Percent Yield Calculator (Actual ÷ Theoretical Yield)
Percent yield of a reaction from the actual and theoretical yields, with the amount not recovered.
Inputs
- Actual yield0.392(77%)
- Loss0.115(23%)
- Working
- 0.392 ÷ 0.5072 × 100 = 77.29%
- Actual yield
- 0.392 g
- Theoretical yield
- 0.5072 g
- Not recovered
- 0.1152 g
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What each Percent Yield input means
- Actual yield
- The mass of product you actually recovered, dried and weighed, in grams.
- Theoretical yield
- The mass the balanced equation predicts from the limiting reagent, in grams.
Percent Yield formula
percent yield = actual yield ÷ theoretical yield × 100% · theoretical yield = the mass the limiting reactant could produce by stoichiometry · not recovered = theoretical − actual
The theoretical yield is the most product a reaction could give if every molecule of the limiting reactant converted and every molecule of product was collected; it comes from the balanced equation and the moles of the limiting reactant. The actual yield is what was weighed at the end. Their ratio, as a percentage, is the percent yield — the number that says how well the reaction and the work-up went, since side reactions, incomplete conversion and losses in transfer and filtration all reduce it. Both yields must be in the same unit, usually grams of the same product. The page divides, shows the working, and reports the mass that did not make it into the flask.
Percent Yield worked example
OpenStax's copper example: 1.274 g of copper sulfate with excess zinc could produce 0.5072 g of copper by stoichiometry, and 0.392 g was actually obtained. Percent yield = 0.392 ÷ 0.5072 × 100 = 77.3%. The 0.115 g not recovered is copper still in solution, lost in filtering, or never formed.
More Percent Yield examples
A synthesis predicted to give 50 g yields 42 g: 42 ÷ 50 × 100 = 84.00%, with 8 g not recovered.
A product weighed at 5.2 g against a theoretical 4.8 g: 108.33%. A yield above 100% almost always means the product is still wet or impure — dry it to constant mass and weigh again.
Where the Percent Yield calculator can give the wrong answer
- The theoretical yield must come from the limiting reactant. Using the reactant in excess overstates it and understates the percent yield.
- Over 100% means the product is not pure — wet, contaminated with reactant or a by-product, or weighed with the filter paper. The page flags it rather than calling it a good result.
- Both yields in the same unit: grams with grams, or moles with moles. Grams of product against moles of reactant is a stoichiometry error, not a yield.
- Percent yield is a property of one run, not of the reaction. A published 'literature yield' is what a careful chemist got once; yours will differ.
- Recovered starting material is not product. Yield counts the product only, unless the report says 'based on recovered starting material'.
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Percent Yield FAQ
- Divide what you obtained by what the stoichiometry says you could have obtained, and multiply by 100. 8.2 g of product against a theoretical 10.0 g is 82%. Enter both masses and the page shows the division and the shortfall.
- From the balanced equation: convert the limiting reactant's mass to moles, multiply by the mole ratio of product to that reactant, and convert back to grams with the product's molar mass. 2 mol of H₂ with excess O₂ gives 2 mol of H₂O = 36.03 g. The molar-mass calculator does the conversions.
- It depends on the reaction. Simple precipitations and acid–base reactions often give over 90%; multi-step organic syntheses are happy with 60–80% per step, and some difficult steps are run at 20–30%. Compare with the literature for that reaction, not a universal bar.
- Not truly. A figure above 100% means the weighed product includes something that is not product — solvent, water, unreacted starting material or a by-product — or the theoretical yield was worked out from the wrong reactant. Dry the product and recheck the limiting reactant.
- Convert each reactant to moles and compare with the equation's ratio. For 2H₂ + O₂ → 2H₂O with 10 g of H₂ (4.96 mol) and 64 g of O₂ (2.00 mol): 2 mol of O₂ needs 4 mol of H₂, which is available, so O₂ runs out first. It limits the product to 4 mol of water, 72.06 g — the theoretical yield.
- Some product is lost in transfers and filtering, some reactant takes side reactions, and some reactions stop at equilibrium short of completion. OpenStax's copper example lost 0.115 g of a possible 0.507 g to such steps.
- The theoretical yield is what the balanced equation says the limiting reagent can make; the actual yield is what you recover and weigh. Their ratio, times 100, is the percent yield.