Percent Yield Calculator

Divide what you got by what the equation allowed, or let the second mode work out that allowance for you. Give it a balanced equation and the amounts you actually have, and it identifies the limiting reactant, prices the theoretical yield, tells you how much of every other reactant is left in the flask, and turns your result into a percentage.

Put this calculator on your website for free

Copy one snippet and give your visitors a working Percent Yield Calculator.

How this percent yield calculator works

The first mode is the straightforward one: divide what you got by what the equation allowed and multiply by a hundred. The second mode does the harder half of the job, which is working out what the equation allowed in the first place. Give it a balanced equation and how much of each reactant you actually have, and it identifies the limiting reactant, prices the theoretical yield from it, tells you how much of everything else is left sitting in the flask, and then turns your actual yield into a percentage.

Molar masses come from the formulas you type, summed from the IUPAC standard atomic weights, so there is nothing to look up. The formulas parse the way real ones are written, brackets and hydrates included.

The formula

percent yield = actual ÷ theoretical × 100
moles = grams ÷ molar mass
limiting reactant = the smallest value of (moles ÷ coefficient)
theoretical yield = (moles ÷ coefficient)limiting × coefficientproduct × molar massproduct
atom economy = mass of product ÷ mass of all reactants × 100

The second line is the bridge every stoichiometry problem crosses: a balanced equation counts molecules, a balance measures grams, and moles are the only place those two meet. The third line is the one people get wrong. Moles divided by coefficient is the comparison that matters, not moles and certainly not grams, because a reactant that needs three of itself per reaction runs out three times faster than its mole count suggests.

Worked example

The aspirin synthesis, which is most people's first real lab. Salicylic acid plus acetic anhydride gives aspirin plus acetic acid, all coefficients 1. You weigh out 2.00 g of salicylic acid (C7H6O3, 138.12 g/mol) and add 5.41 g of acetic anhydride (C4H6O3, 102.09 g/mol).

In moles: 0.01448 and 0.05299. Divide each by its coefficient of 1 and the salicylic acid is smaller, so salicylic acid is limiting and the anhydride is in nearly fourfold excess, which is exactly what the protocol intends: it is cheap, it doubles as the solvent, and pushing an equilibrium is what excess reagent is for. Theoretical yield is 0.01448 mol of aspirin (C9H8O4, 180.16 g/mol) = 2.609 g.

Get 1.85 g out of the recrystallisation and that is a 70.9% yield, which is a good day for that reaction. The missing 0.76 g is not destroyed: most of it is still dissolved in the mother liquor you poured away, which is the price of recrystallising to get something pure enough to melt sharply.

Moles divided by coefficient, not moles

Here is the mistake that survives longest, because it usually works. Given two reactants, the instinct is to convert both to moles and pick the smaller one. That is right whenever the coefficients happen to be equal, which covers a great many textbook problems, and it is wrong the moment they are not.

Take 2 H2 + O2 to 2 H2O with 3 moles of hydrogen and 2 moles of oxygen. Oxygen has fewer moles, so the instinct says oxygen is limiting. Divide by the coefficients and you get 3 ÷ 2 = 1.5 for hydrogen and 2 ÷ 1 = 2 for oxygen, so hydrogen runs out first and half a mole of oxygen is left over. The instinct was wrong, and it was wrong in the direction that overestimates the yield, which is the direction that turns into a puzzled afternoon.

The reason is worth holding onto rather than memorising the rule: the coefficient tells you the rate at which a reactant is consumed relative to the others. Something used two at a time depletes twice as fast per mole present. Dividing by the coefficient converts every reactant onto the same scale, which is "how many times can this reaction run before this particular ingredient is gone", and the smallest answer wins.

A yield above 100% is a diagnosis, not a triumph

Atoms are conserved, so a reaction cannot make more product than its limiting reactant contains. If your percent yield comes out above 100, the extra mass is something that is not your product. In an undergraduate lab it is almost always one of three things: solvent that has not evaporated, water the compound has pulled out of the air, or unreacted starting material carried through the workup. A fourth, less common, is salt left behind from an aqueous wash.

The standard fix is drying to constant mass: dry, weigh, dry again, weigh again, and keep going until two consecutive weighings agree. If the mass stops falling and the yield is still above 100%, the impurity is not volatile and the problem is purification rather than drying. That is a genuinely useful diagnostic sequence, and it is why the number being impossible is more informative than a number that merely looks disappointing.

There is one honourable exception worth knowing. If the reported yield is above 100% because the product was weighed as a salt, a hydrate, or with a counterion the theoretical yield did not account for, then the arithmetic is fine and the theoretical yield was calculated for the wrong compound. Recalculate against what you actually isolated. Getting that right is the difference between a puzzling result and a corrected one.

Percent yield and atom economy answer different questions

Percent yield asks how well you ran the reaction. Atom economy asks whether the reaction was worth running. It is the mass of the product you want as a share of the mass of everything you put in, calculated on the balanced equation with a perfect yield assumed, and it counts the atoms that are destined to leave as byproducts no matter how careful you are.

The two can disagree sharply. A Wittig reaction can run at 95% yield and still have an atom economy near 20%, because a large triphenylphosphine oxide molecule departs carrying most of the mass with it. An addition reaction, where everything you put in ends up in the product, can hit 100% atom economy by construction. Neither number is the whole picture: a high-yield, low-economy route may still be the right choice if it is selective and the byproduct is recyclable. But if you have ever wondered why industrial chemistry cares so much about catalysis and addition chemistry, the answer is the second number, and it is the one that shows up on the disposal invoice.

Sources

Where the numbers on this page come from. We go to the body that publishes the figure, not to another calculator. See how we verify.

Frequently asked questions

How do I calculate percent yield?

Divide the actual yield by the theoretical yield and multiply by 100. If the equation allowed 2.609 g of aspirin and you isolated 1.85 g, that is 70.9%. Both numbers have to measure the same substance in the same unit, either both in grams or both in moles. Comparing grams of product against grams of starting material is not a yield, and it is the most common way this calculation goes wrong.

How do I find the limiting reactant?

Convert every reactant to moles, then divide each by its coefficient from the balanced equation. The smallest result is the limiting reactant. Dividing by the coefficient is the step people skip, and skipping it gives the right answer only when the coefficients happen to be equal. With 3 mol of H2 and 2 mol of O2 in 2 H2 + O2 to 2 H2O, oxygen has fewer moles but hydrogen is limiting, because 3 divided by 2 is smaller than 2 divided by 1.

What is theoretical yield?

The amount of product you would get if the reaction went perfectly and nothing was lost. It is set entirely by the limiting reactant: take its moles divided by its coefficient, multiply by the product's coefficient, then by the product's molar mass. Every other reactant is in excess and has no say in the ceiling, which is why identifying the limiting one correctly is the whole game.

Can percent yield be more than 100%?

The arithmetic can, the chemistry cannot. Atoms are conserved, so anything above 100% means the mass you weighed includes something that is not your product: solvent that has not evaporated, water absorbed from the air, unreacted starting material, or salt from a wash. Dry to constant mass, weighing repeatedly until two readings agree. If it still reads high, the impurity is not volatile and you have a purification problem. There is one honest exception: if you isolated a salt or a hydrate, the theoretical yield was calculated for the wrong compound and should be redone.

What is a good percent yield?

It depends entirely on the reaction, so beware of any single number. Above 90% is excellent, 70 to 90% is a good day for most organic synthesis, and below 50% is common for difficult transformations without meaning anyone did anything wrong. What matters more is that losses multiply across a route rather than adding: three steps at 70% each give 34% overall, which is the arithmetic behind why long synthetic routes are judged on their worst step.

Why is the actual yield always lower than the theoretical?

Because a real workup leaks material at every stage. Product stays dissolved in the mother liquor after recrystallisation, clings to filter paper and glassware, and is lost in transfers. Side reactions consume some starting material, and reversible reactions never fully complete. None of it is destroyed; it is simply somewhere other than your weighing boat, which is why a 70% yield on a recrystallised product is often the price of purity rather than a sign of poor technique.

What is atom economy and how is it different from yield?

Percent yield asks how well you ran the reaction; atom economy asks whether the reaction was worth running. It is the mass of the desired product as a share of the mass of everything you put in, assuming a perfect yield, so it counts the atoms destined to leave as byproducts. A Wittig reaction can run at 95% yield with an atom economy near 20% because a large phosphine oxide departs with most of the mass. An addition reaction can be 100% atom economical by construction. The two answer different questions and both are worth knowing.

Do I need a balanced equation to use this?

For the limiting reactant mode, yes: the coefficients are what turn moles into a comparison, and getting them wrong changes which reactant runs out first. For the plain percent yield mode, no, because you are only dividing one number by another. If you have the amounts but not the equation, balance it first; an unbalanced equation will give you a confidently wrong limiting reactant and a theoretical yield to match.

Related calculators