Mole calculations are at the heart of IGCSE Chemistry (Cambridge 0620). Almost every paper includes at least one question where you must find moles from mass, from gas volume, or from concentration, or use moles to find mass or concentration. Getting the formulae, units, and working right is essential for full marks. This revision guide covers what the mole is, the three main mole formulae you need, relative formula mass (Mr), and worked examples so you can tackle any similar question in the exam.

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What is the mole?

The mole is the unit of amount of substance in chemistry. One mole of any substance contains 6.02 × 10²³ particles (atoms, molecules, or ions). This number is called the Avogadro constant. So when we say “1 mole of carbon atoms” or “1 mole of water molecules”, we mean 6.02 × 10²³ carbon atoms or 6.02 × 10²³ water molecules.

The mole links the microscopic (particles) to the macroscopic (mass and volume) in a simple way. For any substance, the mass of one mole (in grams) is equal to its relative formula mass, Mr (for compounds) or relative atomic mass, Ar (for elements). For example, the Ar of carbon is 12, so one mole of carbon atoms has a mass of 12 g. The Mr of water (H₂O) is (2×1) + 16 = 18, so one mole of water has a mass of 18 g. You use this idea every time you use the formula moles = mass ÷ Mr.

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Relative formula mass (Mr)

Relative formula mass, Mr, is the sum of the relative atomic masses (Ar) of all the atoms in the formula of a compound. The Ar values are given in the Periodic Table (e.g. H = 1, C = 12, O = 16, Na = 23, Cl = 35.5). You must multiply each Ar by the number of that atom in the formula, then add up.

Example: Find the Mr of sodium carbonate, Na₂CO₃.
Na: 2 × 23 = 46
C: 1 × 12 = 12
O: 3 × 16 = 48
Mr = 46 + 12 + 48 = 106.

Example: Find the Mr of sulfuric acid, H₂SO₄.
H: 2 × 1 = 2
S: 1 × 32 = 32
O: 4 × 16 = 64
Mr = 2 + 32 + 64 = 98.

Always show this working in exams. A common error is to use the atomic number instead of the mass number or Ar — the Ar is the larger number (decimal) in the Periodic Table.

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Moles from mass: moles = mass ÷ Mr

The most used relationship is moles = mass ÷ Mr, where mass is in grams (g) and Mr is the relative formula mass.

Worked example 1: How many moles are there in 21.2 g of sodium carbonate, Na₂CO₃?
Mr(Na₂CO₃) = (2×23) + 12 + (3×16) = 106.
moles = mass ÷ Mr = 21.2 ÷ 106 = 0.2 mol.

Worked example 2: What is the mass of 0.5 mol of water, H₂O?
Mr(H₂O) = (2×1) + 16 = 18.
mass = moles × Mr = 0.5 × 18 = 9 g.

Always write the formula (moles = mass ÷ Mr or mass = moles × Mr), substitute with numbers, and give the unit (mol or g) in your answer.

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Moles from volume of gas: moles = volume ÷ 24

At room temperature and pressure (r.t.p.), one mole of any gas occupies a volume of 24 dm³ (24 000 cm³). So moles = volume ÷ 24, where volume must be in dm³. If the volume is given in cm³, convert to dm³ first by dividing by 1000.

Worked example 3: How many moles of gas are there in 1.2 dm³ at r.t.p.?
moles = volume ÷ 24 = 1.2 ÷ 24 = 0.05 mol.

Worked example 4: A volume of 600 cm³ of oxygen at r.t.p. is collected. How many moles is that?
600 cm³ = 600 ÷ 1000 = 0.6 dm³.
moles = 0.6 ÷ 24 = 0.025 mol.

Forgetting to convert cm³ to dm³ is a very common cause of lost marks. Always check the unit of volume before using moles = V ÷ 24.

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Concentration: moles = concentration × volume

Concentration of a solution is often given in mol/dm³ (moles per cubic decimetre). The relationship is moles = concentration × volume, where volume is in dm³. So concentration = moles ÷ volume (in dm³).

Worked example 5: What is the concentration of a solution containing 0.1 mol of sodium chloride in 0.5 dm³ of solution?
concentration = moles ÷ volume = 0.1 ÷ 0.5 = 0.2 mol/dm³.

Worked example 6: How many moles of solute are there in 250 cm³ of a 0.4 mol/dm³ solution?
250 cm³ = 0.25 dm³.
moles = concentration × volume = 0.4 × 0.25 = 0.1 mol.

If volume is in cm³, convert to dm³ (÷ 1000) before using these formulae. Always give the unit mol/dm³ for concentration.

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Combining the formulae

Many questions combine two steps: for example, find moles from mass, then use the equation to find moles of another substance, then find its mass or concentration.

Worked example 7: What mass of oxygen is produced when 24.5 g of potassium chlorate, KClO₃, decomposes completely? (Equation: 2KClO₃ → 2KCl + 3O₂.)
Mr(KClO₃) = 39 + 35.5 + (3×16) = 122.5.
moles of KClO₃ = 24.5 ÷ 122.5 = 0.2 mol.
From the equation, 2 mol KClO₃ produce 3 mol O₂, so 0.2 mol KClO₃ produce (3/2) × 0.2 = 0.3 mol O₂.
Mr(O₂) = 2×16 = 32.
mass of O₂ = 0.3 × 32 = 9.6 g.

Show each step clearly; marks are often given for correct method even if the final number is wrong.

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Exam tips and command words

• Calculate: Use the correct formula; show substitution; give the answer with the correct unit (mol, g, dm³, mol/dm³).
• Show your working: Write “moles = mass ÷ Mr”, then substitute, then give the numerical answer and unit.
• Convert units: Mass in g; volume in dm³ for moles = V÷24 and for concentration (convert cm³ to dm³ by ÷ 1000).
• State: For “state the value of the Avogadro constant”, give 6.02 × 10²³ (per mole).

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Common mistakes to avoid

• Using volume in cm³ in moles = V ÷ 24 without converting to dm³ (÷ 1000).
• Using atomic number instead of Ar when finding Mr.
• Giving the numerical answer without the unit (e.g. “0.2” instead of “0.2 mol”).
• Using the wrong formula (e.g. mass ÷ volume instead of mass ÷ Mr for moles).

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Revision checklist

• Calculate Mr for a compound from its formula; give the unit (no unit for Mr).
• Calculate moles from mass (moles = mass ÷ Mr) and mass from moles (mass = moles × Mr).
• Calculate moles from gas volume at r.t.p. (moles = V ÷ 24, V in dm³); convert cm³ to dm³.
• Calculate concentration (mol/dm³) from moles and volume, and moles from concentration and volume (volume in dm³).
• Combine mole calculations with a balanced equation to find mass or moles of another substance.

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Next steps

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