What is the formula for percentage yield?

Actual yield divided by theoretical yield, times a hundred. Both masses must be of the same substance and in the same units, and the actual yield should be of dried, purified product — weighing a damp sample is the commonest way to produce a nonsense figure. The formula itself is trivial; almost all the difficulty in a percent yield question lives in working out the theoretical value it divides by.

How do I calculate the theoretical yield?

Four steps from the limiting reagent. Convert its mass to moles by dividing by its molar mass. Divide by its coefficient in the balanced equation to get moles of reaction. Multiply by the product coefficient to get moles of product. Multiply by the product molar mass to get grams. Written out: (mass ÷ molar mass) × (product coefficient ÷ reactant coefficient) × product molar mass. It is the ceiling the equation permits, not an estimate of what a good day looks like.

How do I find the limiting reagent?

Convert each reactant to moles, divide each by its own coefficient in the balanced equation, and compare. The smallest result is the limiting reagent. What does not work is comparing masses, and what also does not work — though it is widely taught as a shortcut — is comparing moles directly. Ten grams of a reactant at molar mass 100 with coefficient 1 gives 0.100 mol of reaction; six grams at molar mass 50 with coefficient 3 gives 0.040. The second has more moles and still runs out first.

Can percentage yield be more than 100%?

The number can be, but the chemistry cannot. Mass is conserved, so a reaction cannot produce more product than its equation allows — a figure above 100% always means the material on the balance is not all product. Residual solvent is the usual explanation, especially after recrystallisation from something with a high boiling point. Unreacted starting material, inorganic salts carried through the workup, and absorbed water account for most of the rest. Dry to constant mass and weigh again.

Why is my yield so low?

Usually the workup rather than the reaction. Material stays behind in the flask, dissolves in the mother liquor and is poured away, clings to filter paper, or is lost at each transfer between vessels. Recrystallisation is the biggest single contributor and it is deliberate — it trades yield for purity, which is often the right trade. Beyond that, a competing side reaction consumes starting material, and an equilibrium that does not lie far enough to the right caps the conversion regardless of technique.

What counts as a good percentage yield?

It depends entirely on the reaction, and the number alone says little. Above 90% is excellent for a simple, well-behaved transformation; 70 to 90% is a good result for most preparative work; below 40% would be poor for a straightforward reaction and perfectly respectable for a difficult one. The comparison worth making is against the literature value for the same reaction rather than against 100%. And 65% of pure material beats 85% of something that still needs purifying.

What is the difference between percentage yield and atom economy?

Percent yield measures how well the reaction was executed; atom economy measures how good the reaction is in principle. Atom economy is the mass of desired product divided by the total mass of all reactants, and it is fixed the moment the equation is balanced — before anything is weighed out. Aspirin from salicylic acid and acetic anhydride has an atom economy of 75%, so a perfect 100% yield would still send a quarter of the reactant mass out as acetic acid. Only a different route improves atom economy; no amount of care at the bench will.

How do yields combine across several steps?

They multiply, and the result is far worse than intuition suggests. Five steps at 80% each give 0.8 to the fifth power, which is 32.8% overall — not the 80% the individual figures might imply. Ten steps at 90% each leave only 34.9%. This is why the step count of a synthesis is reported as prominently as the yields, and why a shorter route at lower per-step yield often beats a longer one at higher yields. Averaging the step yields is meaningless; the product is what matters.

Should I work in moles or grams?

Either, as long as you are consistent — percent yield is a ratio so the units cancel. Working in moles is often cleaner because the mole ratio from the equation applies directly and there is one fewer conversion to get wrong. Working in grams matches what the balance actually reads. What you must not do is divide a mass by a mole count, which is a surprisingly common slip and produces a number with units of grams per mole rather than a percentage.

Does the excess reactant affect the calculation at all?

Not the theoretical yield, no. Once the limiting reagent is identified, everything downstream depends on it alone — doubling the excess reagent changes nothing about how much product the equation permits. It can affect the actual yield, by pushing an equilibrium further towards the product or by driving a slow reaction closer to completion, which is exactly why chemists add an excess in the first place. But it never raises the ceiling, only how close you get to it.

Percentage Yield Calculator

Percent yield is the actual mass you isolated divided by the mass the balanced equation allows, times a hundred. That division takes a second. The theoretical yield underneath it is where the work is, and where nearly every wrong answer comes from. Getting it right means finding the limiting reagen...

percent yield = actual ÷ theoretical × 100

That division is the easy part. The theoretical yield is where the work is — it comes from the limiting reagent, the mole ratio in the balanced equation, and two molar masses. This page will compute it, or take it directly if you already have it.

THE REACTION

REACTANT A

coefficient in the balanced equation

needed to find the limiting reagent

coefficient

PRODUCT

MOLAR MASS (G/MOL)

COEFFICIENT

ACTUAL YIELD (G) — WHAT YOU ISOLATED

PERCENTAGE YIELD

80.5%

2.100 g recovered against a theoretical 2.609 g, set by salicylic acid as the limiting reagent. 0.5087 g did not survive the reaction and workup.

THEORETICAL

2.609 g

ACTUAL

2.100 g

LOST

0.5087 g

The working, in full. 2.000 g of salicylic acid ÷ 138.12 g/mol = 0.014480 mol. Divided by its coefficient of 1 that is 0.014480 mol of reaction, and multiplied by the product coefficient of 1 gives 0.014480 mol of product. At 180.16 g/mol that is 2.609 g — the most the balanced equation permits.

Atom economy for this equation is 75.0%. That is a different measure and it answers a different question: not how well the reaction was run, but how much of the reactant mass the equation even allows into the product. It is fixed the moment the equation is balanced. Even a flawless 100% yield here would put 25.0% of the reactant mass into by-products, and no technique at the bench can improve on that — only a different route can.

Where the missing mass usually goes. A yield below 100% rarely means the reaction stopped early. Losses accumulate through the workup: material left in the reaction flask, product dissolved in the mother liquor after recrystallisation, transfer losses on filter paper and glassware, and a competing side reaction consuming starting material. Recrystallisation in particular trades yield for purity by design, and a chemist reporting 65% after two recrystallisations may have run a better reaction than one reporting 85% crude.

THE YIELD

The bar is a ceiling, not a target. It is the mass the balanced equation permits from the limiting reagent, which no reaction can exceed and almost none reaches. The empty portion is what the workup took.

Live calculation · updates as you type

Created with❤️byeaglecalculator.com

HOW TO USE

1
Enter the mass and molar mass of each reactant, along with its coefficient from the balanced equation. The coefficients matter — they are what turn moles of reactant into moles of reaction.
2
Add the second reactant if you have one. Without it there is nothing to compare, so the first is limiting by default and no atom economy can be calculated.
3
Give the product its molar mass and coefficient. The page then works out which reactant runs out first and how much product the equation permits.
4
Enter what you actually isolated, dried to constant mass. If the figure comes out above 100%, the sample is not dry or not pure — that is what the number is telling you.

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Last updated: August 2, 2026 · actual ÷ theoretical × 100, with the theoretical yield from the limiting reagent · Above 100% means impurity, not success.