### Percent Yield

```PERCENT YIELD
Brought to you by Coach Cox
WHAT IS PERCENT YIELD?
• Theoretical Yield – the maximum amount of product that can be produced
from a given amount of reactant. This is determined mathematically using
stoichiometry calculations
• Actual Yield – the amount of product actually produced when a chemical
reaction is carried out in the lab
• Percent Yield – ratio of the actual yield to the theoretical yield expressed as
a percent (%)
PERCENT YIELD FORMULA
• Percent Yield =
Actual Yield
x 100
Theoretical Yield
It is on your formula chart! 
PERCENT YIELD EXAMPLE 1
•
In lab, you burned 3.2 g of Mg and produced 4.8 g of MgO according to the given
reaction. Your calculations showed that you should have been able to make 5.3 g of
MgO. What is your percent yield?
Mg(s) + O2(g)  MgO(s)
•
Actual Yield:
4.8 g MgO
•
Theoretical Yield:
5.3 g MgO
•
Percent yield = Actual Yield
x 100
Theoretical Yield
4.8 g MgO x 100 =
5.3 g MgO
90.6% Yield MgO
PERCENT YIELD EXAMPLE 2
•
During a lab experiment, 24.8 g of calcium carbonate, CaCO 3, is decomposed and 13.1
grams of calcium oxide, CaO, is produced.
CaCO3(s)  CaO(s) + CO2(g)
a. What is the actual yield of CaO?
_________________
b. What is the theoretical yield of CaO?
_________________
c. What is the percent yield of CaO?
_________________
a. What is the actual yield of CaO?
13.1 g CaO
b. What is the theoretical yield of CaO?
13.90 g CaO
24.8 g CaCO 3 x 1 mol CaCO3
x 1 mol CaO
100.086 g CaCO 3
1 mol CaCO3
c. What is the percent yield of CaO?
13.1 g CaO x 100 =
13.9 g CaO
x 56.077 g CaO = 13.895 g CaO
1 mol CaO
94.2% Yield CaO
94.2% Yield CaO
PERCENT YIELD EXAMPLE 3
•
During a lab experiment, 1.87 g of aluminum reacts with an excess of copper (II) sulfate,
(CuSO4), and 4.65 g or copper is produced. What is the percent yield?
2 Al + 3 CuSO4  Al2(SO4)3 + 3 Cu
•
Actual Yield:
4.65 g Cu
•
Theoretical Yield:
6.61 g Cu
1.87 g Al x 1 mol Al
x 3 mol Cu
26.98 g Al
•
x 63.546 g Cu = 6.61 g Sb
2 mol Al
Percent yield = Actual Yield
1 mol Cu
x 100
Theoretical Yield
4.65 g MgO
6.61 g MgO
x 100
=
70.3% Yield Cu
PERCENT YIELD EXAMPLE 4
•
An ore of antimony, (Sb 2S3), reacts with excess iron according to the given reaction. If 15
g of Sb 2S3 reacts and the percent yield of Sb is 91.5%, what mass of Sb is actually
produced?
Sb2S3(s) + 3 Fe(s)  2 Sb(s) + 3 FeS(s)
•
Actual Yield:
? g Sb
•
Theoretical Yield:
10.75 g Sb
15 g Sb2S3 x 1 mol Sb2S3
x 2 mol Sb
339.718 g Sb2S3
•
Percent yield = Actual Yield
x 121.76 g Sb = 10.75 g Sb
1 mol Sb2S3
1 mol Sb
x 100
Theoretical Yield
? g MgO
x 100 =
10.75 g MgO
Actual Yield = 9.84 g Sb
91.5% Yield Sb
```