Stoichiometric Calculations

Chemistry· Unit 1: Atomic structure and stoichiometry· 25 min read

1. Core Mole Calculations★★☆☆☆⏱ 8 min

📘 Definition

Amount of substance

$n$

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

Example:

1 mole of water contains $6.02 \times 1 0^{23}$ 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 $n = c V$ .

2. Empirical and Molecular Formulas★★☆☆☆⏱ 9 min

📘 Definition

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).

3. Reacting Quantity Calculations★★★☆☆⏱ 8 min

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

📐 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 $n (C O_{2}) = 0.0625$ mol.
6
Calculate mass of carbon dioxide:
7
$M(CO_2) = 12.0 + (2 \times 16.0) = 44.0 \text{ g mol}^{-1}; \quad m(CO_2) = 0.0625 \times 44.0 = 2.75 \text{ g}$

4. Percentage Yield and Purity★★★☆☆⏱ 7 min

📘 Definition

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: $C a C O_{3} + 2 H C l \to C a C l_{2} + C O_{2} + H_{2} O$ . 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
$\text{Percentage purity} = \frac{0.793}{1.00} \times 100\% = 79\% \text{ (2 s.f.)}$

5. Common Pitfalls

Wrong move:

Forgetting to convert volume from cm³ to dm³ for solution calculations

Why:

Concentration is given in mol dm⁻³, so using cm³ gives an answer 1000 times too small

Correct move:

Divide volume in cm³ by 1000 to convert to dm³ before substituting into $n = c V$

Wrong move:

Using an unbalanced equation to get the mole ratio

Why:

An unbalanced equation gives the wrong stoichiometric ratio, leading to an incorrect answer

Correct move:

Always balance the equation before starting any calculation, and double-check the balancing

Wrong move:

Leaving the answer as the empirical formula instead of finding the molecular formula

Why:

The empirical formula is only the simplest ratio, not the actual formula of the compound

Correct move:

Always compare the empirical formula mass to the given relative molecular mass and multiply by the correct multiple

Wrong move:

Confusing molar volume at r.t.p. and s.t.p.

Why:

CIE uses 24.0 dm³ mol⁻¹ for r.t.p., not 22.4 dm³ mol⁻¹ which is for s.t.p.

Correct move:

Check the question conditions, and refer to the Data Booklet for the correct $V_{m}$ value

Wrong move:

Rounding intermediate values too early leading to inaccurate final answers

Why:

Rounding intermediate steps to 2 significant figures introduces large errors into the final result

Correct move:

Keep at least one extra significant figure in intermediate steps, only round the final answer

6. Quick Reference Cheatsheet

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

7. Frequently Asked

Do I need to remember Avogadro's constant and molar volume?

No, CIE provides all constants ( $N_{A} = 6.02 \times 1 0^{23}$ mol⁻¹, $V_{m} = 24.0$ dm³ mol⁻¹ at r.t.p.) in the Data Booklet for all exams.

How many significant figures should I use for my answer?

Use the same number of significant figures as the least precise value given in the question, usually 2 or 3 significant figures, unless stated otherwise.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
Empirical formula calculation
2023 · 21
Yield and purity problem
2021 · 11
Gas volume stoichiometry

Going deeper

syllabusCIE 9701 2025-2026 SyllabusOfficial syllabus document

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.