Chemistry · Lesson 4

The mole: chemistry's "dozen"

JAMBWAECNECO

What the syllabus asks

JAMB: Stoichiometry and chemical combination (mole concept, Avogadro constant, formulae, equations, laws of chemical combination, molar volume)

WAEC: The mole concept, stoichiometry and volumetric analysis

NECO: The mole concept, stoichiometry and volumetric analysis

What you will learn

Explain why chemists count particles in moles, and define the mole and the Avogadro constant.
Calculate relative molecular masses and molar masses.
Convert between mass, moles, number of particles and gas volume.
Work out empirical and molecular formulae and percentage composition.
Use balanced equations for reacting masses, limiting reagents and gas volumes.
Calculate concentrations and carry out titration calculations.
State and apply the laws of chemical combination.

Key points

1 mol = 6.02 × 10²³ particles (the Avogadro constant).
n = m/M; molar mass in g mol⁻¹ = Mᵣ.
Empirical formula: mass → ÷ Aᵣ → ÷ smallest → whole numbers. Molecular formula = (empirical)ₙ with n = Mᵣ ÷ empirical mass.
Balanced equations give mole ratios: mass → moles → ratio → moles → mass.
The limiting reagent runs out first and fixes the yield; % yield = actual/theoretical × 100.
Gases: 22.4 dm³ mol⁻¹ at s.t.p.; volume ratio = mole ratio (Gay-Lussac).
C = n/V (mol dm⁻³); g dm⁻³ = mol dm⁻³ × M; C₁V₁ = C₂V₂.
Titration: CₐVₐ/C_bV_b = nₐ/n_b.
Laws: conservation of mass, definite proportions, multiple proportions, reciprocal proportions.

Start here: Why chemists count in moles

Go to any market in Nigeria and you will see that people rarely count tiny things one by one. Nobody counts grains of rice; we buy rice by the mudu or by the bag. Eggs are sold by the crate (30 eggs). Sachet water comes in a bag of 20. Grouping makes counting easy.

Atoms and molecules are unimaginably small. A single drop of water contains about 1.7 × 10²¹ water molecules, far more than the number of grains of sand on all the beaches in Nigeria. Counting them one by one is impossible, so chemists use a very large "crate" called the mole.

Definition. One mole is the amount of substance that contains as many elementary particles (atoms, molecules, ions or electrons) as there are atoms in exactly 12 g of carbon-12. That number is the Avogadro constant: L = 6.02 × 10²³ mol⁻¹.

So, just as "a crate" always means 30 eggs, "a mole" always means 6.02 × 10²³ particles, whatever the particles are. One mole of water molecules is 6.02 × 10²³ molecules; one mole of sodium ions is 6.02 × 10²³ ions.

Fig. 4.1: A mole is just a (very large) counting unit.

Worked example 4.1

How many molecules are in 2.5 mol of carbon dioxide?

Each mole contains 6.02 × 10²³ molecules.
2.5 × 6.02 × 10²³ = 1.505 × 10²⁴.

1.505 × 10²⁴ molecules

Worked example 4.2

How many moles is 3.01 × 10²² atoms of iron?

Divide by the number in one mole: 3.01 × 10²²/6.02 × 10²³.
= 0.05 mol.

0.05 mol

Worked example 4.3 (atoms inside molecules)

How many atoms are in 0.5 mol of ammonia, NH₃?

Molecules = 0.5 × 6.02 × 10²³ = 3.01 × 10²³.
Each NH₃ molecule has 4 atoms (1 N + 3 H).
Atoms = 4 × 3.01 × 10²³ = 1.204 × 10²⁴.

1.204 × 10²⁴ atoms

⚠️ Common mistake: answering with the number of molecules when the question asks for atoms or ions. Underline the last word of the question.

Try it yourself. How many ions are in 1 mol of NaCl?

Show solution

Each NaCl unit has 2 ions (Na⁺ and Cl⁻), so 2 × 6.02 × 10²³ = 1.204 × 10²⁴ ions.

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Practice questions

1. How many moles are in 22 g of carbon(IV) oxide? (C = 12, O = 16)

0.5 mol
2.0 mol
0.25 mol
1.0 mol

Show answer

A: 0.5 mol
Molar mass CO₂ = 44 g/mol; 22/44 = 0.5 mol.

2. What volume does 2 moles of a gas occupy at s.t.p.?

22.4 dm³
11.2 dm³
24.0 dm³
44.8 dm³

Show answer

D: 44.8 dm³
1 mol = 22.4 dm³ at s.t.p.; 2 × 22.4 = 44.8 dm³.

3. The number of moles of NaOH in 250 cm³ of a 0.2 mol/dm³ solution is

0.8
50
0.05
0.5

Show answer

C: 0.05
n = C × V = 0.2 × 0.25 = 0.05 mol.

4. In 2H₂ + O₂ → 2H₂O, how many moles of oxygen react with 6 moles of hydrogen?

Show answer

B: 3
Ratio H₂ : O₂ = 2 : 1, so 6 mol H₂ needs 3 mol O₂.

5. What mass of calcium oxide is obtained by heating 50 g of CaCO₃? (Ca = 40, C = 12, O = 16)

28 g
56 g
22 g
44 g

Show answer

A: 28 g
CaCO₃ (100 g/mol) → CaO (56). 50 g = 0.5 mol → 0.5 × 56 = 28 g.

Key words

Mole (mol): The amount of substance that contains 6.02 × 10²³ particles.
Avogadro constant: 6.02 × 10²³ mol⁻¹, the number of particles in one mole.
Relative atomic mass (Aᵣ): The mass of an atom compared with 1/12 of the mass of a carbon-12 atom; no units.
Relative molecular mass (Mᵣ): The sum of the Aᵣ values of all the atoms in a formula.
Molar mass (M): The mass of one mole, in g mol⁻¹.
Empirical formula: The simplest whole-number ratio of atoms in a compound.
Molecular formula: The actual number of each kind of atom in one molecule.
Stoichiometry: The calculation of quantities in chemical reactions.
Limiting reagent: The reactant that is used up first; it controls how much product forms.
s.t.p.: Standard temperature and pressure: 0 °C (273 K) and 1 atm.
Molar volume: The volume of one mole of gas: 22.4 dm³ at s.t.p.
Concentration: The amount of solute in a given volume of solution (mol dm⁻³ or g dm⁻³).

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