What is the mole in Chemistry and why is it important in stoichiometry?

In Chemistry, a mole is a fundamental unit used to measure the amount of substance. It is important in stoichiometry because it allows chemists to count particles like atoms, molecules, or ions in a given sample by relating mass to the number of particles. This is crucial for balancing chemical equations and calculating reactant and product quantities in chemical reactions, which is a key aspect of stoichiometry.

How is the Avogadro Constant defined and what is its significance?

The Avogadro Constant, approximately 6.022 x 10^23 mol^-1, defines the number of particles in one mole of a substance. It is significant because it provides a bridge between the macroscopic scale of grams and the microscopic scale of atoms and molecules, allowing chemists to convert between mass and number of particles, which is essential for stoichiometric calculations.

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: 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 a fundamental step in stoichiometry for determining how much of a substance is involved in a chemical reaction.

What is the relationship between the mole, molar mass, and Avogadro's number in stoichiometry?

The relationship between the mole, molar mass, and Avogadro's number is central to stoichiometry. One mole of any substance contains Avogadro's number of particles and has a mass equal to its molar mass. This relationship allows chemists to convert between mass, moles, and number of particles, facilitating the quantitative analysis of chemical reactions.

Why is understanding the concept of the mole crucial for Cambridge IGCSE Chemistry students?

Understanding the concept of the mole is crucial for Cambridge IGCSE Chemistry students because it is foundational for mastering stoichiometry, which involves calculating reactants and products in chemical reactions. The mole concept helps students grasp how chemical equations are balanced and how to predict the outcomes of reactions, which are essential skills for success in the IGCSE Chemistry curriculum.

Why is "Using $\text{cm}^{3}$ in $c = n/V$" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Question gives cm^3. How to avoid it: Convert: V(dm^3) = V(cm^3) / 1000. Source: 0620/42 — every series

Why is "Mixing mass with moles in stoichiometric ratios" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Forgetting that ratios in equations are MOLES, not grams. How to avoid it: Convert mass → moles first; apply ratio; then convert back to mass if needed.

Why is "Writing $N_{A}$ as $6.022 \times 10^{22}$" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Misremembering the exponent. How to avoid it: N_A = 6.02 × 10^23. Big number.

Why is "Quoting concentration in g/dm³ when mol/dm³ is required" a common mistake in The Mole and the Avagadro Constant?

Why it happens: Mixing concentration units. How to avoid it: Read the question. mol/dm³ if 'in moles', else g/dm³.

StoichiometryCambridge IGCSE Chemistry 0620 Extended

The Mole and the Avagadro Constant

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The Mole and the Avogadro Constant — Cambridge IGCSE 0620 Chemistry Extended (2026)

$n = m/M_r$ , $n_{\text{gas}} = V/24\,\text{dm}^{3}$ at r.t.p., concentration $c = n/V$ . The mole is the bridge between equation coefficients and lab quantities.

What you’ll learn

Mapped to the Cambridge IGCSE 0620 syllabus (2026-2028).

3.3 — Define the mole and Avogadro's constant.
3.3 — Convert between mass, moles, $M_r$ , gas volume, and concentration.
3.3 — Use mole ratios in balanced equations.
3.3 — Identify the limiting reagent in a reaction (Extended).

What is a mole?

$6.02 \times 10^{23}$ particles. Defined as the amount in $12\,\text{g}$ of $^{12}$ C.

Mole. The amount of substance containing the same number of particles as there are atoms in $12\,\text{g}$ of $^{12}\text{C}$ . That number is Avogadro's constant: $N_{A} = 6.02 \times 10^{23}\,\text{mol}^{-1}.$

Why $12\,\text{g}$ of $^{12}\text{C}$ ? It's a definition. The point is: 1 mole of any substance has $6.02 \times 10^{23}$ particles.

Worked. How many atoms in $0.5\,\text{mol}$ of iron?

$0.5 \times 6.02 \times 10^{23} = 3.01 \times 10^{23}$ atoms.

Mole and mass. The mass of 1 mole of an element (in grams) is numerically equal to its $A_r$ . So 1 mole of carbon = $12\,\text{g}$ ; 1 mole of iron = $56\,\text{g}$ .

For compounds, 1 mole has a mass equal to $M_r$ in grams. 1 mol of $\text{H}_{2}\text{O}$ = $18\,\text{g}$ .

Conversion: mass ↔ moles. $n = \dfrac{m}{M_r}, \quad m = n \times M_r.$

Always convert TO moles first — the mole is the hub that links mass, gas volume, particles and concentration.

Worked. Convert $36\,\text{g}$ of water to moles.

$n = 36/18 = 2\,\text{mol}$ .

Worked. Convert $0.25\,\text{mol}$ of CO₂ to mass.

$m = 0.25 \times 44 = 11\,\text{g}$ .

$1\,\text{mol} = 6.02 \times 10^{23}$ particles.
Mass of 1 mol = $A_r$ or $M_r$ in grams.
$n = m/M_r$ .
Conversion is the most-tested skill.

Gas volume and concentration

1 mol of gas at r.t.p. = $24\,\text{dm}^{3}$ . Concentration = mol/dm³.

Gas volume at r.t.p. Room temperature and pressure (about $20\degree\text{C}$ , $1\,\text{atm}$ ): 1 mole of any gas occupies $24\,\text{dm}^{3}$ ( $24{,}000\,\text{cm}^{3}$ ). $n_{\text{gas}} = \dfrac{V}{24\,\text{dm}^{3}/\text{mol}}.$

Worked. What volume does $0.5\,\text{mol}$ of CO₂ occupy at r.t.p.?

$V = 0.5 \times 24 = 12\,\text{dm}^{3}$ .

Concentration of solutions. $c = \dfrac{n}{V}, \quad n = c \times V,$ where $V$ is in $\text{dm}^{3}$ ( $1\,\text{dm}^{3} = 1000\,\text{cm}^{3} = 1\,\text{L}$ ). Units: $\text{mol/dm}^{3}$ (sometimes written $M$ ).

Worked. $250\,\text{cm}^{3}$ of $0.4\,\text{mol/dm}^{3}$ HCl. Find moles of HCl.

$V = 250/1000 = 0.25\,\text{dm}^{3}$ .
$n = 0.4 \times 0.25 = 0.1\,\text{mol}$ .

Cambridge tip. Watch the volume unit — convert $\text{cm}^{3}$ to $\text{dm}^{3}$ by dividing by $1000$ .

1 mol gas at r.t.p. = $24\,\text{dm}^{3}$ .
$n = V/24$ (with $V$ in dm³).
Concentration: $c = n/V$ , units mol/dm³.
Convert cm³ → dm³ by dividing by 1000.

Using mole ratios from balanced equations

Coefficients in a balanced equation are mole ratios. Use them to find quantities of products from quantities of reactants.

Steps for stoichiometric calculations.

Write the BALANCED equation.
Convert known quantity to moles.
Apply the mole ratio from the equation.
Convert the answer back to the required units (mass or volume).

Never skip the mole ratio — moving straight from one mass to another loses marks.

Worked. What mass of magnesium oxide forms from $4.8\,\text{g}$ of magnesium?

Equation: $2\text{Mg} + \text{O}_{2} \to 2\text{MgO}$ .
Moles of Mg: $4.8/24 = 0.2\,\text{mol}$ .
Mole ratio Mg : MgO = $2 : 2 = 1 : 1$ .
Moles of MgO: $0.2\,\text{mol}$ .
Mass of MgO: $0.2 \times 40 = 8\,\text{g}$ .

Worked. Volume of CO₂ from burning $0.5\,\text{mol}$ of methane?

Equation: $\text{CH}_{4} + 2\text{O}_{2} \to \text{CO}_{2} + 2\text{H}_{2}\text{O}$ .
Mole ratio CH₄ : CO₂ = $1 : 1$ .
Moles of CO₂: $0.5\,\text{mol}$ .
Volume at r.t.p.: $0.5 \times 24 = 12\,\text{dm}^{3}$ .

Limiting reagent (Extended). When two reactants are mixed in non-stoichiometric amounts, one runs out first — the limiting reagent. The product yield is determined by the limiting reagent.

Worked. $0.4\,\text{mol}$ Mg reacted with $0.1\,\text{mol}$ O₂. Equation: $2\text{Mg} + \text{O}_{2} \to 2\text{MgO}$ .

Mg available: $0.4\,\text{mol}$ . To use it all, you'd need $0.2\,\text{mol}$ O₂.
O₂ available: only $0.1\,\text{mol}$ .
O₂ is LIMITING.
Product: $0.1 \times 2 = 0.2\,\text{mol}$ MgO.
$0.2\,\text{mol}$ Mg leftover (unreacted).

Coefficients = mole ratios.
Find moles → use ratio → convert to required units.
Limiting reagent = runs out first.
Product yield set by limiting reagent.

How it’s examined

Mole calculations appear every Paper 4 (8-12 marks) and most Paper 2s (3-5 marks). Cambridge stacks $n = m/M_r$ , gas volume, concentration, and stoichiometry into multi-step problems. Examiner reports flag dm³/cm³ unit confusion and forgetting the mole ratio (just using mass directly).

Sources: Cambridge IGCSE Chemistry 0620 syllabus 2026-2028 (3.3); 0620/42 Oct/Nov 2024 — Q8 (mole + limiting reagent); 0620 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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Step-by-step worked examples — The Mole and the Avagadro Constant

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

1Mass ↔ moles
Core• n=m/M
Question
Find the moles in $25.0\,\text{g}$ of $\text{CaCO}_{3}$ . ( $M_{r} = 100$ .)
Step-by-step solution
1. Step 1
  $n = m / M_{r}$ .
  $n = 25.0 / 100 = 0.250\,\text{mol}$
Answer
$0.250\,\text{mol}$
2Use stoichiometric ratios
Extended• Adapted from 0620/42 May/Jun 2024 Q11• stoichiometry
Question
How many grams of $\text{CaO}$ form when $50\,\text{g}$ of $\text{CaCO}_{3}$ decomposes fully? $\text{CaCO}_{3} \to \text{CaO} + \text{CO}_{2}$ . ( $M_{r}$ : $\text{CaCO}_{3} = 100$ , $\text{CaO} = 56$ .)
Step-by-step solution
1. Step 1
  Moles of $\text{CaCO}_{3}$ .
  $n = 50/100 = 0.5$
2. Step 2
  1:1 ratio → 0.5 mol of CaO.
3. Step 3
  Mass.
  $m = n M_{r} = 0.5 \times 56 = 28\,\text{g}$
Answer
$28\,\text{g}$
3Concentration of a solution
Extended• c=n/V
Question
$0.10\,\text{mol}$ of NaOH is dissolved to make $250\,\text{cm}^{3}$ of solution. Find the concentration in $\text{mol/dm}^{3}$ .
Step-by-step solution
1. Step 1
  Convert volume to dm³.
  $V = 250/1000 = 0.250\,\text{dm}^{3}$
2. Step 2
  $c = n/V$ .
  $c = 0.10/0.250 = 0.40\,\text{mol/dm}^{3}$
Answer
$0.40\,\text{mol/dm}^{3}$
4Volume of a gas at room conditions
Extended• gas volume
Question
Find the volume occupied by $0.5\,\text{mol}$ of $\text{CO}_{2}$ at room temperature & pressure (rtp). $V_{m} = 24\,\text{dm}^{3}/\text{mol}$ .
Step-by-step solution
1. Step 1
  $V = n V_{m}$ .
  $V = 0.5 \times 24 = 12\,\text{dm}^{3}$
Answer
$12\,\text{dm}^{3}$
5Titration calculation
Extended• titration
Question
$25.0\,\text{cm}^{3}$ of NaOH solution is exactly neutralised by $20.0\,\text{cm}^{3}$ of $0.100\,\text{mol/dm}^{3}$ HCl. Find the concentration of NaOH. $\text{NaOH} + \text{HCl} \to \text{NaCl} + \text{H}_{2}\text{O}$ .
Step-by-step solution
1. Step 1
  Moles of HCl.
  $n_{\text{HCl}} = c V = 0.100 \times 0.020 = 0.00200$
2. Step 2
  1:1 → moles of NaOH.
  $n_{\text{NaOH}} = 0.00200$
3. Step 3
  Concentration.
  $c = 0.00200 / 0.0250 = 0.0800\,\text{mol/dm}^{3}$
Answer
$0.0800\,\text{mol/dm}^{3}$

Key Formulae — The Mole and the Avagadro Constant

The formulae you need to memorise for the mole and the avagadro constant on the Cambridge IGCSE 0620 paper, with every variable defined in plain English and a note on when to use it.

Moles from mass
$n = \dfrac{m}{M_{r}}$
$n$
moles
$m$
mass in grams
When to use
Convert mass to moles.
Concentration
$c = \dfrac{n}{V}$
$c$
concentration in mol/dm³
$V$
volume of solution in dm³
When to use
Solutions in stoichiometry.
Molar gas volume
$V = n V_{m},\ \ V_{m} = 24\,\text{dm}^{3}/\text{mol at rtp}$
When to use
Gas volumes at room temperature and pressure.

Key Definitions and Keywords — The Mole and the Avagadro Constant

Definitions to memorise and the exact keywords mark schemes credit for the mole and the avagadro constant answers — sharpened from recent examiner reports for the 2026 0620 sitting.

Mole
Examiner keyword
The amount of substance containing $6.02 \times 10^{23}$ particles (Avogadro's number).
Avogadro constant ( $N_{A}$ )
Examiner keyword
$6.02 \times 10^{23}$ particles per mole.
Room temperature and pressure (rtp)
Examiner keyword
$25\degree\text{C}$ and $1\,\text{atm}$ . Molar gas volume $\approx 24\,\text{dm}^{3}$ .
Limiting reagent
Examiner keyword
The reactant fully used up first, fixing the maximum amount of product.

Common Mistakes and Misconceptions — The Mole and the Avagadro Constant

The traps other students keep falling into on the mole and the avagadro constant questions — taken from recent Cambridge IGCSE 0620 examiner reports and mark schemes — and how to avoid them.

✕Using $\text{cm}^{3}$ in $c = n/V$
0620/42 — every series
Why it happens
Question gives $\text{cm}^{3}$ .
How to avoid it
Convert: $V(\text{dm}^{3}) = V(\text{cm}^{3}) / 1000$ .
✕Mixing mass with moles in stoichiometric ratios
Why it happens
Forgetting that ratios in equations are MOLES, not grams.
How to avoid it
Convert mass → moles first; apply ratio; then convert back to mass if needed.
✕Writing $N_{A}$ as $6.022 \times 10^{22}$
Why it happens
Misremembering the exponent.
How to avoid it
$N_{A} = 6.02 \times 10^{23}$ . Big number.
✕Quoting concentration in g/dm³ when mol/dm³ is required
Why it happens
Mixing concentration units.
How to avoid it
Read the question. mol/dm³ if 'in moles', else g/dm³.

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