Percent Yield Calculator

Percent yield = (actual yield ÷ theoretical yield) × 100

Y

The percent yield itself: what you got as a fraction of what the reaction could have given you, written as a percentage. It is a ratio of two amounts of the same substance, so it has no unit

m(actual)

The actual yield: the mass of purified product you weighed out, in grams. This is the number from the balance at the end of the work-up

m(theoretical)

The theoretical yield: the mass the balanced equation says is possible, worked out from the limiting reagent. It is never measured, only calculated

Use it at the end of a synthesis, once you have weighed the product and know what the formula predicted. The formula is only a division, so the page earns its place by filling in whichever of the three numbers you are missing, and by catching the case where two of them contradict a third. It is also the page for a quick sanity check on a recipe: a literature yield that claims 95% over four steps is claiming more than 0.95 to the fourth power, which the table below makes visible. If you need the theoretical yield itself, that comes from the limiting reagent and the mole ratio, on its own page.

1. A synthesis that gave back most of what it promised

   1. Divide the actual yield by the theoretical yield: 8.5 ÷ 10 = 0.85
   2. Multiply by 100 to turn the fraction into a percentage: 0.85 × 100 = 85%
   3. Subtract to see what stayed behind in the flask: 10 − 8.5 = 1.5 g
   4. The shortfall row reports 1.5 g, which is the number to check against your weighing records

   This is the opening state of the page, and 85% is an ordinary preparative yield — good enough to publish, far from quantitative. Notice that the shortfall row is the useful one when the reaction is on a small scale: 15% of a gram is hard to picture, while 1.5 grams is something you can look for in the filtrate.

2. Working backwards from a yield target

   1. The percentage and one mass are given, so the missing mass is the thing to find
   2. Turn 92% back into a fraction: 92 ÷ 100 = 0.92
   3. Multiply by the theoretical yield: 0.92 × 5 = 4.6 g
   4. The shortfall follows: 5 − 4.6 = 0.4 g
   5. Only two of the three fields are filled in, and the page does not compare anything

   Only the theoretical yield and the percentage are entered here, and that is a complete question: two of the three quantities always pin the third. This is the direction a problem sheet usually asks it in, and it is also how a lab sets a specification — a 92% recovery is a target, and the mass that should be on the balance is what the target means.

3. A yield above 100%, where the shortfall goes negative

   1. Divide: 1.53 ÷ 1.5 = 1.02
   2. As a percentage that is 102% — above 100, and the page reports it rather than refusing
   3. Subtract anyway: 1.5 − 1.53 = −0.03 g
   4. The minus sign is not an overflow: it says you weighed out more than the equation allows

   A yield over 100% does not break the arithmetic, and the page deliberately does not clamp it — the yield field accepts up to 1000%. In practice it means the product is not dry, or the sample carries unreacted starting material or a side product. The negative shortfall is the honest way to print that: you have 0.03 grams more than the theory accounts for, and the explanation is in the sample, not in the equation.

Both inputs are masses of the product, so a problem that gives you a reagent's mass has to be converted to product first — that is the theoretical yield calculation, not this one. Because the two numbers are masses of the same substance, the percentage is independent of the unit; a ratio taken in grams equals the same ratio taken in moles. A theoretical yield is only as good as the stoichiometry behind it, so a wrong limiting reagent or a wrong mole ratio propagates straight through and the percentage looks perfectly reasonable. The one-percent agreement check catches contradictions between the three fields, not errors inside a single one. And a high yield is not a purity claim: a sample can be 102% of theory and still be the wrong compound.

How do I calculate percent yield?

Divide the actual yield by the theoretical yield and multiply by 100. Weighing 8.5 grams of product where the equation allowed 10 grams gives 8.5 ÷ 10 = 0.85, which is 85%. The theoretical yield is never measured — it comes from the limiting reagent and the balanced equation. If you have the percentage and one of the two masses instead, this page works out the missing mass for you, so the same division covers all three directions.

Why is my percent yield more than 100%?

Because the product weighs more than the equation says it can. The usual cause is that it is not dry: residual solvent or water is being weighed as product. The other cause is contamination — unreacted starting material, a side product, or a desiccant from the drying step. It is almost never a genuine surplus, since a reaction cannot make more product than its limiting reagent allows. Dry the sample and weigh it again before reading anything into a number above 100%.

What is the difference between theoretical and actual yield?

The theoretical yield is what the stoichiometry predicts: a calculation from the limiting reagent, its molar mass and the balanced equation. The actual yield is what you have in your hand after the work-up, from the balance. The theoretical yield is therefore an upper bound, and the actual yield is always at or below it in a clean experiment. Keeping the two apart matters because they are wrong for different reasons — one for a bad equation, the other for a bad day at the bench.

Do both masses have to be in the same unit?

They have to be the same substance, and the percentage is the same whichever mass unit you use, because it is a ratio of two like quantities. Each field has its own gram, milligram or kilogram selector, so you can enter 10 grams against 8500 milligrams and the page will convert before dividing. What you cannot do is mix substances: the yield of a reaction compares two masses of the product, never the mass of a reagent against the mass of a product.

How do I work out the overall yield of several steps?

Multiply the per-step yields together, not average them. Three steps at 90% each give 0.9 × 0.9 × 0.9 = 0.729, or 72.9% overall, and ten steps at 90% leave only 34.87%. That decay is why a long synthesis with an impressive yield at every stage can still end with a disappointing amount of material, and it is the reason the table on this page lists several step counts side by side rather than one number.

• SI Units — Amount of Substance: the mole — National Institute of Standards and Technology (NIST)
• Standard Atomic Weights — abridged to four significant figures — Commission on Isotopic Abundances and Atomic Weights (CIAAW), IUPAC
• IUPAC Gold Book — amount of substance — International Union of Pure and Applied Chemistry (IUPAC)