The Mole and Stoichiometry

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1Summary 重點摘要

The mole is the chemist's counting unit: one mole contains 6.02×10²³ particles. Mass divided by molar mass gives moles; following the coefficient ratios of a balanced equation then yields the amounts of all species, and gas volumes can be found at 24 dm³ per mole at room temperature and pressure.

2Vocabulary cards Vocabulary

3Step by step Steps

1. Step 1: Why chemists invented the moleParticles are far too small and too numerous to count individually, so chemists bundle 6.02×10²³ particles into one mole, the unit of amount of substance. For example, a single drop of water contains hundreds of quintillions of molecules — awkward to describe — but '0.05 mole of water molecules' is clean and precise. This number is the Avogadro constant, in mol⁻¹. Remember that the mole counts particles regardless of their size, and answers should name the particle: one mole of oxygen molecules and one mole of oxygen atoms contain the same number of particles but different masses.
2. Step 2: Molar mass turns mass into molesThe molar mass is the mass of one mole of a substance, numerically equal to its relative atomic or molecular mass, in g mol⁻¹; the formula n = m ÷ M converts mass to moles. For example, water has relative molecular mass 18, so M = 18 g mol⁻¹ and 36 g of water is exactly 2 mol; magnesium has relative atomic mass 24, so 24 g of magnesium is 1 mol. Before calculating, check the formula: carbon dioxide's molar mass is 12 + (16 × 2) = 44 g mol⁻¹ — never forget to multiply by the number of atoms.
3. Step 3: Converting between moles and particlesMultiply moles by the Avogadro constant to get the number of particles; divide particles by the constant to return to moles. For example, 0.5 mol of carbon dioxide contains 3.01×10²³ molecules, and each molecule has 3 atoms, so the total number of atoms is 9.03×10²³ — a chained follow-up that appears often in the DSE. Watch whether the question asks for molecules or atoms: they differ by the factor 'atoms per molecule'. Reading the wording carefully before writing saves needless marks.
4. Step 4: Molar volume and gas volume calculationsAt room temperature and pressure (r.t.p.), one mole of any gas occupies about 24 dm³, i.e. 24000 cm³ — the molar volume of gas. Use V = n × 24 with volume in dm³. For example, 0.25 mol of oxygen occupies 6 dm³ at r.t.p.; conversely, 12 dm³ of hydrogen collected means 0.5 mol. This shortcut applies only to gases: particles in solids and liquids are packed closely, so the rule cannot be used — one of the commonest exam errors. Always confirm the state of matter first.
5. Step 5: Molarity of solutionsMolarity states how many moles of solute are dissolved per dm³ of solution, in mol dm⁻³; n = c × V gives the moles of solute, with volume in dm³. For example, 100 cm³ (0.1 dm³) of 0.5 mol dm⁻³ sodium hydroxide solution contains 0.05 mol of NaOH; if a lab record is in cm³, divide by 1000 before substituting. The same formula finds the mass of solute needed to prepare a solution, and titration calculations hang entirely on this relation — practise it until it is automatic.
6. Step 6: Reading mole ratios from balanced equationsThe coefficients of a balanced equation are the mole ratios — the heart of stoichiometry. Take the complete combustion of methane: CH₄ + 2O₂ → CO₂ + 2H₂O means 1 mol of methane needs 2 mol of oxygen and forms 1 mol of carbon dioxide and 2 mol of water. For example, burning 0.2 mol of methane completely needs 0.4 mol of oxygen by the 1 : 2 ratio and yields 0.2 mol of carbon dioxide. Balance the equation before any calculation, or every ratio will be wrong; check that atom counts match on both sides as a habit.
7. Step 7: The master route of stoichiometryEvery stoichiometric problem follows the same route: divide the known mass by the molar mass to get moles, follow the coefficient ratios to the target species, then multiply by its molar mass for mass, or by 24 for gas volume at r.t.p. For example, to find the mass of carbon dioxide from burning 16 g of methane: 16 ÷ 16 = 1 mol of CH₄, a 1 : 1 ratio gives 1 mol of CO₂, and 1 × 44 = 44 g. Memorise this 'mass → moles → ratio → moles → mass' flow and even unfamiliar contexts can be dismantled step by step.
8. Step 8: Empirical formula from percentage by massPercentage composition leads to the empirical formula, the simplest whole-number ratio of atoms: divide each element's percentage by its relative atomic mass, then reduce to the smallest ratio. For example, a compound containing 40% carbon, 6.7% hydrogen and 53.3% oxygen gives 3.33, 6.7 and 3.33 after dividing by 12, 1 and 16 — a ratio of 1 : 2 : 1, so the empirical formula is CH₂O. If the relative molecular mass is 60, the molecular formula is C₂H₄O₂, which is ethanoic acid. Such questions recur in DSE Paper 1 and the steps never change.

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8Recap Recap

• One mole contains 6.02×10²³ particles (the Avogadro constant); the mole is the chemist's counting unit.
• n = m ÷ M: dividing mass by molar mass (g mol⁻¹) gives moles — the first step of stoichiometry.
• At room temperature and pressure the molar volume of gas is 24 dm³ mol⁻¹, valid for gases only.
• Molarity is in mol dm⁻³; n = c × V with volume in dm³.
• Master route: mass → moles → coefficient ratios of the equation → moles of target → mass or volume.