Percent Yield and Stoichiometry in Chemical Reactions

Percent Yield and Stoichiometry

Percent Yield

Percent yield is a measure of how efficiently a chemical reaction produces the desired product. It compares the actual amount of product obtained to the theoretical maximum predicted by stoichiometry.

• Definition: Percent yield is the ratio of actual yield to theoretical yield, expressed as a percentage.

• Formula:

$\text{Percent Yield} = \left( \frac{\text{actual yield}}{\text{theoretical yield}} \right) \times 100\%$

• Interpretation: A higher percent yield indicates a more efficient reaction.

• Classification:

  • >90%: Excellent

  • ~80%: Very Good

  • >70%: Good

  • >40%: Poor

Actual Yield vs. Theoretical Yield

The actual yield is the measured amount of product obtained from a reaction, while the theoretical yield is the maximum possible amount calculated from the balanced chemical equation.

• Actual yield is usually less than theoretical yield due to side reactions, incomplete reactions, or loss during recovery.

• Units: Yields are typically measured in grams or moles.

Stoichiometry in Percent Yield Calculations

Stoichiometry is used to determine the theoretical yield based on the balanced chemical equation and the amounts of reactants.

• Map out the stoichiometric relationships between reactants and products.

• Convert given quantities to moles.

• Use mole-to-mole ratios from the balanced equation to find the moles of desired product.

• Convert moles of product to grams if needed.

• Calculate percent yield using the actual and theoretical yields.

Example Calculation

Example: Consider the reaction:

$2\text{C}_4\text{H}_{10} + 15\text{O}_2 \rightarrow 12\text{CO}_2 + 6\text{H}_2\text{O}$

• If 2.6 g of C4H10 reacts with excess O2 to produce 1.25 g of water, calculate the percent yield of water.

1. Map out the stoichiometry for H2O.

2. Convert 2.6 g C4H10 to moles.

3. Use mole-to-mole ratio to find moles of H2O.

4. Convert moles of H2O to grams.

5. Calculate percent yield:

   • Actual yield = 1.25 g H2O

   • Theoretical yield = (calculated from stoichiometry)

   • Percent yield = $\left( \frac{1.25}{\text{theoretical yield}} \right) \times 100\%$

Example Answer: 69.48%

Practice Problems

• Problem 1: What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 45.5 g Zn with 70.1 g CuSO4?

  • Reaction: Zn (s) + CuSO4 (aq) → Cu (s) + ZnSO4 (aq)

  • Percent yield: 69.48%

• Problem 2: Ammonia reacts with hypochlorite ion to produce hydrazine. How many grams of hydrazine are produced from 115.0 g NH3 if the reaction has an 81.51% yield?

  • Reaction: 2 NH3 + OCl- → N2H4 + Cl- + H2O

  • Percent yield: 69.48%

• Problem 3: The reduction of iron(III) oxide creates the following reaction:

  • Reaction: Fe2O3 (s) + 3 H2 (g) → 2 Fe (s) + 3 H2O (g)

  • If the reaction only went to 75% completion, how many moles of Fe2O3 are required to produce 0.850 moles of Fe?

  • Percent yield: 9.48%

Summary Table: Percent Yield Classifications

Percent Yield (%)	Classification
>90	Excellent
~80	Very Good
>70	Good
>40	Poor

Additional info: The notes provide step-by-step guidance for percent yield calculations, including mapping stoichiometry, converting units, and using mole ratios. Practice problems reinforce the concepts with real chemical equations and yield calculations.