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Percentage Yield
Percentage Yield
N2 + 3H2  2NH3
Based on this equation and your
knowledge, if you reacted 1 mol N2 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?!
Percentage Yield
N2 + 3H2  2NH3
Where did the 0.1 mol go?!
-Experimental design
Competing reactions (products reacting with
reactants)
Experimenter techniques
Impure reactants
.
.
.
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
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
mN2
= 7.5g
Percentage Yield
N2 + 3H2  2NH3
Given:
Limiting reactant = N2
mN2
= 7.5g  n = m
M
= 7.5g
28.02g/mol
= 0.2677mol
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?
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
1 mol N2 0.2677mol N2
x = 0.5353 mol NH3
CONVERT BACK TO GRAMS
mNH3
= n x M
= (0.5353 mol) x (17.04 g/mol)
= 9.121512g
Percentage Yield
N2 + 3H2  2NH3PROBLEM:
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
= 1.72g x 100%
9.12g
= 18.9%
Therefore the percentage yield is 19%
The actual yield is
ALWAYS smaller
than the theoretical
yield
Percentage Purity
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)
Percentage Purity
PROBLEM:
You have a 13.9g sample of impure iron pyrite, The
sample is heated in air to produce iron (III) oxide, Fe2O3,
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 = ?
Percentage PurityPROBLEM:
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 Fe2S
expected to have produced 8.02g Fe2O3!!!
Percentage PurityPROBLEM:
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
= n x M
= (0.100 mol) x (120 g/mol)
= 12.0g FeS2
Percentage PurityPROBLEM:
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%

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18 percentage yield

  • 2. Percentage Yield N2 + 3H2  2NH3 Based on this equation and your knowledge, if you reacted 1 mol N2 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 mN2 = 7.5g
  • 6. Percentage Yield N2 + 3H2  2NH3 Given: Limiting reactant = N2 mN2 = 7.5g  n = m M = 7.5g 28.02g/mol = 0.2677mol 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?
  • 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 1 mol N2 0.2677mol N2 x = 0.5353 mol NH3 CONVERT BACK TO GRAMS mNH3 = n x M = (0.5353 mol) x (17.04 g/mol) = 9.121512g
  • 8. Percentage Yield N2 + 3H2  2NH3PROBLEM: 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 = 1.72g x 100% 9.12g = 18.9% Therefore the percentage yield is 19% The actual yield is ALWAYS smaller than the theoretical yield
  • 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, Fe2O3, 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. Percentage PurityPROBLEM: 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 Fe2S expected to have produced 8.02g Fe2O3!!!
  • 13. Percentage PurityPROBLEM: 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 = n x M = (0.100 mol) x (120 g/mol) = 12.0g FeS2
  • 14. Percentage PurityPROBLEM: 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%

Editor's Notes

  1. FeS2 is called iron pyrite. Alan, it’s because this question is ASKINg you how pure iron pyrite is. So Why would you solve for something else?