Percent Yield in Chemical Reactions: Concepts, Calculations, and Practice

Quantities in Chemical Reactions

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction, indicating how much product is actually obtained compared to the theoretical maximum possible. It is a key concept in stoichiometry and laboratory chemistry.

• Definition: Percent yield determines how successful a chemist is in creating their desired product.

• Formula:

$\text{Percent Yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$

• Actual Yield: The amount of pure product actually obtained from the experiment or reaction in a laboratory.

• Theoretical Yield: The maximum amount of product that could be formed from the given amounts of reactants, calculated using stoichiometry.

• Percent Yield Interpretation:

  • 90-99%: Excellent

  • 80-89%: Very Good

  • 70-79%: Good

  • 60-69%: Fair

  • <60%: Poor

Calculating Percent Yield: Step-by-Step

1. Map out the problem: Write the balanced chemical equation and identify the given and unknown quantities.

2. Convert the given quantity to moles of the given substance.

3. Use mole ratios from the balanced equation to convert moles of the given substance to moles of the desired (unknown) substance.

4. Convert moles of the unknown to grams (if necessary) to find the theoretical yield.

5. Calculate percent yield using the actual and theoretical yields in the formula above.

Example Problem

Consider the reaction:

$2\ \text{Ca}_3(\text{PO}_4)_2\ (s) + 6\ \text{SiO}_2\ (s) + 10\ \text{C}\ (s) \rightarrow 6\ \text{CaSiO}_3\ (s) + \text{P}_4\ (g) + 10\ \text{CO}\ (g)$

If 2.49 g of Ca3(PO4)2 reacts to give 1.25 g of P4, what is the percent yield of P4?

1. Convert 2.49 g Ca3(PO4)2 to moles: $2.49\ \text{g} \times \frac{1\ \text{mol}}{310.18\ \text{g}} = 0.00803\ \text{mol}$

2. Use the mole ratio to find moles of P4: $0.00803\ \text{mol} \times \frac{1\ \text{mol P}_4}{2\ \text{mol Ca}_3(\text{PO}_4)_2} = 0.00402\ \text{mol P}_4$

3. Convert moles of P4 to grams: $0.00402\ \text{mol} \times 123.88\ \text{g/mol} = 0.498\ \text{g}$

4. Calculate percent yield: $\text{Percent Yield} = \frac{1.25\ \text{g}}{0.498\ \text{g}} \times 100\% = 251\%$ Note: This value is over 100%, which suggests a calculation or experimental error, or that the actual yield was measured incorrectly.

Practice Problems

• Problem 1: What is the percent yield for a reaction in which 22.1 g Cu is isolated by reacting 4.50 g Zn with 70.1 g CuSO4? Equation: $\text{Zn} (s) + \text{CuSO}_4 (aq) \rightarrow \text{Cu} (s) + \text{ZnSO}_4 (aq)$ Answer: 63.1%

• Problem 2: Ammonia (NH3) reacts with hydrazine (N2H4) to produce hydrazine (N2H4). How many grams of hydrazine are produced from 119.0 g NH3 if the reaction has an 81.5% yield? Equation: $\text{NH}_3 + \text{OCl}^- \rightarrow \text{N}_2\text{H}_4 + \text{Cl}^- + \text{H}_2\text{O}$ Answer: 52.2 g

• Problem 3: The reduction of iron(III) oxide creates the following reaction: $\text{Fe}_2\text{O}_3 (s) + 3\text{H}_2 (g) \rightarrow 2\text{Fe} (s) + 3\text{H}_2\text{O} (g)$ If the above reaction only went to 77% completion, how many moles of Fe2O3 were required to produce 8.55 mol Fe? Answer: 2.27 mol

Summary Table: Key Terms and Formulas

Term	Definition	Formula
Actual Yield	Amount of product actually obtained from a reaction	Measured experimentally
Theoretical Yield	Maximum possible amount of product, calculated from stoichiometry	Calculated from balanced equation
Percent Yield	Efficiency of a reaction, comparing actual to theoretical yield	$\text{Percent Yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%$