Percentage Yield Problems
1. In an experiment, 5.00 g of silver nitrate is added to a solution of sodium bromide. 5.03 g of silver bromide is produced. The balanced equation is shown below:
AgNO3 (aq) + NaBr (aq) → AgBr (s) + NaNO3 (aq)
Fill in the table below and provide your calculations for the theoretical yield and percentage yield of AgBr (s).
Balanced Eq’n / AgNO3 (aq) + NaBr (aq) → AgBr (s) + NaNO3 (aq)Given mass (g) / 5.00 g / 5.03 g
Molar mass (g/mol) / 169.88 g/mol / 187.77 g/mol
a) Find # of moles of reactant (AgNO3 in this case)
nAgNO3 = m
M
= 5.00 g
169.88 g/mol
= 0.0294 mol
b) Perform a mole ratio conversion. Since 1 mol AgNO3 = 1 mol AgBr, then:
nAgBr = 0.0294 mol AgNO3 x 1 mol AgBr
1 mol AgNO3
= 0.0294 mol
c) Find mass (theoretical yield) of AgBr
m = 0.0294 mol x 187.77 g/molThe theoretical yield of AgBr is 5.527 g
= 5.527 g
d) Calculate percentage yield of AgBr
percentage yield = actual yieldx 100%The % yield of AgBr is 91%
theoretical yield
= 5.03 g x 100%
5.527 g
= 91%
2. 45.0 g of sodium chloride is reacted with iron(II) sulphide. You end up retrieving 20.0 g of sodium sulphide. Fill in the table below and find the theoretical yield and percentage yield of Na2S. The balanced equation is shown below:
2NaCl (aq) + FeS (aq) → Na2S (s) + FeCl2 (aq)
Balanced Eq’n / 2NaCl (aq) + FeS (aq) → Na2S (s) + FeCl2 (aq)Given mass (g) / 45.0 g / 20.0 g
Molar mass (g/mol) / 58.44 g/mol / 78.04 g/mol
a) Find # of moles of reactant (NaCl in this case)
nNaCl = m
M
= 45.0 g
58.44 g/mol
= 0.77 mol
b) Perform a mole ratio conversion. Since 2 mol NaCl = 1 mol Na2S, then:
nNa2S = 0.77 mol NaCl x 1 mol Na2S
2 mol NaCl
= 0.385 mol
c) Find mass (theoretical yield) of Na2S
m = 0.385 mol x 78.04 g/molThe theoretical yield of Na2S is 30.18 g
= 30.18 g
d) Calculate percentage yield of Na2S
percentage yield = actual yieldx 100%The % yield of Na2S is 66.3%
theoretical yield
= 20.0 g x 100%
30.18 g
= 66.3%
3. A student produces Aspirin, C9H8O4 (s) and acetic acid, HC2H3O2 (aq), from a reaction of salicyclic acid, C7H6O3 (s) with acetic anhydride, C4H6O3 (aq). The balanced equation is shown below:
C7H6O3 (s) + C4H6O3 (aq) → C9H8O4 (s) + HC2H3O2 (aq)
Fill in the table and find the theoretical yield and percentage yield for aspirin if 213.0 g of salicyclic acid is used and 189.3 g of aspirin is produced.
Balanced Eq’n / C7H6O3 (s) + C4H6O3 (aq) → C9H8O4 (s) + HC2H3O2 (aq)Given mass (g) / 213.0 g / 189.3 g
Molar mass (g/mol) / 138.13 g/mol / 180.17 g/mol
a) Find # of moles of reactant (C7H6O3 in this case)
nC7H6O3 = m
M
= 213.0 g
138.13 g/mol
= 1.542 mol
b) Perform a mole ratio conversion. Since 1 mol C7H6O3 = 1 mol C9H8O4, then:
nC9H8O4 = 1.542 mol C7H6O3 x 1 mol C9H8O4
1 mol C7H6O3
= 1.542 mol
c) Find mass (theoretical yield) of C9H8O4
m = 1.542 mol x 180.17 g/molThe theoretical yield of C9H8O4 is 277.83 g
= 277.83 g
d) Calculate percentage yield of C9H8O4
percentage yield = actual yieldx 100%The % yield of C9H8O4 is 68.1%
theoretical yield
= 189.3 g x 100%
277.83 g
= 68.1%
4. List some possible reasons that can affect the percentage yield of a product in a chemical reaction. How do these reasons you list affect the yield? What are some ways to improve the yield? (Check out Section 2.12 in your textbook) – IMPORTANT!
*Be prepared to answer a similar question on the test. Read 2.12 and use your work in the formal lab to help you answer this question.*