The mole concept and stoichiometry: a step-by-step guide

The mole concept is the backbone of chemical calculations. Almost every calculation in senior secondary Chemistry (reacting masses, gas volumes, concentration, titration and electrolysis) depends on it. Students who understand moles find these questions straightforward; students who don't, find them terrifying. This guide builds the idea from the beginning.

What is a mole?

A mole is a counting unit, like a "dozen" but much bigger. One mole of anything contains 6.02 × 10²³ particles. This number is called the Avogadro constant (L or Nₐ).

• 1 mole of carbon atoms = 6.02 × 10²³ carbon atoms
• 1 mole of water molecules = 6.02 × 10²³ water molecules

Chemists use moles because atoms are far too small to count individually, but we can weigh them in bulk.

Molar mass

The molar mass of a substance is the mass of one mole of it, in grams per mole (g/mol). It equals the relative atomic or molecular mass in grams.

To find it, add up the relative atomic masses of all the atoms in the formula.

Example: molar mass of sulphuric acid, H₂SO₄ (H = 1, S = 32, O = 16): (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98 g/mol

Example: molar mass of calcium carbonate, CaCO₃ (Ca = 40, C = 12, O = 16): 40 + 12 + (3 × 16) = 100 g/mol

The four key mole formulas

s.t.p. means standard temperature and pressure (0 °C and 1 atm). At s.t.p., one mole of any gas occupies 22.4 dm³. Use the molar volume stated in your question; some questions use room conditions (24 dm³).

Worked example 1: Mass to moles

How many moles are in 4.9 g of H₂SO₄? (H = 1, S = 32, O = 16)

Molar mass = 98 g/mol. Moles = 4.9 ÷ 98 = 0.05 mol

Worked example 2: Moles to particles

How many molecules are in 0.25 mol of water?

Number of molecules = 0.25 × 6.02 × 10²³ = 1.505 × 10²³ molecules

Worked example 3: Gas volume

What volume does 8.8 g of carbon(IV) oxide occupy at s.t.p.? (C = 12, O = 16; molar volume = 22.4 dm³)

Molar mass of CO₂ = 12 + 32 = 44 g/mol. Moles = 8.8 ÷ 44 = 0.2 mol. Volume = 0.2 × 22.4 = 4.48 dm³

Worked example 4: Concentration

What is the concentration of a solution containing 4 g of sodium hydroxide in 500 cm³ of solution? (Na = 23, O = 16, H = 1)

Molar mass of NaOH = 23 + 16 + 1 = 40 g/mol. Moles = 4 ÷ 40 = 0.1 mol. Volume = 500 cm³ = 0.5 dm³ (divide cm³ by 1000). Concentration = 0.1 ÷ 0.5 = 0.2 mol/dm³

Stoichiometry: using balanced equations

Stoichiometry is calculating the amounts of substances in a reaction. A balanced equation tells you the mole ratio in which substances react.

For example:

CaCO₃ → CaO + CO₂

This says 1 mole of CaCO₃ produces 1 mole of CaO and 1 mole of CO₂.

The three-step method

1. Convert what you're given into moles.
2. Use the mole ratio from the balanced equation.
3. Convert moles of the substance you want into the units asked for.

Worked example 5: Reacting masses

What mass of calcium oxide is produced when 25 g of calcium carbonate is heated strongly? (Ca = 40, C = 12, O = 16)

CaCO₃ → CaO + CO₂

Step 1: moles of CaCO₃ = 25 ÷ 100 = 0.25 mol Step 2: ratio CaCO₃ : CaO = 1 : 1, so moles of CaO = 0.25 mol Step 3: molar mass of CaO = 40 + 16 = 56 g/mol; mass = 0.25 × 56 = 14 g

Worked example 6: A ratio that isn't 1 : 1

What mass of oxygen is needed to burn 6 g of magnesium completely? (Mg = 24, O = 16)

2Mg + O₂ → 2MgO

Step 1: moles of Mg = 6 ÷ 24 = 0.25 mol Step 2: ratio Mg : O₂ = 2 : 1, so moles of O₂ = 0.25 ÷ 2 = 0.125 mol Step 3: molar mass of O₂ = 32 g/mol; mass = 0.125 × 32 = 4 g

Worked example 7: Gas volume from a reaction

What volume of hydrogen at s.t.p. is produced when 2.4 g of magnesium reacts with excess hydrochloric acid? (Mg = 24; molar volume = 22.4 dm³)

Mg + 2HCl → MgCl₂ + H₂

Step 1: moles of Mg = 2.4 ÷ 24 = 0.1 mol Step 2: ratio Mg : H₂ = 1 : 1, so moles of H₂ = 0.1 mol Step 3: volume = 0.1 × 22.4 = 2.24 dm³

Worked example 8: Titration

25.0 cm³ of sodium hydroxide solution is neutralised by 20.0 cm³ of 0.10 mol/dm³ hydrochloric acid. Find the concentration of the sodium hydroxide.

NaOH + HCl → NaCl + H₂O

Step 1: moles of HCl = 0.10 × (20.0 ÷ 1000) = 0.002 mol Step 2: ratio NaOH : HCl = 1 : 1, so moles of NaOH = 0.002 mol Step 3: concentration = 0.002 ÷ (25.0 ÷ 1000) = 0.08 mol/dm³

Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

A compound contains 40% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Find its empirical formula. (C = 12, H = 1, O = 16)

Empirical formula: CH₂O

Common mistakes

• Using an unbalanced equation. Always balance first.
• Ignoring the mole ratio and assuming everything reacts 1 : 1.
• Forgetting to convert cm³ to dm³ (divide by 1000) in concentration and titration questions.
• Using atomic mass for a diatomic gas: the molar mass of O₂ is 32, not 16; for H₂ it's 2, not 1.
• Rounding too early. Keep full values until the final answer.

Practise until it's routine

The three-step method (moles, ratio, convert) works for almost every stoichiometry question. Levia's Chemistry levels include mole and stoichiometry practice at increasing difficulty, with a step-by-step explanation for every calculation.