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O-Level Chemistry · Secondary 3 and 4

Mole Concept and Stoichiometry

The mole concept is how chemists count particles by weighing them. It links the mass of a substance you can measure on a balance to the number of particles that actually react, so it sits at the heart of the chemical calculations section of O-Level Chemistry. Once you are fluent with moles, titration results, reacting masses, gas volumes, percentage yield and energy change questions all become the same kind of calculation, which is why this topic repays careful practice early in upper secondary.

What the syllabus expects

  • Calculate relative molecular mass and relative formula mass from relative atomic masses
  • Calculate the number of moles from a mass and a molar mass, and from a gas volume at room temperature and pressure
  • Use the mole ratio in a balanced equation to find reacting masses and gas volumes
  • Calculate concentrations in mol/dm³ and g/dm³, and use titration results to find an unknown concentration
  • Identify the limiting reactant and calculate percentage yield and percentage purity
  • Deduce empirical and molecular formulae from composition data

Moles, relative atomic mass and molar mass

One mole of any substance contains 6.02 × 10²³ particles, the Avogadro constant. The particles may be atoms, molecules or formula units, so always be clear which you mean. Relative atomic mass, Ar, compares the mass of an atom with one twelfth of the mass of a carbon-12 atom. Relative molecular mass, Mr, is the sum of the Ar values of all the atoms in a molecule. For CO₂, Mr = 12 + 2 × 16 = 44. For ionic compounds the same sum is called the relative formula mass, so for CaCO₃ it is 40 + 12 + 3 × 16 = 100. The molar mass is the Mr written in g/mol. The key formula is: number of moles = mass in grams ÷ molar mass. So 11 g of carbon dioxide is 11 ÷ 44 = 0.25 mol. Rearranged, mass = moles × molar mass, so 0.25 mol of CaCO₃ has a mass of 25 g.

Gas volumes and concentration of solutions

One mole of any gas occupies 24 dm³ at room temperature and pressure (r.t.p.), whatever the gas. So moles of gas = volume in dm³ ÷ 24, and 1 dm³ = 1000 cm³. For example, 600 cm³ of oxygen is 0.600 ÷ 24 = 0.0250 mol. This rule applies only to gases, never to liquids or solutions. For solutions, concentration in mol/dm³ = moles of solute ÷ volume of solution in dm³. So 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide contains 0.100 × 25.0/1000 = 0.00250 mol. To change mol/dm³ into g/dm³, multiply by the molar mass: 0.100 mol/dm³ NaOH is 0.100 × 40 = 4.00 g/dm³. In a titration you know the volume and concentration of one solution, so you can find its moles, use the mole ratio from the equation, then divide by the volume of the other solution to find its concentration.

Reacting masses and the limiting reactant

A balanced equation gives the ratio in which particles react, and therefore the ratio of moles. In 2Mg + O₂ gives 2MgO, 2 mol of magnesium makes 2 mol of magnesium oxide. If you burn 4.8 g of magnesium, that is 4.8 ÷ 24 = 0.20 mol, which makes 0.20 mol of MgO, and 0.20 × 40 = 8.0 g. The method is always the same: change what you know into moles, use the mole ratio, then change moles into what is asked for. When amounts of two reactants are given, one of them usually runs out first. This is the limiting reactant, and it decides how much product forms. The other reactant is in excess. To find the limiting reactant, work out the moles of each, then compare with the ratio in the equation, not just the raw numbers.

Percentage yield and percentage purity

The theoretical yield is the mass of product calculated from the limiting reactant. The actual yield is what you really collect, which is usually less because of incomplete reactions, side reactions and losses during filtering or transferring. Percentage yield = actual yield ÷ theoretical yield × 100%. A value above 100% is a warning sign: the product may still be wet or impure, or the theoretical yield was worked out wrongly. Percentage purity compares the mass of the pure substance in a sample with the mass of the whole sample. Percentage purity = mass of pure substance ÷ mass of impure sample × 100%. You usually find the mass of the pure substance from a reaction, for example from the volume of gas it gives off or from a titration.

Empirical and molecular formulae

The empirical formula shows the simplest whole-number ratio of atoms in a compound. The molecular formula shows the actual number of each atom in one molecule. To find the empirical formula, divide the mass or percentage of each element by its Ar, then divide every answer by the smallest. For a compound with 40.0% carbon, 6.7% hydrogen and 53.3% oxygen: C is 40.0 ÷ 12 = 3.33, H is 6.7 ÷ 1 = 6.7, O is 53.3 ÷ 16 = 3.33. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O, which has a mass of 30. If the Mr is 60, the molecular formula has 60 ÷ 30 = 2 empirical units, so it is C₂H₄O₂. If a ratio comes out as 1.5 or 1.33, multiply all the numbers by 2 or 3 rather than rounding.

Worked example

5.00 g of calcium carbonate is added to 50.0 cm³ of 1.00 mol/dm³ hydrochloric acid. CaCO₃ + 2HCl gives CaCl₂ + H₂O + CO₂. Which reactant is limiting, and what volume of carbon dioxide forms at r.t.p.? (Ar: Ca = 40, C = 12, O = 16)

  1. Mr of CaCO₃ = 40 + 12 + 3 × 16 = 100, so moles of CaCO₃ = 5.00 ÷ 100 = 0.0500 mol.
  2. Moles of HCl = 1.00 × 50.0/1000 = 0.0500 mol.
  3. The equation needs 2 mol of HCl for every 1 mol of CaCO₃, so 0.0500 mol of CaCO₃ would need 0.100 mol of HCl.
  4. Only 0.0500 mol of HCl is present, so the hydrochloric acid is the limiting reactant and some calcium carbonate is left over.
  5. From the ratio 2HCl to 1CO₂, moles of CO₂ = 0.0500 ÷ 2 = 0.0250 mol.
  6. Volume of CO₂ = 0.0250 × 24 = 0.600 dm³.

Answer: Hydrochloric acid is limiting, and 0.600 dm³ (600 cm³) of carbon dioxide forms at r.t.p.

Common mistakes

  • Using a volume in cm³ directly in moles = concentration × volume. Divide by 1000 first, so 25.0 cm³ becomes 0.0250 dm³.
  • Assuming a 1 : 1 ratio without reading the equation. In CaCO₃ + 2HCl, the acid reacts in a 2 : 1 ratio, so ignoring the 2 halves or doubles your answer.
  • Using 16 as the Mr of oxygen gas or 35.5 for chlorine gas. The gases are diatomic, so O₂ is 32 and Cl₂ is 71.
  • Applying the 24 dm³ rule to water or to a solution. It works only for gases at room temperature and pressure.
  • Rounding a ratio such as 1 : 1.5 to 1 : 2 when finding an empirical formula. Multiply every number by 2 to get 2 : 3 instead.
  • Choosing the limiting reactant by comparing masses in grams. Compare moles against the ratio in the equation.

Exam tips

  • Write a short label for every line, such as 'moles of HCl' or 'mole ratio 2 : 1', so that a marker can award method marks even if a later number slips.
  • Keep at least four significant figures in your working and round only the final answer, usually to three significant figures.
  • When two amounts are given in the question, expect a limiting reactant step and do it before calculating any product.
  • Check the units the question asks for. Convert 0.600 dm³ to 600 cm³ or mol/dm³ to g/dm³ only at the end.

Practice questions

1. Easy: How many moles of water are in 9.0 g of water? (Ar: H = 1, O = 16)

Show answer

0.50 mol. Mr of H₂O = 2 × 1 + 16 = 18, so moles = 9.0 ÷ 18 = 0.50 mol.

2. Medium: 20.0 cm³ of 0.150 mol/dm³ sulfuric acid exactly neutralises 25.0 cm³ of sodium hydroxide solution. H₂SO₄ + 2NaOH gives Na₂SO₄ + 2H₂O. Find the concentration of the sodium hydroxide.

Show answer

0.240 mol/dm³. Moles of H₂SO₄ = 0.150 × 20.0/1000 = 0.00300 mol. The ratio is 1 : 2, so moles of NaOH = 0.00600 mol. Concentration = 0.00600 ÷ 0.0250 = 0.240 mol/dm³.

3. Harder: 8.0 g of iron(III) oxide is reduced by carbon monoxide. Fe₂O₃ + 3CO gives 2Fe + 3CO₂. The student collects 4.2 g of iron. Calculate the percentage yield. (Ar: Fe = 56, O = 16)

Show answer

75%. Mr of Fe₂O₃ = 2 × 56 + 3 × 16 = 160, so moles = 8.0 ÷ 160 = 0.050 mol. This gives 2 × 0.050 = 0.10 mol of Fe, which is 0.10 × 56 = 5.6 g in theory. Percentage yield = 4.2 ÷ 5.6 × 100% = 75%.

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