Percent Yield and Limiting Reagent Calculator
Enter the mass, molar mass and balanced-equation coefficient of two reactants and of the product. The calculator picks the limiting reagent and compares your actual yield to the theoretical one.
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Formula
For aA + bB → pP:
- Moles:
n = mass ÷ molar mass - Limiting extent:
ξ = min(nA ÷ a, nB ÷ b); the reactant with the smaller value is the limiting reagent - Theoretical yield:
ξ × p × molar mass of P - Percent yield:
actual yield ÷ theoretical yield × 100 - Excess left over:
(n excess - ξ × coefficient of excess) × molar mass of excess
Worked example
2 H2 + O2 → 2 H2O with 4.00 g H2 (2.016 g/mol), 32.0 g O2 (31.998 g/mol) and 30.0 g of water collected (18.015 g/mol):
nH2 = 1.9841 mol,nO2 = 1.0001 molξ = min(1.9841 ÷ 2, 1.0001 ÷ 1) = 0.9921, so H2 is limiting- Theoretical yield:
0.9921 × 2 × 18.015 =35.744 g - Percent yield:
30.0 ÷ 35.744 × 100 =83.93 % - O2 left over: 0.256 g
Frequently asked questions
How do I calculate percent yield?
Percent yield = actual yield ÷ theoretical yield × 100. If the theoretical yield is 15 g and you collect 12 g, the percent yield is 12 ÷ 15 × 100 = 80 %.
How do I find the limiting reagent?
Divide each reactant's moles by its coefficient; the smaller value limits the reaction. For 2 H2 + O2 → 2 H2O with 4.00 g H2 (1.984 mol) and 32.0 g O2 (1.000 mol): 1.984 ÷ 2 = 0.992 and 1.000 ÷ 1 = 1.000, so H2 is limiting.
How do I get the theoretical yield?
Take the limiting extent (0.992 mol for the H2 example), multiply by the product coefficient (2) and the product's molar mass (18.015 g/mol): 0.992 × 2 × 18.015 = 35.74 g of water. Collecting 30.0 g is a 83.9 % yield.
Why is my percent yield over 100 %?
A yield above 100 % usually means the product is wet or contaminated with solvent or by-products, or a mass was misread. This calculator still shows the number so you can see how far off it is.