1. Percentage Yield

2. Percentage Yield
N2 + 3H2  2NH3
Based on this equation and your
knowledge, if you reacted 1 mol N 2 with 3
mol H2, you will get 2 mol NH3
BUT if you carried out this experiment, you may
find out that you only make 1.9 mol. Where did
the 0.1 mol go?!

3. Percentage Yield
N2 + 3H2  2NH3
Where did the 0.1 mol go?!
-Experimental design
Competing reactions (products reacting with
reactants)
Experimenter techniques
Impure reactants
.
.
.

4. Percentage Yield
N2 + 3H2  2NH3
Formula to find percentage yield:
Percentage yield = Actual yield x 100%
Theoretical Yield
This formula is similar to:
Test score = Marks earned x 100%
Maximum possible marks

5. Percentage Yield
N2 + 3H2  2NH3
PROBLEM:
When 7.5g of nitrogen reacts with sufficient hydrogen, the actual
yield of ammonia is 1.72g. What is the percentage yield of the
reaction?
Given:
Limiting reactant = N2
mN = 7.5g
2

6. Percentage Yield
N2 + 3H2  2NH3
When 7.5g of nitrogen reacts with sufficient hydrogen, the actual
yield of ammonia is 1.72g. What is the percentage yield of the
reaction?
Given:
Limiting reactant = N2
mN = 7.5g  n = m
2
M
= 7.5g
28.02g/mol
= 0.2677mol

7. Percentage Yield
N2 + 3H2  2NH3
When 7.5g of nitrogen reacts with sufficient hydrogen, the actual
yield of ammonia is 1.72g. What is the percentage yield of the
reaction?
Given:
Limiting reactant = N2
nN2 = 0.2677mol
STEP 1: SOLVE for the theoretical yield
2 mol NH3 = x CONVERT BACK TO GRAMS
mNH3 = nxM
1 mol N2 0.2677mol N2
= (0.5353 mol) x (17.04 g/mol)
x = 0.5353 mol NH3 = 9.121512g

8. Percentage Yield
PROBLEM:
N2 + 3H2  2NH3
When 7.5g of nitrogen reacts with sufficient hydrogen, the
theoretical yield of ammonia is 9.12g. If 1.72g of ammonia is
obtained by experiment, what is the percentage yield of the
reaction? Given:
Actual yield = 1.72g
Theoretical yield = 9.12g
STEP 2: SOLVE for the percentage yield
Percentage yield = Actual yield x 100%
Theoretical yield
The actual yield is = 1.72g x 100%
ALWAYS smaller
than the theoretical 9.12g
yield = 18.9%
Therefore the percentage yield is 18.9%

9. Percentage Purity

10. Percentage Purity
Percentage purity: How much of a sample, by
mass, is actually of a specific compound or
element
Sample = pure compound or element + impurities
Percentage purity = theoretical mass (g) x 100%
sample size (g)

11. Percentage Purity
PROBLEM:
You have a 13.9g sample of impure iron pyrite, The
sample is heated in air to produce iron (III) oxide, Fe 2O3,
and sulfur dioxide SO2
4FeS2 + 11O2  2Fe2O3 + 8SO2
If you obtain 8.02g of iron (III) oxide, what was the
percentage purity of iron pyrite in the original sample?
Given:
Actual yield of Fe2O3= 8.02g
Sample mass = 13.9g
Theoretical mass = ?

12. PROBLEM:
Percentage Purity
You have a 13.9g sample of impure iron pyrite and you obtain 8.02g of iron (III) oxide
4FeS2 + 11O2  2Fe2O3 + 8SO2
What was the percentage purity of iron pyrite?
Given:
Actual yield of Fe2O3= 8.02g
Sample mass = 13.9g
Theoretical mass = ?
STEP 1:
USE YOUR STOICHIOMETRY SKILLS to find the mass of Fe 2S
expected to have produced 8.02g Fe2O3!!!

13. PROBLEM: Percentage Purity
You have a 13.9g sample of impure iron pyrite and you obtain 8.02g of iron (III) oxide
4FeS2 + 11O2  2Fe2O3 + 8SO2
Given:
Actual yield of Fe2O3= 8.02g
Sample mass = 13.9g
Theoretical mass = ?
STEP 1: Stoichiometry
x = 4 mol FeS2
0.0502 mol Fe2O3 2 mol Fe2O3
x = 0.100 mol FeS2
CONVERT BACK TO GRAMS
mFeS2 = nxM
= (0.100 mol) x (120 g/mol)
= 12.0g FeS2

14. PROBLEM: Percentage Purity
You have a 13.9g sample of impure iron pyrite and you obtain 8.02g of iron (III) oxide
4FeS2 + 11O2  2Fe2O3 + 8SO2
Given:
Actual yield of Fe2O3= 8.02g
Sample mass = 13.9g
Theoretical mass = ?
STEP 2: Solve for the percentage purity
Percentage purity = theoretical mass (g) x 100%
sample size (g)
Percentage purity = 12.0g x 100%
13.9g
= 86.3%
Therefore the percent purity of the iron pyrite sample
is 86.3%