HSC Exam Practice

Sample response. Molar mass is the mass in grams of one mole of a substance; units are g mol⁻¹. Because one mole of a compound contains one mole of each type of atom (in the ratio given by the formula), the total mass of one mole is simply the sum of the masses of all those atoms. Since the atomic mass of each element is the mass in grams of one mole of that atom, multiplying each atomic mass by the subscript in the formula and summing gives the molar mass of the compound.

Marking notes. 1 mark for definition of molar mass (mass per mole) with correct units; 1 mark for stating that it equals the mass of Avogadro’s number of the compound’s formula units; 1 mark for explaining the additive relationship with atomic masses and subscripts.

Sample response. Concordant titres are two or more consecutive titres that agree within ±0.10 mL; they confirm the endpoint is being judged consistently. A rough titre is a fast, approximate titration performed first to find the approximate endpoint volume; it is excluded from the concordant average because it is performed quickly without the careful dropwise addition required near the endpoint, and therefore has greater uncertainty than the careful titres.

Marking notes. 1 mark for defining concordant titres (within ±0.10 mL / ±0.1 mL); 1 mark for correctly explaining the purpose of the rough titre (approximate endpoint); 1 mark for explaining why it is excluded (performed quickly / less precise).

Sample response. Percentage purity must be applied first because only the pure substance participates in the chemical reaction. The formula is: m(pure) = m(sample) × (% purity ÷ 100). If purity is applied after the stoichiometry, the mole calculation uses the total sample mass including impurities, producing a moles value that is too high. This leads to an overestimate of the theoretical yield (or an incorrect calculated purity if working backwards).

Marking notes. 1 mark for stating that only the pure substance reacts; 1 mark for the correct formula m(pure) = m(sample) × (% purity ÷ 100); 1 mark for explaining the consequence of applying purity after stoichiometry (moles too high; yield or purity answer incorrect).

Sample response. Divide each given amount by its stoichiometric coefficient: Al: 0.400 ÷ 4 = 0.100; O₂: 0.600 ÷ 3 = 0.200. Since 0.100 < 0.200, Al is the limiting reagent. It is consumed first and controls the amount of Al₂O₃ produced.

Marking notes. 1 mark for dividing each amount by the correct coefficient; 1 mark for comparing the resulting values; 1 mark for correctly identifying Al as the limiting reagent with a reason (smaller ratio / consumed first).

Sample response. n(Na₂CO₃) = c × V = 0.100 × 0.250 = 0.0250 mol. m(Na₂CO₃) = n × MM = 0.0250 × 105.99 = 2.65 g. Weigh exactly 2.65 g of anhydrous Na₂CO₃ on an analytical balance. Dissolve in ~100 mL of distilled water in the 250 mL volumetric flask. Once dissolved, carefully make up to the 250.0 mL graduation mark with distilled water; stopper and invert 20 times to ensure uniform mixing.

Marking notes. 1 mark for correct mass (2.65 g); 1 mark for specifying use of a volumetric flask (not a measuring cylinder); 1 mark for making up to the graduation mark with distilled water; 1 mark for inverting/mixing to ensure uniform concentration.

Sample response. A systematic error shifts all results in the same direction by the same amount (or proportion) in every trial; it cannot be reduced by repeating the experiment. Example: a burette that consistently reads 0.10 mL too high due to a calibration error will make every titre appear 0.10 mL larger than the true value. A random error causes results to scatter unpredictably around the true value, affecting different trials differently; it can be reduced by taking more measurements and averaging. Example: inconsistent judgement of the endpoint colour change (adding one drop too many or too few in different trials).

Marking notes. 1 mark for a correct definition of systematic error (consistent directional offset); 1 mark for a valid titration systematic-error example (calibration, indicator colour, always using same side of meniscus); 1 mark for a correct definition of random error (unpredictable scatter, reduced by averaging); 1 mark for a valid titration random-error example (endpoint judgement, parallax varying, room vibration).

Part (a) — 3 marks. V(H₂SO₄) = 28.4 mL = 0.02840 L. n(H₂SO₄) = 0.0500 × 0.02840 = 1.420 × 10⁻³ mol. Ratio 2NaOH : 1H₂SO₄; n(NaOH) = 2 × 1.420 × 10⁻³ = 2.840 × 10⁻³ mol. Award: 1 mark n(H₂SO₄); 1 mark correct ratio applied; 1 mark n(NaOH).

Part (b) — 2 marks. c(NaOH) = n ÷ V = 2.840 × 10⁻³ ÷ 0.02500 = 0.1136 mol L⁻¹. n(NaOH) in 500.0 mL = 0.1136 × 0.5000 = 5.680 × 10⁻² mol. Award 1 mark each.

Part (c) — 2 marks. m(NaOH) = 5.680 × 10⁻² × 39.997 = 2.272 g. % purity = (2.272 ÷ 3.00) × 100 = 75.7%. Award 1 mark each.

Part (d) — 2 marks. The calculated purity would be lower than the true NaOH purity. The titration measures the total alkalinity of the solution; Na₂CO₃ also reacts with H₂SO₄ and neutralises acid. Since the method counts all base (including carbonate) as NaOH, it underestimates the NaOH content: the true amount of NaOH is even less than calculated, so the NaOH purity is overstated. Equivalently, the mass of “NaOH” calculated using a 1:1 back-calculation misidentifies carbonate as NaOH, inflating the apparent moles and therefore mass of NaOH — but the actual conversion assumes NaOH only, so the reported purity as NaOH is too low because the carbonate does not contribute fully on a per-gram basis. [Accept: Na₂CO₃ requires 2 mol HCl per mole but has a higher MM than NaOH, so each gram of Na₂CO₃ neutralises fewer moles of acid per gram than NaOH — the calculated mass of “NaOH” will be less than if the sample were pure NaOH, giving a lower reported purity.] Award 1 mark for the direction (lower); 1 mark for a valid chemical explanation.

Sample response. The claim is partially correct but overstated. In theory, 100% yield would require that every mole of limiting reagent is converted into product with zero loss — an ideal scenario that is never fully achievable in practice for several interconnected reasons.

At the theoretical level, many reactions are reversible equilibria; the forward reaction does not proceed to completion but instead reaches an equilibrium where both reactants and products coexist. In such cases, even with perfect technique, the yield is fundamentally limited by the equilibrium position. For example, in the industrial production of ammonia (Haber process — a Module 2 context for gas stoichiometry), the yield at atmospheric pressure and room temperature is very low because the reaction is reversible: N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Industrial conditions (high pressure, moderate temperature, catalyst) improve but cannot reach 100%.

At the practical level, yield is further reduced by product recovery losses (filtering, washing, transferring), side reactions that consume the limiting reagent to form unintended products, and incomplete mixing or insufficient reaction time. A student titration, for instance, can achieve very high percentage yield of NaCl from a neutralisation reaction because it is irreversible and the product is fully soluble, but even here, evaporation losses when recovering the solid reduce the actual yield.

In the specific case of an acid–base neutralisation (e.g. NaOH + HCl → NaCl + H₂O), which is irreversible and proceeds essentially to completion, a well-designed laboratory procedure can achieve yields above 95% with careful technique, approaching (but not reaching) 100%. By contrast, an industrial esterification reaction (carboxylic acid + alcohol ⇌ ester + water) is reversible and typically achieves 60–70% yield even under optimised conditions.

In summary, the claim is valid for irreversible reactions under good laboratory conditions, but fails for reversible reactions where the theoretical maximum is fixed by equilibrium. The absolute upper limit is always 100%, set by the conservation of mass — any result above this indicates a calculation or measurement error.

Marking criteria (7 marks). 1 = identifies the claim as partially correct / overstated with a clear evaluative stance. 1 = correctly defines theoretical yield as the maximum from the limiting reagent (conservation of mass basis). 1 = identifies reversible reactions as a fundamental theoretical limitation on yield, with a named example. 1 = identifies at least two distinct practical factors that reduce actual yield below theoretical (recovery loss, side reactions, incomplete reaction, etc.). 1 = names a relevant industrial or Module 2 example correctly (blast furnace iron smelting, Haber process, acid precipitation gravimetric, any neutralisation). 1 = identifies a scenario where high yield is achievable (irreversible reaction under good conditions) and contrasts with one where it is not (reversible / equilibrium reaction). 1 = reaches an explicit, justified evaluative conclusion that integrates theoretical and practical limits and references conservation of mass.