The formula
- Percent yield
percent yield = (actual yield / theoretical yield) × 100%- Theoretical yield from the limiting reactant
theoretical yield = n(limiting) × (mole ratio) × M(product)- Solved for the actual yield
actual yield = theoretical yield × (percent yield / 100)
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
actual yield | Mass of product actually isolated from the run | g |
theoretical yield | Mass of product the balanced equation allows from the limiting reactant | g |
percent yield | Fraction of the possible product that was recovered | % |
n(limiting) | Moles of the limiting reactant available | mol |
mole ratio | Coefficient of the product over the coefficient of the limiting reactant | dimensionless |
M(product) | Molar mass of the product | g/mol |
When it applies
- The theoretical yield came from the limiting reactant. Basing it on a reactant that is in excess inflates the denominator and deflates the percentage.
- Both yields are the same substance in the same units. Comparing grams of product with moles of reactant is the usual source of a nonsense answer.
- The equation is balanced before the mole ratio is taken, since the ratio is read straight off the coefficients.
- A percent yield above 100% signals a measurement problem, most often product still wet with solvent or contaminated with unreacted starting material.
Worked example
Problem. Ammonia is made by N₂(g) + 3H₂(g) → 2NH₃(g). A run starts with 25.0 g of N₂ and excess hydrogen, and isolates 24.0 g of ammonia. What is the percent yield?
- Hydrogen is in excess, so nitrogen is limiting. Convert it to moles: M(N₂) = 28.014 g/mol, so n = 25.0 / 28.014 = 0.89241 mol.
- Apply the mole ratio from the balanced equation, 2 mol NH₃ per 1 mol N₂: n(NH₃) = 2 × 0.89241 = 1.78482 mol.
- Convert to mass with M(NH₃) = 17.031 g/mol: theoretical yield = 1.78482 × 17.031 = 30.397 g, which is 30.4 g.
- Divide and scale: percent yield = (24.0 / 30.397) × 100% = 78.95%, which is 79.0% to three significant figures.
Answer. A theoretical yield of 30.4 g and a percent yield of 79.0%.
Common mistakes
- Computing the theoretical yield from whichever reactant is mentioned first instead of from the limiting one.
- Dividing theoretical by actual. The actual yield is the numerator, and inverting it turns a 79% run into a 127% impossibility.
- Skipping the mole ratio and converting moles of reactant straight into grams of product, which silently assumes a 1:1 reaction.
- Rounding the intermediate moles hard and carrying that through. Keep the extra digits until the final percentage.
Related formulas
- Empirical formula:
%X = (mass X / mass compound) × 100% - Limiting reactant:
compare n(X) / coefficient(X) across reactants; the smallest value is limiting - Molar mass:
M = Σ (atoms of each element × atomic mass)