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STOICHIOMETRY
Chapter 3
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.
You should note the distinction
between the terms mass and
weight in precise usage.
Using the Law of Conservation of Mass


By weighing substances before and after
chemical change, Lavoisier (1743-1794)
demonstrated the law of conservation of
mass which states that:
the total mass remains constant during a
chemical change (chemical reaction).
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
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:
Chemical Equations
From the law of conservation of mass:
Total mass of substances before reaction
= Total mass of substance after reaction
Therefore:
That is:
A chemical equation is a symbolic
representation of a chemical reaction in terms
of chemical formulas.
eg.
2Na + Cl2  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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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:
(g) ; (I) ; (s) ; (aq)
eg.
2Na(s) + Cl2(g)  2NaCl(s)
Consider the burning of natural gas:
CH4 + O2
You can also indicate the conditions under
which a reaction takes place, as well as the
presence of a catalyst.
CO2 + H2O
eg.
2H2O2(aq)
Pt
2H2O(l) + O2(g)
Consider the burning or propane gas
C3H8 + O2
CO2 + H2O
To balance the equation, you select coefficients that
will make the numbers of atoms of each element equal
on both sides of the equation.
C3H8 + O2
CO2 + H2O
C3H8 + O2
CO2 + H2O
Balancing Simple Equations
Balance the following equations:
CS2 + O2
CO2 + SO2
C2H5OH + O2
Now have
C3H8 + O2
CO2 + H2O
BALANCED
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)
CO2 + H2O
NH3 + O2
NO + H2O
NH3 + O2
NO + H2O
Atomic mass unit - amu
Today based on carbon-12
Mass of 12C = 12 amu
1 amu =
1
12
mass of 12C
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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Alternate Example
Fractional Abundance
The fractional abundance of an isotope is the
fraction of the total number of atoms that is
composed of a particular isotope
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?
Percentage Composition
The mass percentages sums to 100
ie.
1.48 + 23.6 + 22.6 + 52.3 = 99.98
= 100 (SIG FIG’S)
Percentage composition of a compound = the
percentage by mass contributed by each
element in the substance
The percentage of element B Is defined as:
Mass % B =
Atomic weight =
Therefore the element is :
(Atoms of B)(AW of B)
x 100%
mass of the whole
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,
C12H22O11. How many grams of carbon are in
61.8 g of sucrose?
Lead(II)chromate = PbCrO4
Ar(Pb) = 207.2 g/mol
Ar(Cr) = 51.9961 g/mol
Ar(O) = 15.9994 g/mol
%Pb =
Mr(PbCrO4)
=
Consider 1 mol of
PbCrO4
%Cr =
%O =
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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
b) methylamine = CH3NH2
Calculate the formula weight of the following
compounds from their formulae:
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 (NA)
The term mole, just like a dozen or gross,
refers to a particular number of things.
one dozen
one gross
one mole
=
=
=
12
144
6.0221367 x 1023
Avogadro’s number (NA)
When using the term mole it is important to
specify what is being referred to!
eg : 1 mol of CH3COOH molecules contains
2 moles of C atoms etc.
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The molar mass of a substance is the mass of
one mole of the substance
Calculating the Mass of an Atom or
Molecule
What is the mass of a nitric acid molecule?
nitric acid =
For all substances, the molar mass in grams
per mole is numerically equal to the formula
weight in atomic mass units.
Example
Mole Calculations
Converting Grams of Substance to Moles
Because molar mass is the mass per mole we
have the formula:
Mass
or
Molar Mass =
Mr =
m
Moles
n
Calcite is a mineral composed of calcium
carbonate, CaCO3. A sample of calcite
composed of pure CaCO3 weighs 23.6 g. How
many moles of calcium carbonate is this?
Remember that mass must be expressed in
grams here!
Example
Converting Moles of Substance to Grams
A sample of nitric acid contains 0.253 mol
HNO3. How many grams is this?
Mr(HNO3) = 63.013 g.mol-1
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?
(see previous example)
n = 0.253 mol
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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 = H2O2
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 = C2H2
benzene = C6H6
both have the empirical formula =
Example
Empirical formula from Composition
Consider the following flow-diagram:
Percent composition
Mass Composition
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?
Number of moles of
each element
Divide by smallest number of
moles to find the molar ratios
Multiply by appropriate number to
get whole number subscripts
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.
Ratio C : H =
 Empirical Formula =
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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Example
Stoichiometry
Determining the Molecular Formula from the
Percentage Composition and Molecular Weight.
Stoichiometry is the calculation of the
quantities of reactants and products involved in
a chemical reaction
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
 formula weight =
Empirical Formula =
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:
N2(g) + 3H2(g)
Example
2NH3(g)
Molecular Interpretation:
1 molecule N2 + 3 molecules H2
Molar interpretation:
1 mol N2 + 3 mol H2
Mass interpretation:
28.0 g N2 + 3 x 2.02 g H2
Amounts of Substance in a Chemical Reaction
2 molecules NH3
2 mol NH3
2 x 17.0 g NH3
Find the amount of water produced when 3.68
mol of NH3 is consumed according to the
following equation:
4NH3 + 5O2
4 NO + 6H2O
We assume that sufficient O2 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
Use molar mass of A
Moles of substance A
Use coefficients of A & B in
balanced eqn
Moles of substance B
Example
Calculate the mass of sulfur dioxide (SO2)
produced when 3.84 mol O2 is reacted with
FeS2 according to the equation:
4FeS2 + 11O2
3.84 mol
2Fe2O3 + 8SO2
m=?
2.79 mol
Use molar mass of B
Mass of substance B
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Another Example
One of the most spectacular reactions of
aluminium, the thermite reaction, is with iron
oxide, Fe2O3, by which metallic iron is made.
The equation is :
2Al(s) + Fe2O3(s)
Al2O3(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 Fe2O3 that must be used for the operation?
Calculate also how many grams of aluminium are
needed.
Strategy:
2Al(s) + Fe2O3(s)
mass of Fe
mol of Fe
mol of Fe2O3
Al2O3(s) + 2Fe(l)
Limiting Reactants
mol of Fe
mol of Fe2O3
mass of Fe2O3
The Cheese Sandwich Analogy
The limiting reactant (or limiting reagent) is
the reactant that is entirely consumed when a
reaction goes to completion.
The moles of product are always determined
by the starting moles of the limiting reactant.
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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Example
Example
When 100.0 g mercury is reacted with 100.0 g
bromine to form mercuric bromide, which is
the limiting reagent?
Hg + Br2
HgBr2
More Challenging!!
Iron can be obtained by reacting the ore
heamatite (Fe2O3) with coke (C). The latter is
converted to CO2. As manager of a blast
furnace you are told that you have 20.5 Mg of
Fe2O3 and 2.84 Mg of coke on hand.
(a) Which should you order first - another
shipment of iron ore or coke?
(b) How many megagrams of iron can you
make with the materials you have?
Thus the limiting reagent is
2Fe2O3 + 3C
4Fe + 3CO2
(b) How many megagrams of iron can you
make with the materials you have?
Recommended tut questions
2.17
2.18
2.26
2.27
3.1
3.5
3.21
3.27
3.33
3.35
3.59
3.63
3.96
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