Worksheet on percent yield problems with definitions, examples, and practice questions for calculating theoretical and actual yields in chemical reactions.

Problem Analysis and Solution

The provided document outlines the concept of percent yield in chemical reactions and provides several problems to practice calculating theoretical yield, actual yield, and percent yield. Below, I will solve each problem step by step.

---

#### 1. Percent Yield Calculation for Aspirin Synthesis

Problem Statement:
A student adds 200.0 g of \( C_7H_6O_3 \) (salicylic acid) to an excess of \( C_4H_6O_3 \) (acetic anhydride), producing \( C_9H_8O_4 \) (aspirin) and \( C_2H_4O_2 \) (acetic acid). Calculate the percent yield if 231 g of aspirin (\( C_9H_8O_4 \)) is produced in an experiment.

Balanced Equation:
\[ C_7H_6O_3 + C_4H_6O_3 \rightarrow C_9H_8O_4 + C_2H_4O_2 \]

Step 1: Determine the molar mass of \( C_7H_6O_3 \) (salicylic acid).
- Molar mass of \( C_7H_6O_3 \):
\[
(7 \times 12.01) + (6 \times 1.01) + (3 \times 16.00) = 84.09 + 6.06 + 48.00 = 138.15 \, \text{g/mol}
\]

Step 2: Calculate the moles of \( C_7H_6O_3 \) used.
\[
\text{Moles of } C_7H_6O_3 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{200.0 \, \text{g}}{138.15 \, \text{g/mol}} \approx 1.448 \, \text{mol}
\]

Step 3: Use stoichiometry to find the theoretical yield of \( C_9H_8O_4 \) (aspirin).
- From the balanced equation, 1 mole of \( C_7H_6O_3 \) produces 1 mole of \( C_9H_8O_4 \).
- Therefore, 1.448 mol of \( C_7H_6O_3 \) will produce 1.448 mol of \( C_9H_8O_4 \).

Step 4: Determine the molar mass of \( C_9H_8O_4 \) (aspirin).
- Molar mass of \( C_9H_8O_4 \):
\[
(9 \times 12.01) + (8 \times 1.01) + (4 \times 16.00) = 108.09 + 8.08 + 64.00 = 180.17 \, \text{g/mol}
\]

Step 5: Calculate the theoretical mass of \( C_9H_8O_4 \).
\[
\text{Theoretical mass of } C_9H_8O_4 = \text{Moles} \times \text{Molar Mass} = 1.448 \, \text{mol} \times 180.17 \, \text{g/mol} \approx 261.0 \, \text{g}
\]

Step 6: Calculate the percent yield.
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{231 \, \text{g}}{261.0 \, \text{g}} \right) \times 100 \approx 88.51\%
\]

Final Answer:
\[
\boxed{88.51\%}
\]

---

#### 2. Percent Yield Calculation for Toluene Nitration

Problem Statement:
According to the following equation, calculate the percentage yield if 550.0 g of toluene (\( C_7H_8 \)) added to an excess of nitric acid (\( HNO_3 \)) provides 305 g of the p-nitrotoluene (\( C_7H_7NO_2 \)) product in a lab experiment.
\[ C_7H_8 + HNO_3 \rightarrow C_7H_7NO_2 + H_2O \]

Step 1: Determine the molar mass of \( C_7H_8 \) (toluene).
- Molar mass of \( C_7H_8 \):
\[
(7 \times 12.01) + (8 \times 1.01) = 84.07 + 8.08 = 92.15 \, \text{g/mol}
\]

Step 2: Calculate the moles of \( C_7H_8 \) used.
\[
\text{Moles of } C_7H_8 = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{550.0 \, \text{g}}{92.15 \, \text{g/mol}} \approx 5.97 \, \text{mol}
\]

Step 3: Use stoichiometry to find the theoretical yield of \( C_7H_7NO_2 \).
- From the balanced equation, 1 mole of \( C_7H_8 \) produces 1 mole of \( C_7H_7NO_2 \).
- Therefore, 5.97 mol of \( C_7H_8 \) will produce 5.97 mol of \( C_7H_7NO_2 \).

Step 4: Determine the molar mass of \( C_7H_7NO_2 \) (p-nitrotoluene).
- Molar mass of \( C_7H_7NO_2 \):
\[
(7 \times 12.01) + (7 \times 1.01) + (1 \times 14.01) + (2 \times 16.00) = 84.07 + 7.07 + 14.01 + 32.00 = 137.15 \, \text{g/mol}
\]

Step 5: Calculate the theoretical mass of \( C_7H_7NO_2 \).
\[
\text{Theoretical mass of } C_7H_7NO_2 = \text{Moles} \times \text{Molar Mass} = 5.97 \, \text{mol} \times 137.15 \, \text{g/mol} \approx 818.0 \, \text{g}
\]

Step 6: Calculate the percent yield.
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{305 \, \text{g}}{818.0 \, \text{g}} \right) \times 100 \approx 37.30\%
\]

Final Answer:
\[
\boxed{37.30\%}
\]

---

#### 3. Mass of Copper Produced from Aluminum Reaction

Problem Statement:
Aluminum reacts with an aqueous solution containing excess copper(II) sulfate. If 1.85 g of Al reacts and the percentage yield of Cu is 56.6%, what mass of Cu is produced?

Balanced Equation:
\[ 2Al + 3CuSO_4 \rightarrow 3Cu + Al_2(SO_4)_3 \]

Step 1: Determine the molar mass of Al.
- Molar mass of Al:
\[
26.98 \, \text{g/mol}
\]

Step 2: Calculate the moles of Al used.
\[
\text{Moles of Al} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{1.85 \, \text{g}}{26.98 \, \text{g/mol}} \approx 0.0686 \, \text{mol}
\]

Step 3: Use stoichiometry to find the theoretical yield of Cu.
- From the balanced equation, 2 moles of Al produce 3 moles of Cu.
- Therefore, 0.0686 mol of Al will produce:
\[
\text{Moles of Cu} = 0.0686 \, \text{mol Al} \times \frac{3 \, \text{mol Cu}}{2 \, \text{mol Al}} = 0.1029 \, \text{mol Cu}
\]

Step 4: Determine the molar mass of Cu.
- Molar mass of Cu:
\[
63.55 \, \text{g/mol}
\]

Step 5: Calculate the theoretical mass of Cu.
\[
\text{Theoretical mass of Cu} = \text{Moles} \times \text{Molar Mass} = 0.1029 \, \text{mol} \times 63.55 \, \text{g/mol} \approx 6.52 \, \text{g}
\]

Step 6: Calculate the actual mass of Cu produced using the percent yield.
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
\]
\[
56.6\% = \left( \frac{\text{Actual Yield}}{6.52 \, \text{g}} \right) \times 100
\]
\[
\text{Actual Yield} = \frac{56.6}{100} \times 6.52 \, \text{g} \approx 3.69 \, \text{g}
\]

Final Answer:
\[
\boxed{3.69 \, \text{g}}
\]

---

#### 4. Percent Yield Calculation for Methane Combustion

Problem Statement:
The combustion of methane (\( CH_4 \)) produces carbon dioxide and water. Assume that 2.0 mol of \( CH_4 \) burned in the presence of excess air. What is the percentage yield if in an experiment the reaction produces 87.0 g of \( CO_2 \)?

Balanced Equation:
\[ CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \]

Step 1: Determine the molar mass of \( CO_2 \).
- Molar mass of \( CO_2 \):
\[
12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \, \text{g/mol}
\]

Step 2: Use stoichiometry to find the theoretical yield of \( CO_2 \).
- From the balanced equation, 1 mole of \( CH_4 \) produces 1 mole of \( CO_2 \).
- Therefore, 2.0 mol of \( CH_4 \) will produce 2.0 mol of \( CO_2 \).

Step 3: Calculate the theoretical mass of \( CO_2 \).
\[
\text{Theoretical mass of } CO_2 = \text{Moles} \times \text{Molar Mass} = 2.0 \, \text{mol} \times 44.01 \, \text{g/mol} = 88.02 \, \text{g}
\]

Step 4: Calculate the percent yield.
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{87.0 \, \text{g}}{88.02 \, \text{g}} \right) \times 100 \approx 98.84\%
\]

Final Answer:
\[
\boxed{98.84\%}
\]

---

#### 5. Percent Yield Calculation for Lithium-Copper Phosphate Reaction

Problem Statement:
15.3 grams of Lithium is dropped into a solution containing excess copper(II) phosphate. When the reaction is completed, 1.25 grams of copper is formed. What is the percent yield?

Balanced Equation:
\[ 2Li + Cu_3(PO_4)_2 \rightarrow 3Cu + Li_3PO_4 \]

Step 1: Determine the molar mass of Li.
- Molar mass of Li:
\[
6.94 \, \text{g/mol}
\]

Step 2: Calculate the moles of Li used.
\[
\text{Moles of Li} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{15.3 \, \text{g}}{6.94 \, \text{g/mol}} \approx 2.20 \, \text{mol}
\]

Step 3: Use stoichiometry to find the theoretical yield of Cu.
- From the balanced equation, 2 moles of Li produce 3 moles of Cu.
- Therefore, 2.20 mol of Li will produce:
\[
\text{Moles of Cu} = 2.20 \, \text{mol Li} \times \frac{3 \, \text{mol Cu}}{2 \, \text{mol Li}} = 3.30 \, \text{mol Cu}
\]

Step 4: Determine the molar mass of Cu.
- Molar mass of Cu:
\[
63.55 \, \text{g/mol}
\]

Step 5: Calculate the theoretical mass of Cu.
\[
\text{Theoretical mass of Cu} = \text{Moles} \times \text{Molar Mass} = 3.30 \, \text{mol} \times 63.55 \, \text{g/mol} \approx 209.7 \, \text{g}
\]

Step 6: Calculate the percent yield.
\[
\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{1.25 \, \text{g}}{209.7 \, \text{g}} \right) \times 100 \approx 0.596\%
\]

Final Answer:
\[
\boxed{0.596\%}
\]

---

Final Answers:

1. \(\boxed{88.51\%}\)
2. \(\boxed{37.30\%}\)
3. \(\boxed{3.69 \, \text{g}}\)
4. \(\boxed{98.84\%}\)
5. \(\boxed{0.596\%}\)