STOICHIOMETRY
Stoichiometry means using balanced chemical equations to calculate the amount of
substance produced or required in chemical reactions.
Interpreting Balanced Equations
Elemental hydrogen is said to be the fuel of the future. It burns cleanly in air
producing only water according to the following balanced chemical equation.
O2

2H2O
1 mole of O2

2 moles of H2O
2H2
+
Meaning:
2 moles of H2 +
Since:
mm H2 = 2.0 g,
Therefore:
2  (2.0 g H2) + 1  (32.0 g O2) 
mm O2 = 32.0 g
mm H2O = 18.0 g
2  (18.0 g H2O)
In the lab or in the workplace we are often required to determine the mass or number of
moles of a substance involved in a chemical reaction. This is easily accomplished
whenever a balanced equation can be written. Four basic types of calculations which you
will encounter as listed.
a) (moles given  moles required), e.g., 0.25 mol O2  ? mol H2O
b) (moles given  mass required), e.g., 0.25 mol O2  ? g H2O
c) (mass given  moles required), e.g., 64 g O2
 ? mol H2O
d) (mass given  mass required),
 ? g H2 O
e.g., 64 g O2
General (Expedient) Procedure for Solving Stoichiometry Problems
1. Write a balanced chemical equation for the reaction and
above it write the information given and required in the problem
below it write the molar masses of the compounds involved (only if required).
Note that you only need to calculate molar masses of compounds if data is given or
required in mass units.
2. From the wording in the problem, write a short logical statement linking the given and
required components.
Above this, from the balanced equation, write another short logical statement linking
the same 2 components.
3. Solve for the unknown by ratio and proportion. Check units, sig figs, and
reasonableness of your answer.
1
Solve each of the 4 problems stated above using the described method.
a)
0.25 mol O2  ? mol H2O
0.25 mol
+
O2

2H2
1
(given & req’d info)
x mol
2H2O
mol O2 produces 2 mol H2O
 0.25 mol O2 
 = 0.5 mol H2O
 1 mol O2 
0.25 mol O2 produce x mol H2O = 2 mol H2O  
b) 0.25 mol O2  ? g H2O
0.25 mol
+
O2

2H2
1
mol O2 produces (2  18.0 g) H2O
 0.25 mol O2 
 = 9.0 g H2O
 1 mol O2 
x g H2O = (2  18.0 g) H2O  
0.25 mol O2 produce
c)
xg
(given & req’d info)
2H2O
2  (18.0 g)
 ? mol H2O
64.0 g O2
2H2
64.0 g
+
O2

(32.0 g)
x mol
2H2O
(given & req’d info)
32.0 g O2 produce 2 mol H2O
 64.0 g O2 
 = 4 mol H2O
 32.0 g O2 
64.0 g O2 produce x mol H2O = 2 mol H2O  
d)
64 g O2
 ? g H2 O
2H2
64.0 g
+
O2

(32.0 g)
xg
(given & req’d info)
2H2O
2  (18.0 g)
32.0 g O2 produce 36.0 g H2O
64.0 g O2 produce
 64.0 g O2 
 = 72.0 g H2O
 32.0 g O2 
x g H2O = 36.0 g H2O  
2
Solving Percentage Yield Problems
Up to this point we have been calculating the theoretical yield, i.e., the yield
obtained assuming that all the starting materials react completely and no product is lost.
The balanced equation always gives the theoretical yield.
In many industrial processes and laboratory reactions the actual yield obtained is
significantly less than the theoretical yield. The relationship between the actual yield and
the theoretical yield is called the percentage yield (% yield).
 actual yield 
% yield = 
  100%
 theoretical yield 
Often, either accidentally or deliberately, one of the reagents in a reaction is present
in excess while another reagent is the limiting reagent, i.e., there is not enough of it to use
up all the reagent which is in excess. Only the quantity of limiting reagent can be used to
determine the theoretical yield in these cases.
Example of % yield problem:
When a student applied an electric current to water, 100 grams of water was converted into
hydrogen and oxygen, and 80.0 grams of oxygen was recovered. Calculate the percentage
yield of the experiment.
100 g
2H2O
2  (18.0 g)

2H2
+
80.0 g (actual)
x g (theoretical)
O2
32.0 g
36.0 g H2O produce (theoretically) 32.0 g O2
100 g H2O produce (theoretically)
 actual yield 
% yield = 
  100 %
 theoretical yield 
 100 g H 2 O 
x g O2 = 32.0 g O2  
 = 88.9 g O2
 36.0 g H 2 O 
=
 80.0 g O2 

  100% = 90.0% yield.
 88.9 g O2 
In this problem, there is only one reagent (H2O) so it is the limiting reagent and the
% theoretical yield is calculated based on the quantity of this reagent.
3
STOICHIOMETRY, YIELD, AND LIMITING REAGENT PROBLEMS
1. Under appropriate reaction conditions copper and oxygen react to form copper II
oxide. In an experiment with 45.3 g copper and an excess of oxygen, a percent yield
of 96.7 % was obtained. What is the actual yield of copper II oxide in grams?
Ans: 56.7 g
2. The percent yield for the reaction : PCl3 + Cl2  PCl5 is 74.3 %.
What mass of product in grams, can be obtained from the reaction of 50.0 g Cl2 and
an excess of PCl3 ?
Ans: 108 g
3. The % yield for the reaction: 2NO + O2  2NO2
is 35.6 %. What mass of
product can be obtained from the reaction of 25.0 g of NO and an excess of O2 ?
Ans: 13.6 g
4. Aluminum and monatomic sulfur react at elevated temperatures to form aluminum
sulfide. In an experiment, 125 g of product was produced from 75.0 g aluminum and
300 g of sulfur.
a) What is the theoretical yield of product? Ans: 209 g
b) What is the percent yield of product?
Ans: 60.0%
5. Aluminum and oxygen react to form aluminum oxide. In an experiment, 125 g of
product was produced from 75.0 g aluminum.
a) What is the theoretical yield of aluminum oxide? Ans: 142 g
b) What is the percent yield of product?
Ans: 88.2%
6. 3.000 g TiCl3 was produced from 3.513 g of TiCl4 and an excess of Ti.
What is the % yield of TiCl3 ?
Ans: 78.80%
7. 5.000 g of Ag2S was produced from 5.000 g Ag and an excess of S.
What is the % yield of Ag2S ?
Ans: 87.1%
8. Zinc and monatomic sulfur react to produce zinc sulfide. In an experiment with
37.0 g of zinc and an excess of sulfur, a percent yield of 93.7% zinc sulfide was
obtained. What is the actual yield of zinc sulfide in grams for the experiment?
Ans: 42.9 g
4