# How to calculate percent yield

Canonical: https://duckyhelper.com/learn/chemistry/percent-yield/
Updated: 2026-10-01

Percent yield compares how much product you actually made with the most you could have made. Percent yield = actual yield ÷ theoretical yield × 100%. The theoretical yield comes from stoichiometry, starting from the limiting reactant. The actual yield is what you measured in the lab. Real yields are usually below 100% because of side reactions, product lost in transfers, or reactions that do not finish.

## Key ideas

$$
\text{percent yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\%
$$

- **Actual yield** is the mass of product you collected and weighed, in grams (g). It comes from the lab or is given in the problem.
- **Theoretical yield** is the mass of product the balanced equation predicts if every bit of the limiting reactant turns into product, in grams (g). You calculate it.
- Both yields must be in the same unit, so the units cancel and the answer is a percent.

Most problems have two parts. First, use [stoichiometry](https://duckyhelper.com/learn/chemistry/stoichiometry/) to find the theoretical yield. Then plug both yields into the formula. If two reactant amounts are given, find the [limiting reactant](https://duckyhelper.com/learn/chemistry/limiting-reactant/) first.

**What a percent yield tells you**

| Result | What it usually means |
| --- | --- |
| Below 100% | Normal. Some product was lost, some reactant formed side products, or the reaction did not finish. |
| Exactly 100% | Every bit of limiting reactant became product and none was lost. Rare in real labs. |
| Above 100% | Something is wrong. The product is often still wet or mixed with impurities, or there is a math error. |

> **Note: Atomic masses used on this page**
>
> H 1.008, C 12.01, O 16.00, Ca 40.08 (g/mol).

## Worked examples

**Example 1: both yields given**

Problem: A reaction has a theoretical yield of 25.0 g. A student collects 21.3 g of product. What is the percent yield?

1. Actual yield is what was collected: 21.3 g. Theoretical yield is 25.0 g.
2. Divide and multiply by 100%.

   $$
   \frac{21.3\ \text{g}}{25.0\ \text{g}} \times 100\% = 85.2\%
   $$

Answer: 85.2%

**Example 2: find the theoretical yield first**

Problem: Heating limestone makes lime: \(\mathrm{CaCO_3 \rightarrow CaO + CO_2}\). A student heats 50.0 g of \(\mathrm{CaCO_3}\) and collects 25.9 g of CaO. What is the percent yield?

1. Molar masses.

   $$
   M_{\mathrm{CaCO_3}} = 40.08 + 12.01 + 3(16.00) = 100.09\ \text{g/mol} \qquad M_{\mathrm{CaO}} = 40.08 + 16.00 = 56.08\ \text{g/mol}
   $$
2. Grams of \(\mathrm{CaCO_3}\) to moles. The ratio to CaO is 1 to 1.

   $$
   \frac{50.0\ \text{g}}{100.09\ \text{g/mol}} = 0.49955\ \text{mol CaO}
   $$
3. Theoretical yield in grams. Keep an extra digit for now.

   $$
   0.49955\ \text{mol} \times 56.08\ \text{g/mol} = 28.01\ \text{g}
   $$
4. Percent yield.

   $$
   \frac{25.9\ \text{g}}{28.01\ \text{g}} \times 100\% = 92.47\%
   $$
5. The data have 3 significant figures, so round to 3.

Answer: 92.5%

**Example 3: working backward**

Problem: A reaction usually gives a 75.0% yield. How many grams of product must you be able to make in theory to collect 15.0 g?

1. Rearrange the formula to solve for theoretical yield.

   $$
   \text{theoretical} = \frac{\text{actual}}{\text{percent yield}} \times 100\%
   $$
2. Plug in.

   $$
   \frac{15.0\ \text{g}}{75.0\%} \times 100\% = 20.0\ \text{g}
   $$

Answer: 20.0 g theoretical yield

## Common mistakes and how to fix them

- **Dividing the wrong way.** Fix: actual (what you got) goes on top. If you get more than 100%, check that first.
- **Using the mass of a reactant as the theoretical yield.** Fix: the theoretical yield is a mass of product, found with the mole ratio.
- **Starting from the excess reactant.** Fix: the theoretical yield always comes from the limiting reactant.
- **Rounding the theoretical yield too early.** In practice problem 3, rounding 35.75 g to 35.8 g gives 88.0% instead of 88.1%. Fix: keep one extra digit until the final step.

**Practice problems**

1. A student expects 5.00 g of product and collects 4.50 g. What is the percent yield?
   A. 0.900%
   B. 90.0%
   C. 111%
   D. 4.50%

   Answer: 90.0%. \(4.50 \div 5.00 \times 100\% = 90.0\%\). 111% comes from dividing the wrong way.

2. The theoretical yield of a reaction is 18.0 g and the percent yield is 65.0%. What is the actual yield?

   Answer: 11.7 g. Actual = percent yield × theoretical ÷ 100% = \(0.650 \times 18.0 = 11.7\) g.

3. \(\mathrm{2\,H_2 + O_2 \rightarrow 2\,H_2O}\). Burning 4.00 g of \(\mathrm{H_2}\) in excess oxygen gives 31.5 g of water. What is the percent yield?

   Answer: 88.1%. \(4.00 \div 2.016 = 1.984\) mol \(\mathrm{H_2}\), which makes 1.984 mol water (ratio 2 to 2). Theoretical yield: \(1.984 \times 18.02 = 35.75\) g. Percent yield: \(31.5 \div 35.75 \times 100\% = 88.1\%\).

4. A student reports a percent yield of 108%. What is the most likely reason?
   A. Some product spilled
   B. The product was still wet when weighed
   C. The reaction did not finish
   D. A side reaction used up some reactant

   Answer: The product was still wet when weighed. Spills, side reactions and unfinished reactions all lower the yield. Only extra mass, such as water or impurities left in the product, can push it above 100%.

5. Which amount is used to calculate the theoretical yield?
   A. The excess reactant
   B. The limiting reactant
   C. The actual yield
   D. The total mass of all reactants

   Answer: The limiting reactant. The limiting reactant runs out first, so it sets the most product that can form.

## Frequently asked questions

### Can percent yield be more than 100%?

Not for a pure, dry product. A result over 100% means extra mass got weighed, usually water that has not dried off or impurities mixed in, or that a calculation went wrong. Check the math first, then think about the lab.

### Why is percent yield almost never 100%?

Some product sticks to glassware or filter paper, some reactant forms side products, and many reactions stop before every reactant is used. Chemists still aim high, because a low yield wastes materials and money.

### What is the difference between percent yield and percent error?

Percent yield compares the product you made with the most you could make. Percent error compares any measured value with an accepted value: \(\lvert\text{measured} - \text{accepted}\rvert \div \text{accepted} \times 100\%\). They use different formulas and answer different questions.

## Sources

- [OpenStax Chemistry 2e, 4.4 Reaction Yields](https://openstax.org/books/chemistry-2e/pages/4-4-reaction-yields), accessed 2026-10-01

## Related

- [How to find the limiting reactant](https://duckyhelper.com/learn/chemistry/limiting-reactant/)
- [Stoichiometry: how much reacts and how much forms](https://duckyhelper.com/learn/chemistry/stoichiometry/)
- [Unit conversion and significant figures](https://duckyhelper.com/learn/physics/unit-conversion-and-significant-figures/)
- [Chemistry study guides](https://duckyhelper.com/learn/chemistry/)

## Try asking Ducky

- "My percent yield came out over 100%. Look at my data table and tell me where it went wrong."
- "Check my theoretical yield for the lab, did I start from the limiting reactant?"
- "Explain why rounding early changed my answer from 88.1% to 88.0%."

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