Percent Yield Calculator
A 5.10 g product against a theoretical 5.603 g is a percent yield of 91.02 percent. Divide the actual mass by the theoretical mass and multiply by 100, where the theoretical mass comes from the limiting reactant and the mole ratio.
| Percent yield | |
| Moles of limiting reactant | |
| Moles of product the equation allows | |
| Theoretical yield | |
| Actual yield | |
| Short of theoretical by | |
Molar masses are yours to enter, so any balanced equation works. Values are kept to four figures until the last step, which is why a small difference in the theoretical yield can move the percent yield. All of the work happens in your browser.
Enter the limiting reactant's mass and molar mass to get the theoretical yield, then add the mass you collected and the percent yield falls out.
Key takeaways
- Percent yield = actual yield / theoretical yield x 100, so 5.10 / 5.603 = 91.02 percent
- The theoretical yield comes from the limiting reactant, never from the reactant in excess
- 10.0 g of CaCO3 gives 0.09991 mol, which is 0.09991 x 56.08 = 5.603 g of CaO
- A result above 100 percent means wet product, an impurity or a weighing error
- Percent yield, percent recovery and atom economy answer three different questions
On this page
The percent yield formula
Percent yield compares what a reaction produced in practice with what the balanced equation said it could produce. Divide the actual mass by the theoretical mass and multiply by 100. The units cancel as long as both masses use the same one, so grams over grams and moles over moles both work.
Two numbers have to be settled before the division is possible. The theoretical yield is the mass of product that the limiting reactant could make if every molecule of it reacted and nothing was lost. The actual yield is the mass you weighed, scraped out of the flask and dried.
The IUPAC Gold Book defines chemical yield as the fraction of the amount of a compound obtained after a specified reaction or separation, which is the same idea written in one line. A yield of 0.9102 is 91.02 percent, and a yield of 1.02 is 102 percent, a result that always needs an explanation rather than a celebration.
Finding the theoretical yield from the limiting reactant
The theoretical yield comes from the reactant that runs out first, not from the one you have most of. Four steps get there, and skipping any of them is where most wrong answers start.
- Balance the equation, so the mole ratio between reactant and product is known.
- Convert the mass of each reactant to moles by dividing by its molar mass.
- Divide each reactant's moles by its coefficient and pick the smallest result. That reactant is the limiting one.
- Multiply the limiting moles by the product to reactant ratio, then by the product's molar mass to get grams.
Take the decomposition of calcium carbonate, CaCO3 going to CaO and CO2. The equation balances one to one, so one mole of carbonate gives one mole of oxide. Ten grams of carbonate is 10.0 / 100.09 = 0.09991 moles, which can make 0.09991 x 56.08 = 5.603 grams of calcium oxide. That 5.603 grams is the theoretical yield, and it is the number the percent yield gets divided by.
Molar masses here use the standard atomic weights from the Commission on Isotopic Abundances and Atomic Weights: calcium 40.078, carbon 12.011 and oxygen 15.999, so CaCO3 is 40.078 + 12.011 + 3 x 15.999 = 100.086, which rounds to 100.09. CaO is 40.078 + 15.999 = 56.077, which rounds to 56.08.
Worked example: 10.0 g of calcium carbonate
A crucible holds 10.0 grams of calcium carbonate. After heating and cooling, the crucible holds 5.10 grams of calcium oxide. The table walks the arithmetic from the weighed reactant to the percent yield.
| Step | Arithmetic | Result |
|---|---|---|
| Moles of reactant | 10.0 / 100.09 | 0.09991 mol |
| Moles of product, 1:1 ratio | 0.09991 x 1 | 0.09991 mol |
| Theoretical mass of CaO | 0.09991 x 56.08 | 5.603 g |
| Actual mass weighed | 5.10 g | |
| Percent yield | 5.10 / 5.603 x 100 | 91.02 percent |
The missing 0.503 grams is 9 percent of the product, and it went somewhere real: dust lost when the powder was transferred, a little carbonate that never reached temperature, or product that stayed stuck to the crucible. A yield in the low nineties is typical for a school lab run on a hot plate.
Note which mass went into the division. The 10.0 grams of reactant is not part of the percent yield at all, and neither is the carbon dioxide that left as a gas. Only the product the equation predicted and the product you weighed matter.
Worked example: aspirin from salicylic acid
The same four steps handle an organic synthesis. Salicylic acid reacts with acetic anhydride to give acetylsalicylic acid, the aspirin in the bottle, and the balanced ratio between the acid and the product is one to one. Starting from 25.0 grams of salicylic acid at 138.12 grams per mole gives 25.0 / 138.12 = 0.1810 moles.
One mole of salicylic acid makes one mole of aspirin at 180.16 grams per mole, so the theoretical yield is 0.1810 x 180.16 = 32.61 grams. A student who recovers 28.4 grams after recrystallization has 28.4 / 32.61 x 100 = 87.09 percent.
| Case | Reactant | Theoretical yield | Actual yield | Percent yield |
|---|---|---|---|---|
| Calcium oxide | 10.0 g CaCO3 | 5.603 g | 5.10 g | 91.02% |
| Aspirin | 25.0 g salicylic acid | 32.61 g | 28.4 g | 87.09% |
| Water from hydrogen | 5.00 g H2 | 44.69 g | 40.0 g | 89.50% |
The third row shows why the mole ratio matters. Hydrogen burns as 2 H2 plus O2 going to 2 H2O, so one mole of H2 makes one mole of water: 5.00 / 2.016 = 2.4802 moles of H2 give 2.4802 x 18.02 = 44.69 grams of water. Had the ratio been read as 1 to 2, the theoretical yield would have doubled and the percent yield would have halved.
Yields over 100 percent and what they mean
A percent yield above 100 is not a discovery, it is a sign that the product weighed more than it should. The usual causes are all measurable.
- The product is still wet. Water counts on the balance like product, and a few drops can add a tenth of a gram.
- The product carries an impurity, unreacted starting material or a side product that crystallized with it.
- The product is a hydrate, so it includes water in its crystal that the formula you used did not count.
- The mass came from a different, heavier compound than the one the equation describes.
The calcium oxide case shows the size of the effect. If the crucible had come out at 5.75 grams instead of 5.10, the calculation would read 5.75 / 5.603 x 100 = 102.62 percent. Nothing about the reaction changed, only the weighing, and the answer announced it.
Drying to constant mass is the fix, along with weighing the empty vessel first and checking that the product matches the expected appearance. A yield of 102 percent reported without comment is the kind of number a reviewer treats as evidence that the method, not the chemistry, needs work.
Percent yield, percent recovery and atom economy
Three numbers get used for how well a reaction went, and they are not interchangeable. The table separates them, because a paper that mixes them up reports a good process as a bad one.
| Measure | What is divided by what | Answers |
|---|---|---|
| Percent yield | actual product / theoretical product x 100 | How much of the possible product did I collect? |
| Percent recovery | mass recovered / mass started with x 100 | How much of my material survived the workup? |
| Atom economy | molar mass of product / molar mass of all reactants x 100 | How much of the reactant mass ends up in the product by design? |
| Percent error | measured minus accepted, over accepted, x 100 | How far is a measurement from the reference value? |
Percent recovery can beat percent yield and often does. Recrystallizing 28.4 grams of crude aspirin and getting 24.0 grams back is an 84.5 percent recovery, but the yield against the theoretical 32.61 grams is still 73.6 percent for that second step. Both numbers are true and they answer different questions.
Atom economy ignores the bench entirely. A reaction with 100 percent atom economy can still deliver a 5 percent yield if the conditions are wrong, and a reaction with 40 percent atom economy can run at 95 percent yield. For a measured value against a published constant, the right tool is percent error, which this site covers separately.
Five mistakes that change the answer
Most wrong percent yields come from five habits, and each one has a specific symptom.
- Using the excess reactant. The theoretical yield comes from the reactant that runs out first. If 10 grams of one reagent and 50 grams of the other are mixed, the limiting one is decided by moles divided by coefficients, not by which pile looks smaller.
- Dropping the mole ratio. A 2 to 1 equation halves or doubles the product count. Read the coefficients before multiplying by the molar mass.
- Dividing grams by grams of a different compound. Actual and theoretical masses must describe the same substance, otherwise the ratio is meaningless.
- Rounding moles too early. Keep four figures until the last step. Rounding 0.09991 to 0.1 turns the theoretical yield from 5.603 into 5.608 grams, which moves the answer from 91.02 to 90.94 percent.
- Weighing wet product. Dry to constant mass, then weigh. A kitchen scale that reads to 0.1 gram cannot settle a yield better than about 2 percent on a 5 gram product, so use a balance that reads to 0.01 gram or better.
Related tools pick up the steps either side of this one. Grams to moles handles the mass to mole conversion, molar mass builds the formula weight, mole fraction works out the share each component holds in a mixture, and molarity converts between moles and solution volume. This page answers the question those pages leave open: how close did the reaction come to what the equation promised.
Frequently asked questions
How do you calculate percent yield?
Divide the actual yield by the theoretical yield and multiply by 100. A 5.10 gram product against a theoretical 5.603 grams is 5.10 / 5.603 x 100 = 91.02 percent.
What is the difference between actual and theoretical yield?
Theoretical yield is what the limiting reactant could make according to the balanced equation. Actual yield is the mass you collected and weighed.
How do I find the theoretical yield?
Convert the limiting reactant to moles, multiply by the product to reactant ratio from the equation, then multiply by the product's molar mass. Ten grams of CaCO3 gives 0.09991 mol, which is 0.09991 x 56.08 = 5.603 grams of CaO.
Why is my percent yield over 100?
The product weighs more than the equation allows, which points to wet product, an impurity, a hydrate, or a weighing error. A 5.75 gram result against 5.603 theoretical is 102.62 percent.
Can percent yield be negative?
No. Both masses are positive, so the result runs from 0 upward. A negative number means the actual yield was entered as a negative value by mistake.
What counts as a good percent yield?
Above 90 percent is strong for a small lab synthesis, 70 to 90 percent is normal once recrystallization is involved, and below 50 percent usually means a step lost material or the reaction did not go to completion.
Is percent yield the same as percent recovery?
No. Yield compares product to what the equation predicted, while recovery compares the mass you got back to the mass you started with. Recovering 24.0 of 28.4 grams is an 84.5 percent recovery.
Does the mole ratio change the answer?
Yes, and it is the most common reason two students get different numbers from the same data. In 2 H2 + O2 going to 2 H2O the ratio of H2 to water is 1 to 1, so 2.4802 moles of H2 give 2.4802 moles of water.
How many moles are in 10 grams of calcium carbonate?
10.0 / 100.09 = 0.09991 moles, using calcium 40.078, carbon 12.011 and three oxygens at 15.999.
What does a 0.01 gram balance do to the answer?
It sets the smallest step you can see. On a 5 gram product, 0.01 gram is about 0.2 percent of the mass, so a yield quoted to two decimals assumes the weighing was at least that good.
Why does rounding early change the yield?
Because the error is multiplied by the product's molar mass. Rounding 0.09991 to 0.1 gives 5.608 grams instead of 5.603, and the percent yield drops from 91.02 to 90.94.
Can two products have different percent yields in one reaction?
Yes. Each product has its own theoretical mass and its own actual mass, so a reaction can give 91 percent of one product and 40 percent of another when side reactions compete.
What is atom economy?
The molar mass of the desired product divided by the molar mass of all reactants, times 100. It measures how much of the reactant mass the equation puts into the product, while percent yield measures how much you collected at the bench.
Does the calculator work for a reaction with a 2 to 1 ratio?
Yes. Set the product per reactant ratio to 2 and the theoretical moles double. For 5.00 grams of hydrogen making water the ratio is 1, so 2.4802 moles of H2 give 2.4802 moles of H2O.
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