What is the mole in Chemistry and why is it important for Cambridge International A Levels?

The mole is a fundamental unit in Chemistry that represents a specific number of particles, typically atoms or molecules. It is defined as the amount of substance containing the same number of entities as there are atoms in 12 grams of carbon-12. This number is known as Avogadro's constant, approximately 6.022 x 10^23. Understanding the mole is crucial for Cambridge International A Levels as it allows students to perform calculations involving chemical reactions, enabling them to determine the amounts of reactants and products involved.

How is Avogadro's constant used in stoichiometry calculations?

Avogadro's constant, approximately 6.022 x 10^23, is used in stoichiometry to convert between the number of moles and the number of particles (atoms, molecules, ions) in a substance. For example, if you have 1 mole of a substance, it contains 6.022 x 10^23 particles. This conversion is essential for solving problems in stoichiometry, such as determining the number of molecules in a given mass of a compound, which is a key skill in the Cambridge International A Levels Chemistry curriculum.

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

To calculate the number of moles from a given mass, you 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 by summing the atomic masses of all the atoms in a molecule. This calculation is fundamental in stoichiometry and is frequently used in Cambridge International A Levels Chemistry to relate mass to the amount of substance.

What is the relationship between the mole and molar volume at standard temperature and pressure (STP)?

At standard temperature and pressure (STP), which is 0°C and 1 atm, one mole of any ideal gas occupies a volume of 22.4 liters. This relationship is known as the molar volume of a gas. It is a crucial concept in stoichiometry, allowing students to convert between the volume of a gas and the number of moles, which is particularly relevant for Cambridge International A Levels Chemistry when dealing with gas reactions.

Why is understanding the mole concept essential for mastering stoichiometry in AS-Level Physical Chemistry?

Understanding the mole concept is essential for mastering stoichiometry because it provides a bridge between the atomic scale and the macroscopic scale. It allows students to quantify substances in chemical reactions, enabling them to calculate reactant and product amounts accurately. This understanding is vital for Cambridge International A Levels Chemistry, as it underpins the ability to solve complex chemical equations and predict the outcomes of reactions, which are key components of the AS-Level Physical Chemistry curriculum.

Why is "Using Ar instead of Mr (e.g. M(O₂) = 16 not 32)." a common mistake in The mole and the Avogadro constant?

Why it happens: Forgetting the element exists as a molecule. How to avoid it: Use the molar mass of the actual species in the question (O₂ = 32.0). Source: 9701 Examiner Reports

Why is "Counting molecules when asked for atoms or ions." a common mistake in The mole and the Avogadro constant?

Why it happens: Not multiplying by the formula subscripts. How to avoid it: Multiply by the number of that particle per formula unit (e.g. 3 O²⁻ per Al₂O₃; 9 O atoms per CuSO₄·5H₂O).

Why is "Giving formula units instead of 'total ions'." a common mistake in The mole and the Avogadro constant?

Why it happens: Stopping at moles of compound. How to avoid it: For NaCl, total ions = 2 × moles of NaCl × L.

Why is "Assuming equal masses contain equal numbers of particles." a common mistake in The mole and the Avogadro constant?

Why it happens: Treating grams as if they counted particles directly. How to avoid it: Convert to moles first — the lighter the molar mass, the more particles per gram. Source: 9701 Examiner Reports 2022-2024

Why is "Treating the Avogadro constant as a mass." a common mistake in The mole and the Avogadro constant?

Why it happens: Confusing 'number per mole' with 'grams per mole'. How to avoid it: L counts particles (mol⁻¹); molar mass M carries the grams (g mol⁻¹) — different quantities.

Atoms, molecules and stoichiometry(AS-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

The mole and the Avogadro constant

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The Mole and the Avogadro Constant — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

The mole as a counting unit for particles, the Avogadro constant, molar mass, and the two equations ( $n = m/M$ and $N = nL$ ) that convert between mass, moles and number of particles.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

2.2.1 — Define and use the term mole in terms of the Avogadro constant.

What is a mole?

A mole is just a count — a fixed, huge number of particles, like 'a dozen' but for atoms.

Atoms are far too small to count one by one, so chemists count them in moles.

Definition. One mole is the amount of substance that contains the same number of particles as there are atoms in exactly 12 g of carbon-12.

That number is the Avogadro constant: $L = 6.02 \times 10^{23}\,\text{mol}^{-1}$

So one mole of anything contains $6.02 \times 10^{23}$ of that particle:

1 mol of He = $6.02 \times 10^{23}$ helium atoms.
1 mol of $\text{H}_2\text{O}$ = $6.02 \times 10^{23}$ water molecules.
1 mol of $\text{Na}^+$ = $6.02 \times 10^{23}$ sodium ions.

Always say what you are counting. "1 mole of oxygen" is ambiguous — 1 mol of $\text{O}$ atoms is different from 1 mol of $\text{O}_2$ molecules.

Mole = amount with as many particles as atoms in 12 g of C-12.
$L = 6.02 \times 10^{23}\,\text{mol}^{-1}$ .
Specify atoms / molecules / ions.

Molar mass and moles from mass

Molar mass is the mass of one mole in grams — numerically the same as Ar or Mr. Use n = m/M.

Molar mass ( $M$ ) is the mass of one mole of a substance, in g mol⁻¹. Its number equals the $A_r$ (for an element) or $M_r$ (for a compound):

$M(\text{C}) = 12.0\,\text{g mol}^{-1}$
$M(\text{H}_2\text{O}) = 18.0\,\text{g mol}^{-1}$
$M(\text{CaCO}_3) = 100.1\,\text{g mol}^{-1}$

The key equation: $n = \frac{m}{M}\quad\text{(moles} = \frac{\text{mass in g}}{\text{molar mass}}\text{)}$

A simple triangle ( $m$ on top, $n$ and $M$ below) helps you rearrange: $m = nM$ and $M = m/n$ .

Worked. How many moles in $25.0\,\text{g}$ of $\text{CaCO}_3$ ? $n = \frac{25.0}{100.1} = 0.250\,\text{mol}$

Molar mass = mass of 1 mole, g mol⁻¹ (= $A_r$ / $M_r$ numerically).
$n = m/M$ ; rearrange to $m = nM$ .
Watch the difference: relative mass (no units) vs molar mass (g mol⁻¹).

Counting particles with the Avogadro constant

Multiply moles by L to get the number of particles — then scale up if a formula unit contains several of them.

To go from moles to an actual number of particles: $N = n \times L$

Worked — molecules. Number of molecules in $0.250\,\text{mol}$ of $\text{CO}_2$ : $N = 0.250 \times 6.02 \times 10^{23} = 1.51 \times 10^{23}\,\text{molecules}$

Worked — ions inside a compound. Number of oxide ions in $0.100\,\text{mol}$ of $\text{Al}_2\text{O}_3$ :

Each formula unit has 3 $\text{O}^{2-}$ ions. $N(\text{O}^{2-}) = 0.100 \times 3 \times 6.02 \times 10^{23} = 1.81 \times 10^{23}$

Worked — atoms in a molecule. Number of hydrogen atoms in $2\,\text{mol}$ of $\text{NH}_3$ : $N(\text{H}) = 2 \times 3 \times 6.02 \times 10^{23} = 3.61 \times 10^{24}$

$N = nL$ .
Multiply by how many of that particle are in each formula unit.
Read the question: molecules vs atoms vs ions.

How it’s examined

The mole underpins every quantitative question. Paper 1 favours 'number of atoms/ions' MCQs (where the formula subscript trap lives); Papers 2 and 5 use $n=m/M$ as the first step of nearly all calculations. Examiner reports flag using the wrong relative mass and forgetting to multiply by the number of ions per formula unit.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 2.2; 9701 Examiner Reports 2022-2024; 9701/11 & 9701/21 May/Jun 2024 question papers and mark schemes. Last reviewed 2026-05-20.

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Step-by-step worked examples — The mole and the Avogadro constant

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

1Moles from mass (2 marks)
Getting started• n=m/M
Question
Calculate the amount, in moles, of $\text{NaOH}$ in $10.0\,\text{g}$ . ( $M_r$ NaOH = 40.0)
Step-by-step solution
1. Step 1
  Use $n = m/M$ .
  $n = \frac{10.0}{40.0}$
2. Step 2
  Evaluate.
  $n = 0.250\,\text{mol}$
Answer
$0.250\,\text{mol}$ .
Examiner tip
Always quote the unit 'mol'. Use the molar mass of the whole formula (40.0), not of Na alone.
2Number of molecules (2 marks)
Getting started• N=nL
Question
How many molecules are in $0.50\,\text{mol}$ of methane, $\text{CH}_4$ ? ( $L = 6.02 \times 10^{23}\,\text{mol}^{-1}$ )
Step-by-step solution
1. Step 1
  $N = nL$ .
  $N = 0.50 \times 6.02 \times 10^{23}$
2. Step 2
  Evaluate.
  $N = 3.01 \times 10^{23}$
Answer
$3.01 \times 10^{23}$ molecules.
3Counting ions in a compound (3 marks)
Building confidence• ions, N=nL
Question
Calculate the total number of ions in $5.85\,\text{g}$ of sodium chloride, NaCl. ( $M_r$ = 58.5)
Step-by-step solution
1. Step 1
  Moles of NaCl.
  $n = \frac{5.85}{58.5} = 0.100\,\text{mol}$
2. Step 2
  Each formula unit gives 2 ions (1 Na⁺ + 1 Cl⁻), so moles of ions $= 0.200$ .
  $N = 0.200 \times 6.02 \times 10^{23}$
3. Step 3
  Evaluate.
  $N = 1.20 \times 10^{23}$
Answer
$1.20 \times 10^{23}$ ions (total).
Examiner tip
Read carefully: 'total number of ions' means count both Na⁺ and Cl⁻ — double the formula units.
4Mass from a number of atoms (3 marks)
Building confidence• m=nM
Question
What is the mass of $3.01 \times 10^{22}$ atoms of carbon? ( $A_r$ C = 12.0)
Step-by-step solution
1. Step 1
  Moles from N: $n = N/L$ .
  $n = \frac{3.01 \times 10^{22}}{6.02 \times 10^{23}} = 0.0500\,\text{mol}$
2. Step 2
  Mass: $m = nM$ .
  $m = 0.0500 \times 12.0 = 0.600\,\text{g}$
Answer
$0.600\,\text{g}$ .
5Counting a specified atom in a compound (4 marks)
Stretch• atoms, hydrate, N=nL
Question
Calculate the number of oxygen atoms in $8.00\,\text{g}$ of $\text{CuSO}_4{\cdot}5\text{H}_2\text{O}$ . ( $M_r$ = 249.6)
Step-by-step solution
1. Step 1
  Moles of the hydrate.
  $n = \frac{8.00}{249.6} = 0.03205\,\text{mol}$
2. Step 2
  Oxygen atoms per formula unit: 4 in $\text{SO}_4$ + 5 in the waters = 9.
  $n(\text{O}) = 9 \times 0.03205 = 0.2885\,\text{mol}$
3. Step 3
  Convert to a count of atoms.
  $N = 0.2885 \times 6.02 \times 10^{23} = 1.74 \times 10^{23}$
Answer
$1.74 \times 10^{23}$ oxygen atoms.
Examiner tip
Don't forget the 5 oxygens in the water of crystallisation — they count too.
6Equal masses, unequal atom counts (3 marks)
Stretch• comparison, atoms
Question
Which contains more atoms in total: $16\,\text{g}$ of $\text{O}_2$ or $16\,\text{g}$ of $\text{CH}_4$ ? ( $M_r$ : O₂ 32.0, CH₄ 16.0)
Step-by-step solution
1. Step 1
  Moles of each substance.
  $n(\text{O}_2) = \frac{16}{32.0} = 0.50;\quad n(\text{CH}_4) = \frac{16}{16.0} = 1.0$
2. Step 2
  Atoms per molecule: O₂ has 2; CH₄ has 5.
  $n_{atoms}(\text{O}_2) = 1.0;\quad n_{atoms}(\text{CH}_4) = 5.0$
3. Step 3
  Compare — CH₄ has five times as many atoms.
Answer
$\text{CH}_4$ has far more atoms (5.0 mol vs 1.0 mol of atoms).
Examiner tip
Equal masses do NOT mean equal numbers of particles — the lighter molar mass and the atoms-per-molecule both matter.

Key Formulae — The mole and the Avogadro constant

The formulae you need to memorise for the mole and the avogadro constant on the Cambridge International A Level 9701 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 of substance / mol
$m$
mass / g
$M$
molar mass / g mol⁻¹
When to use
First step of almost every stoichiometry calculation; rearrange to $m = nM$ or $M = m/n$ .
Number of particles
$N = n \times L$
$N$
number of particles (atoms, molecules, ions, electrons)
$L$
Avogadro constant, 6.02 × 10²³ mol⁻¹
When to use
Converting moles to a count of particles; rearrange to $n = N/L$ . Multiply by atoms/ions per formula unit for a specific particle.

Key Definitions and Keywords — The mole and the Avogadro constant

Definitions to memorise and the exact keywords mark schemes credit for the mole and the avogadro constant answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Mole
Examiner keyword
The amount of substance that contains the same number of particles as there are atoms in 12 g of carbon-12 (i.e. 6.02 × 10²³ particles).
Avogadro constant (L or Nₐ)
Examiner keyword
The number of particles in one mole, 6.02 × 10²³ mol⁻¹.
Molar mass (M)
Examiner keyword
The mass of one mole of a substance, in g mol⁻¹; numerically equal to Ar or Mr.
Amount of substance
Examiner keyword
The quantity measured in moles (symbol n) — the chemist's way of counting particles by weighing.
Formula unit
The simplest group of ions/atoms shown by the formula of a giant or ionic structure (e.g. one NaCl unit = one Na⁺ and one Cl⁻).

Common Mistakes and Misconceptions — The mole and the Avogadro constant

The traps other students keep falling into on the mole and the avogadro constant questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Using Ar instead of Mr (e.g. M(O₂) = 16 not 32).
9701 Examiner Reports
Why it happens
Forgetting the element exists as a molecule.
How to avoid it
Use the molar mass of the actual species in the question (O₂ = 32.0).
✕Counting molecules when asked for atoms or ions.
Why it happens
Not multiplying by the formula subscripts.
How to avoid it
Multiply by the number of that particle per formula unit (e.g. 3 O²⁻ per Al₂O₃; 9 O atoms per CuSO₄·5H₂O).
✕Giving formula units instead of 'total ions'.
Why it happens
Stopping at moles of compound.
How to avoid it
For NaCl, total ions = 2 × moles of NaCl × L.
✕Assuming equal masses contain equal numbers of particles.
9701 Examiner Reports 2022-2024
Why it happens
Treating grams as if they counted particles directly.
How to avoid it
Convert to moles first — the lighter the molar mass, the more particles per gram.
✕Treating the Avogadro constant as a mass.
Why it happens
Confusing 'number per mole' with 'grams per mole'.
How to avoid it
L counts particles (mol⁻¹); molar mass M carries the grams (g mol⁻¹) — different quantities.

Practice questions

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

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The mole and the Avogadro constant

Short Study Notes in the form of Slides

Short Study Notes — The mole and the Avogadro constant

Detailed Notes

What you’ll learn

How it’s examined

1Moles from mass (2 marks)

2Number of molecules (2 marks)

3Counting ions in a compound (3 marks)

4Mass from a number of atoms (3 marks)

5Counting a specified atom in a compound (4 marks)

6Equal masses, unequal atom counts (3 marks)

Moles from mass

Number of particles

Mole

Avogadro constant (L or Nₐ)

Molar mass (M)

Amount of substance

Formula unit

✕Using Ar instead of Mr (e.g. M(O₂) = 16 not 32).

✕Counting molecules when asked for atoms or ions.

✕Giving formula units instead of 'total ions'.

✕Assuming equal masses contain equal numbers of particles.

✕Treating the Avogadro constant as a mass.

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

The mole and the Avogadro constant — frequently asked questions