Stoichiometry and the Mole Concept
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Chem Lectures
10. Stoichiometry and the Mole Concept
Learning Outcomes
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
State the symbols of the elements and formulae of the compounds mentioned in the syllabus.
Deduce the formulae of simple compounds from the relative numbers of atoms present and vice versa.
Deduce the formulae of ionic compounds from the charges on the ions present and vice versa.
Interpret chemical equations with state symbols.
Construct chemical equations, with state symbols, including ionic equations.
Define relative atomic mass, Ar.
Define relative molecular mass, Mr, and calculate relative molecular mass (and relative formula mass) as the sum of
relative atomic masses.
Calculate the percentage mass of an element in a compound when given appropriate information.
Calculate empirical and molecular formulae from relevant data.
Calculate stoichiometric reacting masses and volumes of gases (one mole of gas occupies 24 dm3 at room
temperature and pressure); calculations involving the idea of limiting reactants may be set (The gas laws and the
calculations of gaseous volumes at different temperatures and pressures are not required).
Apply the concept of solution concentration (in mol/dm3 or g/dm3) to process the results of volumetric experiments and
to solve simple problems (appropriate guidance will be provided where unfamiliar reactions are involved).
Calculate % yield and % purity.
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Stoichiometry and the Mole Concept
1. Calculating Mr

Relative atomic mass, Ar, of an element is the average mass of one atom of the element
when compared with 1/12 of the mass of an atom of carbon-12.

Ar has no units.
Table 1.1 Common elements and their Ar
Element
Symbol
Ar
Hydrogen
H
1.0
Helium
He
4.0
Carbon
C
12.0
Nitrogen
N
14.0
Oxygen
O
16.0
Chlorine
35.5
Cl
Iron
Fe
55.8
Checkpoint 1
1. Naturally-occurring chlorine consists of two isotopes, chlorine-35 and chlorine 37. In a mixture
of chlorine, it was found to be made up of 25% chlorine-37 and 75% chlorine-35. Calculate the
relative atomic mass of naturally-occurring chlorine.
Worked example:
Ar of chlorine = 37 0.25 + 35 × 0.75
= 35.5
2. The relative atomic mass of chlorine is 35.5. A sample of naturally-occurring chlorine comprises
the 37Cl and 35Cl isotopes. Calculate the relative abundance of each isotope.
Let the relative abundance of 37Cl = x
 Relative abundance of 35Cl = 1  x
Ar of chlorine = 35.5 = 37x + 35(1  x)
35.5 = 2x + 35
0.5 = 2x
x = 0.25
Relative abundance of 37Cl = 25%
Relative abundance of 35Cl = 1 - 0.25 = 0.75 or 75%
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Stoichiometry and the Mole Concept
3. The metal copper consists of 69% of
of copper.
63Cu
and 31% of
65Cu.
Calculate the relative atomic mass
Ar of copper = 63 0.69 + 65 × 0.31
= 63.6

Relative molecular mass, Mr, of a molecule is the average mass of one molecule of a
substance when compared with 1/12 of the mass of one atom of carbon-12.

Relative formula mass, Mr, is the average mass of one formula unit of an ionic compound
when compared with 1/12 of the mass of one atom of carbon-12.

Mr has no units.
Checkpoint 2
1. Calculate the relative molecular mass of water.
Worked example:
Step 1: Determine the chemical formula for water, if not given.
H2O
Step 2: Calculate the Mr.
Mr of water = 2 × 1.0 + 1 × 16.0
= 18.0
2. Calculate the relative molecular mass/ relative formula mass of the following compounds:
a) ammonia
b) oxygen gas
c) carbon dioxide
d) glucose, C6H12O6
e) iron(III) hydroxide
f) sodium hydroxide
g) hydrochloric acid
h) sulfuric acid
i) magnesium nitrate
j) sodium carbonate
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Stoichiometry and the Mole Concept
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Stoichiometry and the Mole Concept
2. Calculating % Composition by Mass

% of an element =
Ar of the element
Mr of the compound
100%
Checkpoint 3
1. Calculate the percentage composition, by mass, of the elements in sodium chloride.
Worked example:
Step 1: Determine the chemical formula of the compound, if not given.
NaCl
Step 2: Calculate the Mr of the compound.
Mr of NaCl = 23.0 + 35.5
= 58.5
Step 3: Calculate the % composition of the elements.
% composition of Na in NaCl =
23.0
58.5
100%
= 39.3 %
% composition of Cl in NaCl =
35.5
58.5
100%
= 60.7 %
2. Calculate the percentage composition by mass of the elements in the following compounds:
a) iron(III) oxide
b) sodium carbonate
c) copper(II) sulfate
d) potassium manganate(VIII), KMnO4
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Stoichiometry and the Mole Concept
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Stoichiometry and the Mole Concept
3. % Mass of an Element/ Substance in a Compound

% mass of an element =
Ar of the element
Mr of the compound
mass of sample
Checkpoint 4
1. Calculate the mass of copper in 32 g of copper(II) sulfate.
Worked example:
Step 1: Determine the chemical formula, if not given.
CuSO4
Step 2: Calculate the Mr of the compound.
Mr of CuSO4 = 63.5 + 32.1 + 16.0 × 4
= 159.6
Step 3: Calculate the mass of copper.
Mass of copper =
63.5
159.6
32
= 12.7 g
2. Calculate the mass of oxygen in 100 g of water.
3. Calculate the mass of carbon in 200 g of ethanoic acid, CH3COOH.
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Stoichiometry and the Mole Concept
4. Calculate the mass of water in 100 g of hydrated copper(II) sulfate, CuSO4.5H2O.
5. Calculate the % by mass of water in 200 g of hydrated iron(III) chloride, FeCl3.6H2O.
6. A sample of hydrated iron(II) sulfate has the formula FeSO4.xH2O. The composition of water in
the compound is 45.3%. Calculate the value of x.
4. The Mole Concept
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Stoichiometry and the Mole Concept

A mole is a term used to describe or measure the amount of particles (e.g., atoms, ions,
electrons, molecules).

The SI unit for amount of substance is mol.

One mole is the amount of substance which contains 6.02 × 1023 of particles.
1 mol = 6.02 × 1023 of particles
Example,
1 mol of H2(g) contains 6.02 × 1023 of H2 molecules.
1 mol of H2O(l) contains 6.02 × 1023 of H2O molecules.
1 mol of Fe(s) contains 6.02 × 1023 of Fe atoms.
1 mol of He(g) contains 6.02 × 1023 of He atoms.

The Avogadro's constant is 6.02 × 1023 mol-1, i.e., 6.02 × 1023 of particles in 1 mole of any
substance.

The number of moles of a substance, is also known as the amount of a substance. We can use
n to represent number of moles, i.e., nCO2  number of moles of CO2  amount of CO2.
[Note: n cannot be used to represent number of particles/ions/molecules/atoms!]

Number of moles of a substance is related to the number of particles with the following equation:
number of moles =
number of particles
6.02 × 1023 (in mol-1 )
Checkpoint 5
1. How many moles of water molecules are there in 3 × 1024 molecules of H2O?
Worked example:
nwater molecules =
3 × 1024
6.02 × 1023
= 4.98 mol
2. How many molecules are there in 0.25 mol of carbon dioxide?
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Stoichiometry and the Mole Concept
3. Calculate the number of moles of particles in 2.4 × 1023 atoms of hydrogen, H.
4. Calculate the number of particles in:
a) 3 moles of carbon atoms.
b) 1 mole of ammonia.
c) 20 moles of sodium chloride.
5. How many (a) molecules, and (b) atoms, are present in 1 mol of hydrogen gas?
5. What is Molar Mass, M?
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Stoichiometry and the Mole Concept

Molar mass is the mass of one mole of any substance. The units of molar mass are g/mol or
g mol-1.
Example,
2H
1O
Molar mass of H2O = 2 × 1.0 + 1 × 16.0 = 18.0 g/mol
Ar of H
Ar of O
Molar mass of NaCl = 23.0 + 35.5 = 58.5 g/mol
Checkpoint 6
1. Calculate the molar mass of the following substances:
a) carbon dioxide
b) nitrogen molecules
c) nitrogen atoms
d) copper(II) oxide
e) calcium carbonate
f) copper(II) sulfate pentahydrate, CuSO4.5H2O

Number of moles of a substance is related to the molar mass with the following equation:
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Stoichiometry and the Mole Concept
n, number of moles =
mass (in g)
molar mass (in g/mol)
Example,
Calculate the number of moles of carbon atoms in 12 g of carbon.
nC atoms =
12
12.0
= 1.00 mol

We can link the previous equations together, giving us:
n, number of moles =
mass (in g)
molar mass (in g/mol)
number of particles
6.02 × 1023 (in mol-1 )
Checkpoint 7
1. Calculate the number of moles of each of the following substances:
a) 10 g of hydrogen gas.
b) 40 g of water.
c) 54 g of sulfur dioxide.
d) 43 g of potassium manganate(VII), KMnO4.
2. Calculate the mass of the following substances:
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Stoichiometry and the Mole Concept
a)
b)
c)
d)
3 mol of oxygen gas.
5 mol of nitrogen dioxide.
3.5 mol of silicon(IV) oxide, SiO2.
5 mol of aluminium oxide, Al2O3.
3. Calculate the mass for each of the following substances:
a) 2.0 × 1026 atoms of helium.
b) 6.0 × 1025 sodium ions.
c) 2.0 × 1026 iodine atoms.
d) 2.0 × 1026 iodine molecules.
4. Calculate the number of particles in each of the following substances:
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Stoichiometry and the Mole Concept
a)
b)
c)
d)
100 g of chlorine gas.
150 g of lithium.
200 g of sulfuric acid.
250 g of hydrogen chloride.
6. Chemical Formulae

The molecular formula of a compound shows the actual number and kinds of atoms present.

The structural formula of a compound shows how the atoms are bonded in the molecule.

The empirical formula shows the simplest whole number ratio of the different atoms present.
Compound
Molecular Formula
Structural Formula
Water
H2O
H2O
Butane
C4H10
C2H5
Hydrogen peroxide
H2O2
HO
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Empirical Formula
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Stoichiometry and the Mole Concept

The empirical formula can be calculated using data from experiments.
Examples,
(1) 0.48 g of magnesium was burnt in oxygen to produce 0.80 g of magnesium oxide.
Determine the empirical formula of magnesium oxide.
Step 1: Determine the mass of each of the substances which were used to produce
magnesium oxide.
Mass of Mg = 0.48 g
Mass of magnesium oxide = 0.80 g
 Mass of O = 0.80  0.48 g
= 0.32 g
Step 2: Calculate empirical formula as follows:
Mass/ g
n/ mol
Mole ratio
(this means to divide
by the smallest n
throughout)
Simplest mole ratio
Mg
0.48
O
0.32
0.48
0.0198
24.3
0.0198
1
0.0198
0.32
0.0200
16.0
0.0200
1.01≈1
0.0198
1
1
Empirical formula of magnesium oxide is MgO.
(2) A gaseous oxide of nitrogen contains 30.4% nitrogen and 69.6% oxygen by mass.
Determine its empirical formula.
% mass
% mass
molar mass
Mole ratio
Simplest mole ratio
N
30.4
30.4
2.17
14.0
2.17
1
2.17
1
O
69.6
69.6
4.35
16.0
4.35
2.00 ≈ 2
2.17
2
Empirical formula of the oxide of nitrogen is NO2.
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Stoichiometry and the Mole Concept
Checkpoint 8
Determine the empirical formulae for the following:
a) A compound consisting of 24.0 g of carbon and 4.0 g of hydrogen.
b) A compound consisting of 3.5 g of nitrogen and 8.0 g of oxygen.
c) A 6.20 g sample of aluminium oxide, analysed and found to contain 3.28 g of aluminium.
d) A compound with the composition of 43.4% sodium, 11.3% carbon and 45.3% oxygen by mass.
e) 17.4 g of a compound of chlorine and oxygen, which contains 3.2 g of oxygen.
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Stoichiometry and the Mole Concept

The molecular formula can be derived from the empirical formula if we are given either (i) molar
mass, or (ii) relative molecular mass of the compound.
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Stoichiometry and the Mole Concept
Example,
A gaseous oxide of nitrogen contains 30.4% nitrogen and 69.6% oxygen by mass. The molar
mass of the oxide of nitrogen is 92.0 g/mol. Determine its molecular formula.
Step 1: Calculate the empirical formula.
% mass
% mass
molar mass
Mole ratio
Simplest mole ratio
N
30.4
30.4
2.17
14.0
2.17
1
2.17
1
O
69.6
69.6
4.35
16.0
4.35
2.00 ≈ 2
2.17
2
Empirical formula of the oxide of nitrogen is NO2.
Step 2: Calculate the molecular formula.
Let the molecular formula be (NO2)n
14.0n + (16.0 × 2)n = 92.0
46n = 92.0
n = 2.0
Molecular formula of the oxide of nitrogen is N2O4.
Checkpoint 9
1. The empirical formula of a compound is CH2O. Its molar mass is 90.0 g/mol. Determine its
molecular formula.
2. A compound has 39% by mass of carbon, 9.7% by mass of hydrogen and 51.6% by mass of
oxygen. Its relative molecular mass is 62.0. Determine its molecular formula.
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Stoichiometry and the Mole Concept
7. Molar Gas Volume

One mole of any gas occupies a volume of 24 dm3 (or 24000 cm3) at room temperature and
pressure (r.t.p.). [Note: r.t.p. = 25 oC (298 K) and 1 atm, 1.01 × 105 N m-2]

Avogadro's Law states that equal volumes of gases under the same conditions of temperature
and pressure will contain the same number of particles, e.g., 100 cm3 of H2(g) contains the
same number of particles as 100 cm3 of H2O(g), at r.t.p.

Number of moles of a substance is related to the molar gas volume with the following equation:
n, number of moles =
volume of gas (in dm3 )
24 (in dm3 mol-1 )
Example,
Calculate the number of moles of 48 dm3 of hydrogen gas.
nhydrogen gas =
48
24
= 2.00 mol
Checkpoint 10
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Stoichiometry and the Mole Concept
1. Calculate the volume, at r.t.p., of the following:
a) 0.5 mol of hydrogen, H2.
b) 1 mol of oxygen, O2.
c) 3 mol of carbon monoxide.
2. Calculate the number of moles, at r.t.p., of the following:
a) 50 dm3 of carbon dioxide.
b) 4000 cm3 of nitrogen, N2.
c) 64000 cm3 of hydrogen chloride vapour.

We can link the previous equations together, giving us:
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Stoichiometry and the Mole Concept
mass (in g)
n, number of moles =
molar mass (in g/mol)
number of particles
volume of gas (in dm3 )
6.02 × 1023 (in mol-1 )
24 (in dm3 mol-1 )
Checkpoint 11
1. A sample of sulfur trioxide, SO3, has a volume of 6 dm3 at r.t.p.
a) Calculate the amount of SO3 molecules in the sample.
b) Calculate the number of SO3 molecules in the sample.
c) Calculate the mass of SO3 in the sample.
2. A hot air balloon was filled with 100 g of gas X. This mass of gas would have occupied
600000 cm3 at r.t.p. if not contained within the hot air balloon. Determine the identity of gas X.
8. Concentration
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Stoichiometry and the Mole Concept

Concentration of a solution tells us the amount of solute per unit volume of a solution. It has the
units of g/dm3 (g dm-3) or mol/dm3 (mol dm-3).
concentration =
number of moles of solute (in mol)
mass of solute (in g)
3
volume of solution (in dm )
volume of solution (in dm3 )
Example,
What is the concentration of an aqueous solution, in mol dm-3, of NaOH if 50.0 g of NaOH
pellets was dissolved in 1500 cm3 of water?
Step 1: Convert the mass to number of moles (or amount).
nNaOH =
50.0
23.0 + 16.0 + 1.0
= 1.25 mol
Step 2: Calculate concentration. Note the units required. If not mentioned, always calculate
in mol dm-3.
Concentration of NaOH(aq) =
1.25
1500
1000
= 0.833 mol dm-3
very impt to convert from cm3 to dm3
(1 dm3 = 1000 cm3)
[To convert concentration from mol dm-3 to g dm-3, relate to this following equation,
n=
mass
molar mass
 mass = n × molar mass, so 0.833 × (23.0 + 16.0 + 1.0) = 33.3 g dm-3]
Checkpoint 12
1. Calculate the concentration of the following solutions, in mol dm-3:
a) 40.0 g of Na2CO3 dissolved in 500 cm3 of water.
b) 50.0 g of KOH dissolved in 1200 cm3 of water.
c) 1000 cm3 of HCl bubbled (and completely dissolved) in 3000 cm3 of water.
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Stoichiometry and the Mole Concept
2. Calculate the amount of H2SO4 in 1200 cm3 of 0.5 mol dm-3 dilute sulfuric acid.
3. Calculate the mass of Ca(OH)2 dissolved in 5000 cm3 of 0.05 mol dm-3 limewater.
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Stoichiometry and the Mole Concept
4. What is the mass of KOH which must be dissolved in 800 cm3 of water to obtain a 0.5 mol dm-3
KOH solution?
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Stoichiometry and the Mole Concept
9. Calculations from Equations



We correctly read a chemical equation as follows:
2NaOH
+
H2SO4

Na2SO4
+
2H2O
2 mol of
NaOH
reacts with
1 mol of
H2SO4
to give
1 mol of
Na2SO4
and
2 mol of H2O
2H2
+
O2

2H2O
2 mol of H2
2 molecules
of H2
reacts with
reacts with
1 mol of O2
1 molecule of O2
to give
to give
2 mol of H2O
2 molecules of
H2O
We DO NOT read a chemical equation as follows:
2NaOH
+
H2SO4

Na2SO4
+
2H2O
2 g of
NaOH
reacts with
1 g of
H2SO4
to give
1 g of
Na2SO4
and
2 g of H2O
2 g of
NaOH =
0.05 mol
does not
react with
1 g of
H2SO4 =
0.01 mol
to give
1 g of
Na2SO4 =
0.007 mol
and
2 g of H2O =
0.111 mol
WRONG
WRONG
WRONG
WRONG
WRONG
WRONG
WRONG
Hence, for chemical calculations, we always work with number of moles (or amount), mol.
Example 1,
Calculate the mass of water produced during the complete combustion of 100 g of hydrogen gas.
Step 1: Write the balanced chemical equation for the reaction.
2H2 + O2  2H2O
Step 2: Work with whatever numbers you are provided with. In this case, hydrogen gas.
Because the chemical equation shows molar equivalents, we will need to convert the mass of
hydrogen gas to the number of moles of hydrogen gas.
nhydrogen gas =
100
2
= 50 mol
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Stoichiometry and the Mole Concept
Step 3: Use the balanced chemical equation to determine the number of moles of water
produced. This is also known as determining the mole ratio.
2 mol H2 reacts with 1 mol O2 to produce 2 mol H2O, or we can write this as:
2H2O22H2O
nH2 O =
50
2
2
= 50 mol
Step 4: Calculate the mass of water produced.
Mass of water produced = 50 × (2 × 1.0 + 16.0)
= 900 g
Example 2,
Hydrogen peroxide decomposes according to the following equation:
2H2O2  2H2O + O2
Calculate the volume of oxygen gas evolved at r.t.p. when 50 cm3 of 1.0 mol dm-3 H2O2 is
decomposed.
Step 1: The chemical equation, in this case, has already been given to you.
Step 2: Determine the number of moles of H2O2 decomposed.
nH2 O2 decomposed =
50
1000
1.0
= 0.0500 mol
Step 3: Determine the mole ratio to determine the number of moles of oxygen gas evolved.
2H2O2H2OO2
noxygen gas =
0.0500
2
1
= 0.0250 mol
Step 4: Calculate the volume of oxygen gas evolved.
Volume of oxygen gas evolved = 0.0250 × 24
= 0.600 dm3
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Stoichiometry and the Mole Concept
Checkpoint 13
1. Calculate the volume of hydrogen gas produced at room temperature and pressure when 1.3 g
of zinc react with excess dilute sulphuric acid.
2. Copper(II) oxide reacts with ammonia gas according to the equation:
3CuO(s) + 2NH3(g)  3Cu(s) + 3H2O(l) + N2(g)
In an experiment, 4.0 g of copper(II) oxide was reacted completely with ammonia. Calculate
a) the amount of CuO that reacted.
b) the amount of NH3 that reacted.
c) the volume of NH3 that reacted.
d) the mass of H2O produced in the reaction.
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Stoichiometry and the Mole Concept
3. Lead (II) oxide reacts with hydrogen gas according to the equation:
PbO(s) + H2(g)  Pb(l) + H2O(g)
A steady stream of hydrogen (measured at r.t.p.) was passed across 67 g of heated lead(II)
oxide.
Calculate:
a) the mass of lead produced.
b) the volume of hydrogen gas required to completely reduce all the lead(II) oxide.
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Stoichiometry and the Mole Concept
4.
Zinc reacts with silver ions according to the equation:
Zn(s) + 2Ag+(aq)  Zn2+(aq) + 2Ag(s)
Calculate the mass of silver produced when 0.4 mol of zinc is added to an aqueous solution
containing excess silver ions.
5. 100 g of sodium hydrogen carbonate was heated, and decomposed according to the following
equation:
2NaHCO3(s)  Na2CO3(s) + CO2(g) + H2O(g)
Calculate:
a) the mass of sodium carbonate produced.
b) the volume of carbon dioxide evolved (measured at r.t.p.).
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Stoichiometry and the Mole Concept
6. Ammonia gas is produced when ammonium chloride and calcium hydroxide are mixed and heated.
The equation for the reaction is:
2NH4Cl(s) + Ca(OH)2(s)  CaCl2(s) + 2NH3(g) + 2H2O(l)
If 0.4 g of ammonium chloride is used, calculate:
a) the minimum mass of calcium hydroxide required to completely react away all the
ammonium chloride.
b) the maximum volume of ammonia produced at r.t.p.
7. Calculate the mass of silver produced when 0.48 g of magnesium reacts completely with an
aqueous solution of silver nitrate.
Mg(s) + 2Ag+(aq)  Mg2+(aq) + 2Ag(s)
10. Calculation from Equations - Volumes
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Stoichiometry and the Mole Concept

The volume of a gas is a reaction is proportional to the number of moles. This only applies for
gases.

This means that the mole ratio of gaseous reactants and products as indicated by a chemical
equation is equivalent to the ratio of their volumes.

2N2
+
3H2

2NH3(g)
2 mol of N2
2 molecules
of N2
2 volumes of
N2
200 cm3 of N2
reacts with
reacts with
to give
to give
reacts with
3 mol of H2
3 molecules of
H2
3 volumes of H2
reacts with
300 cm3 of H2
to give
2 mol of NH3
2 molecules of
NH3
2 volumes of
NH3
200 cm3 of
NH3
to give
We CANNOT use volumes to determine product when the product (or reactant) is not in the
gaseous state.
2H2(g)
+
O2(g)

2H2O(l)
2 mol of H2
2 molecules
of H2
2 volumes of
H2
reacts with
reacts with
1 mol of O2
1 molecule of O2
to give
to give
2 mol of H2O
2 molecules of H2O
OK
OK
reacts with
1 volume of O2
but does
not give
WRONG
reacts with
100 cm3 of O2
but does
not give
2 volumes of liquid
H2O because water
here is in the liquid
state, NOT gaseous
state
200 cm3 of liquid
H2O because water
here is in the liquid
state, NOT gaseous
state.
200 cm3 of H2
WRONG
Example,
The Haber process involves the synthesis of ammonia gas from nitrogen and hydrogen.
Calculate the volume of nitrogen required to react completely with 500 cm3 of hydrogen.
Calculate the volume of ammonia produced.
Step 1: Write the chemical equation and notice that they are all gases.
2N2(g) + 3H2(g)  2NH3(g)
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Stoichiometry and the Mole Concept
Step 2: Determine the mole ratio.
2N23H22NH3
Step 3: Determine the number of moles of nitrogen and ammonia.
Volume of nitrogen required =
500
3
2
= 333 cm3
Volume of ammonia gas produced =
500
3
2
= 333 cm3
Checkpoint 14
1. What volume of sulphur dioxide reacts with 80 cm3 of hydrogen sulphide, at r.t.p., according to
the equation below?
2H2S(g) + SO2(g)  3S(s) + 2H2O(l)
2. Methane (CH4) burns in oxygen to give carbon dioxide and steam. If 10 dm3 of methane burns,
calculate
a) the volume of oxygen required.
b) the respective volumes of the products.
11. Calculations from Equations - With Limiting Reagents
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Stoichiometry and the Mole Concept

The number of moles of a product formed during a chemical reaction is determined by the
number of moles of the reactant which is not in excess. This reactant is known as the limiting
reagent.
Example,
150 g of magnesium is combusted in a container with 50000 cm3 of oxygen (measured at r.t.p.)
to form magnesium oxide. Calculate the mass of magnesium oxide formed.
Step 1: Write the balanced chemical equation.
2Mg + O2  2MgO
Step 2: Notice that values are given to allow you to count all the number of moles of reactants.
This means that there will likely be a limiting reagent. In fact, almost all stoichiometry questions
from this point onwards could involve a limiting reagent. To determine the limiting reagent, we
must first calculate the number of moles of each reactant.
nmagnesium =
150
24.3
= 6.17 mol
noxygen =
50000
1000
24
= 2.08 mol
Step 3: Determine the mole ratio, and determine the limiting reagent.
2MgO22MgO.
Thought processes that should be going on in your mind
 This means that 2 mol of Mg will react with 1 mol of O2 to produce 2 mol of MgO.
 We have 6.17 mol of Mg. This means that it will react with 3.09 mol of O2.
 However, we only have 2.08 mol of O2, which means that not all the Mg will be reacted away.
 The reactant which prevents all the reactants from reacting completely is the limiting reagent.
 Hence, oxygen is the limiting reagent.
Oxygen is the limiting reagent.
Step 4: Calculate the product. Because oxygen limits the reaction, the number of moles of
product which will be formed will only be as much as the limiting reagent, after comparing
stoichiometry.
nMgO formed =
2.08
1
2
= 4.16 mol
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Mass of MgO formed = 4.16 × (24.3 + 16.0)
= 168 g
Checkpoint 15
1. In the manufacture of calcium carbide
CaO(s) + 3C(s)  CaC2(s) + CO(g)
Calculate the maximum mass of calcium carbide that can be obtained from 160 g of CaO and
180 g of coke (carbon).
2. A mixture of 125 cm3 of oxygen and 50.0 cm3 of hydrogen at room temperature is exploded in a
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Stoichiometry and the Mole Concept
suitable apparatus. After the reaction, the apparatus is allowed to cool to room temperature.
Give the name(s) of the gas(es) remaining and also its(their) volume(s).
3. 24 cm3 methane burnt completely in 106 cm3 of oxygen. After reaction, the products were
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Stoichiometry and the Mole Concept
cooled to the original temperature.
a) Write the equation for this reaction.
b) What was the volume of oxygen used up (reacted)?
c) What was the volume of carbon dioxide produced?
d) What was the total volume of gas that remained?
12. Calculating Percentage Purity and Percentage Yield
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Stoichiometry and the Mole Concept

In a real lab chemical reaction, the number of moles of product formed is almost never the
theoretical amount.
Start with:
Theoretically
should get:
But actually
will get:

Mg(s)
1 mol
-
O2(g)
excess
excess
+
-
excess

MgO(s)
1 mol
<1 mol
(because of incomplete
reaction, impure reactants, loss
of product during transfer, etc)
Hence, we sometimes will need to calculate:
a) Percentage purity
Percentage purity =
mass of pure substance
mass of impure substance
100%
[Note: We cannot count number of moles of an impure substance, e.g., molar mass of
water is 18.0 g/mol, but molar mass of impure water could be >18 g/mol, and we do not
know what it is!]
b) Percentage yield
Percentage yield =
actual mass of pure product
theoretical mass of pure product
100%
OR
Percentage yield =
actual number of moles of pure product
theoretical number of moles of pure product
100%
Example 1: Percentage Purity
100 g of calcium carbonate was reacted with excess dilute sulfuric acid to form 20 g of carbon
dioxide. What is the percentage purity of the sample of calcium carbonate used?
Step 1: Write the balanced chemical equation.
CaCO3 + H2SO4  CaSO4 + H2O + CO2
Step 2: Because the initial reactant, calcium carbonate is NOT pure, we cannot convert its
mass into number of moles. However, we know that it forms 20 g of carbon dioxide. The
product formed is pure. We can convert its mass into number of moles.
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Stoichiometry and the Mole Concept
ncarbon dioxide =
20
12.0 + 16.0 × 2
= 0.455 mol
Step 3: We can use the number of moles of pure product formed to determine the number of
moles of the initial pure reactant.
CaCO3CO2
ncalcium carbonate = 0.455 mol
mass of calcium carbonate = 0.455 × (40.1 + 12.0 + 16.0 × 3)
= 45.5 g
Step 4: Calculate % purity.
% purity of calcium carbonate =
45.5
100
100%
= 45.5%
Example 2: Percentage Yield
100 g of calcium carbonate was reacted with excess dilute sulfuric acid to form 30 g of carbon
dioxide. What is the percentage yield of carbon dioxide?
Step 1: Write the balanced chemical equation.
CaCO3 + H2SO4  CaSO4 + H2O + CO2
Step 2: Percentage yield means that given 100 g of calcium carbonate, we should get X g of
carbon dioxide. However, we only obtained Y g. So we will need to calculate how much of the
product we could theoretically obtain. We assume that the reactants are 100% pure in this
case.
ncalcium carbonate =
100
40.1 + 12.0 + 16.0 × 3
= 0.999 mol
CaCO3CO2
ncarbon dioxide = 0.999 mol
Step 3: We can now find percentage yield. We will need to convert mass to mole first for
comparison, or mole to mass.
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Stoichiometry and the Mole Concept
ncarbon dioxide produced =
30
12.0 + 16.0 × 2
= 0.682 mol
% yield of carbon dioxide =
0.682
0.999
100%
= 68.2 %
Checkpoint 16
1. When 4.3 g of copper was heated in oxygen, 3.2 g of copper(II) oxide was obtained. Calculate
the percentage purity of copper.
2. Zinc reacts with sulfur according to the equation:
Zn(s) + S(s)  ZnS(s)
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Stoichiometry and the Mole Concept
In an experiment, 6.5 g of zinc was reacted with sulfur to make zinc sulfide, ZnS. 9.0 g of ZnS
was obtained. Calculate the percentage yield.
13. Determining Equations From Empirical Data

We can use the mass of reactants in an experiment to determine the chemical equation for a
reaction.
Example,
In an experiment, 6.2 g of phosphorus (P4) reacted with 4.8 g of oxygen (O2) to produce an
oxide of phosphorus. Determine the equation for the reaction.
Step 1: Convert all the values to moles because we cannot work with mass in a chemical
equation.
nphosphorus =
6.2
31.0 × 4
= 0.0500 mol
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Stoichiometry and the Mole Concept
noxygen =
4.8
16.0 × 2
= 0.150 mol
Step 2: Compare the mole ratios to determine the equation.
0.0500 mol of P4 reacted with 0.150 mol of O2.
 1 mol of P4 will react with 3 mol of O2.
 the oxide of phosphorus formed will contain all the P and O  P4O6
P4 + 3O2  P4O6
Checkpoint 17
1. In an experiment, 11.2 g of iron was found to react with 4.8 g of oxygen to form an oxide of iron.
a) Calculate the number of moles of iron atoms reacted.
b) Calculate the number of moles of oxygen molecules reacted.
c) What is the simplest whole number ratio of iron atoms to oxygen molecules?
d) Find the equation for the reaction.
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Stoichiometry and the Mole Concept
2. 6.7 g of potassium nitrate (KNO3) was heated to give 5.7 g of potassium nitrite (KNO2) and
0.816 dm3 of oxygen (measured at r.t.p). Find the equation for the reaction.
14. Calculations Involving Titrations

Titration allows us to find the concentration of a solution by reacting it with a solution of known
concentration, known as the standard.
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Stoichiometry and the Mole Concept

The calculations are similar to calculations involving concentrations.
Example 1,
If 25 cm3 of aqueous NaOH is required to completely neutralise 20 cm3 of 0.5 mol dm-3 H2SO4,
what is the concentration of the aqueous NaOH?
2NaOH + H2SO4  Na2SO4 + 2H2O
nH2 SO4
20
0.5
1000
= 0.0100 mol
2NaOHH2SO4
nNaOH
0.0100
2
1
= 0.0200 mol
Concentration of NaOH =
0.0200
25
1000
= 0.800 mol dm-3
Example 2,
A 250 cm3 solution of dilute hydrochloric acid was prepared from 10 cm3 of a stock solution of
dilute hydrochloric acid of unknown concentration. 25 cm3 of the prepared acid was completely
neutralised by 12.5 cm3 of 0.5 mol dm-3 aqueous NaOH. What is the concentration of the stock
solution?
NaOH + HCl  NaCl + H2O
nNaOH
12.5
0.5
1000
= 0.00625 mol
NaOHHCl
nHCl in 25 cm3 of the prepared acid = 0.00625 mol
nHCl in the stock solution =
0.00625
25
250
= 0.0625 mol
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Stoichiometry and the Mole Concept
concentration of stock solution =
0.0625
10
1000
= 6.25 mol dm-3
Checkpoint 18
1. 32 cm3 of 0.1 mol dm-3 NaOH(aq) reacts with 25 cm3 of HCl(aq) in a titration. Calculate the
concentration of HCl in (a) mol dm-3, and (b) g dm-3.
2. 18 cm3 of 0.2 mol dm-3 H2SO4(aq) reacts with 24 cm3 of KOH(aq). Calculate the concentration
of the KOH in (a) mol dm-3, and (b) g dm-3.
3. The mass concentration of an aqueous solution of NaOH is 6.0 g dm-3. 20 cm3 of the NaOH
solution reacted completely with 40 cm3 of dilute HNO3. Calculate:
a) the concentration of NaOH in mol dm-3.
b) the amount of NaOH used in the reaction.
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Stoichiometry and the Mole Concept
c) the amount of HNO3 used in the reaction.
d) the concentration of HNO3 in mol dm-3.
e) the concentration of HNO3 in g dm-3.
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