Q4 notebook

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

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.
Assignment Categories
Notebook
Test/Projects
Labs/Quizzes
Work
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
1

Unit 1 (22 days)
Chapter 1 Introduction to Chemistry
Honors Chemistry
2017/2018 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
5 days
Unit 3 (28 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
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
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
Unit 4 (8 days)
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
2

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
3

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.
4

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.
5

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 Cristopher __________________________
Crisostomo
5th
PERIOD ________________________
PARENT NAME Genarina ____________________________
Tavarez
PARENT NUMBER _________________________
862-899-6625
SIGNATURE ____________________________
DATE ____________________________________
6

7

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
Inference
Observation
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
Hypothesis
Observation
Controlled experiment
Identify sources of information and assess their reliability according to the
strict standards of scientific investigation.
8
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
mixture
physical property
intensive property
volume
solution
chemical property
element
vapor
compound
extensive property
Chapter 3
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
Experimental value
International System of Units (SI)
Percent error
Significant figures
Dimensional analysis
Accepted value
9

The Learning Goal for this assignment is:
Determine appropriate and consistent standards of measurement for the data to be collected in a survey or
experiment
Notes Section
kilo 1000
Hecto 100
Deca 10
Base Meter,Gram,Liter=1
Deci .1
Centi .01
Milli .001
10

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
.004 8000
b) 4 mL = ______L c) 8 g = _______mg
d) 9 mg =_______g .009
e) 2 mL = ______L .002 f) 6 kg = _____g 6000
g) 4 cm =_______m .04 h) 12 mg = ______ .012 g i) 6.5 cm 3 = _______L .0065
j) 7.02 mL =_____cm 7.02 3 k) .03 hg = _______ 30
dg l) 6035 mm _____cm 603.5
m) .32 m = _______cm 320
n) 38.2 g = .0382 _____kg
11

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.
6 cereal bars is 222g while 1kg would be 1000g since kilo means thousand
3. Wanda needs to move 110 kg of rocks. She can carry l0 hg each trip. How many trips must she
make? Explain your answer.
110 trips since 110kg is 110,000g and 10hg is 1000g so she would need to make 110 trips 110000/10000=110
4. Dr. O is playing in her garden again She needs 1 kg of potting soil for her plants. She has 750 g.
How much more does she need? Explain your answer.
1000-750=250 she needs 250 to complete the 1000g's
5. Weather satellites orbit Earth at an altitude of 1,400,000 meters. What is this altitude in kilometers?
1400
6. Which unit would you use to measure the capacity? Write milliliter or liter.
a) a bucket __________
liter
b) a thimble __________
milliliter
c) a water storage tank__________
liter
d) a carton of juice__________
liter
7. Circle the more reasonable measure:
a) length of an ant 5mm or 5cm
b) length of an automobile 5 m or 50 m
KHDBDCM
c) distance from NY to LA 450 km or 4,500 km
d) height of a dining table 75 mm or 75 cm
8. Will a tablecloth that is 155 cm long cover a table that is 1.6 m long? Explain your answer.
no 155cm would be 1.55 so it would be covering everything except .05
9. A dollar bill is 15.6 cm long. If 200 dollar bills were laid end to end, how many meters long would
the line be?
31.2 meters long if 200 bills were laid end to end
10. The ceiling in Jan’s living room is 2.5 m high. She has a hanging lamp that hangs down 41 cm.
Her husband is exactly 2 m tall. Will he hit his head on the hanging lamp? Why or why not?
no he wont hit his head since 41cm is .41m there will be 9cm of space
12

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
b. volume
c. mass
_____ 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
1
_____ 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
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 are related to temperature
________________________________________________________________________________
17. balance, second, mass __________________________________________________________
a balance is used to find the mass in something
________________________________________________________________________________
18. kilogram, liter, cubic centimeter __________________________________________________
cm cubed can be turned into ml
________________________________________________________________________________
19. time, second, distance __________________________________________________________
seconds are a measurement of time
________________________________________________________________________________
20. decimeter, kilometer, Kelvin _____________________________________________________
the base of both of them is meter
________________________________________________________________________________
13

1. How many meters are in one kilometer? 1000m __________
2. What part of a liter is one milliliter? __________
.001
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?__________
1kg
5. What part of a meter is a decimeter? __________
.1m
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 ______________ standard ten tenth .
7. The system of measurement used worldwide in science is _______________
si ten
.
8. SI is based on units of _______________ Metric prefixes .
9. The system of measurement that was based on units of ten was the _______________ si ten
system.
10. In SI, _______________ ,Metric prefixes 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 _______________ standard ten tenth .
14

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? __________________________________
15

The Learning Goal for this assignment is:
Notes Section
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
15

16
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
17

Convert each number from Scientific Notation to real numbers:
1. 7.485 X 10³ 7485
6. 1.683 X 10⁰
1
2. 8.842 X 10² 884.2
7. 3.622 10⁰
1
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 6.633 x 10^-4
16. 1,937,000
1.937x 10^6
12. 0.0004445 4.445x 10^-4
17. 34,570
3.457x 10^4
13. 0.002182 2.182x10^-3
18. 0.00003948
3.948x 10^-5
14. 469.5 4.695x 10^2
19. 894,500
8.945x 10^5
15. 727,400 7.274x10^5
20. 678.3
6.783x 10^2
18

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
19

Significant Figures Rules
There are three rules on determining how many significant figures are in a
number:
1. Non-zero digits are always significant.
1 2 3 4 5 6 7 8 9
2. Any zeros between two significant digits are significant.
101 8405 15.057 8450
3. A final zero or trailing zeros in the DECIMAL PORTION ONLY are
significant.
15.240 .5220 .10010 .006040
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.
20

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.
21

How Many Significant Digits for Each Number?
1) 2359 = 4______
2) 2.445 x 10−⁵= ______ 4
3) 2.93 x 10⁴= ______ 3
4) 1.30 x 10−⁷= ______ 3
5) 2604 = ______ 4
6) 9160 = ______ 3
7) 0.0800 = ______ 3
8) 0.84 = ______ 2
9) 0.0080 = ______ 2
10) 0.00040 = ______ 2
11) 0.0520 = ______ 3
12) 0.060 = ______ 2
13) 6.90 x 10−¹= ______ 3
14) 7.200 x 10⁵= ______ 2
15) 5.566 x 10−²= ______ 4
16) 3.88 x 10⁸= ______ 3
17) 3004 = ______ 4
18) 0.021 = ______ 2
19) 240 = ______ 2
20) 500 = ______ 1
22

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 = _______ 55.36 55.361
2) 87.54 - 3.3 = _______ 84.2
3) 99.1498 + 6.5397 + 9.7 = _______ 11534
84.24
115.3895
4) 5.868 - 5.1 = _______ .8 .768
5) 59.9233 + 86.21 + 99.396 = _______ 245.53
6) 7.7 + 26.756 = _______ 34.5
7) 66.8 + 2.3 + 4.8516 = _______ 74.0
8) 9.7419 + 43.545 = 53.287 _______
9) 4.8976 + 48.4644 + 1.514 = _______ 54.876
245.5293
34.456
73.951
53.2869
54.876
10) 4.335 + 35.85 = _______ 40.19 40.185
11) 9.448 - 1.7 = _______ 7.7 7.748
12) 75.826 - 8.6555 = _______ 67.171 67.1705
13) 57.2 + 23.814 = _______ 81.0
14) 77.684 - 4.394 = _______ 73.290
15) 26.4496 + 3.339 = _______ 29.789
16) 9.6848 + 29.85 = 39.53 _______
17) 63.11 + 2.5412 + 4.93 = 70.58 _______
18) 11.2471 + 75.4 = _______ 86.6
19) 73.745 - 8.755 = 64.990 _______
20) 6.5238 + 1.7 + 27.79 = _______ 36.0
81.014
73.29
29.7886
39.5348
70.5812
86.6471
64.99
36.0138
23

24
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 = __________
20000
2) 720 ÷ 7.7 = __________
94
3) 929 x 0.3 = __________
300
4) 300 ÷ 44.31 = __________
7
5) 608 ÷ 9.8 = __________
62
.005
6) 0.06 x 0.079 = __________
7) 0.008 x 72.91 x 7000 = __________
8) 73.94 x 67 x 780 = __________
3,800,000
9) 0.62 x 0.097 x 40 = __________
2.4
10) 600 x 10 x 5030 = __________
30,000,000
11) 5200 ÷ 4.46 = __________
1,100
12) 0.0052 x 0.4 x 107 = __________
.2
13) 0.099 x 8.8 = __________
.87
14) 0.0095 x 5.2 = __________
.049
15) 8000 ÷ 4.62 = __________
1,000
16) 0.6 x 0.8 = __________
.5
4000
17) 2.84 x 0.66 = __________
1.9
18) 0.5 x 0.09 = __________
.05
19) 8100 ÷ 34.84 = __________
230
20) 8.24 x 6.9 x 8100 = __________
460,000

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.
25

1qt=32 oz 1gal = 4qts 1.00 qt = 946 mL 1L = 1000mL
2. Convert 8135.6 mL to quarts
8135.6
1qt
=
8135.6
8.6qts
8.6000
1
946mL
946
3. Convert 115.2 oz to mL
115.2oz
1qt
946ml
=
108979.2
3405.6ml
1
32oz
1qt
32
3406ml
4. Convert 2.3 g to Liters
2.3g
4qt
946mL
1L
=
8703.2
8.7032
8.7
1
1g
1qt
1000mL
1000
5. Convert 8.42 L to oz
8.42L
1
1000mL
1L
1qt
946mL
32oz
1qt
=
169440
946
284.8202959831
284
Go to http://science.widener.edu/svb/tutorial/ chose #7 “Converting Volume” and do 5 more in the
space provided.
1. Convert _________ 8.42L to _________ qts
8.42L 1000mL
1qt 8420
=
8..9006342495
8.90
1L
946mL
946
2. Convert _________ 5.39L to _________ oz
5.39L 1000mL 1qt
1L
946mL
=
32oz 172480
1qt
946
183.3255813953
182
3. Convert _________ 1.2G to _________ L
1.2g 4qt
946ml
=
1L 4540.8
4.5408
4.5
1g
1qt
1000ml
1000
4. Convert _________ 1513.6mL to _________ oz
1513.6mL 1qt
32oz 48435.2
946ml
1qt
5. Convert _________ 7.38L to _________ oz
7.38L
=
=
946
1000 1qt 32 236160
51.2
51.200
249.6405919662
250
1L 946ml 1qt 946
26

Chapter 7
Unit 3
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.
27

Students will be able to predict shapes of various compounds.
Atom
Electron
Element
Compound
Molecule
empirical formula
Chapter 9
Chemical Names and Formulas
The students will learn how the periodic table helps them
determine the names and formulas of ions and compounds.
28

The Learning Goal for this assignment is:
Distinguish between bonding forces holding compounds together
and other attractive forces. incuding hydrogen bonding and van der
vass 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:
29

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.
11 e- lose 1 electron 10 e-
Positive charge is
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.
30

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.
31

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.
https://www.youtube.com/watch?v=zpaHPXVR8WU
Writing Ionic Compounds
An easy technique for creating a neutral combination of two charged ions is called the criss-cross
technique. When writing a formula for an ionic compound the charges from each ion are simply
switched to become the subscript values written to designate the number of atoms present in a
compound. See the example below.
Naming Binary Ionic Compounds
Naming Ionic Compounds
Learning to name ionic compounds is both easy and hard depending on the complexity of the
compound. Before we start, though, I just wanted to review a few terms. Remember that positively
charged ions are called cations. Negatively charged ions are called anions. An ionic compound is a
compound held together by ionic bonds. A simple binary compound is just what it seems - a simple
compound with two elements in it.
Binary compounds are easy to name. The cation is always named first and gets its name from the
name of the element. For example, K + is called a potassium ion. An anion also takes its name from its
element, but it adds the suffix -ide to it. So, Cl - is called a chloride ion; O 2- is an oxide ion.
Take the binary compound NaCl. The Na + is a sodium cation. The Cl - is a chlorine anion, which gets
the suffix -ide added to it. When you put them together, it becomes sodium chloride.
32

Here are some examples for you:
Zn 2+ is zinc. S 2- is sulfide. Put them together for zinc sulfide (ZnS).
Here's another:
K2O is potassium oxide
Naming Ionic Compounds Containing Transition Metals
A transition metal is a metal that can use the inner shell before using the outer shell to bond. These
are the elements in the middle of the periodic table - things like zinc, iron and copper. Naming
polyatomic ionic compounds that have transition metals in them is also fairly easy. It follows the same
naming rules as the simple binary compounds, but with an extra rule added in. So, you still name the
cation first, followed by the anion with the suffix -ide added to the end of it.
The new rule is that transition metals form more than one ion, so this has to be accounted for in the
naming. We do this by using Roman numerals to denote which ion it is. The Roman numeral will
equal the charge on the ion. For instance, Fe 2+ is iron (II). Fe 3+ is iron (III).
When compounds are formed with these metals, the different ions still have to be accounted for. If I
told you the compound was iron chloride, that wouldn't give you the full story. You wouldn't know if it
was iron (II) or iron (III), which means you don't know how many chlorine atoms are in the compound
to bond with the iron, since two chlorines would be needed for iron (II) and three for iron (III). If I
instead told you that the compound was iron (II) chloride, you would know that it was Fe 2+ in there,
which means you have two chlorine atoms bonding with it. The formula would be FeCl2. If I said it
was iron (III) chloride, the formula would be FeCl3.
Naming Polyatomic Ionic Compounds
A polyatomic ionic compound is a compound made up of a polyatomic ion, which is two or more
atoms bonded together, and a metal. Naming polyatomic ions is harder, but doable. First, name the
cation, which is just the name of the element. Next, name the anion. This gets trickier.
Notes Section:
33

1
Name of
Cation
Calcium ion
Name of
Anion
Chloride ion
Formula
of Cation
Formula of
Anion
Formula of
Compound
Name of Compound
2
3
Iron(III) ion
Phosphide
ion
Na 1+ S 2-
4
Al 3+ Br 1-
5
Lithium Sulfide
6
Platinum (IV) Oxide
7 Magnesium
ion
8
Carbonate
ion
Ca 2+ NO3 1-
9
HgS
10
Th3(PO4)2
Write the formulas for these ionic compounds
11. Chromium (IV) oxide
12. Chromium (III) oxide
13. Aluminum oxide
14. Nickel (II) bromide
15. Silver (I) sulfide
16. Magnesium chloride
17. Nickel (II) sulfate
18. Iron (III) phosphate
19. Potassium dichromate
20. Lead (IV) hydroxide
34

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.
The students will learn how molecular bonding is different
than ionic bonding and electrons affect the shape of a
molecule and its properties.
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.
35

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:
Bonds between non-metals is dependant on electrons between both of the elements. not all elements will
bond fully like with flourine will take part of others electrons.
36

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.
37

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.
Add any left over bonds to make double or triple
bonds.
Then, if needed, use any lone pairs to make
double or triple bonds so that all atoms meet
the Octet Rule.
See Step 4 for total number of bonds.
Step 9
Find the formal charges for the atoms in the compound.
Arrange atoms so that all formal charges
are as close to 0 as possible.
Some central atoms do not meet the octet rule.
Boron can sometimes have only 6 electrons and
some elements in Periods 3—7 may exceed the
octet rule.
38

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-
39

Orbitals Equation Lone​ ​Pairs Angle
Name
sp AX2 NONE 180
LINEAR
SP2 AX3 NONE 120 TRILINEAR​ ​PLANER
SP2 AX2E 1 116
BENT
SP3 AX4 NONE 109.5
TETRAHEDREAL
SP3 AX3E 1 107
TRIG​ ​PER
SP3 AX2E2 2 104.5
BENT
SP3D AX5 NONE 100/90
TRIG​ ​BI-PER
SP3D AX3E2 2 90
T-SHAPED
SP3D2 AX6 NONE 90
OCTOHEDREAL
SP3D2 AX4E2 2 90
SQUARE​ ​PLANAR
40

Name
Date
Example:
Xenon Hexafluoride
Chemical Name
XeF 6
Chemical Formula
Octahedral
MUSTANG LABORATORIES
EXPERIMENT 6 – Molecular Geometry
In this lab you will use the website https://phet.colorado.edu/sims/html/moleculeshapes/latest/molecule-shapes_en.html.
You will construct the following 10 molecular
compounds using the rules of VSEPR. Use the snip it tool to get pictures of the model
from the side and from the top showing its form. Place these pictures in the appropriate
boxes. You can either do the diagram by hand and take a picture or you can construct
it in this program. Then you will fill in the rest of the form with the relevant information.
I have given you an example.
H 3 O + O 3 SF 6
PF 5 CO 2 OF 2
COCl 2 XeF 4 ClF 3
POF 3
When you have finished save this as a pdf and put it in the drop box in Angel.
sp 3 d 2
Molecular Geometry
None
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
41

Carbon Dioxide
Chemical Name
CO 2
Chemical Formula
Linear
sp
Molecular Geometry
none
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
Phosgene
Chemical Name
COCL 2
Chemical Formula
Trigonal Planar
sp 3
Molecular Geometry
none
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
42

Oxygen difluoride
Chemical Name
OF 2
Chemical Formula
Bent
Molecular Geometry
sp 2
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
Phosphoryl fluoride
Chemical Name
POF 3
Chemical Formula
Tetrahedral
sp 3
Molecular Geometry
None
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
43

Hydronium
Chemical Name
H 3 O +
Chemical Formula
Trigonal Pyramidal
Molecular Geometry
Sp 2 1
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
Trioxide
Chemical Name
O 3
Chemical Formula
Bent
Molecular Geometry
sp 1
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
44

PhosphorousPentaFlouride
Chemical Name
PF 5
Chemical Formula
Trigonal Bipyramidal
sp 3 d
Molecular Geometry
none
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
Chemical Name
ClF 3
Chemical Formula
T-Shaped
Molecular Geometry
sp2 22
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
45

Chemical Name
SF 6
Chemical Formula
Octahedral
Molecular Geometry
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
Xenon TretraFlouride
Chemical Name
XeF4
Chemical Formula
Square Planar
Molecular Geometry
sp 3 2
Side View
Orbitals
Lone Pairs
Molecular Drawing
With Angles
Top View
46

Finding Bond Angles, Shapes, and Hybridizations
Sometimes people have a hard time with the whole VSEPR thing. In this helpdesk section we'll
discuss what VSEPR means, what it's all about, and how you can use a great big flow chart to figure
out the bond angles, shapes, and hybridizations of various covalent compounds.
This whole thing assumes, by the way, that you know how to draw Lewis structures. If you don't, click
here.
What is VSEPR?
VSEPR stands for Valence Shell Electron Pair Repulsion. It's a complicated acronym, but it means
something that's not difficult to understand. Basically, the idea is that covalent bonds and lone pair
electrons like to stay as far apart from each other as possible under all conditions. This is because
covalent bonds consist of electrons, and electrons don't like to hang around next to each other much
because they have the same charge.
This VSEPR thing explains why molecules have their shapes. If carbon has four atoms stuck to it (as
in methane), these four atoms want to get as far away from each other as they can. This isn't
because the atoms necessarily hate each other, it's because the electrons in the bonds hate each
other. That's the idea behind VSEPR.
What is hybridization (AP Chemistry)?
Now, one problem with the whole VSEPR thing is that if you have four things stuck to carbon, for
example, there are no orbitals that want to get 109.5 degrees apart from each other (109.5 degrees
corresponds to the geometric maximum distance the atoms can get apart). After all, s-orbitals go in a
complete sphere (360 degrees) and p-orbitals are 90 degrees apart.
What happens instead of using s- or p- orbitals is that when covalent bonds are formed, the s- and p-
orbitals mix to form something called hybrid orbitals. "Hybrid" just means "mixture of two different
things", and that's exactly what a hybrid orbital is. When three p-orbitals with 90 degree angles
combine with one s-orbital with 360 degrees, they average to form four sp3 orbitals with 109.5 degree
bond angles. Depending on the numbers of s- and p-orbitals that mix, you can get a bunch of
different bond angles.
Common shapes you should know
There are a whole bunch of common shapes you need to know to accurately think of covalent
molecules. Here they are:
Tetrahedral: Tetrahedral molecules look like pyramids with four faces. Each point on the pyramid
corresponds to an atom that's attached to the central atom. Bond angles are 109.5 degrees.
Trigonal pyramidal: It's like a tetrahedral molecule, except flatter. It looks kind of like a squished
pyramid because one of the atoms in the pyramid is replaced with a lone pair. Bond angles are 107.5
degrees (it's less than tetrahedral molecules because the lone pair shoves the other atoms closer to
each other).
Trigonal planar: It looks like the hood ornament of a Mercedes automobile, or like a peace sign with
that bottom-most line gone. The bond angles are 120 degrees.
47

Bent: They look, well, bent. Bond angles can be either 116 degrees for molecules with one lone pair
or 104.5 degrees for molecules with two lone pairs.
Linear: The atoms in the molecule are in a straight line. This can be either because there are only
two atoms in the molecule (in which case there is no bond angle, as there need to be three atoms to
get a bond angle) or because the three atoms are lined up in a straight line (corresponding to a 180
degree bond angle).
There are other types, but we won't worry about them until college or AP Chemistry.
Using a flow chart to figure out what shape, and bond angle an atom has
Take a look at this flow chart. I'll explain how to use it to find all the stuff above at the end.
Complicated, huh? Here's how to use it:
1. Draw the Lewis structure for the molecule. This vital if you're going to get the answer right.
2. Count the number of "things" on the atom you're interested in. Let's say that you're looking at
methane, CH4. If you want to find the bond angles, shape, and hybridization for carbon, count
the number of things that are stuck to it.
Now, the vague term "things" refers to atoms and lone pairs. IT DOES NOT REFER TO THE
NUMBER OF BONDS! When you look at methane, there are four atoms stuck to it, so you'd go down
the line that says "four" toward the green boxes on this chart.
People get confused with multiple bonds. Take carbon dioxide, for example. There are four bonds
(carbon is double-bonded to each oxygen) but only two oxygen atoms bonded to carbon. In this
48

case, we count two things stuck to carbon, because we only count the atoms, NOT the number of
bonds.
Likewise, with ammonia there are four things. Three of the things on nitrogen are hydrogen atoms
and the fourth is a lone pair. For the purposes of VSEPR, lone pairs count exactly the same as
atoms, because they consist of negative charge, too.
3. Count the number of lone pairs that are on the atom you're interested in. IMPORTANT: This
does NOT mean to count the number of lone pairs on all of the atoms in the molecule. Lone
pairs on other atoms aren't important - what's important is only what's directly stuck to the atom
you're interested in.
We mentioned above that methane has four things stuck to it. Since all four things are hydrogen
atoms, we moved toward the green boxes on the flow chart. When we get to our second question,
we find that there are no lone pairs on carbon, so our answer is zero. When we go down the line that
says "zero" from that box, we find that methane is sp3 hybridized, with a 109.5 degree bond angle
and tetrahedral shape.
And, hey, that's what we were looking for!
Some sample problems:
What are the shapes, bond angles, and hybridizations of the following molecules? Use the flow chart
and instructions above to figure it out.
1. carbon tetrabromide
2. phosphorus trichloride
3. oxygen

4. the chlorine atom in hydrochloric acid (HCl)
5. boron trichloride
6. CH2O
7. sulfur difluoride
8. either carbon atom in C2H2
1. sp 3 , tetrahedral, 109.5 degrees.
2. sp 3 , trigonal pyramidal, 107.5 degrees.
3. sp 2 , linear, no bond angle
4. sp 3 , linear, no bond angle
5. sp 2 , trigonal planar, 120 degrees
6. sp 2 , trigonal planar, 120 degrees
7. sp 3 , bent, 104.5 degrees
8. sp, linear, 180 degrees
50

Unit 4
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
51

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
52

https://www.bbc.co.uk/education/guides/zvvwxnb/revision
53

54

Introduction to Organic Chemistry
To understand life as we know it, we must first understand a little bit of organic chemistry. Organic
molecules contain both carbon and hydrogen. Though many organic chemicals also contain other
elements, it is the carbon-hydrogen bond that defines them as organic. Organic chemistry defines life.
Just as there are millions of different types of living organisms on this planet, there are millions of
different organic molecules, each with different chemical and physical properties. There are organic
chemicals that make up your hair, your skin, your fingernails, and so on. The diversity of organic
chemicals is due to the versatility of the carbon atom. Why is carbon such a special element? Let's
look at its chemistry in a little more detail.
The uniqueness of carbon
Carbon (C) appears in the second row of the periodic table and has four bonding electrons in its
valence shell (see our Periodic Table module for more information). Similar to other non-metals,
carbon needs eight electrons to satisfy its valence shell. Carbon therefore forms four bonds with other
atoms (each bond consisting of one of carbon's electrons and one of the bonding atom's electrons).
Every valence electron participates in bonding; thus, a carbon atom's bonds will be distributed evenly
over the atom's surface. These bonds form a tetrahedron (a pyramid with a spike at the top), as
illustrated below:
Carbon forms 4 bonds
Organic chemicals get their diversity from the many different ways carbon can bond to other atoms.
The simplest organic chemicals, called hydrocarbons, contain only carbon and hydrogen atoms; the
simplest hydrocarbon (called methane) contains a single carbon atom bonded to four hydrogen
atoms:
Methane - a carbon atom bonded to 4 hydrogen atoms
But carbon can bond to other carbon atoms in addition to hydrogen, as illustrated in the molecule
ethane below:
Ethane - a carbon-carbon bond
In fact, the uniqueness of carbon comes from the fact that it can bond to itself in many different ways.
Carbon atoms can form long chains:
55

anched chains:
Hexane - a 6-carbon chain
rings:
Isohexane - a branched-carbon chain
Cyclohexane - a ringed hydrocarbon
There appears to be almost no limit to the number of different structures that carbon can form. To add to the
complexity of organic chemistry, neighboring carbon atoms can form double and triple bonds in addition to
single carbon-carbon bonds:
Single bonding Double bonding Triple bonding
Keep in mind that each carbon atom forms four bonds. As the number of bonds between any two
carbon atoms increases, the number of hydrogen atoms in the molecule decreases (as can be seen
in the figures above).
Simple hydrocarbons
The simplest hydrocarbons are those that contain only carbon and hydrogen. These simple
hydrocarbons come in three varieties depending on the type of carbon-carbon bonds that occur in the
molecule.
Alkanes
Alkanes are the first class of simple hydrocarbons and contain only carbon-carbon single bonds. The
alkanes are named by combining a prefix that describes the number of carbon atoms in the molecule
with the root ending "ane". The names and prefixes for the first ten alkanes are given in the following
table.
56

Carbon Prefix Alkane Name Chemical Structural Formula
atoms
Formula
1 Meth Methane CH4 CH4
2 Eth Ethane C2H6 CH3CH3
3 Prop Propane C3H8 CH3CH2CH3
4 But Butane C4H10 CH3CH2CH2CH3
5 Pent Pentane C5H12 CH3CH2CH2CH2CH3
6 Hex Hexane C6H14 …
7 Hept Heptane C7H16
8 Oct Octane C8H18
9 Non Nonane C9H20
10 Dec Decane C10H22
The chemical formula for any alkane is given by the expression CnH2n+2. The structural formula,
shown for the first five alkanes in the table, shows each carbon atom and the elements that are
attached to it. This structural formula is important when we begin to discuss more complex
hydrocarbons. The simple alkanes share many properties in common. All enter into combustion
reactions with oxygen to produce carbon dioxide and water vapor. In other words, many alkanes are
flammable. This makes them good fuels. For example, methane is the main component of natural
gas, and butane is common lighter fluid.
CH4 + 2O2 → CO2 + H2O
The chemical reaction between a fuel (for example wood) and an oxidation agent.
Alkenes
The second class of simple hydrocarbons, the alkenes, consists of molecules that contain at least
one double-bonded carbon pair. Alkenes follow the same naming convention used for alkanes. A
prefix (to describe the number of carbon atoms) is combined with the ending "ene" to denote an
alkene. Ethene, for example is the two-carbon molecule that contains one double bond. The chemical
formula for the simple alkenes follows the expression CnH2n. Because one of the carbon pairs is
double bonded, simple alkenes have two fewer hydrogen atoms than alkanes.
Ethene
Alkynes
Alkynes are the third class of simple hydrocarbons and are molecules that contain at least one triplebonded
carbon pair. Like the alkanes and alkenes, alkynes are named by combining a prefix with the
ending "yne" to denote the triple bond. The chemical formula for the simple alkynes follows the
expression CnH2n-2.
Ethyne
57

Isomers
Because carbon can bond in so many different ways, a single molecule can have different bonding
configurations. Consider the two molecules illustrated here:
C6H14
CH3CH2CH2CH2CH2CH3
C6H14
CH3
I
CH2CH2CH2CH2CH3
Both molecules have identical chemical formulas (shown in the left column); however, their structural
formulas (and thus some chemical properties) are different. These two molecules are called isomers.
Isomers are molecules that have the same chemical formula but different structural formulas.
Functional groups
In addition to carbon and hydrogen, hydrocarbons can also contain other elements. In fact, many
common groups of atoms can occur within organic molecules, these groups of atoms are called
functional groups. One good example is the hydroxyl functional group. The hydroxyl group consists of
a single oxygen atom bound to a single hydrogen atom (OH - ). The group of hydrocarbons that contain
a hydroxyl functional group is called alcohols. The alcohols are named in a similar fashion to the
simple hydrocarbons, a prefix is attached to a root ending (in this case "anol") that designates the
alcohol. The existence of the functional group completely changes the chemical properties of the
molecule. Ethane, the two-carbon alkane, is a gas at room temperature; ethanol, the two-carbon
alcohol, is a liquid.
Ethanol
Ethanol, common drinking alcohol, is the active ingredient in "alcoholic" beverages such as beer and
wine.
Summary
The chemical basis of all living organisms is linked to the way that carbon bonds with other atoms.
This introduction to organic chemistry explains the many ways that carbon and hydrogen form bonds.
Basic hydrocarbon nomenclature is described, including alkanes, alkenes, alkynes, and isomers.
Functional groups of atoms within organic molecules are discussed.
58

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
59

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
60

Learning Goal for this section:
al equations are used in stoichiometric calculations and how to calculate amounts
hemical equation
The Mole: Its History and Use
Simply put, the mole represents a number. Just as the term dozen refers to the number twelve, the
mole represents the number 6.02 x 10 23 . (If you're confused by the form of this number refer to our
The Metric System module).
Now that's a big number! While a dozen eggs will make a nice omelet, a mole of eggs will fill all of the
oceans on earth more than 30 million times over. Think about it: It would take 10 billion chickens
laying 10 eggs per day more than 10 billion years to lay a mole of eggs. So why would we ever use
such a big number? Certainly the local donut store is not going to "supersize" your dozen by giving
you a mole of jelly-filled treats.
The mole is used when we're talking about numbers of atoms and molecules. Atoms and molecules
are very tiny things. A drop of water the size of the period at the end of this sentence would contain
10 trillion water molecules. Instead of talking about trillions and quadrillions of molecules (and more),
it's much simpler to use the mole.
History of the mole
The number of objects in one mole, that is, 6.02 x 10 23 , is commonly referred to as Avogadro's
number. Amedeo Avogadro was an Italian physics professor who proposed in 1811 that equal
volumes of different gases at the same temperature contain equal numbers of molecules. About fifty
years later, an Italian scientist named Stanislao Cannizzaro used Avogadro's hypothesis to develop a
set of atomic weights for the known elements by comparing the masses of equal volumes of gas.
Building on this work, an Austrian high school teacher named Johann Josef Loschmidt calculated the
size of a molecule of air in 1865, and thus developed an estimate for the number of molecules in a
given volume of air. While these early estimates have since been refined, they led to the concept of
the mole - that is, the theory that in a defined mass of an element (its atomic weight) there is a
precise number of atoms: Avogadro's number.
Molar mass
A sample of any element with a mass equal to that element's atomic weight (in grams) will contain
precisely one mole of atoms (6.02 x 10 23 atoms). For example, helium has an atomic weight of 4.00
amu. Therefore, 4.00 grams of helium will contain one mole of helium atoms. You can also work with
fractions (or multiples) of moles:
Mole/weight relationship
examples using helium
moles
helium
# helium
atoms
grams
helium
¼ 1.505 X 10 23 1g
½ 3.01 X 10 23 2g
1 6.02 X 10 23 4g
2 1.204 X 10 24 8g
10 6.02 X 10 24 40g
61

Other atomic weights are listed on the periodic table (see our Periodic Table module). For each
element listed, measuring out a quantity of the element equal to its atomic weight in grams will yield
6.02 x 10 23 atoms of that element.
The atomic weight of an element identifies both the mass of one mole of that element and the total
number of protons and neutrons in an atom of that element. How can that be? Let's look at hydrogen.
One mole of hydrogen atoms will weigh 1.01 grams.
A hydrogen atom
with its single electron
Each hydrogen atom consists of one proton surrounded by one electron. But remember, the electron
weighs so little that it does not contribute much to an atom's weight. Ignoring the weight of hydrogen's
electrons, we can say that one mole of protons (H nuclei) weighs approximately one gram. Since
protons and neutrons have about the same mass, a mole of either of these particles will weigh about
one gram. For example, in one mole of helium, there are two moles of protons and two moles of
neutrons - four grams of particles.
Molecular weight
If you stand on a scale with a friend, the scale will register the combined weight of both you and your
friend. When atoms form molecules, the atoms bond together, and the molecule's weight is the
combined weight of all of its parts.
For example, every water molecule (H2O) has two atoms of hydrogen and one atom of oxygen. One
mole of water molecules will contain two moles of hydrogen and one mole of oxygen.
2 moles
hydrogen
Mole/weight relationships
of water and its parts
1 mole
oxygen
1 mole
water
+ =
A bottle filled with exactly 18.02 g water will contain 6.02 x 10 23 water molecules. The concept of
fractions and multiples described above also applies to molecules: 9.01 g of water would contain 1/2
mole, or 3.01 x 10 23 molecules. You can calculate the molecular weight of any compound simply by
summing the weights of atoms that make up that compound.
62

Example: Converting between mass and moles
Knowing a substance’s molar mass is useful, because the molar mass acts as a conversion factor
between the mass of a sample and the number of moles in that sample (Equation 1). For converting
between the number of moles in a sample and the number of molecules in the sample, Avogadro’s
number acts as the conversion factor, as shown in Equation 2 below.
Equation 1
Sample mass (g) = Moles in sample (mol)
Sample's molar mass (g/mol)
Equation 2
Moles in sample (mol) x Avogadro's number (number/mol) = Number of sample molecules
To understand how molar mass and Avogadro’s number act as conversion factors, we can turn to an
example using a popular drink: How many CO2 molecules are in a standard bottle of carbonated
soda?
Thanks to molar mass and Avogadro’s number, figuring this out doesn’t require counting each
individual CO2 molecule! Instead, we can start by determining the mass of CO2 in this sample. In an
experiment, a scientist compared the mass of a standard 16-ounce (454 milliliters) bottle of soda
before it was opened, and then after it had been shaken and left open so that the CO2 fizzed out of
the liquid. The difference between the masses was 2.2 grams—the sample mass of CO2 (for this
example, we’re going to assume that all the CO2 has fizzled out). Before we can calculate the number
of CO2 molecules in 2.2 grams, we first have to calculate the number of moles in 2.2 grams of CO2
using molar mass as the conversion factor (see Equation 1 above):
Equation 3
Now that we’ve figured out that there are 0.050 moles in 2.2 grams of CO2, we can use Avogadro’s
number to calculate the number of CO2 molecules (see Equation 2 above):
Equation 4
While scientists today commonly use the concept of the mole to interconvert number of particles and
mass of elements and compounds, the concept started with 19th-century chemists who were puzzling
out the nature of atoms, gas particles, and those particles’ relationship with gas volume.
63

Find Molecular Mass 1- 10
1. NaBr
94g
2. PbSO4
303g
3. Zn(C2H3O2)2
183g
4. Na3PO4
141g
5. (NH4)2CO3
86g
6. Glucose
7. Iron(II) Phosphate
8. Ammonium Sulfide
9. Calcium Hydroxide
10. Silver(I) Fluoride
11. How many moles is 51g of NaBr?
.54 mol
12. How many moles is 60g of PbSO4?
.20mol
13. How many moles is 150g of Zn(C2H3O2)2?
.82mol
14. How many moles is 578g of Na3PO4?
4.1mol
15. How many moles is 500g of (NH4)2CO3?
5.8mol
1. 103 g/mol 6. 180 g/mol 11. .50 mol
2. 303 g/mol 7. 358 g/mol 12. .20 mol
3. 74 g/mol 8. 68 g/mol 13. 2.0 mol
4. 164 g/mol 9. 183 g/mol 14. 3.5 mol
5. 96 g/mol 10. 127 g/mol 15. 5.2 mol
64

The Learning Goal for this assignment is:
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
65

Step 3
Always leave hydrogen and oxygen for last. This means that you will need to balance the carbon
atoms first.
Step 4
66
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.
67

1) ___ NaNO3 + ___ PbO ___ Pb(NO3)2 + ___ Na2O
2) ___ AgI + ___ Fe2(CO3)3 ___ FeI3 + ___ Ag2CO3
3) ___ C2H4O2 + ___ O2 ___ CO2 + ___ H2O
4) ___ ZnSO4 + ___ Li2CO3 ___ ZnCO3 + ___ Li2SO4
5) ___ V2O5 + ___ CaS ___ CaO + ___ V2S5
68

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
69

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:
70

The Learning Goal for this assignment is:
we will learn how chemical reactions obey the law of conservation of mass and how
we can predict the products of a chemical reaction
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:
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:
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:
A+BC->AC+B or BA+C
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
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.
71

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:
AB+CD->AD+CB
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
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+B OH-> BA+H2O
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:
Org+O2->CO2+H2O
Organic
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.
72

List what type the following reactions are:
1) NaOH + KNO3 --> NaNO3 + KOH
double displacement
2) CH4 + 2 O2 --> CO2 + 2 H2O
3) 2 Fe + 6 NaBr --> 2 FeBr3 + 6 Na
Combustion
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
decomposition
73

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) __Ag2SO4 + __NaNO3 →
2) __NaI + __CaSO4 →
3) __HNO3 + __Ca(OH)2 →
4) __CaCO3 →
5) __AlCl3 + __(NH4)PO4 →
6) __Pb + __Fe(NO3)3 →
7) __C3H6 + __O2 →
8) __Na + __CaSO4 →
74

Stoichiometry
Cristopher Crisostomo
Pd.5
Chemistry
Molecular mass: ​the sum of the atomic weights of each constituent element multiplied by the
number of atoms of that element in the molecular formula
Steps:
Find Molecular mass
Find the conversion units from grams to moles and to moles to moles
Convert one molecule to the other for example from O​2​ to CO​2​ by using the conversion units
Multiply the totals and then divide
Mole To Mole
Mass to Mass
106

Gas StoiChiometry
30.0 g of chlorine gas bubbled through liquid sulfur to produce liquid disulfur dichloride.How
much product is produced in grams
_Cl​2​+2S=_S​2​Cl​2
1Mol of Cl​2​ = 1Mol of S​2​Cl​2
71 g Cl​2​=1Mol of Cl​2
135 g S​2​Cl​2​ = 1Mol of S​2​Cl​2
Limiting Reagents
Is like a regular mole to mole or mass to mass just that you have to do it twice to see which
reactant runs out first
Percentile yields
Percent yield = Actual/Theoretical
Actual yield = Theoretical*percentage
107

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
108

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
Solvent
Aqueous solution
Solute
109

Name:
Name:
Grade:
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 Monday Feb 26.
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 Monday February 26 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 Wednesday Feb 28.
This page is due by the end of class on Monday February 26.
This Project is due by the end of class on Wednesday February 28.
Jeopardy (https://jeopardylabs.com)
Password:
Edit Link:
Play Link:
Kahoot (https://getkahoot.com)
Username: Dreadnought213
Email:
Password: Mustang2134
110

Chapter 13 notes
Differentiate among the four states of matter
The Nature of Gases
Key Q’s
What are the three assumptions of the kinetic theory as it applies to gases?
Single particles of gas are considered to be small, hard spheres of an insignificant volume.
The movement of the particles in a gas is random, constant, and rapid
All collisions between particles in gases are perfectly elastic
How does kinetic theory explain gas pressure?
Gas pressure is the result of billions of moving particles in a gas simultaneously colliding with an object
What is the relationship between the template in kelvins and the average kinetic energy of particles?
The Kelvin of a substance is directly proportional to the average kinetic energy of the particles of the substance
Definitions
Kinetic energy Energy because of motion
Kinetic Theory All matter consists of particles that are constantly moving
Gas pressure Force exerted by a gas per unit surface area of an object
Vacuum Empty space with no particles and no pressure
Atmospheric pressure Collisions of molecules and atoms in air with objects
Barometer Device to measure Atmospheric pressure
Pascal(Pa) SI unit of pressure
Standard atmosphere(atm) Pressure required to support 760mm of mercury in a mercury barometer
Moving bodies exert a force when they collide with other bodies
Like a liquid a gas will always fill the container no matter the size or shape
Uncontained gas can spread through space limitlessly because particles travel straight until they collide
with something like another particle.
An elastic collision is when particles collide transferring kinetic energy without losing any
If no particles are present then no collisions can occur so consequently no pressure
Air exerts pressure on earth since gravity holds it in the atmosphere
Higher altitudes results in lower atmospheric pressure
Particles in a collection of anything varies in kinetic energy
In a wide collection of particles most of the particles have a kinetic energy somewhere in the middle of
the range
Particles of all substances, regardless of state, have the same kinetic energy in varying temperatures
111

The Nature of Liquids
Key Q’s
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
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
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
Under what conditions does boiling occur
When a liquid is heated to a temperature at which the particles throughout the liquid have enough kinetic
energy to vaporize, the liquid begins to boil.
Definitions
Vaporization Conversion of a liquid into a gas or vapor
Evaporation Vaporization of but on the surface of a liquid that is not boiling
Vapor pressure Measure of the force exerted by a gas above a liquid
Boiling point Temperature at which the vapor pressure is equal to the external pressure on the liquid
Normal boiling point The boiling point of a liquid at a pressure of 101.3kPa
According to Kinetic theory liquids and gases both have kinetic energy
The ability of gases and liquids to flow allows them to change into the shape of a container
Particles in liquids are attracted to each other rather than gases that aren’t
Liquids are denser than gases and have a definite volume because of the attractaction
Increasing pressure on a liquid has no effect on the volume
Liquids and solids are known as condensed states of matter because pressure doesn’t affect volume
During evaporation most molecules don’t have enough kinetic energy to escape into the state of a gas
Some of the particles that evaporate and become a gas collide with something and return back into the
liquid they evaporated from
Heating a liquid leads to faster evaporation because heat increases kinetic energy which allows them to
overcome the attractive forces
Since in evaporation the the particles with the most kinetic energy escape first and since the particles
take their heat with them evaporation is technically a cooling process
Overtime the number of particles turning into vapor increases and some return to being liquid
112

The Nature of Solids
Key Q’s
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
What determines the shape of a crystal
The shape of a crystal reflects the arrangement of the particles within the solid
Definitions
Melting point The temperature at which a solid changes into a liquid
Freezing point The temperature at which a liquid changes into a solid
Crystal Particles arranged in an orderly, repeating, three dimensional patterns
Unit cell Smallest group of particles that within a crystal that retains the geometric shape of the crystal
Allotropes Two or more different molecular forms of the same element in the same physical state
Amorphous solid Solid that lacks internal structure
Glass Transparent fusion product of inorganic substances that have cooled to a rigid state without crystallizing
The particles in liquids are capable of movement but the particles of solids aren’t
Most solids are packed tightly together and are difficult to compress
When you heat solids the organization of the particles in solids break down
At the melting point the kinetic energy of the solids is able to overcome the attractions that hold them
together
When the freezing temperature and melting temperature then the liquid and solid phases are in
equilibrium
Ionic solids have higher melting points because of the strong forces holding them together compared to
molecular solids that have that have relatively low melting points
The angles at which the faces of a crystal intersect are always the same for a given substance and are
characteristics of that specific substance
Crystals are classified in seven groups that differ in terms of angles between faces and the number of
edges of equal length on each face
A lattice is a repeating array of any of the fourteen kinds of unit cells
Each crystal system can be composed of one to four kinds of cell units
Solids can come in multiple forms such as carbon
Allotropes are composed of the same atoms of the same element but different properties because of their
different structures
few elements have allotropes: phosphorus,sulfur,oxygen,boron,carbon,and antimony have allotropes
When a crystalline solid is shattered the fragments have the same surface angles as the original
113

114
The Learning Goal for this is:
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
Gas Laws
Introduction to Gases
Of the three common states of matter, the gaseous state was most easily described by early
scientists. As early as 1662, Robert Boyle showed how the volume of a gas, any gas, changed as the
pressure applied to it was changed. Soon thereafter, the effects of temperature and the quantity of
gas on the volume were discovered. The result of all these studies was a set of fundamental
mathematical equations known as the gas laws that applied equally well to any gas, whether pure
oxygen, nitrogen, or a mixture of the two. Through careful studies of gases reacting with one another,
Amedeo Avogadro later concluded that equal volumes of different gases must contain the same
number of molecules. For example, 10.0 L of oxygen contained the same number of oxygen
molecules as there were nitrogen molecules in 10.0 L of nitrogen.
As time passed, it became clear that one mole of any gas contained the same number of molecules,
6.02 × 10 23 molecules to be exact, a number known today as Avogadro's number. One mole of
anything is Avogadro's number of that thing—atoms, molecules, people, or dollars—and that would
be a lot of dollars!
Boyle's Law
Pressure is the amount of force exerted on one unit of area. The example of an ocean diver should
make the concept clearer: The greater the depth the diver reaches, the greater the pressure due to
the weight of the overlying water. Pressure is not unique to liquids but can be transmitted by gases
and solids, too. At the surface of the Earth, the weight of the overlying air exerts a pressure equal to
that generated by a column of mercury 760 mm high. The two most common units of pressure in
chemical studies are atmosphere and millimeters of mercury.
1 atm = 760 mm Hg = 760 torr.
However, the standard international unit for pressure is the Pascal.
1 atm = 101325 Pa
1 atm = 101.3 kPa
The English scientist Robert Boyle performed a series of experiments involving pressure and, in
1662, arrived at a general law—that the volume of a gas varies inversely with pressure.
P1V1 = P2V2
This formulation has become established as Boyle's law. Of course, the relationship is valid only if the
temperature remains constant.
As an example of the use of this law, consider an elastic balloon holding 5 L of air at the normal
atmospheric pressure of 760 mm Hg. If an approaching storm causes the pressure to fall to 735 mm
Hg, the balloon expands. The product of the initial pressure and volume is equal to the product of the
final pressure and volume while the temperature is constant.

It is important that you realize that pressure and volume vary inversely; therefore, an increase in
either one necessitates a proportional decrease in the other.
Convert a pressure of 611 mm Hg to atmospheres.
1. If a gas at 1.13 atm pressure occupies 732 milliliters, what pressure is needed to reduce the
volume to 500 milliliters?
Charles' Law
In 1787, the French inventor Jacques Charles, while investigating the inflation of his man‐carrying
hydrogen balloon, discovered that the volume of a gas varied directly with temperature. This relation
can be written as
V1 = V2
and is called Charles' law. For this law to be valid, the pressure must be held constant, and the
temperature must be expressed on the absolute temperature or Kelvin scale.
T1
Because the volume of a gas decreases with falling temperature, scientists realized that a natural
zero‐point for temperature could be defined as the temperature at which the volume of a gas
theoretically becomes zero. At a temperature of absolute zero, the volume of an ideal gas would be
zero. The absolute temperature scale was devised by the English physicist Kelvin, so temperatures
on this scale are called Kelvin (K) temperatures. The relationship of the Kelvin scale to the common
Celsius scale must be memorized by every chemistry student:
T2
K = °C + 273
Therefore, at normal pressure, water freezes at 273 K (0°C), which is called the freezing point, and
boils at 373 K (100°C). Room temperature is approximately 293 K (20°C). Both temperature scales
are used in tables of chemical values, and many simple errors arise from not noticing which scale is
presented.
Use Charles' law to calculate the final volume of a gas that occupies 400 ml at 20°C and is
subsequently heated to 300°C. Begin by converting both temperatures to the absolute scale:
T1 = 20°C = 293.15 K
T2 = 300°C = 573.15 K
115

Then substitute them into the constant ratio of Charles' law:
When using Charles' law, remember that volume and Kelvin temperature vary directly; therefore, an
increase in either requires a proportional increase in the other.
2. A gas occupying 660 ml at a laboratory temperature of 20°C was refrigerated until it shrank to 125
ml. What is the temperature in degrees Celsius of the chilled gas?
Gay-Lussac's Law
This relationship between temperature and pressure is known as Gay-Lussac's law. It states that if
the volume of a container is held constant as the temperature of a gas increases, the pressure inside
the container will also increase. As with the other gas laws, this one can be represented in the form of
an equation:
P1 = P2
T2
Recall that we use 1s and 2s to indicate the quantities before (1s) and after (2s) a change has taken
place. Also, note that the units for pressure do not matter, as long as they are the same throughout
the entire equation. The units for temperature must be Kelvins or the equation will not work. This is
because the Kelvin scale is an absolute scale - it doesn't go negative. Finally, this equation only
works for an ideal gas. Most gases that surround you and me behave very much like ideal gases, so
we can use this equation as an approximation for the gases we encounter.
T2
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3. The pressure of nitrogen gas in a light bulb is 60 kPa at 25°C. Calculate the pressure of the gas
when the temperature inside the bulb rises to 167°C after the bulb is lighted up?
Combined gas law
The combined gas law makes use of the relationships shared by pressure, volume, and temperature:
the variables found in other gas laws, such as Boyle's law, Charles' law and Gay-Lussac's law. Let's
review the basic principles of these three laws.
Imagine you are a diver, and you begin your dive with lungs full of air. As
you go deeper under water, the pressure you experience in your lungs
increases. When this happens, the air inside your lungs gets squished, so
the volume decreases. This is an example of Boyle's law in action, which
states that the higher the pressure (P), the lower the volume (V), as
shown in this image. Here, k is any constant number.
Have you ever tried putting a balloon in the refrigerator and notice that it
shrinks? As the temperature of the refrigerated balloon decreases, the
volume of the gas inside the balloon also decreases. When you take the
balloon out of the refrigerator, it reverts to its original size, so the opposite is
also true; when the temperature increases, the volume also increases. The
shrinking balloon serves as a demonstration of Charles' law, which states
that the higher the temperature (T), the higher the volume (V).
Imagine yourself driving down a road, which can cause the
temperature to increase within your tires. As a result, the air inside the
tires expands, and the pressure increases. This is an example of Gay-
Lussac's law, which shows the relationship between pressure (P) and
temperature (T) when the volume remains constant; as the
temperature increases, the pressure also increases.
When we put Boyle's law, Charles' law, and Gay-Lussac's law together, we come up with the
combined gas law, which shows that:
Pressure is inversely proportional to volume, or higher volume equals lower pressure.
Pressure is directly proportional to temperature, or higher temperature equals higher pressure.
Volume is directly proportional to temperature, or higher temperature equals higher volume.
Let's take a look at the formula for the combined gas law. Here, PV / T = k shows how pressure,
volume and temperature relate to each other, where k is a constant number.
The formula for the combined gas law can be adjusted to compare two sets of conditions in one
substance. In the equation, the figures for pressure (P), volume (V), and temperature (T) with
subscripts of one represent the initial condition, and those with the subscripts of two represent the
final condition.
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P1V1 = P2V2
T1
It is important to note that the temperature should always be in Kelvin, so if the given units are in
Celsius, then those should be converted to Kelvin by adding 273.
450 mL of a gas occupies a container that has a temperature of 28°C and a pressure of 788 mmHg.
What is the temperature if the volume is reduced to 50 mL at 760 mmHg? As the unit of
measurement is in Celsius, remember to convert it to Kelvin.
T2
4. A closed gas system initially has pressure and temperature of 1160torr and 542K with the volume
unknown. If the same closed system has values of 912torr, 9720mL and 686K, what was the initial
volume in L?
Ideal Gas Equation
The relations known as Boyle's law, Charles' law, and Avogadro's law can be combined into an
exceedingly useful formula called the Ideal Gas Equation,
where R denotes the gas constant:
PV = nRT
The temperature is, as always in gas equations, measured in Kelvin.
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This formula is strictly valid only for ideal gases—those in which the molecules are far enough apart
so that intermolecular forces can be neglected. At high pressures, such forces cause significant
departure from the Ideal Gas Equation, and more complicated equations have been devised to treat
such cases. The Ideal Gas Equation, however, gives useful results for most gases at pressures less
than 100 atmospheres.
This is the value stated in the carbon dioxide reaction; you were asked to memorize that 1 mole of
any gas occupies 22.4 liters at STP.
You should be able to use the Ideal Gas Equation to determine any one of the four quantities—
pressure, volume, moles, or temperature—if you are given values for the other three.
One important application is to deduce the molecular mass and formula for a gas. Assume that you
know that the hydrocarbon propylene is, by mass, 85.6% carbon and 14.4% hydrogen. Then the
atomic ratios of the compound are
Therefore, the propylene molecule is some integral multiple of CH2: It can be CH2 or C2H4 or C3H6 or
a yet larger molecule. Measuring the volume of 10 grams of propylene at STP yields 5.322 liters,
which you can use to calculate its molecular mass.
Because the atomic masses of 1 CH2 unit added together is 14.03, the molecule contains three such
units. Consequently, the molecular formula for propylene is C3H6.
5. What is the volume occupied by 1 kilogram of carbon monoxide at 700°C and 0.1 atm?
6. The ozone molecule contains only oxygen atoms. Determine the molecular formula of ozone given
that 2.3 grams occupies 1,073 milliliters at standard temperature and pressure.
1*1.073=2.3*.1*.0821
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The students will learn how the interactions
between water molecules account for the unique
The Learning Goal for this assignment is: 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
Water is essential for all life on earth being present deep underground and in the atmosphere. Water
is simple being made of 3 atoms 2 hydrogen and one oxygen. This simple structure allows it to be
highly polar, one side (the hydrogens) being positive and the other (the oxygen) negative. water is
capable of high surface pressure, low vapor pressure, and a high boiling point because to the
hydrogen bonding between the negative end of one molecule to the positive end of another. attractive
forces between water are balanced everywhere except the surface. at the surface the molecules
experience unbalanced attractive forces surface particles to be pulled in wards which reduces ​surface
tension​ 1​ ( surface area of liquid). All liquids have surface tension but water has the highest causing it
to have a spherical shape compared to other liquids that flatten because of gravity. The tension that
water has can be reduced by adding a ​surfactant​2​(substance that interferes with hydrogen bondings).
Adding a surfactant to water will cause the tension decrease by interfering with the hydrogen bonding
between the molecules. The hydrogen bonding between water molecules are also responsible for low
vapor pressure since the bonds have tom be broken to be able to evaporate. The hydrogen bonding
in water increases the amount of heat needed to break these bonds.
Water in the Solid State
Water in solid state has a lower density than liquid water, on the contrary most liquids increase in
density as molecules pack together causing it to sink in it’s own liquid water’s density is the highest at
4 degrees celsius at 1.0000 yet below that the density decreases again. Because of hydrogen
bonding water molecules are more spaced out and arranged when solid compared to liquid. Ice
capable of becoming an insulator for water that is underneath it keeping it warmer that 0 degrees
celsius. more energy is required to turn a liquid to
15.2 Homogeneous Aqueous Systems
Solutions
Water that is out in nature and even tap water are ​aqueous solutions​ 3​ (water that contains dissolved
substances) and never chemically pure because it dissolves so many things it comes in contact with.
In solutions​ solvents​4​(the dissolving medium in the solvent) are used to dissolve ​solutes​5​(dissolved
particles in a solution). Solutions are homogeneous and stable. some things dissolve better than
others; in water ionic compounds and polar covalent compounds are the most ready to dissolve but
nonpolar covalent will not dissolve in water. water being a liquid has kinetic energy and is always
moving and begins a process called ​solvation ​ 6​ (the process by which positive and negative ionic
solids become surrounded by solvent molecules)when there are solid ionic compounds are present.
some compound have bonds stronger than the attractions of water and are nearly insoluble. polar
solvents dissolve polar and ionic compounds while nonpolar solvents dissolve nonpolar solutes.
Electrolytes and Nonelectrolytes
water being a aqueous solvent is used to find if a substance is an ​electrolyte​ 7​ ( a compound that
conducts an electric current when in an aqueous solution or molten state)or a
nonelectrolyte​ 8​ (Compound that does not conduct electricity in neither a aqueous solution nor a molten
state) there are two electrolytes ​strong electrolytes​ 9​ (solutioremove rn which solute mostly exists as
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ions)and ​weak electrolytes​ 10​ (solution that has few or no ions)that can be determined by placing
soluthold es in water. Electrolytes are essential for living because they are capable of delivering
electrical impulses.
Hydrates
there is water in crystals that is called ​water of hydration​ 11​ ( Water contained within a crystal) a crystal
that contains water of hydration is called a ​hydrate​12​. whenever a crystal is heated to 100 degrees
celsius or above they begin to lose water but since the forces that hold the water molecules in
hydrates aren’t strong they lose and regain water easily. substances that do not contain water are
known as ​anhydrous​ 13​ . although some anhydrous substances can regain water if it comes in contact
with it like copper(II) sulfate will turn blue with water. sometimes a hydrate can have a higher vapor
pressure than water it will lose its water of hydration or ​effloresce​14​. Hydrated ionic compounds that
have low vapor pressure can remove water from moist air these hydrates are called ​hygroscopic​ 15​ .
Certain substances can also remove water from the air to create a dry atmosphere they are called
desiccant​ 16​ . some compounds are so sufficient that they can dissolve completely and cause solutions
they are called ​deliquisent​ 17​ .
15.3 Heterogeneous Aqueous Systems
Suspensions
a ​suspension​ 18​ is a mixture of particles that will settle after a moment. they are different than a
solution because the size of the particle are 1000 times larger than that of a solution. for example clay
and water don’t mix so if you shake them together then the water will become cloudy with the clay
particles but will not stay that way since the clay will eventually settle.
Colloids
One of the things we eat as snacks is a ​colloid​ 19​ , a colloid is a heterogeneous mixture that has
particles ranging in sizes from 1 nm to 1000 nm. they are different from suspensions because of size,
colloids are smaller than suspensions but bigger than solutions. another key difference is the ​tyndall
effect​20​, in which visible light is scattered by colloidal particles. both suspensions and colloids share
this while solutions do not as light does not pass through. colloids also scintillate because of the
radical movement of particles, this movement is called ​brownian motion​ 21​ . ​The final colloid is
Emulsion​ 22​ , colloidal dispersion of a liquid in a liquid. for example oils wont become solutions with
water but will form a colloidal dispersion if soap is added to the water
121

123
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

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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:
The students will learn what properties are used to describe the nature of solutions and how to quantify
the concentration of a solution.
Properties of Solutions
Solution Composition
Solute - substance which is dissolved (if liquid-liquid, the one you have less of is considered the
solute)
Solvent - substance which is doing the dissolving.
Molarity (M) - the number of moles of solute per liter of solution.
Mass percent (weight percent) - percent by mass of solute in the solution.
Molality - the number of moles of solute per kilogram of solvent.
Energies of Solution Formation
Three steps in creating a liquid solution
Step 1 - break up the solute into individual components (expanding the solute).
Step 2 - overcoming intermolecular forces in the solvent to make room for the solute
(expanding the solvent).
Step 3 - allowing the solute and solvent to interact to form the solution.
Enthalpy - a thermodynamic quantity equivalent to the total heat content of a system. It is equal to the
internal energy of the system plus the product of pressure and volume.
Enthalpy (heat) of solution - sum of the energies it takes to dissolve a substance.
Enthalpy (heat) of hydration - represents the enthalpy change associated with the dispersal of a
gaseous solute in water.
Processes which require large amounts of energy tend not to occur.
Factors Affecting Solubility
Since it is the molecular structure that determines polarity, there should be a definite connection
between structure and solubility.
Like dissolves like - the rule says that most typically, polar solutes can dissolve in polar solvents and
non-polar solutes can dissolve in non-polar solvents. But, polar solutes rarely dissolve in non-polar
solvents and vice versa.
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Structure effects: determines what kind of substance will be able to dissolve the sample.
hydrophobic - water-fearing
hydrophilic - water-loving
While pressure has little effect on the solubilities of solids or liquids, it does significantly increase the
solubility of a gas.
Pressure effects: gas solubility increases with increased pressure.
Henry's Law: the amount of a gas dissolved in a solution is directly proportional to the pressure of the
gas above the solution.
P = kC
P represents the partial pressure of the gas, C represents the concentration of the dissolved gas, k is
a constant of the particular solution.
Solubility and Pressure are Directly Related
Dilution is the process of decreasing the concentration of a solute in solution, usually simply by mixing
with more solvent. To dilute a solution means to add more solvent without the addition of more solute.
The resulting solution is thoroughly mixed so as to ensure that all parts of the solution are identical.
Temperature effects - the solubility of most solids increases with temperature.
The Vapor Pressures of Solutions
Volatile - quantified by the tendency of a substance to vaporize. Volatility is directly related to a
substance's vapor pressure. At a given temperature, a substance with higher vapor pressure
vaporizes more readily than a substance with a lower vapor pressure.
Nonvolatile - does not readily form a vapor.
Nonvolatile solutes lowers the vapor pressure of a solvent. The nonvolatile solute decreases the
number of solvent molecules per unit volume. Thus it lowers the number of solvent molecules at the
surface, and it should proportionately lower the escaping tendency of the solvent molecules.
Boiling-Point elevation and Freezing-Point Depression
Colligative properties - depend on the number, and not the identity of the solute particles in an ideal
solution. ex. - freezing-point depression and boiling-point elevation.
Boiling-point elevation - a nonvolatile solute elevates the boiling point of the solvent.
Freezing-point depression - the water in the solution has a lower vapor pressure than that of pure ice.
http://slideplayer.com/slide/10775304/
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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:
endothermic reactions are When energy is added to a substance from the environment
exothermic reactions are When energy is removed from a substance and released off into the
environment

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.
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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:
specific heat is increasing or decreasing the temperature of a gram of a substance by one degree
in celsius.

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:
all of the values must be in the same unit
We use the equation q = mc/\T
there is no need to round the final answer

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.
131

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:
as water reaches 100 degrees Celsius then /\t is also 100

133
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:
134

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,
135

then it is exothermic, meaning the system gives off heat to its surroundings, so q

The Learning Goal for this assignment is:
Thermochemistry Review Questions
Particulate Nature of Matter and Changes of State
Watch the following video and answer the questions.
1. How are solid particles arranged?
Solid particles are arranged very compactly and dont have much room to move.
2. What is supplied to make a solid become a liquid?
Energy is supplied to convert a solid to a liquid
3. What does ice take from the environment when melting?
it takes the surrounding energy from the environment
4. What do molecules need to evaporate?
they need to have enough energy to escape liquid form
5. Do all substance need the same amount of energy to evaporate?
no some substance require less like Ether can evaporate at room temperature
6. Can gases change straight into solids?
Yes if it is cold enough for the environment to aborb enough energy from the gas the gas become a solid
1. Which of the following statements about the atoms and molecules in all three states of matter is
NOT true.
a. Particles of matter are in constant motion
b. Space exists between particles
c. Adding heat increases the size of the particles
d. The particles of matter are too small to be seen with the unaided eye
2. Which statement is correct?
a. All particles in a given substance at a given temperature have equal kinetic energy
b. Heat always flows from a hotter object to a cooler object
c. Heat is a measure of average kinetic energy of a substance
d. Temperature measures the total kinetic energy of a substance
3. In what state does matter have the most kinetic energy?
a. Solid
b. Liquid
c. Gas
d. Plasma
4. In what state does matter have the least kinetic energy?
a. Solid
b. Liquid
c. Gas
d. Plasma
5. What is the best description of the relationship between states of matter and kinetic energy?
a. There is no relationship
b. As kinetic energy increases the attraction between molecules increases
c. As the kinetic energy decreases the attraction between molecules decreases
d. As the kinetic energy increases the attraction between molecules decreases
137

Define the types of reactions. Give the names of the changes of States of Matter and describe the
relative kinetic energy of the particles.
Endothermic Reactions:
Gas to Liquid
Condensation
Liquid to Solid
freezing
Gas to Solid
deposition
Exothermic Reactions:
Solid to Liquid
melting
Liquid to Gas
vaporization evaporation and boiling
Solid to Gas
sublimination
138

Phase Diagram
Information on a Phase Diagram
1. In each colored region the substance is in a single phase
2. The lines describes the equilibrium conditions for the phases on either side of the line
3. The triple point is the conditions required for all three state of matter to exist in equilibrium
139

1. If this compound was at room temperature (25°C), what phase would it most likely be in?
gas
2. At what temperature would all three phases of this substance exist?
350 degrees
3. If this substance is at 45 atm and 100°C what will happen if I raise the temperature to 400°C?
it will subliminate
4. If this substance is at 70 atm and 500°C and the pressure is decreased to 40 atm what phase
change will occur?
it will vaporize
5. Why can’t this substance be boiled at a temperature of 200°C?
because it's liquid phase is at 350 degrees at and it can't be a liquid at 200 degrees
6. If I wanted to, could I drink this substance?
50 atm at 350 degrees is required for it to be a liquid so it is not possible
140

Heating Curve
Information on a Phase Diagram
1. In general the temperature goes up the longer the heating continues.
Q=mc∆T
2. The horizontal parts to the graph is a change of state.
Q=mHf
Q=mHv
141

1. In what part of the curve would a substance have a definite shape and definite volume?
in part 1 of the curve the substance would be frozen and have a definite shape
2. In what part of the curve would substance a have a definite volume but no definite shape?
in part 3 the substance would have a volume but no shape since liquids take up anyshape
3. In what part of the curve would a substance have no definite shape or volume?
in part 5 of the curve the substance would be a gas and have no definite shape or volume
4. What part of the curve represents a mixed solid/liquid phases?
Part 2
5. What part of the curve represents a mixed liquid/vapor phases?
Part 4
6. What is the melting temperature of substance?
the substances melts at 5 degrees
7. What is the boiling temperature of substance?
the boiling temperature of the substance at 55 degrees
8. In what part of the curve would the molecules have the lowest kinetic energy?
the substance would have the least in phase 1 since it would be a solid
9. In what part of the curve would the molecules have the greatest kinetic energy?
it would have the most in phase 5 since it would be a gas
142

The Learning Goal for this assignment is:
Thermochemistry Review Questions
Particulate Nature of Matter and Changes of State
Watch the following video and answer the questions.
1. How are solid particles arranged?
Solid particles are arranged very compactly and dont have much room to move.
2. What is supplied to make a solid become a liquid?
Energy is supplied to convert a solid to a liquid
3. What does ice take from the environment when melting?
it takes the surrounding energy from the environment
4. What do molecules need to evaporate?
they need to have enough energy to escape liquid form
5. Do all substance need the same amount of energy to evaporate?
no some substance require less like Ether can evaporate at room temperature
6. Can gases change straight into solids?
Yes if it is cold enough for the environment to aborb enough energy from the gas the gas become a solid
1. Which of the following statements about the atoms and molecules in all three states of matter is
NOT true.
a. Particles of matter are in constant motion
b. Space exists between particles
c. Adding heat increases the size of the particles
d. The particles of matter are too small to be seen with the unaided eye
2. Which statement is correct?
a. All particles in a given substance at a given temperature have equal kinetic energy
b. Heat always flows from a hotter object to a cooler object
c. Heat is a measure of average kinetic energy of a substance
d. Temperature measures the total kinetic energy of a substance
3. In what state does matter have the most kinetic energy?
a. Solid
b. Liquid
c. Gas
d. Plasma
4. In what state does matter have the least kinetic energy?
a. Solid
b. Liquid
c. Gas
d. Plasma
5. What is the best description of the relationship between states of matter and kinetic energy?
a. There is no relationship
b. As kinetic energy increases the attraction between molecules increases
c. As the kinetic energy decreases the attraction between molecules decreases
d. As the kinetic energy increases the attraction between molecules decreases
143

Define the types of reactions. Give the names of the changes of States of Matter and describe the
relative kinetic energy of the particles.
Endothermic Reactions:
Gas to Liquid
Condensation
Liquid to Solid
freezing
Gas to Solid
deposition
Exothermic Reactions:
Solid to Liquid
melting
Liquid to Gas
vaporization evaporation and boiling
Solid to Gas
sublimination
144

Phase Diagram
Information on a Phase Diagram
1. In each colored region the substance is in a single phase
2. The lines describes the equilibrium conditions for the phases on either side of the line
3. The triple point is the conditions required for all three state of matter to exist in equilibrium
145

1. If this compound was at room temperature (25°C), what phase would it most likely be in?
gas
2. At what temperature would all three phases of this substance exist?
350 degrees
3. If this substance is at 45 atm and 100°C what will happen if I raise the temperature to 400°C?
it will subliminate
4. If this substance is at 70 atm and 500°C and the pressure is decreased to 40 atm what phase
change will occur?
it will vaporize
5. Why can’t this substance be boiled at a temperature of 200°C?
because it's liquid phase is at 350 degrees at and it can't be a liquid at 200 degrees
6. If I wanted to, could I drink this substance?
50 atm at 350 degrees is required for it to be a liquid so it is not possible
146

Heating Curve
Information on a Phase Diagram
1. In general the temperature goes up the longer the heating continues.
Q=mc∆T
2. The horizontal parts to the graph is a change of state.
Q=mHf
Q=mHv
147

1. In what part of the curve would a substance have a definite shape and definite volume?
in part 1 of the curve the substance would be frozen and have a definite shape
2. In what part of the curve would substance a have a definite volume but no definite shape?
in part 3 the substance would have a volume but no shape since liquids take up anyshape
3. In what part of the curve would a substance have no definite shape or volume?
in part 5 of the curve the substance would be a gas and have no definite shape or volume
4. What part of the curve represents a mixed solid/liquid phases?
Part 2
5. What part of the curve represents a mixed liquid/vapor phases?
Part 4
6. What is the melting temperature of substance?
the substances melts at 5 degrees
7. What is the boiling temperature of substance?
the boiling temperature of the substance at 55 degrees
8. In what part of the curve would the molecules have the lowest kinetic energy?
the substance would have the least in phase 1 since it would be a solid
9. In what part of the curve would the molecules have the greatest kinetic energy?
it would have the most in phase 5 since it would be a gas
148

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
149

Reduction
Oxidation reduction reaction
Oxidation number
Half reaction
Electrochemical process
Battery
Cathode
Anode
Electrolysis
150

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.
Acids
Turn blue litmus red
taste sour
acid corrode metal
positively charged hydrogen ions (H)
Bases
turn red litmus blue
taste bitter
negatively hydroxide ions
most hand soaps are bases
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.
151

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¯
152

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 Brønsted and Thomas Lowry in 1923.
153

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:
154

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.
155

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).
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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.
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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
2.82
2. 5.0 x 10¯9
3. 1.0
4. 3.27 x 10¯4
8.30
0.00
3.485
5. 1.00 x 10¯12
6. 0.00010
12.000
4.00
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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
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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
Acid base reaction, taking hydrogens adding hydroxides to get water
Titration
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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:
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Ag+ + Cu ---> Ag + Cu2+ Ag+ -> Ag reduction Cu-> 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:
Oxidation agent is reducing the oxidation number. And a reduction agent does the opposite
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).
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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¯
5+
4+
6+
6+
3+
2+
5+
4+
7+
10. Mn in MnO4 2¯
6+
Notes:
The way to find the number is doing the first rule left to right if there is none on the left then start the one on the
left
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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¯
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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+
2+
0 1-
3) Fe 2+ → Fe 3+
4) I3¯ → I¯
3+
Reduction
Oxidation
Oxidation
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
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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
2¯
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
2¯
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
2¯
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.
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