8 CHEMICAL QUANTITIES
8.1 Relationship between Compounds in Chemical Equations
A balanced chemical equation provides the proportions by which compounds react in the chemical
reaction. Thus, for example, in the combustion of glucose, the balanced chemical reaction
C6H12O6 + 6 O2
6 CO2 + 6 H2O
shows that the combustion of 1 molecule C6H12O6 requires 6 molecules of O2 for complete combustion,
producing 6 molecules CO2 and 6 molecules H2O. The equation shows only the proportions in which the
chemicals react in this equation. Thus, if there are 2 molecules of C6H12O6 , 6 x 2 molecules of O2 will be
required and 6 x 2 molecules each of CO2 and H2O will be produced. In fact, any number of molecules of
glucose, n, will require (6n) molecules O2 to effect a combustion and will produce (6n) molecules each of
CO2 and H2O. These proportions can be used as conversion factors to calculate the quantities of all
chemicals that appear in a chemical equation. This is known as stoichiometry from the Greek stoichion
meaning element and metric meaning measurement.
In stoichiometric calculations, you will always be provided with the quantity of one or more chemicals
appearing in the chemical equation. From this data, you will be required to calculate the quantities of the
other compounds that appear in the equation.
The steps involved are as follow:
Mass of one or
more
compounds in
the reaction
Moles
Divide by
the Molar
mass of the
given
Note:
Moles of other
compounds in
the chemical
equation
Relate the
proportions
by which the
chemicals
react
according to
the chemical
equation
Mass of the other
compounds
Multiply by the
Molar mass of
the other
compounds
1. Always ensure that the equation is balanced first.
2. Proceed with calculations by always working with moles [because reactions are based on
reactions between molecules (and moles are a packet of molecules)].
Example:
Calculate the mass of oxygen required to burn 43.9 g glucose. How much carbon dioxide and
water will be produced?
Step 1: Write the balanced chemical equation
C6H12O6 + 6 O2
6 CO2 + 6 H2O
Step 2: Calculate the number of moles of the given chemical, in this case, glucose
i)
The molecular mass of glucose is:
–1
6 (12.011 g C) + 12 (1.008 g H) + 6 (16.00 g O) = 180.16 g mol
page 61
ii)
nC6 H12 O6 = 43.9 g C6 H12 O6
FG 1 mol C H O IJ = 0.244 mol C H
. gC H O K
H 18016
12
6
6
6
12
6
12 O6
6
Step 3: Calculate the related quantities of the other chemicals by using the proportions expressed in the
balanced chemical equation
i)
Moles of the other compounds in the chemical equation
6 mol O2
= 1.46 mol O2
1 mol C6 H12 O6
nO = 0.244 mol C6 H12 O6
2
nCO = 0.244 mol C6 H12 O6
6 mol CO2
= 1.46 mol CO2
1 mol C6 H12 O6
nH O = 0.244 mol C6 H12 O6
6 mol H 2 O
= 1.46 mol H 2 O
1 mol C6 H12 O6
2
2
ii)
Mass of the other compounds in the chemical equation
Molecular masses:
–1
O2 = 2 (16.00) = 32.00 g mol
CO2 = (12.011) + 2 (16.00) = 44.01 g mol
H2O = 2 (1.008) + (16.00) = 18.02 g mol
mH 2 O
–1
FG 32.00 g O IJ = 46.8 g O
H mol K
F 44.01 g CO IJ = 64.3 g CO
. mol CO G
= 146
H mol K
F 18.02 g H O IJ = 26.3 g H O
. mol H OG
= 146
H mol K
2
. mol O2
mO2 = 146
mCO2
–1
2
2
2
2
2
2
2
Note that mass has been conserved:
Total mass of reactants = 43.9 g C6H12 O6 + 46.8 g O2 = 90.7 g
Total mass of product = 64.3 g CO2 + 26.3 g H2O = 90.6 g
(the small discrepancy is due to round off errors and illustrates the need for using judgement in applying
significant figures).
The point is that if the mass of one substance in a chemical equation is known, it determines the
masses of all other chemicals in the reaction.
page 62
Example:
Calculate the mass of chlorine required to react with 84.453 g iron. How much iron (III)
chloride will be produced?
Step 1: Build the balanced chemical reaction.
Skeleton equation:
Fe (s) + Cl2 (g)
Balanced equation:
2 Fe (s) + 3 Cl2 (g)
FeCl3 (s)
2 FeCl3 (s)
Step 2: Calculate the number of moles of the given material.
n Fe = 84.453 g Fe
FG 1 mol Fe IJ = 1512
.
mol Fe
. g Fe K
H 5585
Step 3: Calculate the number of moles of the other chemicals in the equation.
FG 3 mol Cl IJ = 2.268 mol Cl
H 2 mol Fe K
F 2 mol FeCl IJ = 1512
.
= 1512
mol FeG
H 2 mol Fe K . mol FeCl3
2
.
nCl = 1512
mol Fe
2
n FeCl
2
3
3
Step 4: Calculate the masses
i)
Calculate the molar masses
–1
Cl2 = 2 (35.453) = 70.906 g mol
FeCl3 = (55.85) + 3 (35.453) = 162.21 g mol
ii)
–1
Calculate their masses
FG 70.906 mol Cl IJ = 160.8 g Cl
H mol Cl K
F 162.21 g IJ = 245.3 g FeCl
mol FeCl G
= 1512
.
3
H mol FeCl K
2
mCl = 2.268 mol Cl2
2
mFeCl
3
2
2
3
3
page 63
Example:
How many grams of hydrogen are needed to react completely with 75.4327 g nitrogen? How
many gram of ammonia will be produced?
Step 1: Build the balanced chemical reaction.
Skeleton equation:
N2 (g) + H2 (g)
Balanced equation:
N2 (g) + 3 H2 (g)
NH3 (g)
2 NH3 (g)
Step 2: Calculate the number of moles of the given material.
Molar mass of N2 = 2 (14.0067) = 28.0134 g mol–1
n N 2 = 75.4327 g N 2
FG 1 mol N IJ = 2.69274 mol N
H 28.0134 g N K
2
2
2
Step 3: Calculate the number of moles of the other chemicals in the equation.
FG 3 mol H IJ = 8.07821 mol H
H 1 mol N K
F 2 mol NH IJ = 5.38547 mol NH
= 2.69274 mol N G
H 1 mol N K
2
n H = 2.69274 mol N 2
2
n NH
2
2
3
2
3
3
2
Step 4: Calculate the masses
i)
Calculate the molar masses
H2 = 2 (1.00797) = 2.01594 g mol
–1
–1
NH3 = (14.0067) + 3 (1.00797) = 17.0306 g mol
ii)
Calculate their masses
FG 2.01594 g IJ = 16.2852 g H
H mol H K
F 17.0306 g IJ = 917178
= 5.38547 mol NH G
H mol NH K . g NH
mH = 8.07821 mol H2
2
m NH
3
2
2
3
3
3
page 64
8.2 Limiting Reactants
Example: What mass of aluminum is required in the single displacement reaction of 21.4 g iron (III)
oxide? What mass of products are formed?
Atomic Mass:
Step 1:
Al = 26.98
Fe = 55.85
Write the balanced chemical equation
Fe2O3 (s) + 2Al (s)
Step 2:
O = 16.00
2Fe (s) + Al2O3 (s)
Calculate the number of moles of the given compound
Fe2O3 = 2 x (55.85) + 3 x (16.00) = 159.70
Molar Mass
Al2O3 = 2 x (26.98) + 3 x (16.00) = 101.96
nFe O = 21.4 g Fe2 O3
2 3
Step 3:
mol
= 0.134 mol Fe2 O3
159.70 g Fe2 O3
Calculate the number of moles of the other compounds in the chemical equation
nAl = 0.134 mol Fe2 O3
2 mol Al
= 0.268 mol Al
1 mol Fe2 O3
nFe = 0.134 mol Fe2 O3
2 mol Fe
= 0.268 mol Fe
1 mol Fe2 O3
nAl O = 0.134 mol Fe2 O3
2 3
Step 4:
1 mol Al2 O3
= 0.134 mol Al2 O3
1 mol Fe2 O3
Calculate the mass of the other compounds in the chemical equation
Al2O3 = 2 x (26.98) + 3 x (16.00) = 101.96
Molar Mass
mAl = 0.268 mol Al
26.98 g Al
mol
= 7.23 g
mFe = 0.268 mol Fe
55.85 g Fe
mol
= 15.0 g
mAl O = 0.134 mol Al2 O3
2 3
101.96 g Al2 O3
mol
= 13.7 g
In this example, 7.23 g of Al were required to completely react with the 21.4 g Fe2O3. Suppose however,
that when we were weighing out the reactants, instead of 7.23 g Al we weighed out 7.52 g Al. Clearly we
have more Al than we need – thus when all the Fe2O3 is used up there would remain
(7.52 – 7.23) g Al = 0.29 g Al. The Al is said to be in excess (there is too much Al). Another way of saying
this is that Fe2O3 limits the amount of product. That is, Fe2O3 is the limiting reactant. We will run out of
Fe2O3 before the Al is all used up, so the amount of Fe2O3 will determine the amount of product obtained.
In any stoichiometry problem, it is necessary to determine which reactant is limiting. There is one
definitive clue that you are dealing with a limiting reactant problem:
Whenever you are given the quantities of more than one reactant, the problem is a limiting reactant one.
page 65
Example: When 36.127 g benzene, C6H6 , is burnt in 115.723 g oxygen, how much product is formed?
Atomic Mass:
H = 1.00797
C = 12.011
O = 15.9994
The steps in the solution are identical to the stoichiometry problems solved
previously. There is only one additional step – to determine the limiting reactant
through an extra calculation.
Step 1:
Write the balanced chemical equation
2 C6H6 (l) + 15 O2 (g)
Step 2:
12 CO2 (g) + 6 H2O (l)
Calculate the number of moles of the given compounds
Molar Masses: C6H6 = 6 × (12.011) + 6 × (1.00797) = 78.114
O2 = 2 × (15.9994) = 31.9988
CO2 = (12.011) + 2 × (15.9994) = 44.010
H2O = 2 × (1.00797) + (15.9994) = 18.0153
nC H = 36.127 g C6 H 6
6
6
nO = 115.723 g O2
2
Step 3:
mol
= 0.46249 mol
78.114 g
mol
= 3.61648 mol
31.9988 g
Calculate the number of moles of the other compounds in the chemical equation.
To do this we follow the schematic outlined in the following schematic diagram.
+
2 C6H6 (l)
Moles of C6H6
15 O2 (g)
Moles of O2
12 CO2 (g)
+
6 H2O (l)
Use the moles of the Limiting Reactant
to determine the quantities of produce
Use mole ratios to
determine the
Limiting Reactant
Mole of the Limiting Reactant
Determine the Limiting Reactant. Assume that one of the reactants is
limting. On the basis of the amount of this reactant, calculate the quantity
of the other reactant needed.
For no real reason, let us choose C6H6 . We now ask: On the basis of 0.46249
mol C6H6 , how much oxygen are required?
nO2 = 0.46249 mol C6 H 6
15 mol O2
= 3.4687 mol O2 are required
2 mol C6 H 6
Now we compare the value of O2 required (3.4687 mol), with the amount of
O2 that we actually have (i.e., 3.61648 mol).
page 66
In this problem, we actually have more oxygen than we need. Thus, we have a
basis for judging whether we have established the limiting reactant.
Moles of O2
available
3.61648 mol
Moles of O2
required
3.4687 mol
Since we have more O2 than is actually needed, the C6H6 will be used up before
the oxygen. Therefore, C6H6 is the limiting reactant. The amount of C6H6 initially
present will determine the amount of product obtained.
Step 4:
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
nO = 3.61648 3.4687 = 0.1478 mol
2
nCO = 0.46249 mol C6 H 6
12 mol CO2
= 2.7749 mol CO2
2 mol C6 H 6
nH O = 0.46249 mol C6 H 6
6 mol CO2
= 1.3875 mol H 2 O
2 mol C6 H 6
2
2
Step 5:
Calculate the masses of the other compounds at the end of the reaction.
mO = 0.1478 mol O2
2
31.9988 g O2
= 4.733 g O2
mol O2
mCO = 2.7749 mol CO2
44.010 g CO2
= 122.13 g CO2
mol CO2
mH O = 1.3875 mol H 2 O
18.0153 g H 2 O
= 24.996 g H 2 O
mol H 2 O
2
2
Thus, at the end of the reaction, there will be no C6H6, there will remain 4.733 g O2 , and 122.13 g
CO2 and 24.996 g H2O.
page 67
Example: Calculate the amount of each compound remaining after 5.00 g each of hydrogen and
oxygen are reacted to form water?
Atomic Mass:
Step 1:
H = 1.00797
O = 15.9994
Write the balanced chemical equation
2 H2 (l) + O2 (g)
Step 2:
2 H2O (l)
Calculate the number of moles of the given compounds
Molar Masses: H2 = 2 × (1.008) = 2.016
O2 = 2 × (16.00) = 32.00
H2O = 2 × (1.008) + (16.00) = 18.02
Step 3:
nH 2 = 5.00 g H 2
mol
= 2.48 g H 2
2.016 g H 2
nO2 = 5.00 g O2
mol
= 0.156 g O2
32.00 g O2
Determine the Limiting Reactant.
Assume H2 is the Limiting Reactant. Then the number of moles of O2 required to
react completely with all the H2 is:
nO2 = 2.48 mol H 2
1 mol O2
= 1.24 mol O2
2 mol H 2
Actual moles of O2 = 0.156 mol < 1.24 mol O2 needed.
There is not enough O2 to react with all the H2 . Therefore O2 must be the
Limiting Reactant. The Limiting Reactant will determine how much product
remains.
Step 4:
Calculate the number of moles of the other compounds in the chemical equation
at the end of the reaction based on the Limiting Reactant.
nH = moles H 2 initially present moles H 2 used up = 2.48 0.156 mol O2
2
nH O = 0.156 mol O2
2
Step 5:
2 mol H 2
= 2.17 mol H 2 remains
1 mol O2
2 mol H 2 O
= 0.312 mol H 2 O
1 mol O2
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
mH = 2.17 mol
2
2.016 g
= 4.37 g H 2
mol
mH O = 0.312 mol
2
18.02 g
= 5.62 g H 2 O
mol
page 68
Example: If 20.0 g aluminum is reacted with 15.0 g oxygen, how much aluminum oxide will be obtained?
Atomic Mass:
Step 1:
Al = 26.98
O = 16.00
Write the balanced chemical equation
4 Al (s) + 3 O2 (g)
Step 2:
2 Al2O3 (s)
Calculate the number of moles of the given compounds
Molar Mass
O2 = 2 x (16.00) = 32.00
Al2O3 = 2 x (26.98) + 3 x (16.00) = 101.96
nAl = 20.0 g Al
mol
= 0.741 mol Al
26.98 g Al
nO = 15.0 g O2
mol
= 0.469 mol O2
32.00 g O2
2
Step 3:
Step 3: Determine the Limiting Reactant.
Assume Al is the Limiting Reactant. Then the number of moles of O2 required to
completely react with all the Al is:
nO = 0.741 mol Al
2
3 mol O2
= 0.556 mol O2 are required
4 mol Al
Compare the number of moles O2 that are actually present with the number of
moles O2 that are required”
0.467 mol O2 are given < 0.556 mol O2 are required.
There are not enough moles O2 initially present so O2 must be the Limiting
Reactant.
Step 4:
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
nAl = moles Al initially present moles Al used up = 0.741 0.469 mol O2
nAl O = 0.469 mol O2
2 3
Step 5:
4 mol Al
= 0.116 mol Al remains
3 mol O2
2 mol Al2 O3
= 0.313 mol Al2 O3
3 mol O2
Calculate the mass of the other compounds in the chemical equation
mAl = 0.116 mol Al
26.98 g Al
=3.13 g Al
mol Al
nAl O = 0.313 mol Al2 O3
2 3
101.96 g Al2 O3
= 31.9 g Al2 O3
mol Al2 O3
page 69
8.3 Percent Yield
In the preceding stoichiometric calculations, given a certain mass of reactants, we are asked to calculate
the amount of product we expect to get. This expected value for the amount of product is known as the
theoretical yield as it is achieved by calculations based on a theoretical relationship between moles of
reactants and products described by the balanced chemical equation (i.e., obtained through the
stoichiometric calculations carried out above).
If the reaction is actually carried out in the lab, a certain experimental value for the mass of product will be
obtained. This lab value for the mass is known as the actual yield or the experimental yield.
In an ideal world, real life and theory are one and the same, so that the actual yield and the theoretical
yield should be the same. In real life however, many things can go improperly so that losses may be
incurred, serving to reduce the actual yield (or the product may be contaminated with impurities raising
the apparent mass of product, thereby increasing the apparent actual yield). Thus, chemists usually
report a percent yield:
Percent Yield =
Actual Yield
100%
Theoretical Yield
This is the ratio of what you got to what you should have gotten.
Thus for example, in the preceding problem where 20.0 g Al are reacted with 15.0 g O2 , we calculated
that 31.9 g Al2O3 would be expected. The 31.9 g Al2O3 are the theoretical yield. If the experiment was
actually carried out, and 31.1 g Al2O3 were obtained then:
% yield =
31.1 g
100% = 97.5%
31.9 g
Example: In the following (unbalanced) reaction
NH3 (g) + O2 (g)
NO (g) + H2O (l) ,
16.5 g NH3 is allowed to react with 33.0 g O2 to produce 21.3 g nitrogen monoxide. Calculate
the theoretical yield and the percent yield.
Atomic Mass:
Step 1:
H = 1.008
N = 14.007
Write the balanced chemical equation
4 NH3 (g) + 5 O2 (g)
Step 2:
O = 16.00
4 NO (g) + 6 H2O (l) ,
Calculate the number of moles of the given reactants
O2 = 2 x (16.00) = 32.00
Molar Mass
NH3 = (14.007) + 3 x (1.008) = 17.031
NO = (14.007) + (16.00) = 30.01
H2O = 2 × (1.008) + (16.00) = 18.02
nNH = 16.5 g NH 3
3
nO = 33.0 g O2
2
mol
= 0.969 mol
17.031g
mol
= 1.03 mol
32.00 g
page 70
Step 3:
Determine the Limiting Reactant.
Assume NH3 is the Limiting Reactant. Then the number of moles of O2 required
to completely react with all the NH3 is:
nNH3 = 0.969 mol NH 3
5 mol O2
= 1.21 mol O2 are required
4 mol NH 3
Compare the number of moles O2 that are actually present with the number of
moles O2 that are required”
1.03 mol O2 are given < 1.21 mol O2 are required.
There are not enough moles O2 initially present so O2 must be the Limiting
Reactant.
Step 4:
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
nNH = moles NH 3 initially present moles NH 3 used up = 0.969 1.03 mol O2
3
nNO = 1.03 mol O2
4 mol NH 3
= 0.145 mol NH 3 remains
5 mol O2
4 mol NO
= 0.824 mol NO
5 mol O2
6 mol H 2 O
= 1.24 mol H 2 O
5 mol O2
Calculate the mass of the other compounds in the chemical equation
nH O = 1.03 mol O2
2
Step 5:
Step 6:
mNH3 = 0.145 mol Al
17.031 g
=2.47 g NH 3
mol
mNO = 0.824 mol NO
30.01 g
=24.7 g NO
mol
mH 2O = 1.24 mol H 2 O
18.02 g
= 22.3 g H 2 O
mol
Calculate the % yield
Theoretical yield = 24.8 g NO
Actual yield = 21.3 g NO
% yield =
21.3 g
100% = 86.2%
24.7 g
page 71
Example: Nitrogen gas and hydrogen gas react to produce ammonia gas according to the unbalanced
reaction
N2 ( g ) + H2 ( g )
NH3 ( g ) .
When 25.00 g each of nitrogen and hydrogen were reacted at a certain temperature and
pressure, 25.00 g of ammonia were found to have formed. Calculate the percentage yield.
Atomic Mass:
H = 1.008
N = 14.007
Step 1:
Write the balanced chemical equation
Step 2:
N2 ( g ) + 3 H 2 ( g )
2 NH3 ( g ) .
Calculate the number of moles of the given compounds
H2 = 2 x (1.008) = 2.016
Molar Mass
N2 = 2 x (14.007) = 28.014
NH3 = (14.007) + 3 x (1.008) = 17.031
nH = 25.00 g H 2
mol
= 12.40 mol H 2
2.016 g H 2
nN = 25.00 g N 2
mol
= 0.8924 mol N 2
28.014 g H 2
2
2
Step 3:
Determine the Limiting Reactant.
Assume N2 is the Limiting Reactant. Then the number of moles of H2 required to
completely react with all the N2 is:
n H2 = 0.8924 mol N 2
FG 3 mol H IJ = 2.677 mol H
H 1 mol N K
2
2
are required
2
Compare the number of moles H2 that are actually present with the number of
moles H2 that are required”
12.4 mol H2 are given > 2.677 mol H2 are required.
There are enough moles H2 initially present so N2 is the Limiting Reactant as was
assumed.
Step 4:
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
nH = 12.40 mol H 2 given 2.677 mol H 2 required = 9.72 mol H 2 remain
2
2 mol NH 3
= 1.785 mol NH 3
1 mol N 2
Calculate the mass of the compounds in the chemical equation at the end
nNH = 0.8924 mol N 2
3
Step 5:
mH = 9.72 mol H 2
2
2.016 g H 2
= 19.6 g H 2
mol
mNH = 1.785 mol NH 3
3
Step 6:
17.031 g NH 3
= 30.40 g NH 3
mol
Calculate the % yield
Theoretical yield = 30.40 g NH3 Actual yield = 25.00 g NH3
% yield =
25.00
100% = 82.24%
30.40
page 72
Example: If 49.8 g Fe3O4 is mixed with 10.0 g C and they react according to the following unbalanced
equation,
Fe3O4 (s) + C (s)
Fe (s) + CO (g),
calculate the mass of CO produced if the yield was 76.5%.
Atomic Mass:
Step 1:
C = 12.011
O = 16.00
Write the balanced chemical equation
Fe3O4 (s) + 4 C (s)
Step 2:
Fe = 55.85
3 Fe (s) + 4 CO (g),
Calculate the number of moles of the given compounds
Molar Mass
CO = (12.011) + (16.00) = 28.01
Fe3O4 = 3 x (55.85) + 4 (16.00) = 231.55
nFe O = 49.8 g Fe3O4
3 4
nC = 10.0 g C
Step 3:
mol
= 0.215 mol Fe3O4
231.55 g
mol
= 0.833 mol C
12.011 g
Determine the Limiting Reactant.
Assume Fe3O4 is the Limiting Reactant. Then the number of moles of C required
to completely react with all the Fe3O4 is:
nC = 0.215 mol Fe3O4
FG 4 mol C IJ = 0.860 mol C required
H 1 mol Fe O K
3 4
Compare the number of moles C that are actually present with the number of
moles C that are required”
0.832 mol C are given > 0.860 mol C are required.
There are not enough moles C initially present so C is the Limiting Reactant.
Step 4:
Based on the Limiting Reactant, calculate the number of moles of the other
compounds at the end of the reaction.
nFe3O4 = 0.215 mol Fe3O4 present 0.833 mol C
Step 5:
1 mol Fe3O4
= 0.007 mol Fe3O4 remain
4 mol C
nFe = 0.833 mol C
3 mol Fe
= 0.624 mol Fe
4 mol C
nCO = 0.833 mol C
4 mol CO
= 0.833 mol CO
4 mol C
Calculate the mass of the compounds in the chemical equation at the end
mFe3O4 = 0.007 mol Fe3 O4
mFe = 0.624 mol Fe
mCO = 0.833 mol CO
231.55 g
= 2 g Fe3 O4
mol
55.85 g
= 34.9 g Fe
mol
28.01 g
= 23.3 g CO
mol
page 73
Step 6:
Calculate the actual yield
% yield =
Actual yield
100%
Theoretical yield
Rearranging this equation:
Theoretical yield
100%
23.3 g CO
Actual yield = (76.5%)
= 17.8 g CO
100%
Actual yield = (% yield )
page 74