Percent Yield Calculations Practice Problems

Mastering Percent Yield Calculations: Practice Problems and Solutions

Percent yield is a crucial concept in chemistry, providing a measure of the efficiency of a chemical reaction. It represents the ratio of the actual yield (the amount of product obtained experimentally) to the theoretical yield (the amount of product expected based on stoichiometry), expressed as a percentage. Understanding percent yield calculations is vital for any aspiring chemist, allowing for analysis of experimental procedures and optimization of reaction conditions. This comprehensive guide provides a detailed explanation of percent yield, along with a range of practice problems of varying difficulty, complete with step-by-step solutions. Mastering these problems will solidify your understanding of this essential chemical concept.

Understanding Percent Yield

Before diving into the practice problems, let's reinforce the fundamental concepts. The percent yield is calculated using the following formula:

Percent Yield = (Actual Yield / Theoretical Yield) x 100%

• Actual Yield: This is the actual amount of product obtained from a chemical reaction. It's determined experimentally through techniques like weighing or titration. It's always less than or equal to the theoretical yield.

• Theoretical Yield: This is the maximum amount of product that could be obtained if the reaction proceeded perfectly with 100% efficiency. It is calculated using stoichiometry, based on the balanced chemical equation and the limiting reactant.

The difference between the actual and theoretical yield reflects the losses encountered during the reaction, often due to incomplete reactions, side reactions, or losses during purification. A high percent yield (close to 100%) indicates an efficient reaction, while a low percent yield suggests areas for improvement in the experimental procedure.

Practice Problems: Beginner Level

Let's start with some simpler problems to build your confidence.

Problem 1:

In a synthesis of aspirin (C₉H₈O₄), 2.00 g of salicylic acid (C₇H₆O₃) reacted with excess acetic anhydride. The actual yield of aspirin was 2.50 g. The balanced equation for the reaction is:

C₇H₆O₃(s) + C₄H₆O₃(l) → C₉H₈O₄(s) + CH₃COOH(l)

Calculate the percent yield of aspirin.

Solution 1:

1. Calculate the molar mass of salicylic acid (C₇H₆O₃) and aspirin (C₉H₈O₄):

   • Molar mass of C₇H₆O₃ = (7 x 12.01 g/mol) + (6 x 1.01 g/mol) + (3 x 16.00 g/mol) = 138.12 g/mol
   • Molar mass of C₉H₈O₄ = (9 x 12.01 g/mol) + (8 x 1.01 g/mol) + (4 x 16.00 g/mol) = 180.16 g/mol

2. Calculate the moles of salicylic acid:

   • Moles of C₇H₆O₃ = (2.00 g) / (138.12 g/mol) = 0.0145 mol

3. Determine the theoretical yield of aspirin using stoichiometry:

   • From the balanced equation, 1 mole of salicylic acid produces 1 mole of aspirin.
   • Therefore, 0.0145 mol of salicylic acid will produce 0.0145 mol of aspirin.
   • Theoretical yield of C₉H₈O₄ = (0.0145 mol) x (180.16 g/mol) = 2.61 g

4. Calculate the percent yield:

   • Percent yield = (Actual yield / Theoretical yield) x 100% = (2.50 g / 2.61 g) x 100% = 95.8%

Problem 2:

The reaction of 5.00 g of copper(II) oxide (CuO) with excess hydrogen gas (H₂) produced 3.92 g of copper (Cu). The balanced equation is:

CuO(s) + H₂(g) → Cu(s) + H₂O(l)

What is the percent yield of copper?

Solution 2:

1. Calculate molar masses:

   • Molar mass of CuO = 63.55 g/mol + 16.00 g/mol = 79.55 g/mol
   • Molar mass of Cu = 63.55 g/mol

2. Calculate moles of CuO:

   • Moles of CuO = 5.00 g / 79.55 g/mol = 0.0628 mol

3. Determine theoretical yield of Cu:

   • From the equation, 1 mole of CuO produces 1 mole of Cu.
   • Theoretical yield of Cu = 0.0628 mol x 63.55 g/mol = 3.99 g

4. Calculate percent yield:

   • Percent yield = (3.92 g / 3.99 g) x 100% = 98.2%

Practice Problems: Intermediate Level

These problems incorporate more complex stoichiometric calculations.

Problem 3:

Iron(III) oxide (Fe₂O₃) reacts with carbon monoxide (CO) to produce iron (Fe) and carbon dioxide (CO₂). If 10.0 g of Fe₂O₃ reacts with excess CO, and 5.00 g of Fe is obtained, what is the percent yield of iron? The balanced equation is:

Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)

Solution 3:

1. Calculate molar masses:

   • Molar mass of Fe₂O₃ = (2 x 55.85 g/mol) + (3 x 16.00 g/mol) = 159.70 g/mol
   • Molar mass of Fe = 55.85 g/mol

2. Calculate moles of Fe₂O₃:

   • Moles of Fe₂O₃ = 10.0 g / 159.70 g/mol = 0.0626 mol

3. Determine theoretical yield of Fe:

   • From the equation, 1 mole of Fe₂O₃ produces 2 moles of Fe.
   • Moles of Fe = 0.0626 mol Fe₂O₃ x (2 mol Fe / 1 mol Fe₂O₃) = 0.125 mol Fe
   • Theoretical yield of Fe = 0.125 mol x 55.85 g/mol = 6.98 g

4. Calculate percent yield:

   • Percent yield = (5.00 g / 6.98 g) x 100% = 71.6%

Problem 4:

The reaction of 15.0 g of methane (CH₄) with excess oxygen (O₂) produced 25.0 g of carbon dioxide (CO₂). The balanced equation is:

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)

What was the percent yield of carbon dioxide?

Solution 4:

1. Calculate molar masses:

   • Molar mass of CH₄ = 12.01 g/mol + (4 x 1.01 g/mol) = 16.05 g/mol
   • Molar mass of CO₂ = 12.01 g/mol + (2 x 16.00 g/mol) = 44.01 g/mol

2. Calculate moles of CH₄:

   • Moles of CH₄ = 15.0 g / 16.05 g/mol = 0.935 mol

3. Determine theoretical yield of CO₂:

   • From the equation, 1 mole of CH₄ produces 1 mole of CO₂.
   • Theoretical yield of CO₂ = 0.935 mol x 44.01 g/mol = 41.1 g

4. Calculate percent yield:

   • Percent yield = (25.0 g / 41.1 g) x 100% = 60.8%

Practice Problems: Advanced Level

These problems require a deeper understanding of limiting reactants and stoichiometry.

Problem 5:

20.0 g of sodium chloride (NaCl) reacts with 25.0 g of silver nitrate (AgNO₃) to produce silver chloride (AgCl) and sodium nitrate (NaNO₃). If 20.5 g of AgCl was obtained, what is the percent yield? The balanced equation is:

NaCl(aq) + AgNO₃(aq) → AgCl(s) + NaNO₃(aq)

Solution 5:

1. Calculate molar masses:

   • Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
   • Molar mass of AgNO₃ = 107.87 g/mol + 14.01 g/mol + (3 x 16.00 g/mol) = 169.88 g/mol
   • Molar mass of AgCl = 107.87 g/mol + 35.45 g/mol = 143.32 g/mol

2. Determine the limiting reactant:

   • Moles of NaCl = 20.0 g / 58.44 g/mol = 0.342 mol

   • Moles of AgNO₃ = 25.0 g / 169.88 g/mol = 0.147 mol

   • Since the stoichiometric ratio is 1:1, AgNO₃ is the limiting reactant.

3. Determine theoretical yield of AgCl:

   • Moles of AgCl = 0.147 mol AgNO₃ x (1 mol AgCl / 1 mol AgNO₃) = 0.147 mol AgCl
   • Theoretical yield of AgCl = 0.147 mol x 143.32 g/mol = 21.0 g

4. Calculate percent yield:

   • Percent yield = (20.5 g / 21.0 g) x 100% = 97.6%

Problem 6:

A student performs an experiment to synthesize ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). They start with 10.0 g of N₂ and 5.00 g of H₂. The actual yield of ammonia is 6.00 g. The balanced equation is:

N₂(g) + 3H₂(g) → 2NH₃(g)

What is the percent yield of ammonia?

Solution 6:

1. Calculate molar masses:

   • Molar mass of N₂ = (2 x 14.01 g/mol) = 28.02 g/mol
   • Molar mass of H₂ = (2 x 1.01 g/mol) = 2.02 g/mol
   • Molar mass of NH₃ = 14.01 g/mol + (3 x 1.01 g/mol) = 17.04 g/mol

2. Determine the limiting reactant:

   • Moles of N₂ = 10.0 g / 28.02 g/mol = 0.357 mol

   • Moles of H₂ = 5.00 g / 2.02 g/mol = 2.48 mol

   • From the balanced equation, 1 mole of N₂ requires 3 moles of H₂. 0.357 mol of N₂ would require 1.07 mol of H₂. Since there is more than enough H₂, N₂ is the limiting reactant.

3. Determine theoretical yield of NH₃:

   • Moles of NH₃ = 0.357 mol N₂ x (2 mol NH₃ / 1 mol N₂) = 0.714 mol NH₃
   • Theoretical yield of NH₃ = 0.714 mol x 17.04 g/mol = 12.2 g

4. Calculate percent yield:

   • Percent yield = (6.00 g / 12.2 g) x 100% = 49.2%

Frequently Asked Questions (FAQ)

Q: What factors can cause a low percent yield?

A: Several factors can contribute to a low percent yield, including: incomplete reactions, side reactions forming unwanted byproducts, loss of product during purification or transfer, experimental errors in measurements, and the presence of impurities in the starting materials.

Q: Can the percent yield ever be greater than 100%?

A: While theoretically possible, a percent yield greater than 100% usually indicates errors in the experimental procedure. This could be due to the presence of impurities in the product that increase its measured mass, or inaccuracies in measuring the reactants or products.

Q: How can I improve my percent yield in experiments?

A: Improving percent yield often involves careful optimization of reaction conditions, including temperature, pressure, reaction time, and the use of appropriate catalysts. Careful purification techniques are also crucial to minimize product loss.

Conclusion

Mastering percent yield calculations is a fundamental skill for any chemist. By understanding the concept and practicing with a range of problems, from beginner to advanced levels, you can confidently analyze experimental results and evaluate the efficiency of chemical reactions. Remember that a high percent yield reflects not only accurate calculations but also meticulous experimental technique. Consistent practice is key to developing proficiency in these calculations and strengthening your overall understanding of stoichiometry and chemical reactions. Continue practicing and refine your skills, and you'll find yourself increasingly adept at tackling more complex chemical problems.