr must be +6

4. Cr in Cr 2 O 7


The O is -2 and seven of them makes -14. Since -2 is left over, the two Cr must be +12

What follows is an important point. Each Cr is +6. Each one. We do not speak of Cr 2 being +12. Each Cr atom is considered individually.

5. Fe in Fe 2 O 3

The O is -2 and three of them makes -6. Each Fe must then be +3

6. +2 9. +7

7. +5 10. +6

8. +4

1) Cl 2 + 2e¯ →2Cl¯ 5) ICl 2¯

+ 2e¯ →I¯ + 2Cl¯

2) Sn →Sn 2+ + 2e¯ 6) Sn →SnO 2 and NO 3¯

→NO 2

3) Fe 2+ →Fe 3+ + e¯ 7) HClO →Cl 2 and Co ---> Co 2+

4) I 3¯ → 3I¯ + 2e¯ 8) NO 2 →NO 3¯

and NO 2 →NO

Note that in this last redox equation the NO 2 (actually just the N) is both being reduced AND being oxidized. In the first half-reaction, the N goes from +4

to +5 (oxidation) and in the second, the N goes from +4 to +2 (reduction).

By the way, we could flip the reaction so that NO 3¯

and NO are reacting together to produce only one product, the NO 2 . In that case, the two halfreactions

would be reversed.


Reference Tables for Physical Setting/CHEMISTRY

Table A

Standard Temperature and Pressure

Name Value Unit

Standard Pressure 101.3 kPa kilopascal

1 atm atmosphere

Standard Temperature 273 K kelvin

0°C degree Celsius

Table D

Selected Units

Symbol Name Quantity

m meter length

g gram mass

Pa pascal pressure

K kelvin temperature

Table B

Physical Constants for Water

mol

J

mole

joule

amount of

substance

energy, work,

quantity of heat

Heat of Fusion

Heat of Vaporization

Specific Heat Capacity of H 2

O()

Specific Heat Capacity of H 2

O(s)

Specific Heat Capacity of H 2

O(g)

Table C

Selected Prefixes

Factor Prefix Symbol

10 3 kilo- k

334 J/g

2260 J/g

4.18 J/g•K

2.10 J/g•K

2.01 J/g•K

s second time

min minute time

h hour time

d day time

y year time

L liter volume

ppm parts per million concentration

M

molarity

solution

concentration

u atomic mass unit atomic mass

10 –1 deci- d

10 –2 centi- c

10 –3 milli- m

10 –6 micro- μ

10 –9 nano- n

10 –12 pico- p

R1


Table E

Selected Polyatomic Ions

Formula Name Formula Name

H 3

O +

hydronium

CrO 4

2–

chromate

Hg 2

2+

mercury(I)

Cr 2

O 7

2–

dichromate

NH 4

+

C 2

H 3

O


2 –}

CH 3

COO

CN –

CO 3

2–

HCO


3

C 2

O

2–

4

ClO –

ammonium

acetate

cyanide

carbonate

hydrogen

carbonate

oxalate

hypochlorite

MnO 4


NO


2

NO


3

O

2–

2

OH –

PO 4

3–

SCN –

SO 3

2–

permanganate

nitrite

nitrate

peroxide

hydroxide

phosphate

thiocyanate

sulfite

ClO 2


chlorite

SO 4

2–

sulfate

ClO 3


chlorate

HSO 4


hydrogen sulfate

ClO 4


perchlorate

S 2

O 3

2–

thiosulfate

Table F

Solubility Guidelines for Aqueous Solutions

Ions That Form

Soluble Compounds

Group 1 ions

(Li + , Na + , etc.)

ammonium (NH 4 + )

nitrate (NO 3 – )

acetate (C 2

H 3

O 2 – or

CH 3

COO – )

hydrogen carbonate

(HCO 3 – )

chlorate (ClO 3 – )

halides (Cl – , Br – , I – )

Exceptions

when combined with

Ag + , Pb 2+ , or Hg 2

2+

sulfates (SO 4 2– ) when combined with Ag + ,

Ca 2+ , Sr 2+ , Ba 2+ , or Pb 2+

Ions That Form

Insoluble Compounds*

Exceptions

carbonate (CO 3 2– ) when combined with Group 1

ions or ammonium (NH 4 + )

chromate (CrO 4 2– ) when combined with Group 1

ions, Ca 2+ , Mg 2+ , or

ammonium (NH 4 + )

phosphate (PO 4 3– ) when combined with Group 1

ions or ammonium (NH 4 + )

sulfide (S 2– ) when combined with Group 1

ions or ammonium (NH 4 + )

hydroxide (OH – ) when combined with Group 1

ions, Ca 2+ , Ba 2+ , Sr 2+ , or

ammonium (NH 4 + )

*compounds having very low solubility in H 2 O

R2


150.

140.

Table G

Solubility Curves at Standard Pressure

KI

NaNO 3

130.

120.

KNO 3

110.

100.

Solubility (g solute/100. g H 2

O)

90.

80.

70.

60.

HCl

NH 4

Cl

KCl

50.

40.

30.

NaCl

KClO 3

NH 3

20.

10.

SO 2

0

0 10. 20. 30. 40. 50. 60. 70. 80. 90. 100.

Temperature (°C)

R3


Table H

Vapor Pressure of Four Liquids

200.

propanone

ethanol

150.

water

Vapor Pressure (kPa)

100.

101.3 kPa

ethanoic

acid

50.

0

0 25 50. 75 100. 125

R4


Table I

Heats of Reaction at 101.3 kPa and 298 K

Reaction

ΔH (kJ)*

CH 4

(g) + 2O 2

(g) CO 2

(g) + 2H 2

O() –890.4

C 3

H 8

(g) + 5O 2

(g) 3CO 2

(g) + 4H 2

O() –2219.2

2C 8

H 18

() + 25O 2

(g) 16CO 2

(g) + 18H 2

O() –10943

2CH 3

OH() + 3O 2

(g) 2CO 2

(g) + 4H 2

O() –1452

C 2

H 5

OH() + 3O 2

(g) 2CO 2

(g) + 3H 2

O() –1367

C 6

H 12

O 6

(s) + 6O 2

(g) 6CO 2

(g) + 6H 2

O() –2804

2CO(g) + O 2

(g) 2CO 2

(g) –566.0

C(s) + O 2

(g) CO 2

(g) –393.5

4Al(s) + 3O 2

(g) 2Al 2

O 3

(s) –3351

N 2

(g) + O 2

(g) 2NO(g) +182.6

N 2

(g) + 2O 2

(g) 2NO 2

(g) +66.4

2H 2

(g) + O 2

(g) 2H 2

O(g) –483.6

2H 2

(g) + O 2

(g) 2H 2

O() –571.6

N 2

(g) + 3H 2

(g) 2NH 3

(g) –91.8

2C(s) + 3H 2

(g) C 2

H 6

(g) –84.0

2C(s) + 2H 2

(g) C 2

H 4

(g) +52.4

2C(s) + H 2

(g) C 2

H 2

(g) +227.4

H 2

(g) + I 2

(g) 2HI(g) +53.0

KNO 3

(s) H 2 O K + (aq) + NO 3 – (aq) +34.89

NaOH(s) H 2 O Na + (aq) + OH – (aq) –44.51

NH 4

Cl(s) H 2 O NH 4 + (aq) + Cl – (aq) +14.78

NH 4

NO 3

(s) H 2 O NH 4 + (aq) + NO 3 – (aq) +25.69

NaCl(s) H 2 O Na + (aq) + Cl – (aq) +3.88

LiBr(s) H 2 O Li + (aq) + Br – (aq) –48.83

H + (aq) + OH – (aq) H 2

O() –55.8

*The ΔH values are based on molar quantities represented in the equations.

A minus sign indicates an exothermic reaction.

Most

Active

Least

Active

Table J

Activity Series**

Metals Nonmetals Most

Active

Li

Rb

K

Cs

Ba

Sr

Ca

Na

Mg

Al

Ti

Mn

Zn

Cr

Fe

Co

Ni

Sn

Pb

H 2

Cu

Ag

Au

F 2

Cl 2

Br 2

I 2

**Activity Series is based on the hydrogen

standard. H 2 is not a metal.

Least

Active

R5


Table K

Common Acids

Table N

Selected Radioisotopes

HCl(aq)

Formula

HNO 2

(aq)

HNO 3

(aq)

H 2

SO 3

(aq)

H 2

SO 4

(aq)

H 3

PO 4

(aq)

H 2

CO 3

(aq)

or

CO 2

(aq)

CH 3

COOH(aq)

or

HC 2

H 3

O 2

(aq)

Name

hydrochloric acid

nitrous acid

nitric acid

sulfurous acid

sulfuric acid

phosphoric acid

carbonic acid

ethanoic acid

(acetic acid)

Nuclide Half-Life Decay

Mode

Nuclide

Name

198 Au 2.695 d β – gold-198

14 C 5715 y β – carbon-14

37 Ca 182 ms β + calcium-37

60 Co 5.271 y β – cobalt-60

137 Cs 30.2 y β – cesium-137

53 Fe 8.51 min β + iron-53

220 Fr 27.4 s α francium-220

3 H 12.31 y β – hydrogen-3

131 I 8.021 d β – iodine-131

37 K 1.23 s β + potassium-37

42 K 12.36 h β – potassium-42

Table L

Common Bases

85 Kr 10.73 y β – krypton-85

16 N 7.13 s β – nitrogen-16

Formula

NaOH(aq)

KOH(aq)

Ca(OH) 2

(aq)

NH 3

(aq)

Name

sodium hydroxide

potassium hydroxide

calcium hydroxide

aqueous ammonia

19 Ne 17.22 s β + neon-19

32 P 14.28 d β – phosphorus-32

239 Pu 2.410 × 10 4 y α plutonium-239

226 Ra 1599 y α radium-226

222 Rn 3.823 d α radon-222

90 Sr 29.1 y β – strontium-90

Table M

Common Acid–Base Indicators

Approximate

Indicator pH Range Color

for Color Change

Change

methyl orange 3.1–4.4 red to yellow

bromthymol blue 6.0–7.6 yellow to blue

phenolphthalein 8–9 colorless to pink

litmus 4.5–8.3 red to blue

bromcresol green 3.8–5.4 yellow to blue

thymol blue 8.0–9.6 yellow to blue

99 Tc 2.13 × 10 5 y β – technetium-99

232 Th 1.40 × 10 10 y α thorium-232

233 U 1.592 × 10 5 y α uranium-233

235 U 7.04 × 10 8 y α uranium-235

238 U 4.47 × 10 9 y α uranium-238

Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011,

CRC Press

Source: The Merck Index, 14 th ed., 2006, Merck Publishing Group

R6


Table O

Symbols Used in Nuclear Chemistry

Name Notation Symbol

alpha particle

4

2

He or 4 2 α α

beta particle

0

–1

e or 0

–1 β β–

gamma radiation

0

0

γ γ

neutron

1

0

n n

proton

1

1

H or 1 1 p p

positron

0

+1

e or 0

+1 β β+

Table P

Organic Prefixes

Prefix

meth- 1

eth- 2

prop- 3

but- 4

pent- 5

hex- 6

hept- 7

oct- 8

non- 9

dec- 10

Number of

Carbon Atoms

Table Q

Homologous Series of Hydrocarbons

Name General Examples

Formula Name Structural Formula

R7

alkanes C n

H 2n+2

ethane

alkenes C n

H 2n

ethene

alkynes C n

H 2n–2

ethyne

Note: n = number of carbon atoms

H H

H C C H

H H

H

H

C C

H

H

H C C H


Table R

Organic Functional Groups

Class of

Compound

Functional

Group

General

Formula

Example

halide

(halocarbon)

F (fluoro-)

Cl (chloro-)

Br (bromo-)

I (iodo-)

R X

(X represents

any halogen)

CH 3

CHClCH 3

2-chloropropane

alcohol

OH

R

OH

CH 3

CH 2

CH 2

OH

1-propanol

ether

O

R O R′

CH 3

OCH 2

CH 3

methyl ethyl ether

aldehyde

O

C H

R

O

C H

O

CH 3

CH 2

C H

propanal

ketone

O

C

O

R C R′

O

CH 3

CCH 2

CH 2

CH 3

2-pentanone

organic acid

O

C OH

R

O

C OH

O

CH 3

CH 2

C OH

propanoic acid

ester

O

C O

O

R C O R′

O

CH 3

CH 2

COCH 3

methyl propanoate

amine

N

R

R′

N R′′

CH 3

CH 2

CH 2

NH 2

1-propanamine

amide

O

C NH

R

O R′

C NH

O

CH 3

CH 2

C NH 2

propanamide

Note: R represents a bonded atom or group of atoms.

R8


0

6.941

+1

Li

3

2-1

Na

39.0983

K +1

19

2-8-8-1

85.4678 +1

Rb

Cs

(223)

Fr

87

-18-32-18-8-1

+1

Ra

88

-18-32-18-8-2

39

138.9055

La

57

2-8-18-18-9-2

+2 (227)

Ac

89

-18-32-18-9-2

47.867

Ti

22

2-8-10-2

91.224

Zr

40

2-8-18-10-2

+3 178.49

Hf

72

*18-32-10-2

+3 (261)

Rf

104

+2

+3

+4

+4

+4

50.9415

V

23

2-8-11-2

+2

+3

+4

+5

51.996

Cr

24

2-8-13-1

95.94

Mo

42

2-8-18-13-1

183.84

W

74

-18-32-12-2

+2

+3

+6

+6

+6

54.9380

Mn

25

2-8-13-2

+2

+3

+4

+7

55.845

Fe

26

2-8-14-2

+2

+3 58.9332

Co

27

2-8-15-2

+2

+3

58.693

Ni

28

2-8-16-2

+2

+3 63.546 Cu

2-8-18-1

107.868

Ag

47

2-8-18-18-1

79

+1

+2

+1

65.409

Zn

30

2-8-18-2

10.81

+3 12.011

B

5

2-3

26.98154

Al

13

2-8-3

+2 69.723

Ga

31

2-8-18-3

+3

+3

–4

+2

+4

C

6

2-4

28.0855

Si

14

2-8-4

72.64

Ge

32

2-8-18-4

Pb

–4

+2

+4

+2

+4

74.9216

As

33

2-8-18-5

Sb

–3

+3

15.9994 O

–2 18.9984

8

2-6 2-7

78.96

Se

34

2-8-18-6

127.60

Te

52

2-8-18-18-6

(209)

Po

84

-18-32-18-6

–2

+4

+6

–2

+4

+6

+2

+4

F

79.904

Br

35

2-8-18-7

126.904

l

53

2-8-18-18-7

(210)

At

85

-18-32-18-7

( ? )

Uus

117

4.00260 0

He

2

2

–1 20.180

Ne

10

2-8

0

22.98977

11

2-8-1

1

1.00794 +1

–1

H

1

1

1

37

2-8-18-8-1

–1

+1

+5

–1

+1

+5

+7

83.798

Kr

36

2-8-18-8

131.29

Xe

54

2-8-18-18-8

(222)

Rn

86

-18-32-18-8

0

+2

0

+2

+4

+6

0

132.905

55

2-8-18-18-8-1

Symbol

Relative atomic masses are based

Group on 12 C = 12 (exact)

Group

2

13 14 15 16 17 18

Atomic Number

+1

+1

9.01218 +2

Be

4

2-2

24.305

Mg

12

2-8-2

40.08

Ca

20

2-8-8-2

87.62

Sr

38

2-8-18-8-2

137.33

Ba

56

2-8-18-18-8-2

(226)

+2

+2

+2

+2

3

44.9559

Sc

21

2-8-9-2

88.9059

Y

2-8-18-9-2

+3

+3

4

KEY

92.9064

Nb +3

+5

41

2-8-18-12-1

180.948

Ta

73

-18-32-11-2

(262)

105

5

Periodic Table of the Elements

Atomic Mass

Electron Configuration

+4

Db

+5

6

(266)

Sg

106

12.011 2-4

–4

6

C

+2

+4

(98)

Tc

43

2-8-18-13-2

186.207

Re

75

-18-32-13-2

(272)

Bh

107

7

Group

+4

+6

+7

+4

+6

+7

8

101.07

Ru

44

2-8-18-15-1

190.23

Os

76

-18-32-14-2

(277)

Hs

108

+3

+3

+4

Selected Oxidation States

Note: Numbers in parentheses

are mass numbers of the most

stable or common isotope.

9

102.906

Rh

45

2-8-18-16-1

192.217

Ir

77

-18-32-15-2

(276)

Mt

109

+3

+3

+4

106.42

Pd

46

2-8-18-18

195.08

Pt

78

-18-32-17-1

+2

+4

+2

+4

196.967

Au

-18-32-18-1

(281)

Ds (280) Rg

110

10

29

111

11 12

+1

+3

112.41

Cd

48

2-8-18-18-2

200.59

Hg

80

-18-32-18-2

(285)

Cn

112

+2 114.818

In

+1

+2

49

2-8-18-18-3

204.383

Tl

81

-18-32-18-3

(284)

Uut

113**

+3

+1

+3

118.71

Sn

50

2-8-18-18-4

207.2

82

-18-32-18-4

(289)

Uuq

114

+2

+4

+2

+4

14.0067 –3

–2

N

–1

7

2-5

30.97376

P

15

2-8-5

121.760

51

2-8-18-18-5

208.980

Bi

83

-18-32-18-5

(288)

Uup

115

+1

+2

+3

+4

+5

–3

+3

+5

+5

–3

+3

+5

+3

+5

32.065

S

16

2-8-6

(292)

Uuh

116

–2

+4

+6

35.453

Cl

17

2-8-7

–1

+1

+5

+7

39.948

Ar

18

2-8-8

18

(294)

Uuo

118

140.116

Ce

58

232.038

Th

90

+3

+4

140.908

Pr +3

59

144.24

Nd

60

+4 231.036

Pa +4 238.029 +5

U +3

+4

+5

+6

91

92

+3

(145)

Pm

61

+3

150.36

Sm

62

+2

+3

151.964

Eu

63

+2

+3

157.25

Gd

64

+3

158.925

(237)Np (244) Pu (243) Am (247) Cm +3 (247) Bk +3

+3

+4

+5

+6

93 94

+3

+4

+5

+6

65

+3

+4

+5

+6

95 96 97

Tb

+3

+4

162.500

Dy

66

(251)

+3

164.930

Ho

67

+3

167.259

Er

68

Cf +3 (252) Es (257) Fm

100

98 99

+3

+3

+3

168.934

Tm +3

69

(258)

Md

101

+2

+3

173.04

Yb

70

(259)

No

102

+2

+3

+2

+3

174.9668

Lu

71

(262)

Lr

103

+3

+3

*denotes the presence of (2-8-) for elements 72 and above

**The systematic names and symbols for elements of atomic numbers 113 and above

will be used until the approval of trivial names by IUPAC.

Source: CRC Handbook of Chemistry and Physics, 91 st ed., 2010–2011, CRC Press

9

R9

Period

1

2

3

4

5

6

7


Table S

Properties of Selected Elements

First

Atomic Symbol Name Ionization

Electro- Melting Boiling* Density** Atomic

Number Energy negativity Point Point (g/cm 3 ) Radius

(kJ/mol) (K) (K) (pm)

1 H hydrogen 1312 2.2 14 20. 0.000082 32

2 He helium 2372 — — 4 0.000164 37

3 Li lithium 520. 1.0 454 1615 0.534 130.

4 Be beryllium 900. 1.6 1560. 2744 1.85 99

5 B boron 801 2.0 2348 4273 2.34 84

6 C carbon 1086 2.6 — — .— 75

7 N nitrogen 1402 3.0 63 77 0.001145 71

8 O oxygen 1314 3.4 54 90. 0.001308 64

9 F fluorine 1681 4.0 53 85 0.001553 60.

10 Ne neon 2081 — 24 27 0.000825 62

11 Na sodium 496 0.9 371 1156 0.97 160.

12 Mg magnesium 738 1.3 923 1363 1.74 140.

13 Al aluminum 578 1.6 933 2792 2.70 124

14 Si silicon 787 1.9 1687 3538 2.3296 114

15 P phosphorus (white) 1012 2.2 317 554 1.823 109

16 S sulfur (monoclinic) 1000. 2.6 388 718 2.00 104

17 Cl chlorine 1251 3.2 172 239 0.002898 100.

18 Ar argon 1521 — 84 87 0.001633 101

19 K potassium 419 0.8 337 1032 0.89 200.

20 Ca calcium 590. 1.0 1115 1757 1.54 174

21 Sc scandium 633 1.4 1814 3109 2.99 159

22 Ti titanium 659 1.5 1941 3560. 4.506 148

23 V vanadium 651 1.6 2183 3680. 6.0 144

24 Cr chromium 653 1.7 2180. 2944 7.15 130.

25 Mn manganese 717 1.6 1519 2334 7.3 129

26 Fe iron 762 1.8 1811 3134 7.87 124

27 Co cobalt 760. 1.9 1768 3200. 8.86 118

28 Ni nickel 737 1.9 1728 3186 8.90 117

29 Cu copper 745 1.9 1358 2835 8.96 122

30 Zn zinc 906 1.7 693 1180. 7.134 120.

31 Ga gallium 579 1.8 303 2477 5.91 123

32 Ge germanium 762 2.0 1211 3106 5.3234 120.

33 As arsenic (gray) 944 2.2 1090. — 5.75 120.

34 Se selenium (gray) 941 2.6 494 958 4.809 118

35 Br bromine 1140. 3.0 266 332 3.1028 117

36 Kr krypton 1351 — 116 120. 0.003425 116

37 Rb rubidium 403 0.8 312 961 1.53 215

38 Sr strontium 549 1.0 1050. 1655 2.64 190.

39 Y yttrium 600. 1.2 1795 3618 4.47 176

40 Zr zirconium 640. 1.3 2128 4682 6.52 164

R10


First

Atomic Symbol Name Ionization

Electro- Melting Boiling* Density** Atomic

Number Energy negativity Point Point (g/cm 3 ) Radius

(kJ/mol) (K) (K) (pm)

41 Nb niobium 652 1.6 2750. 5017 8.57 156

42 Mo molybdenum 684 2.2 2896 4912 10.2 146

43 Tc technetium 702 2.1 2430. 4538 11 138

44 Ru ruthenium 710. 2.2 2606 4423 12.1 136

45 Rh rhodium 720. 2.3 2237 3968 12.4 134

46 Pd palladium 804 2.2 1828 3236 12.0 130.

47 Ag silver 731 1.9 1235 2435 10.5 136

48 Cd cadmium 868 1.7 594 1040. 8.69 140.

49 In indium 558 1.8 430. 2345 7.31 142

50 Sn tin (white) 709 2.0 505 2875 7.287 140.

51 Sb antimony (gray) 831 2.1 904 1860. 6.68 140.

52 Te tellurium 869 2.1 723 1261 6.232 137

53 I iodine 1008 2.7 387 457 4.933 136

54 Xe xenon 1170. 2.6 161 165 0.005366 136

55 Cs cesium 376 0.8 302 944 1.873 238

56 Ba barium 503 0.9 1000. 2170. 3.62 206

57 La lanthanum 538 1.1 1193 3737 6.15 194

Elements 58–71 have been omitted.

72 Hf hafnium 659 1.3 2506 4876 13.3 164

73 Ta tantalum 728 1.5 3290. 5731 16.4 158

74 W tungsten 759 1.7 3695 5828 19.3 150.

75 Re rhenium 756 1.9 3458 5869 20.8 141

76 Os osmium 814 2.2 3306 5285 22.587 136

77 Ir iridium 865 2.2 2719 4701 22.562 132

78 Pt platinum 864 2.2 2041 4098 21.5 130.

79 Au gold 890. 2.4 1337 3129 19.3 130.

80 Hg mercury 1007 1.9 234 630. 13.5336 132

81 Tl thallium 589 1.8 577 1746 11.8 144

82 Pb lead 716 1.8 600. 2022 11.3 145

83 Bi bismuth 703 1.9 544 1837 9.79 150.

84 Po polonium 812 2.0 527 1235 9.20 142

85 At astatine — 2.2 575 — — 148

86 Rn radon 1037 — 202 211 0.009074 146

87 Fr francium 393 0.7 300. — — 242

88 Ra radium 509 0.9 969 — 5 211

89 Ac actinium 499 1.1 1323 3471 10. 201

Elements 90 and above have been omitted.

*boiling point at standard pressure

**density of solids and liquids at room temperature and density of gases at 298 K and 101.3 kPa

— no data available

Source: CRC Handbook for Chemistry and Physics, 91 st ed., 2010–2011, CRC Press

R11


Table T

Important Formulas and Equations

d = density

m

Density d = m = mass

V

V = volume

Mole Calculations number of moles =

given mass

gram-formula mass

measured value – accepted value

Percent Error % error = × 100

accepted value

mass of part

Percent Composition % composition by mass = × 100

mass of whole

mass of solute

parts per million = × 1000000

mass of solution

Concentration

molarity =

moles of solute

liter of solution

Combined Gas Law

P

P = pressure

1

V 1

P

= 2

V 2

V = volume

T 1

T 2 T = temperature

M A

= molarity of H + M B

= molarity of OH –

Titration M A

V A

= M B

V B

V A

= volume of acid V B

= volume of base

q = mCΔT q = heat H f

= heat of fusion

Heat q = mH f

m = mass H v

= heat of vaporization

q = mH v

C=specific heat capacity

ΔT = change in temperature

Temperature

K = °C + 273

K = kelvin

°C = degree Celsius

R12


Collier County CHEMISTRY EXAM FORMULA AND RESOURCE PACKET

GENERAL

D m V

[ ExperimentalValue AcceptedVa lue]

% error

x100

AcceptedVa lue

% yield

ExperimentalYield

TheoreticalYield

x100

CONCENTRATIONS

moles of solute

M = Molarity

liters of solution

KEY

P = pressure

V = volume

T = temperature

n = number of moles

m = mass

M = molar mass (grams/mole)

D = density

KE = kinetic energy

Avogadro’s Number = 6.02 x 10 23

GASES, LIQUIDS, SOLUTIONS

m = Molality

M1V1 M2V2

S1

P1

S 2

P 2

ACID/BASE

pH = - log[H + ]

[H + ]=10 -pH

moles of solute

kilograms of solvent


Gas constant

R 8.314 L kPa L atm L mmHg

0.0821 62.4

K mol K mol K mol

1 atm = 760 mmHg = 760 torr = 101.3 kPa

K = o C + 273

o C = K - 273

STP = Standard Temperature and Pressure = 0 o C

and 1 atm

P V

1 1


P2V

2

pOH = - log [OH - ]

[OH - ]= 10 -pOH

pH + pOH = 14

Kw = [H + ] x [OH - ] = 1.0 x 10 -14 M 2

V

T

1

1

P

T

1

1

V


T

P


T

2

2

2

2

Or V1T2 = V2T1

Or P1T2 = P2T1

THERMOCHEMISTRY

ΔH= mCΔT, where ΔT = T f - T

P1V

1

T

1


P2V

2

T

2

Or

P1V1T2=P2V2T1

q = mCΔT

PV

nRT

Specific Heat of Water = 4.18 J/g*˚C or 1.0 cal/g*˚C

Specific Heat of Ice = 2.1 J/g*˚C or 0.5 cal/g*˚C

Specific Heat of Steam = 2.0 J/g*˚C or 0.4 cal/g*˚C

P

Total

P

1

P

2

Rate A

Rate B

...


Molar MassB

Molar MassA

R13


CHEMISTRY EXAM FORMULA AND RESOURCE PACKET

Solubility of Compounds at 25 o C and 1 atm

acetate

bromide

carbonate

chlorate

chloride

hydroxide

iodide

nitrate

oxide

perchlorate

phosphate

sulfate

sulfide

aluminum S S - S S I S S I S I S d

ammonium S S S S S S S S - S S S S

barium S S I S S S S S sS S I I d

calcium S S I S S S S S sS S I sS I

copper(II) S S - S S I S S I S I S I

iron(II) S S I S S I S S I S I S I

iron(III) S S - S S I S S I S I sS d

lithium S S sS S S S S S S S sS S S

magnesium S S I S S I S S I S I S d

potassium S S S S S S S S S S S S S

silver sS I I S I - I S I S I sS I

sodium S S S S S S S S S S S S S

strontium S S I S S S S S S S I I I

zinc S S I S S I S S I S I S I

S=soluble

sS = slightly soluble in water

I = insoluble in water

d = decomposes in water

- = no such compound

R14


CHEMISTRY EXAM FORMULA AND RESOURCE PACKET

Common Polyatomic Ions

1- Charge 2- Charge 3- Charge

Formula Name Formula Name Formula Name

Dihydrogen

Phosphate

Hydrogen

phosphate

Phosphite

Acetate Oxalate Phosphate

Hydrogen

sulfite

Sulfite

Hydrogen

sulfate

Sulfate

Hydrogen

carbonate

Carbonate

Nitrite Chromate 1+ Charge

Nitrate Dichromate Formula Name

Cyanide Silicate Ammonium

Hydroxide

Permanganate

Hypochlorite

Chlorite

Chlorate

Perchlorate

R15


Activity Series of Metals

Name

Symbol

D

Lithium

Li

e

Potassium

K

c

r

Barium

Ba

e

Calcium

Ca

a

Sodium

Na

s

i

Magnesium

Mg

n

Aluminum

Al

g

Zinc

Zn

Iron

Fe

A

c

Nickel

Ni

t

Tin

Sn

i

v

Lead

Pb

i

(Hydrogen)

(H)*

t

Copper

Cu

y

Mercury

Hg

Silver

Ag

Gold

Au

*Metals from Li to Na will replace H from acids and water; from Mg to

Pb they will replace H from acids only.

Decreasing

Activity

Activity Series of Nonmetal (Halogens)

Name

Symbol

Fluorine F 2

Chlorine Cl 2

Bromine Br 2

Iodine I 2

R16

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