Chemistry • Year 11 • Module 2 • Lesson 20
Module 2 Review
Build HSC Band 5–6 extended-response technique on multi-step quantitative reasoning, experimental design, and evaluation of analytical methods.
1. Data + scenario: Analysing a BHP iron ore sample (Band 5–6)
8 marks Band 5–6
Scenario. BHP operates iron ore mines in the Pilbara region of Western Australia. A quality control chemist receives a 50.00 g sample of iron ore that is labelled as containing approximately 70% haematite (Fe2O3, MM = 159.69 g/mol). The sample is reduced in a furnace according to the reaction: Fe2O3(s) + 3CO(g) → 2Fe(l) + 3CO2(g). The chemist collects 24.62 g of iron metal (MM = 55.845 g/mol) from the reduction. The table below shows her calculations at each step.
| Step | Calculation | Value |
|---|---|---|
| Purity correction | m(pure Fe2O3) = 50.00 × 0.70 | 35.00 g |
| n(Fe2O3) | n = 35.00 ÷ 159.69 | 0.2192 mol |
| Mole ratio | Fe2O3 : Fe = 1 : 2; n(Fe) = 0.2192 × 2 | 0.4384 mol |
| Theoretical yield | m(Fe) = 0.4384 × 55.845 | 24.48 g |
| Percentage yield | % yield = (24.62 ÷ 24.48) × 100 | 100.6% |
Illustrative data; MM values: Fe = 55.845, O = 15.999, Fe2O3 = 159.69 g/mol.
Q1. Analyse and evaluate the chemist’s data and calculations above. In your response you must:
- Identify and explain the scientific error in the table that makes the % yield impossible.
- Recalculate the correct theoretical yield and correct % yield, showing full working.
- Explain what a % yield greater than 100% would indicate about the experimental result if it were taken at face value.
- Identify one possible experimental reason why the actual yield (24.62 g) is greater than the theoretical yield calculated from 70% purity, and propose how the chemist could improve accuracy.
- State whether the ore sample is more or less than 70% pure, using your corrected % yield as evidence.
2. Experimental design — determining concentration of a sports drink (Band 5–6)
7 marks Band 5–6
Research question. A sports science student wants to determine the concentration of chloride ions (Cl−) in a commercial sports drink. He claims that gravimetric analysis using silver nitrate (AgNO3) would give a more accurate result than a titration with AgNO3.
Context. The precipitation reaction is: Ag+(aq) + Cl−(aq) → AgCl(s). AgCl is a white, insoluble precipitate (MM = 143.32 g/mol). You have access to: standard AgNO3 solution (0.100 mol/L), an analytical balance (±0.0001 g), a filter funnel and filter paper, a drying oven (60°C), and standard glassware. Time available: one laboratory session (3 hours).
Q2. Design the gravimetric investigation and present it in the format below.
- State your aim and predict the expected result (hypothesis) with reference to the stoichiometric relationship.
- 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 ensure complete precipitation and calculate [Cl−] from the final mass of AgCl.
- Write the stoichiometric calculation pathway from m(AgCl) to c(Cl−) in the sports drink.
- State two limitations of this gravimetric design and assess whether it would be more accurate than a titration, as the student claims.
Q1 — Sample Band 6 response (8 marks), annotated
Error identification: The calculated % yield of 100.6% is impossible in a real experiment [1]. The error arises because the theoretical yield was calculated assuming the ore is exactly 70% pure, but the actual iron recovered (24.62 g) exceeds this theoretical maximum. A % yield above 100% indicates that either the assumed purity is too low (the ore is actually more pure than 70%), or there is an experimental error such as incomplete drying of the iron product (water or slag remaining) or contamination of the collected iron with unreacted ore minerals [1].
Recalculation (showing the error is in the purity assumption): The chemist’s theoretical yield of 24.48 g was calculated from 70% purity. Corrected calculation: n(Fe2O3) = 35.00 ÷ 159.69 = 0.2192 mol; n(Fe) = 0.2192 × 2 = 0.4384 mol; m(Fe) theoretical = 0.4384 × 55.845 = 24.48 g. These steps are correct. The % yield = 24.62 ÷ 24.48 × 100 = 100.6% — this exceeds 100%, which is physically impossible for a pure yield [2 marks for showing full correct recalculation and identifying it still exceeds 100%].
What % yield > 100% would indicate: If taken at face value, a % yield above 100% would imply that more product was obtained than could be produced from the stated starting material — suggesting either the assumed mass of pure reactant was underestimated (purity was higher than 70%), or that the collected product contains impurities (e.g. water, slag, unreduced ore) that artificially inflated its mass. In all real experiments, % yield must be ≤ 100% [1].
Possible reason for excess product — purity higher than 70%: The most likely reason is that the ore sample contains more than 70% Fe2O3 — the purity label was a rough estimate. If purity is, say, 71%, the theoretical yield would be correspondingly larger and the % yield would fall at or below 100%. To improve accuracy, the chemist should determine the purity of the ore independently (e.g. by dissolving a weighed sample and using ICP-OES spectroscopy, or by performing a separate titration to determine iron content) rather than relying on the label [1].
Purity conclusion: Since the actual yield (24.62 g) exceeds the theoretical yield based on 70% purity, the ore must be more than 70% pure. Using back-calculation: n(Fe) actual = 24.62 ÷ 55.845 = 0.4409 mol; n(Fe2O3) = 0.4409 ÷ 2 = 0.2204 mol; m(Fe2O3) = 0.2204 × 159.69 = 35.20 g; % purity = (35.20 ÷ 50.00) × 100 = 70.4% — consistent with the ore being slightly above 70% pure, which is why the calculated % yield slightly exceeds 100% [1].
Marking criteria summary (8 marks): 1 = identifies that % yield > 100% is impossible and explains why; 1 = identifies purity assumption as the source of the error; 2 = correct full recalculation of theoretical yield and % yield showing all steps; 1 = explains what % yield > 100% would mean if taken at face value; 1 = names one experimental reason for the discrepancy and proposes how to improve accuracy; 1 = correctly states ore is more than 70% pure with supporting evidence; 1 = uses precise chemical terminology throughout (theoretical yield, % purity, % yield, stoichiometric ratio, limiting reagent).
Q2 — Sample Band 6 response (7 marks), annotated
Aim and hypothesis: Aim: to determine the concentration of chloride ions in a commercial sports drink using gravimetric analysis (precipitation of AgCl). Hypothesis: when excess AgNO3 is added to a measured volume of sports drink, all Cl− will be precipitated as AgCl; the mass of dried AgCl can be used via stoichiometry (1:1 ratio Ag+:Cl−) to calculate c(Cl−) [1].
Variables: IV = volume of sports drink used (25.0 mL aliquots). DV = mass of dried AgCl precipitate. Controlled: volume of sports drink per trial (pipetted exactly); same brand and batch of sports drink; same drying temperature (60°C, 30 min); same grade of filter paper [1 — IV and DV + two controlled].
Procedure (5 steps): (1) Pipette exactly 25.00 mL of sports drink into a clean 100 mL beaker. (2) Slowly add 30.0 mL of 0.100 mol/L AgNO3 solution (excess) with stirring to ensure complete precipitation of Cl−. Allow to stand 5 minutes. (3) Weigh a dry filter paper on the analytical balance. Set up a gravity filtration funnel over a 100 mL beaker. Carefully pour the contents of the beaker through the filter paper, washing the precipitate 3 times with 5 mL portions of distilled water. (4) Transfer the filter paper + precipitate to a drying oven at 60°C for 30 minutes, then allow to cool in a desiccator for 10 minutes. Weigh the dried filter paper + precipitate and subtract the mass of the dry filter paper. Record m(AgCl). (5) Repeat with two more 25.00 mL aliquots to obtain three values of m(AgCl); calculate the mean [2 — 5 steps with complete precipitation strategy and calculation pathway].
Stoichiometric calculation pathway: n(AgCl) = m(AgCl) ÷ 143.32 → n(Cl−) = n(AgCl) (1:1 ratio) → c(Cl−) = n(Cl−) ÷ V(drink in litres) = n(AgCl) ÷ 0.02500 [1].
Limitations and assessment: Limitation 1: AgCl is slightly soluble in water; washing the precipitate with distilled water dissolves a small amount, reducing the measured mass and giving a slightly low [Cl−]. Improvement: wash with 0.1% HNO3 solution to minimise dissolving. Limitation 2: the sports drink contains other ions that might also precipitate with AgNO3 (e.g. phosphate → Ag3PO4), contaminating the precipitate and inflating its mass. Assessment: the student’s claim that gravimetric analysis is more accurate than titration is partially justified — it does not require a primary standard, so it removes one source of uncertainty. However, the slight solubility of AgCl and potential co-precipitation of other anions introduce errors not present in a well-controlled titration. Both methods require careful technique [1 each limitation; 1 assessment].
Marking criteria summary (7 marks): 1 = aim/hypothesis with reference to 1:1 stoichiometry; 1 = IV, DV, two controlled variables correctly identified; 2 = five numbered steps including excess reagent strategy, washing, drying, and mass measurement; 1 = correct stoichiometric pathway from m(AgCl) to c(Cl−); 1 = one valid limitation with explanation; 1 = second valid limitation plus assessment of student’s claim with nuance.