# Yield and purity calculations

> CIE A-Level Chemistry · 9701
> Source: https://www.owlsprep.com/study/cie-9701-u27-yield-and-purity-calculations/

This module explains how to calculate percentage yield from limiting reactants and percentage purity of crude organic products, two core calculation skills for CIE A-Level 9701 organic chemistry exams.

**Prerequisites:** [Mole and reacting mass stoichiometry calculations](https://www.owlsprep.com/study/cie-9701-u1-mole-and-stoichiometry/)

## Learning objectives

- Distinguish between theoretical yield, actual yield and percentage yield
- Calculate percentage yield for organic synthesis reactions
- Calculate percentage purity of crude organic products

## Key Definitions for Yield

In organic synthesis, reactions rarely produce the maximum possible amount of product. This is due to incomplete reaction, side reactions that form unwanted products, and product loss during purification steps like filtration or distillation.

**Percentage Yield** — A quantitative measure of how much product is obtained compared to the maximum possible amount from the starting reactants

*Notation:* % yield

*Example:* A 75% yield means 75% of the maximum possible product was recovered

- - **Theoretical yield**: The maximum mass of product calculated from the reaction stoichiometry, assuming complete reaction of the limiting reactant and no product loss.
- - **Actual yield**: The mass of pure product you actually measure after isolating and purifying the product from the reaction mixture.
- - **Percentage yield**: Compares actual yield to theoretical yield, to describe how efficient the reaction is.

> **Exam tip:** Always confirm which reactant is limiting before calculating theoretical yield.

## Calculating Percentage Yield

The formula for percentage yield is standard across all CIE exam questions:

$$\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$$

**Worked example:** 0.500 g of ethanol ($C_2H_5OH$) is oxidized to ethanoic acid ($CH_3COOH$). The actual yield of pure ethanoic acid is 0.520 g. Calculate the percentage yield.

1. 1. Write the balanced reaction equation (1:1 mole ratio of ethanol to ethanoic acid):
2. $$C_2H_5OH + 2[O] \rightarrow CH_3COOH + H_2O$$
3. 2. Calculate moles of the starting reactant ethanol:
4. $$n(C_2H_5OH) = \frac{0.500}{46.0} = 0.0109 \text{ mol}$$
5. 3. Theoretical moles of ethanoic acid = 0.0109 mol (from 1:1 ratio)
6. 4. Calculate theoretical mass of ethanoic acid:
7. $$m(CH_3COOH) = 0.0109 \times 60.0 = 0.654 \text{ g}$$
8. 5. Substitute into the percentage yield formula:
9. $$\% \text{ Yield} = \frac{0.520}{0.654} \times 100 = 79.5\%$$

**Check your understanding**

Test your understanding: What is the percentage yield if the theoretical yield is 10.0 g and the actual yield is 8.2 g?

1. What is the correct percentage yield?

   - 8.2%
   - 82%
   - 122%
   - 0.82%

   *Why:* Correct! $\frac{8.2}{10.0} \times 100 = 82\%$, which is a valid value between 0 and 100%.

## Percentage Purity Calculations

After isolating a product from a reaction, the crude product almost always contains impurities. Percentage purity measures how much of the total mass of the crude sample is actually the desired pure product.

**Percentage Purity** — The percentage by mass of the desired pure compound in an impure crude product sample

*Notation:* % purity

*Example:* An 88% pure sample means 88 g of every 100 g of sample is desired product

$$\text{Percentage Purity} = \frac{\text{Mass of pure desired compound}}{\text{Total mass of impure sample}} \times 100\%$$

**Worked example:** 1.50 g of impure sodium ethanoate is reacted with excess hydrochloric acid to produce 0.720 g of ethanoic acid. Assuming only sodium ethanoate reacts, calculate the percentage purity.

1. 1. Write the balanced reaction (1:1 mole ratio of sodium ethanoate to ethanoic acid):
2. $$CH_3COONa + HCl \rightarrow CH_3COOH + NaCl$$
3. 2. Calculate moles of ethanoic acid produced:
4. $$n(CH_3COOH) = \frac{0.720}{60.0} = 0.0120 \text{ mol}$$
5. 3. Moles of pure sodium ethanoate = moles of ethanoic acid = 0.0120 mol
6. 4. Calculate mass of pure sodium ethanoate (molar mass = 82.0 g mol⁻¹):
7. $$m(CH_3COONa) = 0.0120 \times 82.0 = 0.984 \text{ g}$$
8. 5. Calculate percentage purity:
9. $$\% \text{ Purity} = \frac{0.984}{1.50} \times 100 = 65.6\%$$

> **Exam tip:** If you get a yield or purity over 100%, you have made a mistake: check your mole calculations and ratio.

## Common pitfalls

- **Wrong:** Using the mass of excess reactant to calculate theoretical yield
  - Why it fails: Yield is limited by the reactant that runs out first, not the excess reactant
  - Correct: Always calculate moles of all reactants and identify the limiting reactant first
- **Wrong:** Confusing percentage yield with percentage purity
  - Why it fails: Yield measures reaction efficiency, while purity measures sample cleanliness
  - Correct: Read the question carefully to confirm which calculation is required
- **Wrong:** Forgetting to multiply the ratio by 100, leaving a decimal answer
  - Why it fails: Examiners require percentage values, not decimal fractions
  - Correct: Always check your answer is between 0 and 100% for both calculations
- **Wrong:** Using an incorrect molar mass for organic compounds
  - Why it fails: It is easy to miscount carbon/hydrogen atoms in molecular formulas
  - Correct: Double-check the molecular formula and sum atomic masses before calculating moles

## Cheatsheet

| Calculation | Formula |
| --- | --- |
| Percentage Yield | $\frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%$ |
| Percentage Purity | $\frac{\text{Mass of pure compound}}{\text{Total mass of impure sample}} \times 100\%$ |
| Theoretical Yield | Moles of limiting reactant × mole ratio × molar mass of product |

## What's next

Yield and purity calculations are foundational for all organic synthesis topics in CIE 9701, and appear regularly in both Paper 2 and Paper 4, often combined with titration, chromatography, or reaction planning questions. Mastery of these calculations is essential for full marks on longer structured problems, as they are frequently the final step in multi-part questions. Understanding yield and purity also helps you explain why synthetic routes are modified for industrial and laboratory use, a common topic for extended answer questions.

- [Nuclear magnetic resonance spectroscopy](https://www.owlsprep.com/study/cie-9701-u28-overview/)
- [Principles of ¹H NMR](https://www.owlsprep.com/study/cie-9701-u28-principles-of-1h-nmr/)
- [Interpretation of NMR spectra](https://www.owlsprep.com/study/cie-9701-u28-interpretation-of-nmr-spectra/)

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