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Theoretical Yield Calculator

A yield over 100% means wet product, not a good day.

Work out Theoretical Yield. A yield over 100% means wet product, not a good day. Names the misconception directly.

Written and maintained by Mohit PatelLast checked August 4, 2026How we build these
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Theoretical yield

35.744 g

Assuming the limiting reagent converts completely

Moles of limiting reagent1.9841 mol
Mole ratio, product to reagent2 : 2
Moles of product formed1.9841 mol
Theoretical mass35.744 g

Theoretical yield assumes the limiting reagent converts completely. Real yields fall short through incomplete reaction, competing side products, and losses during transfer and purification.

How the Theoretical Yield Calculator works

Theoretical yield from the limiting reagent through the mole ratio to the product, with percent yield when you supply the actual mass. Every step is shown, because the mole ratio is where the arithmetic usually fails and a single number hides that.

Also known as: how much product will i get · why is my percent yield over 100 · limiting reagent product mass · expected yield from grams

Three conversions, and the one that goes wrong

Every theoretical yield is the same three steps: grams of limiting reagent to moles, moles of reagent to moles of product, moles of product to grams. The first and third are divisions and multiplications by molar mass, and they rarely fail. The middle step is the mole ratio, and that is where the answers go wrong.

The ratio comes from the coefficients of the balanced equation, not from the subscripts inside the formulas. In 2H₂ + O₂ → 2H₂O, the ratio of water to hydrogen is 2:2, or 1:1. The subscript 2 in H₂O says each molecule holds two hydrogens; the coefficient 2 says two molecules form. Confusing them is the single most common stoichiometry error, and it survives into professional work.

Which way up the ratio goes is settled by units. You have moles of reagent and want moles of product, so the factor must be moles of product over moles of reagent — product coefficient on top. Writing the ratio as a fraction with the units attached makes the wrong version look obviously wrong, which is the point of doing it.

Why the limiting reagent is the only one that counts

A reaction stops when one reactant runs out, and everything past that point is spectator. The limiting reagent is that reactant, and the excess reagent — however much of it there is — has no influence on how much product forms. Using the excess reagent for the calculation gives a theoretical yield the reaction cannot reach.

Finding the limiting reagent means dividing each reactant's moles by its coefficient and taking the smallest quotient. With 4 g of hydrogen (1.98 mol) and 4 g of oxygen (0.125 mol) reacting as 2H₂ + O₂, the quotients are 0.99 and 0.125, so oxygen limits — despite the two masses being equal. Mass comparisons are useless here; only the mole-per-coefficient comparison decides it.

In practice, the cheap reagent is usually deliberately in excess to drive the expensive one to completion, so the limiting reagent is normally the one you care about anyway. But that is a synthesis decision, not a rule of arithmetic, and the calculation has to be done rather than assumed.

What percent yield actually measures

Percent yield is actual mass over theoretical mass, times a hundred, and it measures the gap between the equation and the bench. Reactions that do not go to completion, side reactions eating starting material, product left in the mother liquor after recrystallisation, and material lost on every glass surface it touches all subtract from it.

The compounding matters more than any single step. A three-step synthesis running at 80% per step ends at 51% overall; at 90% per step it ends at 73%. That arithmetic is why route design in synthetic chemistry cares more about step count than about any individual step's yield, and why a long elegant route often loses to a short crude one.

A yield above 100% is not a good result, it is a measurement fault. The usual cause is a product that has not been dried to constant weight, so solvent is being weighed as product. Residual starting material, an inorganic salt that failed to wash out, and a mis-recorded starting mass account for most of the rest. The right response is to dry and reweigh, not to report it.

Where to go next

The Theoretical Yield question rarely arrives on its own. These are the ones that usually come with it:

Frequently asked questions

How is theoretical yield calculated?

Convert the limiting reagent to moles, multiply by the mole ratio of product to reagent from the balanced equation, then multiply by the product's molar mass. Only the limiting reagent matters; the excess reagent has no say in the answer.

Why is my percent yield above 100%?

The product is not dry, or it contains unreacted starting material or solvent. Yields above 100% are not chemically possible, so treat one as a signal to dry the sample to constant weight and reweigh it.

What is a good percent yield?

It depends on the reaction. Above 90% is excellent for a simple single-step reaction; a multi-step synthesis where each step runs at 80% ends at 51% after three steps, and that is normal.

Why is actual yield always lower than theoretical?

Reactions rarely go to completion, side reactions consume starting material, and product is lost in every transfer, filtration and recrystallisation. The theoretical figure assumes none of that happens.

Do I use the limiting or the excess reagent?

The limiting reagent, always. It runs out first and stops the reaction, so it caps the product no matter how much of the other reagent is present.

Does the mole ratio use the coefficients or the subscripts?

The coefficients from the balanced equation. Subscripts describe what is inside one molecule; coefficients describe how many molecules react, and the ratio between coefficients is what scales the moles.

How do I identify the limiting reagent?

Divide each reactant's moles by its coefficient in the balanced equation and take the smallest result. Comparing masses does not work, and comparing moles without dividing by the coefficients does not work either.

What is atom economy and how is it different?

Percent yield measures how much of the theoretical product you actually recovered. Atom economy measures how much of the reactant mass ends up in the desired product rather than in by-products, and it is fixed by the equation before any experiment happens.

Why does my yield drop with a smaller scale?

Losses that are fixed rather than proportional — film left on glassware, product held in a filter cake, material lost to a transfer — take a much larger share of a 100 mg reaction than of a 100 g one. Milligram-scale chemistry routinely reports lower yields for this reason alone.

Can I use volume instead of mass for a liquid reagent?

Yes, if you convert with the density first. Volume times density gives mass, then mass over molar mass gives moles. Densities are temperature-dependent, so a value measured at 20 °C is slightly off in a warm laboratory.

Does an excess reagent affect the yield at all?

Not the theoretical yield, which depends only on the limiting reagent. It can affect the actual yield by driving an equilibrium reaction further towards products, which is exactly why the cheap reagent is usually the one in excess.

Should percent yield be reported on crude or purified product?

On purified product, and the purification method should be stated with it. A crude yield includes solvent and impurities and is not comparable to anything, which is why it is rarely worth quoting on its own.

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