# Stoichiometric Calculations

> Chemistry · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9701-u1-stoichiometric-calculations/

This sub-topic covers core mole-based stoichiometric calculations, including deriving chemical formulas, reacting quantities, and yield/purity calculations. These methods are foundational for all quantitative chemistry topics in A-Level.

**Prerequisites:** [Basic mole concept and atomic structure](https://www.owlsprep.com/study/cie-9701-u1-mole-concept/)

## Learning objectives

- Calculate amount of substance in moles from mass, gas volume and solution concentration
- Determine empirical and molecular formulas from experimental composition data
- Solve reacting mass and volume stoichiometry problems from balanced equations
- Calculate percentage yield and percentage purity for chemical reactions

## Core Mole Calculations

**Amount of substance** — A measure of the number of specified particles (atoms, ions, molecules) in a sample, measured in moles (mol).

*Notation:* $n$

*Example:* 1 mole of water contains $6.02 \times 10^{23}$ water molecules

Three core relationships are used for all basic mole calculations:

- From mass: $n = \frac{m}{M}$, where $m$ = mass (g), $M$ = molar mass (g mol⁻¹)
- From gas volume (r.t.p.): $n = \frac{V}{V_m}$, where $V$ = volume (dm³), $V_m = 24.0$ dm³ mol⁻¹
- From solution: $n = c \times V$, where $c$ = concentration (mol dm⁻³), $V$ = volume (dm³)

**Worked example:** Calculate the moles of sodium hydroxide (NaOH) in 2.10 g of solid NaOH. (Aᵣ: Na = 23.0, O = 16.0, H = 1.0)

1. Calculate molar mass of NaOH:
2. $$M(NaOH) = 23.0 + 16.0 + 1.0 = 40.0 \text{ g mol}^{-1}$$
3. Substitute into the mole relationship:
4. $$n = \frac{2.10}{40.0} = 0.0525 \text{ mol}$$

> **Exam tip:** Always convert volume from cm³ to dm³ by dividing by 1000 before using $n = cV$.

## Empirical and Molecular Formulas

**Empirical formula** — The simplest whole number ratio of atoms of each element present in a compound.

*Example:* Empirical formula of glucose (C₆H₁₂O₆) is CH₂O

To find the empirical formula from experimental data, follow a standard 4-step method. The molecular formula (actual number of atoms) is found by comparing the empirical formula mass to the known relative molecular mass.

**Worked example:** A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its relative molecular mass is 28. Find the empirical and molecular formula. (Aᵣ: C = 12, H = 1)

1. Step 1: Divide percentages by relative atomic masses:
2. $$C: \frac{85.7}{12} = 7.14; \quad H: \frac{14.3}{1} = 14.3$$
3. Step 2: Divide all values by the smallest result:
4. $$C: \frac{7.14}{7.14} = 1; \quad H: \frac{14.3}{7.14} = 2$$
5. Ratio C:H = 1:2, so empirical formula = CH₂. Calculate empirical formula mass:
6. $$\text{Empirical mass} = 12 + (2 \times 1) = 14$$
7. Find the multiple for molecular formula:
8. $$\text{Multiple} = \frac{28}{14} = 2; \quad \text{Molecular formula} = C_2H_4$$

> **Exam tip:** If you get a ratio like 1:1.5 after step 2, multiply all values by 2 to get whole numbers (2:3 ratio).

## Reacting Quantity Calculations

Stoichiometric calculations use the mole ratio from a balanced chemical equation to find the unknown mass, volume or concentration of a reactant or product.

> **tip**
>
> Always write a balanced equation first. You cannot get the correct mole ratio from an unbalanced equation.

**Worked example:** What mass of carbon dioxide is produced when 1.00 g of methane (CH₄) is completely burned in excess oxygen? (Aᵣ: C = 12.0, H = 1.0, O = 16.0)

1. Write the balanced chemical equation:
2. $$CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$$
3. Calculate moles of methane:
4. $$M(CH_4) = 16.0 \text{ g mol}^{-1}; \quad n(CH_4) = \frac{1.00}{16.0} = 0.0625 \text{ mol}$$
5. From the balanced equation, mole ratio CH₄:CO₂ = 1:1, so $n(CO_2) = 0.0625$ mol.
6. Calculate mass of carbon dioxide:
7. $$M(CO_2) = 12.0 + (2 \times 16.0) = 44.0 \text{ g mol}^{-1}; \quad m(CO_2) = 0.0625 \times 44.0 = 2.75 \text{ g}$$

## Percentage Yield and Purity

**Percentage yield** — Compares the actual mass of product obtained in an experiment to the maximum theoretical mass predicted from the starting reactants.

*Example:* A 100% yield means no product is lost during purification

Two core formulas for these calculations:

$$\text{Percentage yield} = \frac{\text{actual mass of product}}{\text{theoretical mass of product}} \times 100\%$$

$$\text{Percentage purity} = \frac{\text{mass of pure substance}}{\text{mass of impure sample}} \times 100\%$$

**Worked example:** A student reacts 1.00 g of impure calcium carbonate (CaCO₃) with excess HCl and obtains 0.88 g of calcium chloride (CaCl₂). Calculate the percentage purity of CaCO₃. (Aᵣ: Ca = 40, C = 12, O = 16, Cl = 35.5)

1. Balanced equation: $CaCO_3 + 2HCl \rightarrow CaCl_2 + CO_2 + H_2O$. Moles of CaCl₂ produced:
2. $$M(CaCl_2) = 111 \text{ g mol}^{-1}; \quad n(CaCl_2) = \frac{0.88}{111} = 0.00793 \text{ mol}$$
3. Mole ratio CaCO₃:CaCl₂ = 1:1, so moles of pure CaCO₃ = 0.00793 mol. Calculate mass of pure CaCO₃:
4. $$M(CaCO_3) = 100 \text{ g mol}^{-1}; \quad \text{Mass of pure } CaCO_3 = 0.00793 \times 100 = 0.793 \text{ g}$$
5. Calculate percentage purity:
6. $$\text{Percentage purity} = \frac{0.793}{1.00} \times 100\% = 79\% \text{ (2 s.f.)}$$

## Common pitfalls

- **Wrong:** Forgetting to convert volume from cm³ to dm³ for solution calculations
  - Why it fails: Concentration is given in mol dm⁻³, so using cm³ gives an answer 1000 times too small
  - Correct: Divide volume in cm³ by 1000 to convert to dm³ before substituting into $n = cV$
- **Wrong:** Using an unbalanced equation to get the mole ratio
  - Why it fails: An unbalanced equation gives the wrong stoichiometric ratio, leading to an incorrect answer
  - Correct: Always balance the equation before starting any calculation, and double-check the balancing
- **Wrong:** Leaving the answer as the empirical formula instead of finding the molecular formula
  - Why it fails: The empirical formula is only the simplest ratio, not the actual formula of the compound
  - Correct: Always compare the empirical formula mass to the given relative molecular mass and multiply by the correct multiple
- **Wrong:** Confusing molar volume at r.t.p. and s.t.p.
  - Why it fails: CIE uses 24.0 dm³ mol⁻¹ for r.t.p., not 22.4 dm³ mol⁻¹ which is for s.t.p.
  - Correct: Check the question conditions, and refer to the Data Booklet for the correct $V_m$ value
- **Wrong:** Rounding intermediate values too early leading to inaccurate final answers
  - Why it fails: Rounding intermediate steps to 2 significant figures introduces large errors into the final result
  - Correct: Keep at least one extra significant figure in intermediate steps, only round the final answer

## Cheatsheet

| Calculation | Formula |
| --- | --- |
| Moles from mass | $n = \frac{m}{M}$ |
| Moles of solution | $n = cV$ (V in dm³) |
| Moles of gas (r.t.p.) | $n = \frac{V}{24}$ |
| Empirical formula steps | % → ÷Aᵣ → ÷smallest |
| Percentage yield | $\frac{\text{actual}}{\text{theoretical}} \times 100\%$ |
| Percentage purity | $\frac{\text{mass pure}}{\text{mass sample}} \times 100\%$ |
| Molecular formula | $\text{Multiple} = \frac{M_r}{\text{empirical }M_r}$ |

## What's next

Stoichiometric calculations are the foundation of all quantitative chemistry, so mastering this sub-topic is essential for almost every other topic in CIE A-Level Chemistry. You will use the mole concept and balanced equation methods repeatedly in topics like titrations, energetics, equilibrium, and organic synthesis. Building accuracy with these calculations now will help you avoid losing easy marks in exams later. Next, you will explore the structure of the atom, which builds on the understanding of elements and compounds you have developed in stoichiometry.

- [Atomic structure](https://www.owlsprep.com/study/cie-9701-u2-overview/)
- [Atomic nucleus and isotopes](https://www.owlsprep.com/study/cie-9701-u2-atomic-nucleus-and-isotopes/)
- [Electron orbitals and configuration](https://www.owlsprep.com/study/cie-9701-u2-electron-orbitals-and-configuration/)

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