What is the mole concept in stoichiometry and why is it important in Cambridge IGCSE Coordinated Science?

The mole is a fundamental concept in stoichiometry that represents a specific quantity of particles, typically atoms or molecules. One mole is defined as 6.022 x 10^23 particles, known as Avogadro's number. This concept is crucial in Cambridge IGCSE Coordinated Science because it allows students to convert between the mass of a substance and the number of particles, facilitating calculations in chemical reactions and equations.

How do you calculate the number of moles from a given mass in a chemical reaction?

To calculate the number of moles from a given mass, you need to use the formula: number of moles = mass (g) / molar mass (g/mol). The molar mass is the mass of one mole of a substance and can be found on the periodic table as the atomic or molecular weight. This calculation is essential for stoichiometry problems in the Cambridge IGCSE Coordinated Science curriculum.

What is Avogadro's number and how is it used in stoichiometry calculations?

Avogadro's number, 6.022 x 10^23, is the number of atoms, ions, or molecules in one mole of a substance. In stoichiometry, it is used to convert between the number of particles and the amount of substance in moles. This conversion is vital for understanding the relationships between reactants and products in chemical equations, a key aspect of the Cambridge IGCSE Coordinated Science syllabus.

How can the mole concept help in balancing chemical equations?

The mole concept helps in balancing chemical equations by ensuring that the number of moles of each element is the same on both sides of the equation. This reflects the conservation of mass, a fundamental principle in chemistry. By using moles, students can accurately determine the proportions of reactants and products, which is a critical skill in the Cambridge IGCSE Coordinated Science curriculum.

What is the relationship between moles, molar mass, and volume in stoichiometry?

In stoichiometry, the relationship between moles, molar mass, and volume is crucial for solving chemical problems. Moles relate to the number of particles, molar mass connects moles to the mass of a substance, and volume is often used in reactions involving gases. At standard temperature and pressure (STP), one mole of an ideal gas occupies 22.4 liters. Understanding these relationships is essential for mastering stoichiometry in the Cambridge IGCSE Coordinated Science course.

Why is "Forgetting to convert cm³ to dm³ when calculating concentration" a common mistake in The Mole?

Why it happens: Students substitute the cm³ value directly into c = n/V, giving an answer 1000 times too large. How to avoid it: Always divide cm³ by 1000 to get dm³ before using c = n/V. Write '÷ 1000' as a clearly labelled step. Source: 0654 Examiner Report 2023

Why is "Ignoring the mole ratio from the balanced equation" a common mistake in The Mole?

Why it happens: Students assume 1 mol of reactant always gives 1 mol of product. How to avoid it: Always identify the mole ratio from the coefficients in the balanced equation before calculating moles of product.

Why is "Using 22.4 dm³/mol instead of 24 dm³/mol for molar gas volume" a common mistake in The Mole?

Why it happens: 22.4 dm³/mol is the STP value (0 °C); Cambridge IGCSE uses rtp (25 °C) where molar volume = 24 dm³/mol. How to avoid it: For Cambridge IGCSE questions, always use 24 dm³/mol unless the question specifies STP.

StoichiometryCambridge IGCSE Coordinated Sciences 0654

The Mole

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The Mole — Cambridge IGCSE Coordinated Science 0654

The mole is the SI unit for amount of substance — 6 × 10²³ particles. Cambridge Extended tests mole calculations, molar volume of gases, and concentration. These calculation questions are worth significant marks in Paper 4.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

C4.5 — Define the mole and Avogadro's number.
C4.6 — Calculate moles from mass and relative formula mass.
C4.7 — Calculate moles from volume and concentration of a solution.
C4.8 — Use molar volume to calculate moles of a gas.
C4.9 — Perform stoichiometric calculations using moles.

The Mole

The mole connects the number of particles to measurable mass. 1 mole of any substance contains 6.02 × 10²³ particles.

The mole is the SI unit for amount of substance.

Avogadro's number: 1 mole = 6.02 × 10²³ particles (atoms, molecules, ions, etc.)

Molar mass: the mass of one mole of a substance = Mr in grams per mole (g/mol).

Molar mass of H₂O = 18 g/mol (so 1 mole of H₂O weighs 18 g)
Molar mass of NaCl = 58.5 g/mol

The mole triangle:

n = m / Mr where:

n = number of moles

m = mass in grams

Mr = relative formula mass

Examples:

How many moles in 40 g of NaOH (Mr = 40)? n = 40/40 = 1 mol
What mass is 0.5 mol of CO₂ (Mr = 44)? m = 0.5 × 44 = 22 g

1 mole = 6.02 × 10²³ particles. Molar mass = Mr in g/mol.
n = m / Mr. m = n × Mr. Mr = m / n.

Moles in Solutions and Gases

For solutions: moles = concentration × volume. For gases at r.t.p.: moles = volume / 24 dm³.

Moles in solutions:

n = c × V where c = concentration in mol/dm³, V = volume in dm³

1 dm³ = 1000 cm³. Always convert cm³ to dm³ by dividing by 1000.

Examples:

How many moles in 250 cm³ of 0.1 mol/dm³ NaOH? V = 0.25 dm³. n = 0.1 × 0.25 = 0.025 mol
What is the concentration of 0.05 mol NaCl in 200 cm³? V = 0.2 dm³. c = 0.05/0.2 = 0.25 mol/dm³

Molar volume of gases (at r.t.p. — room temperature and pressure):

1 mole of ANY gas occupies 24 dm³ (24,000 cm³) at r.t.p.

n = V / 24 (V in dm³) or n = V / 24000 (V in cm³)

Examples:

How many moles in 12 dm³ of CO₂ at r.t.p.? n = 12/24 = 0.5 mol
What volume does 0.25 mol of O₂ occupy at r.t.p.? V = 0.25 × 24 = 6 dm³

Using moles in reactions (full calculation):

Write balanced equation
Convert given quantity to moles
Use mole ratio from equation
Convert moles of product to required unit (mass, volume, concentration)

n = c × V (solutions). Concentration in mol/dm³, volume in dm³.
1 mole of any gas = 24 dm³ at r.t.p. n = V/24.
Always: given quantity → moles → mole ratio → answer.

How it’s examined

Paper 4: Mole calculations are 3-5 mark structured questions. Typical format: 'Calculate the number of moles of X in Y grams' → 'Calculate the volume of gas produced at r.t.p.' — must show working clearly. Titration calculations use n = c × V for both acid and alkali, then ratio from equation. Examiner reports note that students lose marks by forgetting to convert cm³ to dm³ or by not showing the mole ratio step.

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (C4); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

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Step-by-step worked examples — The Mole

Step-by-step solutions to past-paper-style questions on the mole, written exactly the way a tutor would explain them at the board.

1Moles from mass
Core
Question
Calculate the number of moles in $8\,\text{g}$ of calcium. [Ca = 40]
Step-by-step solution
1. Step 1
  Use the moles formula.
  $n = \frac{m}{M}$
2. Step 2
  Substitute: m = 8 g, M = 40 g/mol.
  $n = \frac{8}{40} = 0.2\,\text{mol}$
Answer
$n = 0.2\,\text{mol}$ of calcium.
2Volume of gas from mole ratio
Extended
Question
What volume of hydrogen gas (at rtp) is produced when $0.5\,\text{mol}$ of zinc reacts with excess hydrochloric acid? $\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$
Step-by-step solution
1. Step 1
  From the equation, the mole ratio Zn : H₂ = 1 : 1. So 0.5 mol Zn produces 0.5 mol H₂.
2. Step 2
  At rtp, 1 mol of any gas occupies 24 dm³.
  $V = n \times 24 = 0.5 \times 24 = 12\,\text{dm}^3$
Answer
$12\,\text{dm}^3$ of hydrogen gas.
Examiner tip
rtp = room temperature and pressure (25 °C, 1 atm). The molar gas volume is 24 dm³/mol at rtp; use 24 000 cm³/mol if the answer is required in cm³.
3Concentration of a NaOH solution
Extended
Question
Calculate the concentration of a solution containing $4\,\text{g}$ of NaOH dissolved in $500\,\text{cm}^3$ of solution. [Na = 23, O = 16, H = 1]
Step-by-step solution
1. Step 1
  Calculate Mr of NaOH.
  $M_r = 23 + 16 + 1 = 40\,\text{g/mol}$
2. Step 2
  Calculate moles of NaOH.
  $n = \frac{4}{40} = 0.1\,\text{mol}$
3. Step 3
  Convert volume to dm³: 500 cm³ = 0.5 dm³.
  $c = \frac{n}{V} = \frac{0.1}{0.5} = 0.2\,\text{mol/dm}^3$
Answer
$c(\text{NaOH}) = 0.2\,\text{mol/dm}^3$
Examiner tip
Always convert cm³ to dm³ by dividing by 1000 before substituting into c = n/V.
4Moles from solution volume (titration)
Extended
Question
$25.0\,\text{cm}^3$ of $0.100\,\text{mol/dm}^3$ sodium hydroxide solution is exactly neutralised by hydrochloric acid. Calculate the moles of NaOH used.
Step-by-step solution
1. Step 1
  Convert volume: 25.0 cm³ = 0.0250 dm³.
  $V = 0.0250\,\text{dm}^3$
2. Step 2
  Apply n = cV.
  $n(\text{NaOH}) = 0.100 \times 0.0250 = 0.00250\,\text{mol}$
Answer
$n(\text{NaOH}) = 2.50 \times 10^{-3}\,\text{mol}$
5Mass of CaCO₃ to produce given moles of CO₂
Extended
Question
In the reaction $2\text{HCl} + \text{CaCO}_3 \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ , calculate the mass of $\text{CaCO}_3$ needed to produce $0.5\,\text{mol}$ of $\text{CO}_2$ . [Ca = 40, C = 12, O = 16]
Step-by-step solution
1. Step 1
  Mole ratio CaCO₃ : CO₂ = 1 : 1 (from the equation). So 0.5 mol CO₂ requires 0.5 mol CaCO₃.
2. Step 2
  Mr of CaCO₃ = 40 + 12 + 48 = 100 g/mol.
3. Step 3
  Calculate mass.
  $m = n \times M = 0.5 \times 100 = 50\,\text{g}$
Answer
$50\,\text{g}$ of calcium carbonate.

Key Formulae — The Mole

The formulae you need to memorise for the mole on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Moles from mass
$n = \frac{m}{M}$
$n$
amount in moles (mol)
$m$
mass in grams (g)
$M$
molar mass in g/mol (numerically equal to Mr)
When to use
Converting between mass and moles for any substance.
Example
8 g Ca ÷ 40 g/mol = 0.2 mol
Moles of gas (at rtp)
$n = \frac{V(\text{dm}^3)}{24}$
$n$
amount in moles (mol)
$V$
volume of gas in dm³
$24$
molar volume at rtp = 24 dm³/mol
When to use
Converting between volume and moles of a gas at room temperature and pressure.
Concentration
$c = \frac{n}{V}$
$c$
concentration in mol/dm³
$n$
amount in moles (mol)
$V$
volume of solution in dm³ (divide cm³ by 1000)
When to use
Finding concentration, or moles in a given volume of solution.
Example
$n = cV$ : 0.1 mol/dm³ × 0.025 dm³ = 0.0025 mol

Key Definitions and Keywords — The Mole

Definitions to memorise and the exact keywords mark schemes credit for the mole answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Mole (mol)
Examiner keyword
The SI unit for amount of substance. One mole contains 6.02 × 10²³ particles (atoms, molecules, ions, or formula units). This number is Avogadro's constant.
Avogadro's constant (Nₐ)
Examiner keyword
$6.02 \times 10^{23}\,\text{mol}^{-1}$ — the number of specified particles in one mole of a substance.
Molar mass (M)
Examiner keyword
The mass of one mole of a substance in g/mol. Numerically equal to the relative formula mass (Mr).
Molar gas volume (at rtp)
Examiner keyword
The volume occupied by one mole of any gas at room temperature and pressure: 24 dm³ (24 000 cm³).
Concentration
Examiner keyword
The amount of solute dissolved per unit volume of solution; units mol/dm³ (or mol/L). Calculated as $c = n/V$ where V is in dm³.
Stoichiometric ratio (mole ratio)
The ratio of the amounts (in moles) of reactants and products as given by the balanced chemical equation coefficients.

Common Mistakes and Misconceptions — The Mole

The traps other students keep falling into on the mole questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to convert cm³ to dm³ when calculating concentration
0654 Examiner Report 2023
Why it happens
Students substitute the cm³ value directly into c = n/V, giving an answer 1000 times too large.
How to avoid it
Always divide cm³ by 1000 to get dm³ before using c = n/V. Write '÷ 1000' as a clearly labelled step.
✕Ignoring the mole ratio from the balanced equation
Why it happens
Students assume 1 mol of reactant always gives 1 mol of product.
How to avoid it
Always identify the mole ratio from the coefficients in the balanced equation before calculating moles of product.
✕Using 22.4 dm³/mol instead of 24 dm³/mol for molar gas volume
Why it happens
22.4 dm³/mol is the STP value (0 °C); Cambridge IGCSE uses rtp (25 °C) where molar volume = 24 dm³/mol.
How to avoid it
For Cambridge IGCSE questions, always use 24 dm³/mol unless the question specifies STP.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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The Mole — frequently asked questions

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The Mole

Short Study Notes in the form of Slides

Short Study Notes — The Mole

Detailed Notes

What you’ll learn

How it’s examined

1Moles from mass

2Volume of gas from mole ratio

3Concentration of a NaOH solution

4Moles from solution volume (titration)

5Mass of CaCO₃ to produce given moles of CO₂

Moles from mass

Moles of gas (at rtp)

Concentration

Mole (mol)

Avogadro's constant (Nₐ)

Molar mass (M)

Molar gas volume (at rtp)

Concentration

Stoichiometric ratio (mole ratio)

✕Forgetting to convert cm³ to dm³ when calculating concentration

✕Ignoring the mole ratio from the balanced equation

✕Using 22.4 dm³/mol instead of 24 dm³/mol for molar gas volume

Practice questions

Past paper style quiz

Assess your understanding

The Mole — frequently asked questions