STOICHIOMETRY

CHEM180/181 2004
STOICHIOMETRY
Law of conservation of mass
• The balance became an
important instrument in
the eighteen century.
• Balances measure mass,
which is the quantity of
matter in a material.
Chapter 3
You should note the distinction
between the terms mass and
weight in precise usage.
• By weighing substances before and after
chemical change, Lavoisier (1743-1794)
demonstrated the law of conservation of
mass which states that:
Using the Law of Conservation of Mass
You heat 2.53 g of metallic mercury in air, which
produces 2.73 g of a red-orange residue. Assume that
the chemical change is the reaction of the metal with
oxygen in air.
Mercury + oxygen
red-orange residue
• the total mass remains constant during a
chemical change (chemical reaction).
What is the mass of oxygen that reacts?
When you strongly heat the red-orange residue,
it decomposes to give back mercury and
releases the oxygen, which you collect.
What is the mass of oxygen you collect?
Solution:
From the law of conservation of mass:
Total mass of substances before reaction
= Total mass of substance after reaction
Therefore:
That is:
Chemical Equations
A chemical equation is a symbolic
representation of a chemical reaction in terms
of chemical formulas.
eg.
2Na + Cl 2 2NaCl
A reactant is a starting substance in a chemical
reaction
A product is a substance that results from a
reaction.
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CHEM180/181 2004
Balancing Chemical Equations
A chemical equation is said to be balanced,
when the numbers of atoms of each element
are equal on either side of the arrow.
The state or phase of a substance may be
indicated with one of the following labels:
eg.
(g) ; (I) ; (s) ; (aq)
2Na(s) + Cl 2 (g) 2NaCl(s)
Consider the burning of natural gas:
CH 4 + O 2
CO 2 + H 2 O
You can also indicate the conditions under
which a reaction takes place, as well as the
presence of a catalyst.
eg. 2H 2 O 2 (aq)
Pt
2H 2 O(l) + O 2 (g)
Consider the burning or propane gas
C 3 H 8 + O 2
CO 2 + H 2 O
To balance the equation, you select coefficients that
will make the numbers of atoms of each element equal
on both sides of the equation.
Balancing Simple Equations
Balance the following equations:
C 3 H 8 + O 2
C 3 H 8 + O 2
CO 2 + H 2 O
CO 2 + H 2 O
CS 2 + O 2 CO 2 + SO 2
C 2 H 5 OH + O 2
CO 2 + H 2 O
Now have
C 3 H 8 + O 2 CO 2 + H 2 O BALANCED
NH 3 + O 2
NH 3 + O 2
NO + H 2 O
NO + H 2 O
Atomic Weights & Molecular Weights
Central to Dalton’s atomic theory was the idea
that each atom had a characteristic mass.
We now know that each isotope has its own
characteristic mass. Dalton actually calculated
the average atomic mass, based on the relative
abundance of each isotope.
Dalton could not weigh individual atoms,
instead he could find the average mass of one
atom relative to the average mass of another.
(using H as a base)
Atomic mass unit - amu
Today based on carbon-12
Mass of 12 C = 12 amu
1
1 amu = 12 mass of 12 C
The atomic weight of an element, is the
average atomic mass for the naturally
occurring element, expressed in atomic mass
units.
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CHEM180/181 2004
Fractional Abundance
The fractional abundance of an isotope is the
fraction of the total number of atoms that is
composed of a particular isotope
Alternate Example
An element has four naturally occurring
isotopes. The mass and percentage
abundance of each isotope are as follows:
Percentage abundance
Mass
1.50 203.973
23.6 205.9745
22.6 206.9745
52.3 207.9766
What is the atomic weight and the name of
the element?
The mass percentages sums to 100
ie. 1.48 + 23.6 + 22.6 + 52.3 = 99.98
= 100 (SIG FIG’S)
Percentage Composition
Percentage composition of a compound = the
percentage by mass contributed by each
element in the substance
The percentage of element B Is defined as:
(Atoms of B)(AW of B)
Mass % B =
mass of the whole
x 100%
Atomic weight = Therefore the element is :
Remember the sum of the percentages of each
element in the substance add up to 100%
Example
Lead(II)chromate is used as a paint pigment
(chrome yellow). What is the percentage
composition of lead(II)chromate?
The chemical name for table sugar is sucrose,
C 12 H 22 O 11 . How many grams of carbon are in
61.8 g of sucrose?
Lead(II)chromate = PbCrO 4
A r (Pb) = 207.2 g/mol
A r (Cr) = 51.9961 g/mol
A r (O) = 15.9994 g/mol
%Pb =
M r (PbCrO 4 )
=
%Cr =
Consider 1 mol of
PbCrO 4
%O =
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CHEM180/181 2004
Mass and Moles of Substance
• We have seen so far that atoms / molecules
react in simple ratios.
• Atoms and molecules are too small to count.
• We need find another way of expressing
such large numbers
In pursuit of this end we are going to consider
the following concepts:
• Molecular weight
• Formula weight
• the Chemical Mole
Formula and Molecular Weight
The formula weight of a substance is the sum
of the atomic weights of all atoms in a formula
unit of the compound.
The molecular weight of a substance is the
sum of the atomic weights of all the atoms in a
molecule of the substance.
Note: the formula weight is
applicable to both molecular and
ionic compounds.
Example
Calculate the formula weight of the following
compounds from their formulae:
b) methylamine = CH 3 NH 2
a) calcium hydroxide
The Mole
Chemists have adopted the mole
concept as a convenient way to
deal with the enormous numbers
of atoms, molecules or ions in
the samples they work with.
Definition of Mole:
A mole (symbol mol) = the amount of matter
that contains as many Objects (atoms,
molecules etc) as the number of atoms in
exactly 12 grams of carbon-12.
The number of atoms in a 12 gram sample of
carbon-12 is called Avogadro’s number (N A )
The term mole, just like a dozen or gross,
refers to a particular number of things.
one dozen = 12
one gross = 144
one mole = 6.0221367 x 10 23
Avogadro’s number (N A )
When using the term mole it is important to
specify what is being referred to!
eg : 1 mol of CH 3 COOH molecules contains
2 moles of C atoms etc.
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CHEM180/181 2004
The molar mass of a substance is the mass of
one mole of the substance
For all substances, the molar mass in grams
per mole is numerically equal to the formula
weight in atomic mass units.
Calculating the Mass of an Atom or
Molecule
What is the mass of a nitric acid molecule?
nitric acid =
Mole Calculations
Because molar mass is the mass per mole we
have the formula:
Molar Mass =
Mass
Moles
or M r =
m
n
Example
Converting Grams of Substance to Moles
Calcite is a mineral composed of calcium
carbonate, CaCO 3 . A sample of calcite
composed of pure CaCO 3 weighs 23.6 g. How
many moles of calcium carbonate is this?
Remember that mass must be expressed in
grams here!
Converting Moles of Substance to Grams
A sample of nitric acid contains 0.253 mol
HNO 3 . How many grams is this?
M r (HNO 3 ) = 63.013 g.mol -1
Example
(see previous example)
Example
Calculating the Number of Molecules in a Given
Mass
The daily requirement of chromium in the
human diet is 1.0x10 -6 g. How many atoms of
chromium does this represent?
n = 0.253 mol
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CHEM180/181 2004
Determining formulae
The percentage composition of a compound
leads directly to its empirical formula.
Recall: An empirical formula for a compound
is the formula of a substance written with the
smallest integer subscripts.
Eg. Consider hydrogen peroxide:
Molecular formula = H 2 O 2
Empirical formula = HO
Compounds with different molecular formulae
can have the same empirical formula, and
such substances will have the same
percentage composition.
Eg. acetylene = C 2 H 2
benzene = C 6 H 6
both have the empirical formula =
Empirical formula from Composition
Consider the following flow-diagram:
Percent composition
Mass Composition
Number of moles of
each element
Example
Determining the Empirical Formula from the
Masses of Elements (Binary Compound)
We have determined the percentage
composition of benzene: 92.3% C and 7.7% H.
What is the empirical formula of benzene?
Divide by smallest number of
moles to find the molar ratios
Multiply by appropriate number to
get whole number subscripts
Ratio C : H = Empirical Formula =
Example
Determining The Empirical Formula from
Percentage Composition. (General)
Sodium pyrophosphate is used in detergent
preparations. The mass percentages of the
elements in this compound are 34.6% Na,
23.3% P and 42.1%O. What is the empirical
formula of sodium pyrophosphate?
Hint: Consider a 100 g sample.
Molecular Formula from Empirical
Formula
The molecular formula of a compound is a
multiple of its empirical formula.
Molecular weight = n x empirical formula weight
where n = number of empirical formula units
in the molecule.
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CHEM180/181 2004
Example
Determining the Molecular Formula from the
Percentage Composition and Molecular Weight.
We have already determined the mass
composition and empirical formula of benzene.
In a separate experiment, the molecular weight
of benzene was determined to be 78.1. What is
the molecular formula of benzene
Empirical Formula = formula weight =
Stoichiometry
Stoichiometry is the calculation of the
quantities of reactants and products involved in
a chemical reaction
It is based on the chemical equation and the
relationship between mass and moles.
We can interpret a chemical equation in terms of
number of molecules (or ions or formula units) or in
terms of number of moles of molecules etc. depending
on our needs.
Remember that because moles can be converted to
mass, we can also produce a mass interpretation of a
chemical equation.
Consider the reaction:
N 2 (g) + 3H 2 (g) 2NH 3 (g)
Molecular Interpretation:
1 molecule N 2 + 3 molecules H 2 2 molecules NH 3
Molar interpretation:
1 mol N 2 + 3 mol H 2 2 mol NH 3
Mass interpretation:
28.0 g N 2 + 3 x 2.02 g H 2 2 x 17.0 g NH 3
Example
Amounts of Substance in a Chemical Reaction
Find the amount of water produced when 3.68
mol of NH 3 is consumed according to the
following equation:
4NH 3 + 5O 2
4 NO + 6H 2 O
We assume that sufficient O 2 is present.
From the equation we know that :
Notice that :
The number of moles involved in a reaction is
proportional to the coefficients in the balanced chemical
equation.
The Steps in a Stoichiometric
Calculation
Mass of substance A
Moles of substance A
Use molar mass of A
Use coefficients of A & B in
balanced eqn
Moles of substance B
Example
Calculate the mass of sulfur dioxide (SO 2 )
produced when 3.84 mol O 2 is reacted with
FeS 2 according to the equation:
4FeS 2 + 11O 2 2Fe 2 O 3 + 8SO 2
3.84 mol m = ?
2.79 mol
Mass of substance B
Use molar mass of B
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CHEM180/181 2004
Another Example
One of the most spectacular reactions of
aluminium, the thermite reaction, is with iron
oxide, Fe 2 O 3 , by which metallic iron is made.
The equation is :
2Al(s) + Fe 2 O 3 (s)
Al 2 O 3 (s) + 2Fe(l)
A certain welding operation, requires that at least 86.0 g
of Fe be produced. What is the minimum mass in grams
of Fe 2 O 3 that must be used for the operation?
Calculate also how many grams of aluminium are
needed.
Strategy:
2Al(s) + Fe 2 O 3 (s)
mass of Fe mol of Fe
mol of Fe mol of Fe 2 O 3
Al 2 O 3 (s) + 2Fe(l)
Limiting Reactants
The limiting reactant (or limiting reagent) is
the reactant that is entirely consumed when a
reaction goes to completion.
mol of Fe 2 O 3 mass of Fe 2 O 3
The moles of product are always determined
by the starting moles of the limiting reactant.
The Cheese Sandwich Analogy
Which is the limiting reactant?
Strategy:
• Use the relationships from the balanced
chemical equation
• You take each reactant in turn and ask how
much product would be obtain, if each were
totally consumed.
• The reactant that gives the smaller amount
of product is the limiting reactant.
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CHEM180/181 2004
Example
Example
More Challenging!!
When 100.0 g mercury is reacted with 100.0 g
bromine to form mercuric bromide, which is
the limiting reagent?
Hg + Br 2 HgBr 2
Iron can be obtained by reacting the ore
heamatite (Fe 2 O 3 ) with coke (C). The latter is
converted to CO 2 . As manager of a blast
furnace you are told that you have 20.5 Mg of
Fe 2 O 3 and 2.84 Mg of coke on hand.
(a) Which should you order first - another
shipment of iron ore or coke?
Thus the limiting reagent is
(b) How many megagrams of iron can you
make with the materials you have?
2Fe 2 O 3 + 3C 4Fe + 3CO 2
(b) How many megagrams of iron can you
make with the materials you have?
Recommended tut questions
2.17 3.1 3.35
2.18 3.5 3.59
2.26 3.21 3.63
2.27 3.27 3.96
3.33
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