Worked examples: The Mole Concept, Chemical Formula and Equation

Cover the solution, attempt each example, then check your working line by line. Every answer shows the formula, the substitution and the evaluation, because that is how method marks are earned. Relative atomic masses are taken from the periodic table: H = 1, C = 12, N = 14, O = 16, Na = 23, Mg = 24, S = 32, Cl = 35.5, Ca = 40, Cu = 64, Zn = 65. These are original questions in SPM style, not past-year papers.

Example 1, Relative molecular mass

Question. Calculate the relative molecular mass of ethanol, C₂H₅OH, and of hydrated sodium carbonate, Na₂CO₃·10H₂O.

Solution. For C₂H₅OH: 2(12) + 6(1) + 16 = 24 + 6 + 16 = 46. (Count all six hydrogen atoms, five in C₂H₅ plus one in OH.) For Na₂CO₃·10H₂O: 2(23) + 12 + 3(16) + 10(18) = 46 + 12 + 48 + 180 = 286.

Common slip: forgetting the ten water molecules in the hydrated salt, giving 106 instead of 286.

Example 2, Moles from mass

Question. How many moles are present in 20 g of sodium hydroxide, NaOH?

Solution. M(NaOH) = 23 + 16 + 1 = 40 g mol⁻¹. Using n = m ÷ M, n = 20 ÷ 40 = 0.5 mol.

Common slip: dividing molar mass by mass (40 ÷ 20) and writing 2 mol. The formula is mass ÷ molar mass, not the reverse.

Example 3, Mass from moles

Question. Calculate the mass of 0.25 mol of calcium carbonate, CaCO₃.

Solution. M(CaCO₃) = 40 + 12 + 3(16) = 100 g mol⁻¹. Using mass = moles × molar mass, mass = 0.25 × 100 = 25 g.

Common slip: using the wrong molar mass, forgetting there are three oxygen atoms and writing 68 instead of 100.

Example 4, Number of particles

Question. How many molecules, and how many atoms, are there in 0.2 mol of carbon dioxide, CO₂?

Solution. number of molecules = moles × the Avogadro constant = 0.2 × 6.02 x 10²³ mol⁻¹ = 1.204 x 10²³ molecules. Each CO₂ molecule has 3 atoms, so number of atoms = 3 × 1.204 x 10²³ = 3.612 x 10²³ atoms.

Common slip: giving the number of atoms when molecules are asked, or the reverse. Read whether the question wants molecules, atoms or ions.

Example 5, Volume of gas

Question. What is the volume, at room conditions, of 0.15 mol of oxygen gas?

Solution. volume = moles × molar volume = 0.15 × 24 dm³ mol⁻¹ = 3.6 dm³.

Common slip: using the STP molar volume of 22.4 dm³ mol⁻¹ when the question clearly states room conditions. Match the molar volume to the stated conditions.

Example 6, Combining mass, moles and gas volume

Question. Calcium carbonate decomposes on heating: CaCO₃ → CaO + CO₂. What volume of carbon dioxide, measured at room conditions, is produced when 5 g of calcium carbonate decomposes completely?

Solution. M(CaCO₃) = 100 g mol⁻¹, so n(CaCO₃) = 5 ÷ 100 = 0.05 mol. The mole ratio CaCO₃ : CO₂ is 1 : 1, so n(CO₂) = 0.05 mol. Volume = 0.05 × 24 dm³ mol⁻¹ = 1.2 dm³.

Common slip: jumping from mass straight to volume without going through moles, or forgetting to balance/read the 1 : 1 ratio.

Example 7, Empirical formula

Question. A compound contains 2.4 g of carbon and 0.6 g of hydrogen. Determine its empirical formula.

Solution. Moles of C = 2.4 ÷ 12 = 0.2 mol; moles of H = 0.6 ÷ 1 = 0.6 mol. Divide by the smaller value (0.2): C = 0.2 ÷ 0.2 = 1; H = 0.6 ÷ 0.2 = 3. The ratio C : H is 1 : 3, so the empirical formula is CH₃.

Common slip: stopping at the mole values (0.2 and 0.6) and writing them as the formula, instead of dividing by the smallest to get the simplest whole-number ratio.

Example 8, Molecular formula

Question. The empirical formula of a compound is CH₂O and its relative molecular mass is 180. Determine its molecular formula.

Solution. Empirical formula mass of CH₂O = 12 + 2(1) + 16 = 30. n = Mr ÷ empirical formula mass = 180 ÷ 30 = 6. Molecular formula = (CH₂O) × 6 = C₆H₁₂O₆.

Common slip: giving the empirical formula CH₂O as the final answer when the molecular formula is asked, or forgetting to multiply every subscript by n.

Example 9, Stoichiometry with a gas

Question. Zinc reacts with dilute sulfuric acid: Zn + H₂SO₄ → ZnSO₄ + H₂. Calculate the volume of hydrogen gas, at room conditions, produced when 3.25 g of zinc reacts completely.

Solution. M(Zn) = 65 g mol⁻¹, so n(Zn) = 3.25 ÷ 65 = 0.05 mol. The mole ratio Zn : H₂ is 1 : 1, so n(H₂) = 0.05 mol. Volume = 0.05 × 24 dm³ mol⁻¹ = 1.2 dm³.

Common slip: using the mass of zinc directly as if it were moles, or reversing the mole ratio. Always convert mass to moles first, then apply the ratio from the balanced equation.

Using these examples

Notice the same discipline in every solution: write the formula, substitute values with units, evaluate, and state the unit in the answer. Once the four conversions, mass, moles, particles and gas volume, feel automatic, move on to the practice questions and mark yourself the same way. A one-to-one teacher can check that your working shows every step, which is exactly where method marks are gained or lost in the SPM Chemistry written papers. Rehearse until you can set out any of these calculations without hesitation, because the same steps underpin the acid–base, redox and energetics chapters that follow.

Example 10, Percentage by mass

Question. Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃.

Solution. M(NH₄NO₃) = 14 + 4(1) + 14 + 3(16) = 14 + 4 + 14 + 48 = 80 g mol⁻¹. There are two nitrogen atoms, with a combined mass of 2 × 14 = 28. Percentage by mass of nitrogen = (28 ÷ 80) × 100 = 35%.

Common slip: counting only one nitrogen atom, or dividing by the mass of nitrogen instead of by the total molar mass. Always place the part over the whole.

Example 11, Balancing and using the mole ratio

Question. Aluminium burns in oxygen to form aluminium oxide. Write the balanced equation and calculate the number of moles of oxygen needed to react completely with 0.4 mol of aluminium.

Solution. The balanced equation is 4Al + 3O₂ → 2Al₂O₃. The mole ratio Al : O₂ is 4 : 3. So moles of O₂ = 0.4 × (3 ÷ 4) = 0.3 mol.

Common slip: leaving the equation unbalanced (Al + O₂ → Al₂O₃) and reading the ratio as 1 : 1, or inverting the 4 : 3 ratio. Balance first, then read the ratio in the correct direction.

From examples to marks

These eleven examples span the whole chapter, and together they drill the same core routine: find the relative mass, convert the given quantity to moles, apply the correct ratio, and convert to the quantity asked, always with units. Practise them until the routine is automatic, then test yourself with the practice questions for this chapter and check your working against the model answers.