HSC Exam Practice

Sample response. Percentage yield = (actual yield ÷ theoretical yield) × 100%. It measures how efficient a chemical reaction was — specifically, what fraction of the theoretically possible product was actually collected. It can never exceed 100% because the theoretical yield is calculated from stoichiometry assuming complete reaction with no losses; in practice, actual yield is always equal to or less than this maximum, since some product is always lost (through incomplete reaction, side reactions, purification losses, etc.). A yield above 100% indicates an experimental error, such as actual and theoretical yield being swapped in the formula, or unreacted impurities inflating the measured product mass.

Marking notes. 1 mark for the correct formula with definitions of actual and theoretical yield; 1 mark for stating yield cannot exceed 100% because actual ≤ theoretical by definition; 1 mark for identifying that a value above 100% indicates error (not efficiency).

Sample response. Percentage purity measures what fraction of a reactant sample consists of the desired substance; it is a property of the starting material and is applied before stoichiometric calculations by calculating the pure mass (m(sample) × purity ÷ 100) and using this in Step 2. Percentage yield measures how efficiently a reaction produced the desired product; it is calculated after stoichiometry by comparing the experimentally measured actual yield to the stoichiometrically calculated theoretical yield. They are different corrections applied at opposite ends of the calculation.

Marking notes. 1 mark for correctly defining percentage purity (fraction of sample that is the desired substance); 1 mark for stating purity is applied before stoichiometry; 1 mark for correctly defining percentage yield (efficiency / actual vs theoretical); 1 mark for stating yield is calculated after stoichiometry.

Sample response. Four reasons: (1) Side reactions — reactants are consumed producing by-products instead of the desired product. (2) Incomplete reaction — not all reactant molecules collide with sufficient energy to react; some reactant remains unreacted. (3) Reversible reaction / equilibrium limitation — the reaction reaches equilibrium before all reactants are consumed. (4) Product loss during workup — product is lost during filtration, evaporation, transfer between containers, or purification steps.

Marking notes. 1 mark per valid reason (maximum 4). Accept: impure reactants (though this is a purity issue not strictly a yield issue); incomplete drying of product. Do not accept: “low temperature” alone without linking to reaction rate or equilibrium.

Sample response. m(pure NaOH) = 30.0 × (94.0 ÷ 100) = 30.0 × 0.940 = 28.2 g.

Marking notes. 1 mark for correct formula/method; 1 mark for correct answer 28.2 g. Accept 28.2 g ± 0.1 g due to rounding.

Sample response. The sample mass includes both the pure reactant and the impurities. Impurities do not take part in the desired reaction, so they contribute no moles of product. If the sample mass is used in n = m ÷ MM (Step 2), the moles of reactant are overestimated because the impurity mass is incorrectly counted as pure reactant. This inflates n(product) via the mole ratio (Step 3) and therefore inflates the calculated theoretical yield (Step 4). The student’s theoretical yield is higher than the true maximum, meaning any subsequent % yield calculation will also be incorrect (lower than the true yield).

Marking notes. 1 mark for identifying that impurities do not react and so their mass should not be included in the moles calculation; 1 mark for explaining that using sample mass overestimates n(reactant) in Step 2; 1 mark for explaining the consequence — the theoretical yield is inflated above the true stoichiometric maximum.

Sample response (a). m(pure Fe₂O₃) = 200 000 × 0.780 = 156 000 g (156 kg). [1 mark]

Sample response (b). Step 2: MM(Fe₂O₃) = 2(55.845) + 3(15.999) = 159.69 g/mol; n(Fe₂O₃) = 156 000 ÷ 159.69 = 976.9 mol. Step 3: Ratio Fe₂O₃:Fe = 1:2; n(Fe) = 976.9 × 2 = 1953.8 mol. Step 4: MM(Fe) = 55.845; m(Fe) = 1953.8 × 55.845 = 109 100 g ≈ 109 kg (theoretical). [1 mark per step; 3 marks total]

Sample response (c). Actual m(Fe) = 109 100 × 0.880 = 96 000 g ≈ 96.0 kg. [1 mark]

Sample response (d). The actual yield would be higher. At 92.0% purity: m(pure) = 200 000 × 0.920 = 184 000 g — more pure Fe₂O₃ is available, so more moles enter the reaction and the theoretical yield is higher; with the same 88.0% yield applied to a larger theoretical mass, the actual yield is greater [1 mark prediction + 1 mark justification].

Sample response (a).

Route A (100% pure Cu): m(pure Cu) = 50.0 g. n(Cu) = 50.0 ÷ 63.546 = 0.7868 mol. Ratio Cu:CuSO₄ = 1:1; n(CuSO₄) = 0.7868 mol. MM(CuSO₄) = 63.546 + 32.06 + 4(15.999) = 159.60; theoretical = 0.7868 × 159.60 = 125.6 g. Actual A = 125.6 × 0.840 = 105.5 g. [3 marks]

Route B (90.0% pure Cu): m(pure Cu) = 50.0 × 0.900 = 45.0 g. n(Cu) = 45.0 ÷ 63.546 = 0.7081 mol. n(CuSO₄) = 0.7081 mol. Theoretical = 0.7081 × 159.60 = 113.0 g. Actual B = 113.0 × 0.960 = 108.5 g. [3 marks]

Sample response (b). Route B produces more CuSO₄ (108.5 g vs 105.5 g) even though it uses impure copper. The higher yield of Route B (96.0% vs 84.0%) more than compensates for the 10% reduction in available moles from the lower purity [1 mark]. This illustrates that purity and yield both multiply together to determine actual output — a large improvement in yield can outweigh a moderate decrease in purity [1 mark quantitative reasoning or general principle].

Marking notes. For (a): 1 mark per correct theoretical yield + 1 mark per correct actual yield for each route (3 marks per route, 6 marks total). For (b): 1 mark for correctly identifying Route B as higher; 1 mark for correctly explaining why (yield improvement outweighs purity reduction, with quantitative or qualitative reasoning).

Sample response. The actual mass of product from any reaction is determined by: actual product = (sample mass × purity ÷ 100) ÷ MM(reactant) × mole ratio × MM(product) × (% yield ÷ 100). Purity and yield both appear as multiplicative factors in this expression; a 1% improvement in purity has the same proportional effect on output as a 1% improvement in yield, all else being equal. The claim that purity improvement is “always” more effective is therefore incorrect as a general rule. A scenario where improving purity is more beneficial: if the starting material has very low purity (e.g. 50%) and the yield is already close to its maximum (e.g. 95%), then improving purity from 50% to 60% produces a larger relative gain in output (20% relative improvement) than any realistic improvement in a yield already near its ceiling. Conversely, a scenario where improving yield is more beneficial: if the starting material is already at 98% purity and the reaction yield is only 40% (e.g. due to an unfavourable reversible equilibrium), then improving yield from 40% to 60% (a 50% relative gain) would deliver far more product than moving purity from 98% to 100% (a 2% relative gain). The two factors interact multiplicatively, not additively. In industrial contexts such as the Haber process for ammonia synthesis, the reactor yield per pass is only about 15–25%, making yield optimisation the dominant concern; the nitrogen and hydrogen feeds are typically very pure. By contrast, in processing of low-grade iron ores (e.g. some Australian hematite deposits at 58–62% Fe₂O₃), upgrading ore purity through beneficiation substantially increases output even when blast furnace yield is already high. In conclusion, the claim is too absolute. Neither purity nor yield is universally more important; the best investment depends on which factor is the larger limiting constraint in a given process. An evidence-based approach requires calculating the proportional gain from improving each factor and choosing accordingly.

Marking criteria (7 marks). 1 = correctly states the quantitative relationship: actual product = f(purity) × f(yield), so both multiply together. 1 = explains that equal relative improvements in purity or yield have equal proportional effects on output. 1 = describes a scenario where improving purity is the more effective strategy (quantitative or qualitative), with correct reasoning. 1 = describes a scenario where improving yield is the more effective strategy (quantitative or qualitative), with correct reasoning. 1 = uses a named industrial or Australian example correctly to support either scenario. 1 = explicitly addresses the word “always” in the claim and concludes it is too absolute. 1 = reaches a clear, evidence-based conclusion that integrates both scenarios and the mathematical relationship. Mark for quality of argument, not length.