Percent Yield Formula
The percent yield formula is percent yield = (actual yield / theoretical yield) x 100. Actual yield is the mass or moles of product you actually collect; theoretical yield is the maximum possible amount predicted by stoichiometry. The same equation rearranges to solve for actual yield or theoretical yield whenever the other two values are known.
The Percent Yield Formula
The percent yield formula is: percent yield = (actual yield / theoretical yield) x 100. Actual yield is the amount of product you physically recover from a reaction, usually measured on a balance. Theoretical yield is the maximum amount of product that stoichiometry predicts is possible, calculated from a balanced equation and the limiting reactant. Both quantities must be expressed in the same unit, grams to grams or moles to moles, before you divide, because the formula is a ratio and any unit mismatch will throw off the result.
This page is a reference for the formula itself: what it says, how it rearranges, and how to plug numbers into it. If you want a calculator that does the arithmetic for you, use the tools for actual yield or theoretical yield directly, or start from the percent yield calculator on the homepage. For the full derivation and a unit-by-unit dimensional analysis walkthrough of why this equation works, see the companion page on the percent yield equation.
Where the Formula Comes From
The formula is fundamentally a ratio of two masses (or two mole counts) that share the same unit. Because actual yield and theoretical yield are measured in the same unit, that unit cancels out of the division, leaving a plain number. Multiplying by 100 converts that plain number into a percentage, which is why percent yield has no unit attached to it, it is a dimensionless comparison of what you got against what was possible. For the step-by-step cancellation of units and a longer treatment of why the equation is built this way, see the percent yield equation page.
Rearranging the Formula for Actual or Theoretical Yield
| Solving for | Rearranged formula | When to use it |
|---|---|---|
| Percent yield | Yp = (actual / theoretical) x 100 | You measured actual yield and calculated theoretical yield, and want the efficiency of the run |
| Actual yield | actual = (Yp / 100) x theoretical | You know the theoretical yield and an expected or literature percent yield, and want to predict the mass that should appear on the balance |
| Theoretical yield | theoretical = actual / (Yp / 100) | You measured actual yield and know the percent yield from a procedure, and want to back-calculate what the theoretical yield should have been |
Solving for percent yield is the everyday case: a lab run is complete, the product has been weighed and dried, and theoretical yield has already been worked out from the stoichiometry of the balanced equation. Dividing actual by theoretical and multiplying by 100 gives a single number that answers the question every lab report asks, how efficient was this reaction.
Solving for actual yield runs the formula in the opposite direction, and it is useful before a reaction even starts. If a published procedure states it typically achieves a certain percent yield, and the theoretical yield for your scale of reaction is already known, you can predict roughly what mass should end up in the flask. This is a planning number, not a guarantee, but it tells you whether the amount you eventually collect is in a reasonable range or whether something went wrong.
Solving for theoretical yield is a checking tool. If you already collected a product, weighed it, and you know from a procedure or from the instructor what percent yield is typically reported, you can work backward to see what theoretical yield the numbers imply. If that implied theoretical yield does not match the value you calculated from your reagent masses, it is a signal that a reagent amount, a molar mass, or the limiting reactant identification was entered incorrectly somewhere upstream.
Five Worked Examples Using the Formula
Each example below states which of the three forms of the formula is used and shows the complete arithmetic.
- Solving for percent yield. Actual yield = 4.0 g, theoretical yield = 5.0 g. Yp = (4.0 / 5.0) x 100 = 0.80 x 100 = 80%.
- Solving for actual yield. Percent yield = 70%, theoretical yield = 12.0 g. actual = (70 / 100) x 12.0 = 0.70 x 12.0 = 8.40 g.
- Solving for theoretical yield. Actual yield = 6.75 g, percent yield = 90.0%. theoretical = 6.75 / (90.0 / 100) = 6.75 / 0.90 = 7.50 g.
- Working in moles instead of grams. Actual yield = 0.0350 mol, theoretical yield = 0.0400 mol. Yp = (0.0350 / 0.0400) x 100 = 0.875 x 100 = 87.5%. The formula does not care whether the quantities are grams or moles, only that both sides of the ratio use the same unit.
- Mixed units requiring a molar-mass conversion first. Actual yield is measured as 3.20 g of a product with a molar mass of 96.00 g/mol, but theoretical yield is already known in moles as 0.0400 mol. Convert the actual mass to moles first: 3.20 g / 96.00 g/mol = 0.03333 mol. Now both values are in moles, so the formula applies directly: Yp = (0.03333 / 0.0400) x 100 = 83.3%. For the reverse conversion, moles to grams, see grams to moles.
Quick-Reference Table for Common Ratios
These are common actual-to-theoretical ratios worked out in advance, useful for a fast sanity check against a similar problem.
| Actual yield | Theoretical yield | Percent yield |
|---|---|---|
| 4 | 5 | 80.0% |
| 7 | 10 | 70.0% |
| 9 | 10 | 90.0% |
| 17 | 20 | 85.0% |
| 23 | 25 | 92.0% |
If your own actual and theoretical values simplify to one of these ratios, you can check your answer against this table before trusting a longer calculation.
Reporting Percent Yield Correctly
Percent yield is normally reported to one decimal place, for example 87.5% rather than 87.53219%, unless an instructor or a specific procedure asks for more precision. The underlying actual and theoretical masses should be reported to the precision your balance actually supports, typically two decimal places in grams, and the final percentage should not carry more significant figures than those measurements justify. For a wider set of solved problems at varying levels of unit complexity, see the worked examples page, and for a bank of practice questions with answers, see the practice problems page.
Frequently asked questions
What is the percent yield formula?
The percent yield formula is percent yield = (actual yield / theoretical yield) x 100. Actual yield is the amount of product actually recovered from a reaction, and theoretical yield is the maximum amount predicted by stoichiometry from the limiting reactant. Both values must be in the same unit, grams or moles, before dividing, since the formula is a simple ratio expressed as a percentage.
How do you rearrange the percent yield formula to find actual yield?
Multiply the percent yield, written as a decimal, by the theoretical yield: actual = (Yp / 100) x theoretical. This form is used when the theoretical yield is already known and a target or expected percent yield is given, letting you predict roughly what mass of product should appear before you even run the reaction or as a check once you do.
How do you rearrange the formula to find theoretical yield?
Divide the actual yield by the percent yield written as a decimal: theoretical = actual / (Yp / 100). This form is used after a product has been collected and weighed, when you want to check whether the theoretical yield implied by a known or expected percent yield matches the theoretical yield you calculated from your original reagent quantities.
Does the formula work in moles as well as grams?
Yes, the formula works identically in moles because it is a plain ratio and does not depend on which unit is used, only that both actual and theoretical yield are in the same unit. Using 0.0350 mol actual against 0.0400 mol theoretical gives Yp = 87.5%, the exact same arithmetic as if both numbers were grams instead.
Why do the units cancel in the percent yield formula?
The units cancel because actual yield and theoretical yield are measured in the same unit, so dividing one by the other leaves a plain, unitless number that is then multiplied by 100 to express it as a percentage. For the full dimensional analysis showing this cancellation step by step, see the percent yield equation page.
What if I have mixed units, grams and moles?
Convert one value so both sides of the ratio share the same unit before applying the formula. If actual yield is in grams and theoretical yield is in moles, convert the grams to moles using the product's molar mass, or convert the moles to grams, then divide as usual. Mixing grams and moles directly in the formula gives an incorrect result.
Can the formula give a negative percent yield?
No, a correctly entered calculation cannot produce a negative percent yield, because mass and moles are always zero or positive quantities. A negative result means a value was entered incorrectly, most often a sign error or a mixed-up subtraction elsewhere in a multi-step problem, not a real chemical outcome, so it should be treated as a data-entry mistake to fix.
Is there a different formula for percent yield in industry versus a teaching lab?
No, the formula is the same in both settings: percent yield = (actual yield / theoretical yield) x 100. Industrial process reports and teaching lab write-ups use identical arithmetic; the only differences are typically scale, the units chosen for reporting, and how many decimal places are expected, not the underlying equation itself.
What is the maximum possible percent yield?
The maximum possible percent yield is 100%, meaning every bit of theoretical product predicted by stoichiometry was actually recovered. Readings above 100% are possible on paper but indicate a measurement or calculation problem, such as leftover solvent or an impure product adding extra mass, rather than a true yield exceeding the stoichiometric maximum. See the homepage for the basic definition.
How many decimal places should percent yield be reported to?
One decimal place is the standard convention, for example 87.5% rather than 87.53219%, unless a specific assignment or procedure requests more precision. The final percentage should not carry more significant figures than the actual and theoretical mass measurements can support, since a balance reading to two decimal places in grams cannot justify five decimal places in the resulting percentage.