Chemistry • Year 11 • Module 2 • Lesson 14
Percentage Yield & Percentage Purity
Build HSC Band 5–6 technique by evaluating industrial trade-off data, designing an investigation, and constructing a multi-step analytical response integrating purity, yield, and economic reasoning.
1. Data + scenario: choosing between two industrial processes for aspirin synthesis (Band 5–6)
8 marks Higher-order
Scenario. A pharmaceutical company produces aspirin (C9H8O4) by reacting salicylic acid (C7H6O3) with acetic anhydride: C7H6O3 + (CH3CO)2O → C9H8O4 + CH3COOH. The company is comparing two manufacturing routes that start with the same nominal 2.00 kg batch of salicylic acid. Their data are summarised below. The company must produce at least 2.20 kg of aspirin per batch to be commercially viable.
| Parameter | Process X | Process Y |
|---|---|---|
| Purity of salicylic acid feed | 88.0% | 97.0% |
| Percentage yield of aspirin | 92.0% | 78.0% |
| Cost of reactant per batch | $120 | $185 |
| Operating time per batch | 3 hours | 1.5 hours |
Molar masses: C7H6O3 = 138.12 g/mol; C9H8O4 = 180.16 g/mol. Illustrative scenario.
Q1. Analyse and evaluate the data above to determine which process the company should adopt. In your response you must:
- For each process, apply the purity correction and complete the full stoichiometric calculation to find the theoretical yield of aspirin (in kg).
- Apply the percentage yield to calculate the actual mass of aspirin produced per batch for each process.
- State which process (or neither) meets the commercial viability threshold of 2.20 kg per batch.
- Evaluate both processes using at least two additional criteria beyond product mass (e.g. cost, efficiency, time, environmental considerations).
- Reach a justified recommendation supported by your quantitative and qualitative analysis.
2. Experimental design, determining the purity of impure NaOH pellets (Band 5–6)
7 marks Higher-order
Research question. A batch of sodium hydroxide (NaOH) has been blended with inert sodium chloride (NaCl) as a filler. NaCl does not react with hydrochloric acid, so a single neutralisation titration against standardised HCl measures only the NaOH present, this is why an inert impurity is chosen, rather than a base such as Na2CO3 that would also consume acid and make the result invalid. A student wants to determine the percentage purity of the NaOH in the batch using a neutralisation reaction with standardised hydrochloric acid (HCl) and an indicator.
Constraints: Standard Year 11 lab equipment including burette, pipette, conical flask, balance (0.01 g precision), standardised 0.1000 mol/L HCl solution, phenolphthalein indicator, and distilled water. The student wants a result accurate to within ±2%.
Q2. Design the titration-based investigation and present it in the format below.
- State a hypothesis in the form: “If the purity of NaOH is X%, then Y mol of HCl will be required to neutralise Z g of the sample.”
- Identify the independent variable, dependent variable, and at least two controlled variables.
- Describe the procedure in at least five numbered steps, including how you will calculate % purity from the titration result.
- Show the calculation structure (formula, not numbers) you will use to convert the titre volume into a % purity value.
- State two sources of error and one improvement to increase the reliability of results.
Q1, Sample Band 6 response (8 marks), annotated
Process X, full working:
m(pure C7H6O3) = 2000 × 0.880 = 1760 g [purity, 1 mark]. n(C7H6O3) = 1760 ÷ 138.12 = 12.74 mol. Ratio 1:1; n(C9H8O4) = 12.74 mol. Theoretical = 12.74 × 180.16 = 2295 g = 2.295 kg [stoichiometry, 1 mark]. Actual (X) = 2295 × 0.920 = 2111 g ≈ 2.11 kg [actual yield, 1 mark].
Process Y, full working:
m(pure) = 2000 × 0.970 = 1940 g [purity, 1 mark]. n = 1940 ÷ 138.12 = 14.04 mol. Theoretical = 14.04 × 180.16 = 2529 g = 2.529 kg [stoichiometry, 1 mark]. Actual (Y) = 2529 × 0.780 = 1973 g ≈ 1.97 kg [actual yield, 1 mark].
Viability check: Neither process meets the 2.20 kg threshold, Process X produces 2.11 kg (below by 0.09 kg) and Process Y produces only 1.97 kg (further below). Both fall short.
Multi-criteria evaluation: Process X produces more aspirin per batch (2.11 kg vs 1.97 kg) and has a lower reactant cost ($120 vs $185), making it more cost-effective per gram of product. However, Process Y is faster (1.5 h vs 3 h), which could allow more batches per day. Cost per gram: Process X = $120 ÷ 2110 g = $0.057/g; Process Y = $185 ÷ 1973 g = $0.094/g, Process X is 65% cheaper per gram produced [multi-criteria, 1 mark].
Recommendation: Process X should be adopted. Although neither process currently meets the viability threshold, Process X comes closer (2.11 vs 2.20 kg) and has a significantly lower cost per gram. The company should invest in improving the yield of Process X (e.g. optimising temperature, using a catalyst, or recovering unreacted material) rather than switching to the more expensive and lower-output Process Y [justified recommendation, 1 mark, annotated as mark 8].
Marking criteria summary (8 marks): 1 = correct purity correction for Process X with working; 1 = correct actual yield for Process X; 1 = correct purity correction for Process Y with working; 1 = correct actual yield for Process Y; 1 = viability comparison against 2.20 kg threshold; 1 = at least two non-mass criteria analysed quantitatively or qualitatively (cost, time, throughput, environmental); 1 = integrated recommendation naming Process X; 1 = uses precise chemical language throughout (purity correction, theoretical yield, actual yield, percentage yield).
Q2, Sample Band 6 response (7 marks), annotated
Hypothesis: If the NaOH pellets are 95% pure, then approximately 0.0475 mol of HCl (47.5 mL of 0.1000 mol/L HCl) will be required to neutralise 0.200 g of the sample, because the ratio NaOH:HCl = 1:1 and the pure mass of NaOH in 0.200 g at 95% purity is 0.190 g = 0.00475 mol [1, hypothesis with X and Y values].
Variables: Independent variable: mass of sample used per titration (set at a fixed value, e.g. 0.200 g). Dependent variable: titre volume of HCl required to reach the endpoint; from this, % purity is calculated. Controlled variables: (1) concentration of HCl (standardised at 0.1000 mol/L); (2) volume of distilled water used to dissolve the sample (25.0 mL); (3) temperature (room temperature, ~22°C); (4) indicator (3 drops phenolphthalein in each flask) [1, variables].
Procedure: (1) Tare a weighing boat on the balance; transfer approximately 0.200 g of NaOH pellets and record the exact mass. (2) Dissolve the pellets in 25.0 mL of distilled water in a conical flask; add 3 drops of phenolphthalein indicator (solution turns pink). (3) Fill burette with standardised 0.1000 mol/L HCl; record initial reading. Titrate slowly until the pink colour just disappears (pale colourless endpoint); record final burette reading. (4) Calculate titre volume V(HCl) = final − initial. (5) Repeat steps 1–4 at least three times; use only concordant titres (±0.10 mL) to calculate an average titre [1, five clear steps including endpoint criterion].
Calculation structure: n(HCl) = c × V(L); n(NaOH) = n(HCl) (1:1 ratio); m(pure NaOH) = n(NaOH) × MM(NaOH) = n(NaOH) × 40.00; % purity = [m(pure NaOH) ÷ m(sample)] × 100 [1, correct formula structure].
Sources of error: (1) NaOH is hygroscopic, so it can absorb water vapour (and a little CO2) from the air between weighing and dissolving; any water taken up increases the weighed sample mass without adding NaOH, lowering the apparent purity. Minimise by weighing quickly in a closed container. (Note: if the impurity were a base such as Na2CO3 rather than inert NaCl, it would also consume HCl and the single-titration method would no longer measure NaOH alone, a two-endpoint method would then be required.) [1, error 1]. (2) Burette reading parallax error: incorrect reading of the meniscus may add up to ±0.05 mL per reading, compounding to ±0.10 mL per titre [1, error 2].
Improvement: Weigh the sample in a sealed weighing vessel and dissolve immediately before titrating; alternatively, repeat titrations until three concordant results within ±0.05 mL are obtained and use the mean, improving precision [1, improvement].
Marking criteria summary (7 marks): 1 = testable hypothesis in the specified form with numerical values; 1 = correct IV, DV, and two controlled variables; 1 = five numbered procedure steps including titration technique and endpoint criterion; 1 = correct formula structure (n = cV; n ratio; m(pure) = n×MM; % = m(pure)÷m(sample)×100); 1 = one valid error specific to this experiment; 1 = second valid error; 1 = one specific improvement linked to a named error.