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 Higher-order
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:
- Explain why the true percentage yield of a reaction cannot exceed 100%, and what an apparent (measured) value of 100.6% therefore tells you.
- Recalculate the theoretical yield and the apparent % yield from the data, showing full working.
- Identify at least two different possible explanations for the apparent yield exceeding 100% (for example: the assumed 70% purity is too low; the collected iron was not fully dry or was contaminated with slag/unreacted ore).
- Explain why a single mass measurement cannot distinguish between these explanations, so the exact ore purity cannot be concluded from this result alone.
- Describe one additional measurement or control that would let the chemist separate the purity question from product-contamination effects.
2. Experimental design, determining concentration of a sports drink (Band 5–6)
7 marks Higher-order
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).
Safety. Silver nitrate is corrosive and stains skin and clothing; wear safety glasses and gloves, use the minimum (microscale) volume, and collect all silver-containing waste in the labelled silver-recovery container, never down the sink. Your design must include these controls.
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
True vs apparent yield: The true yield of a reaction can never exceed 100%, because no more product can form than the limiting reactant allows. The chemist's measured (apparent) value of 100.6% is a real observation, and because it is above 100% it must be a signal that one or more of the assumptions or measurements is off, not evidence that the reaction "over-produced" [1]. The recovered iron (24.62 g) exceeds the theoretical maximum calculated from the assumed 70% purity, so at least one of the following must be true: the assumed purity is too low, OR the collected iron was not fully dry / was contaminated with slag or unreacted ore [1].
Recalculation: The theoretical yield assuming 70% purity: m(Fe2O3) = 0.70 × 50.00 = 35.00 g; 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. Apparent % yield = 24.62 ÷ 24.48 × 100 = 100.6%. Because this exceeds 100%, it cannot be a true yield, so the 70% purity assumption and/or the product mass must be questioned [2 marks for full correct recalculation and identifying it exceeds 100%].
What an apparent % yield > 100% indicates: A measured value above 100% does not mean more product formed than the limiting reactant allows (the true yield is always ≤ 100%). It signals that the assumed mass of pure reactant was underestimated (purity higher than 70%) and/or that the collected product mass was inflated by impurities (water, slag, unreduced ore). The measurement is real; the >100% figure is a diagnostic that an assumption or the product mass is unreliable [1].
Two competing explanations: (1) The ore is actually more than 70% Fe2O3 (the 70% label was approximate). If, for example, the true purity were ~70.4%, the theoretical yield would rise to ~24.62 g and the apparent yield would fall to ~100%. (2) The collected iron was not pure and dry, residual slag, unreacted ore, or moisture would inflate its weighed mass above the true mass of iron produced. Both explanations are fully consistent with the single observation of 24.62 g [1].
Why the purity cannot be concluded from this result: A single product mass cannot separate "higher ore purity" from "contaminated/wet product", both raise the measured mass in exactly the same way. The naive back-calculation (assuming the product is pure: 24.62 ÷ 55.845 = 0.4409 mol Fe → 0.2204 mol Fe2O3 → 35.20 g → 70.4% purity) is only valid if the iron is pure, dry and the reduction went to completion, none of which can be verified here. So 70.4% is at best an upper-bound estimate, not a measured purity [1].
Additional measurement to resolve it: The chemist should determine the iron content of the ore independently, for example by dissolving a separately weighed sub-sample and assaying iron by titration or ICP-OES, and should dry the reduced product to constant mass (and check for slag) before weighing. Only with an independent purity value can the purity and product-contamination effects be separated [1].
Marking criteria summary (8 marks): 1 = distinguishes true yield (≤100%) from an apparent measured value that can read >100%; 2 = correct full recalculation of theoretical yield and apparent % yield; 1 = gives at least two distinct explanations for the apparent >100% result; 1 = explains a single mass cannot distinguish purity from contamination; 1 = states the purity cannot be concluded from one result (70.4% is an upper-bound estimate only); 1 = proposes a valid independent measurement; 1 = precise chemical terminology throughout.
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.