1. Percentage Yield

2. Define the following terms: yield, theoretical yield, actual yield, percentage yield.
Yield: the amount of product
Theoretical yield: the amount of product we expect, based on stoichiometric calculations
Actual yield: amount of product from a procedure or experiment (this is given in the question)
Percent yield Actual yield x 100%
Theoretical Yield
(stoichiometric calculation)

3. The reasons why the actual yield in a chemical reaction often falls short of the theoretical yield.
Not all product is recovered (e.g. spattering)
Reactant impurities (e.g. weigh out 100 g of chemical which has 20 g of junk)
A side reaction occurs (e.g. MgO vs. Mg3N2)
The reaction does not go to completion

4. Sample problem
Q - What is the % yield of H2O if 138 g H2O is produced from 16 g H2 and excess O2?
Step 1: write the balanced chemical equation
2H2 + O2  2H2O
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
16 g H2
143 g =
Step 3: Calculate % yield
actual
theoretical
138 g H2O
143 g H2O =
% yield =
x 100%
x 100%
96.7% =

5. Practice problem
Q - What is the % yield of NH3 if 40.5 g NH3 is produced from 20.0 mol H2 and excess N2?
Step 1: write the balanced chemical equation
N2 + 3H2  2NH3
Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:
2 mol NH3
3 mol H2 x
17.04 g NH3
1 mol NH3 x
# g NH3=
20.0 mol H2
227 g =
Step 3: Calculate % yield
actual
theoretical
40.5 g NH3
227 g NH3 =
% yield =
x 100%
x 100%
17.8% =

6. Challenging question 2H2 + O2  2H2O
What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?
Hint: determine limiting reagent first
1 mol O2
32 g O2 x
2 mol H2O
1 mol O2 x
18.02 g H2O
1 mol H2O x
# g H2O=
60 g O2
68 g =
1 mol H2
2.02 g H2 x
2 mol H2O
2 mol H2 x
18.02 g H2O
1 mol H2O x
# g H2O=
7.0 g H2
62.4 g =
actual
theoretical
58 g H2O
62.4 g H2O =
% yield =
x 100%
x 100%
92.9% =

7. More Percent Yield Questions
The electrolysis of water forms H2 and O H2O  2H2 + O What is the % yield of O2 if 12.3 g of O2 is produced from the decomposition of 14.0 g H2O?
107 g of oxygen is produced by heating 300 grams of potassium chlorate. Calculate % yield. 2KClO3  2KCI + 3O2
What is the % yield of ferrous sulphide if 3.00 moles of Fe reacts with excess sulfur to produce 220 grams of ferrous sulphide? Fe + S  FeS

8. More Percent Yield Questions
Iron pyrites (FeS2) reacts with oxygen according to the following equation:
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of iron pyrites is burned in 200 g of O2, 143 grams of ferric oxide is produced. What is the percent yield of ferric oxide?

9. Q1 The electrolysis of water forms H2 & O2. 2H2O  2H2 + O2
Give the percent yield of O2 if 12.3 g O2 is produced from the decomp. of 14 g H2O?
Actual yield is given: 12.3 g O2
Next, calculate theoretical yield
# g O2=
1 mol H2O
18.02 g H2O x
1 mol O2
2 mol H2O x
32 g O2
1 mol O2 x
14.0 g H2O
12.43 g =
Finally, calculate % yield
actual
theoretical
12.3 g O2
12.43 g O2 =
% yield =
x 100%
x 100%
98.9% =

10. Q2
107 g of oxygen is produced by heating 300 grams of potassium chlorate.
2KClO3  2KCI + 3O2
Actual yield is given: 107 g O2
Next, calculate theoretical yield
# g O2=
1 mol KClO3
g KClO3 x
3 mol O2
2 mol KClO3 x
32 g O2
1 mol O2 x
300 g KClO3
117.5 g =
Finally, calculate % yield
actual
theoretical
107 g O2
117.5 g O2 =
% yield =
x 100%
x 100%
91.1% =

11. Q3
What is % yield of ferrous sulfide if 3 mol Fe produce 220 grams of ferrous sulfide?
Fe + S  FeS
Actual yield is given: 220 g FeS
Next, calculate theoretical yield
1 mol FeS
1 mol Fe x
87.91 g FeS
1 mol FeS x
# g FeS=
3.00 mol Fe
263.7 g =
Finally, calculate % yield
actual
theoretical
220 g O2
263.7 g O2 =
% yield =
x 100%
x 100%
83.4% =

12. First, determine limiting reagent
4FeS2 + 11O2  2Fe2O3 + 8SO2
If 300 g of FeS2 is burned in 200 g of O2, 143 g Fe2O3 results. % yield Fe2O3?
First, determine limiting reagent
# g Fe2O3=
1 mol FeS2
g FeS2 x
2 mol Fe2O3
4 mol FeS2 x
159.7 g Fe2O3
1 mol Fe2O3 x
300 g FeS2
199.7 g Fe2O3 =
1 mol O2
32 g O2 x
2 mol Fe2O3
11 mol O2 x
159.7 g Fe2O3
1 mol Fe2O3 x
200 g O2
g Fe2O3 =
actual
theoretical
143 g Fe2O3
g Fe2O3 =
% yield =
x 100%
x 100%
78.8% =