Mole – IGCSE Chemistry Definition
IGCSE Chemistry definition of mole: the amount of substance containing 6.02 x 10^23 particles. Covers key formulae for mass, gas volume, and solution calculations.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
The mole is the unit for amount of substance in chemistry. It bridges the gap between individual atoms (far too small to count) and the grams you weigh on a balance. Mole calculations are the backbone of stoichiometry on the 0620 syllabus and appear on every Paper 4.
The 0620 definition
The mole (mol) is the amount of substance that contains 6.02 x 10²³ particles. One mole of any substance contains the same number of specified particles — this number is the Avogadro constant.
The three key mole formulae
1. Moles from mass
moles = mass (g) / molar mass (g/mol)
The molar mass in g/mol is numerically equal to the relative atomic mass (Ar) or relative molecular mass (Mr).
Example: moles of 4.0 g of NaOH (Mr = 40) = 4.0 / 40 = 0.1 mol
2. Moles of gas at room temperature and pressure (RTP)
moles = volume (cm³) / 24000
or moles = volume (dm³) / 24
At RTP (20 degrees C, 1 atm), one mole of any gas occupies 24 dm³ (24000 cm³). This is the molar gas volume.
Example: moles in 6000 cm³ of CO₂ = 6000 / 24000 = 0.25 mol
3. Moles from solution concentration
moles = concentration (mol/dm³) x volume (dm³)
Example: moles of HCl in 25 cm³ of 0.1 mol/dm³ solution = 0.1 x 0.025 = 0.0025 mol
Using moles in equation calculations
- Write the balanced equation
- Calculate moles of the substance you know
- Use the mole ratio from the equation to find moles of the substance you want
- Convert back to mass, volume, or concentration as needed
Worked exam question
What mass of magnesium oxide is formed when 2.4 g of magnesium is burned in excess oxygen? (Ar: Mg = 24, O = 16) 2Mg + O₂ → 2MgO (3)
Mark scheme
Moles of Mg = 2.4 / 24 = 0.1 mol [1]; from the equation, 2 mol Mg produces 2 mol MgO, so 0.1 mol Mg produces 0.1 mol MgO [1]; mass of MgO = 0.1 x (24 + 16) = 0.1 x 40 = 4.0 g [1]
Common exam mistakes
- Using Ar instead of Mr for compounds. For NaOH, use Mr = 23 + 16 + 1 = 40, not the Ar of any single element.
- Forgetting to convert cm³ to dm³ in solution calculations. Divide by 1000 first.
- Ignoring the mole ratio. The equation coefficients tell you the ratio — do not assume 1:1.
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