Percent yield calculator
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Result
Calculate the percent yield of a reaction. If you already know the theoretical yield, enter it with the actual yield. Otherwise, enter the mass and molar mass of one or two reactants and the coefficients from the balanced equation: the calculator finds the limiting reactant, the theoretical yield of the product, the amount of excess reactant left over and the percent yield.
Step-by-step solution
Moles of each reactant
Moles of product from the mole ratio
Theoretical yield
Percent yield
Key formulas
- Moles
- n = m/M
- Mole ratio
- use the coefficients of the balanced equation
- Limiting reactant
- the one that runs out first
- Percent yield
- actual / theoretical × 100%
How to calculate percent yield
- Balance the chemical equation and note the coefficients of the reactants and the product.
- Convert the mass of each reactant to moles: n = m/M.
- Use the mole ratio to find how many moles of product each reactant could make. The reactant that gives the least product is the limiting reactant.
- Theoretical yield: multiply those moles of product by the molar mass of the product.
- Percent yield = actual yield/theoretical yield × 100%.
Questions and answers
What is the difference between the actual and the theoretical yield?
The theoretical yield is the mass of product the balanced equation predicts when the limiting reactant reacts completely. The actual yield is what you really collect. Losses during transfer, filtration and purification, side reactions and incomplete reactions make the actual yield lower.
Can the percent yield be above 100%?
Not for a pure, dry product. A result above 100% usually means that the product still contains water or solvent, is contaminated, or that a mass was measured or calculated wrongly.
How do I find the limiting reactant?
Work out how much product each reactant could make on its own, using the mole ratio from the balanced equation. The one that makes the least runs out first and is the limiting reactant; the others are in excess.
Do I have to use grams?
The actual and theoretical yield must be in the same unit, usually grams. With molar masses in g/mol, enter the masses of the reactants in grams.
Can you show a worked example?
For 2 Al + 3 Cl₂ → 2 AlCl₃ with 5.00 g Al (26.98 g/mol) and 10.0 g Cl₂ (70.90 g/mol): 0.1853 mol Al could give 0.1853 mol AlCl₃, but 0.1410 mol Cl₂ gives only 0.0940 mol. Cl₂ is limiting, the theoretical yield is 0.0940 × 133.33 = 12.54 g, and an actual yield of 10.5 g is 83.75%.
Practise it yourself
Practise Stoichiometry
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