Chemistry • Year 11 • Module 2 • Lesson 20
Module 2 Review
Apply the mole concept, stoichiometry, and quantitative analysis skills to data, worked calculations, and real-world chemical contexts.
1. Interpret experimental data — titration results
A student standardised a sodium hydroxide (NaOH) solution using anhydrous oxalic acid (H2C2O4, MM = 90.04 g/mol) dissolved in water to make 250.0 mL of solution. The student weighed 1.125 g of oxalic acid. The equation is: H2C2O4 + 2NaOH → Na2C2O4 + 2H2O. Aliquots of 25.00 mL oxalic acid were titrated with NaOH. The titration results are shown below. 10 marks
| Titration | Initial burette reading (mL) | Final burette reading (mL) | Volume used (mL) |
|---|---|---|---|
| Rough | 0.00 | 18.75 | |
| T1 | 0.00 | 20.10 | |
| T2 | 20.10 | 40.25 | |
| T3 | 0.00 | 20.15 |
1.1 Complete the “Volume used” column in the table above. 2 marks
1.2 Identify the concordant titres and calculate the average volume to be used. Justify why the rough titre is excluded. 2 marks
1.3 Calculate the concentration of the oxalic acid solution. Show full working. 2 marks
1.4 Using your answer to 1.2 and 1.3, calculate the concentration of the NaOH solution. Show full working including the stoichiometric ratio. 3 marks
1.5 Identify one source of error in this procedure that would cause the calculated [NaOH] to be too high, and explain your reasoning. 1 mark
2. Interpret a reaction yield graph — ethanol production from glucose
In an industrial fermentation process, a student monitored the cumulative mass of ethanol (C2H5OH, MM = 46.07 g/mol) produced over 6 days from 500 g of pure glucose (C6H12O6, MM = 180.16 g/mol). The balanced equation is: C6H12O6 → 2C2H5OH + 2CO2. The graph below shows the cumulative mass of ethanol produced each day. 8 marks
Figure 2. Cumulative mass of ethanol produced over 6 days of fermentation of 500 g pure glucose at 30°C. Reaction: C6H12O6 → 2C2H5OH + 2CO2. Illustrative data.
2.1 Calculate the theoretical yield of ethanol from 500 g of pure glucose. Show full working. (MM: C6H12O6 = 180.16, C2H5OH = 46.07) 3 marks
2.2 Using the graph, calculate the percentage yield of ethanol after 6 days. Show your working. 2 marks
2.3 Describe the trend shown in the graph between Day 1 and Day 6 and suggest one scientific reason why the rate of ethanol production decreases after Day 3. 2 marks
2.4 State one reason why the percentage yield of this process is less than 100%. 1 mark
3. Compare gravimetric and volumetric (titration) analysis
Complete the two-column table below. For each feature, write a concise description contrasting the two analytical techniques. 10 marks (1 per cell)
| Feature | Gravimetric analysis | Volumetric (titration) analysis |
|---|---|---|
| What is measured? | ||
| Type of reaction used | ||
| Key equipment | ||
| Need for primary standard? | ||
| One advantage |
4. Predict and justify — a copper sulfate purification scenario
A Year 11 student is given 25.0 g of a copper sulfate sample that is described as “approximately 80% pure CuSO4•5H2O (MM = 249.68 g/mol)”. She reacts this sample with excess sodium hydroxide solution to precipitate copper(II) hydroxide: CuSO4 + 2NaOH → Cu(OH)2 + Na2SO4. She collects and dries the precipitate and obtains 5.20 g of Cu(OH)2 (MM = 97.56 g/mol). 6 marks
4.1 Calculate the theoretical yield of Cu(OH)2 assuming exactly 80% purity. Show all steps. 3 marks
4.2 Calculate the percentage yield of Cu(OH)2 and suggest one reason why the yield is less than 100%. 2 marks
4.3 Back-calculate the actual percentage purity of CuSO4•5H2O in the sample using the 5.20 g of product and 100% yield. Show your reasoning. 1 mark
Q1.1 — Volume used column
Rough: 18.75 mL • T1: 20.10 mL • T2: 40.25 − 20.10 = 20.15 mL • T3: 20.15 mL.
Q1.2 — Concordant titres and average
Concordant titres are T1 (20.10 mL), T2 (20.15 mL), and T3 (20.15 mL) — all within 0.10 mL of each other [1]. Average = (20.10 + 20.15 + 20.15) ÷ 3 = 20.13 mL [1]. The rough titre (18.75 mL) is excluded because it was performed quickly to estimate the endpoint and the operator likely overshot, making it unreliable.
Q1.3 — Concentration of oxalic acid
n(H2C2O4) = 1.125 ÷ 90.04 = 0.01249 mol [1]. c(H2C2O4) = 0.01249 ÷ 0.2500 = 0.04997 ≈ 0.0500 mol/L [1].
Q1.4 — Concentration of NaOH
n(H2C2O4 in aliquot) = 0.0500 × 0.02500 = 1.250 × 10−3 mol [1]. Ratio H2C2O4 : NaOH = 1 : 2; n(NaOH) = 1.250 × 10−3 × 2 = 2.500 × 10−3 mol [1]. c(NaOH) = 2.500 × 10−3 ÷ 0.02013 = 0.1242 mol/L [1].
Q1.5 — Source of error causing [NaOH] too high
Accept any valid answer, for example: if the burette was not rinsed with NaOH solution before filling and retained water, the NaOH would be diluted — more volume would be needed to reach the endpoint, making the titre larger and the calculated [NaOH] higher than the true value. Alternatively: if the endpoint was overshot in the concordant titres (pale pink persisted > 30 s), extra NaOH was added beyond the true equivalence point, inflating the titre and thus the calculated concentration.
Q2.1 — Theoretical yield of ethanol
n(C6H12O6) = 500 ÷ 180.16 = 2.775 mol [1]. Ratio 1:2; n(C2H5OH) = 2.775 × 2 = 5.550 mol [1]. m(C2H5OH) = 5.550 × 46.07 = 255.7 g [1].
Q2.2 — Percentage yield after 6 days
Actual yield from graph = 172 g. % yield = (172 ÷ 255.7) × 100 = 67.3% [2].
Q2.3 — Trend description and explanation
The cumulative mass of ethanol increases rapidly from Day 1 to Day 3, then levels off from Day 4 onwards, approaching a maximum by Day 6 [1]. One reason the rate decreases after Day 3: the build-up of ethanol in the fermentation medium inhibits the yeast enzymes responsible for fermentation (product inhibition), or the glucose substrate is becoming depleted, reducing reaction rate [1].
Q2.4 — Reason for yield less than 100%
Accept any one of: the fermentation is not complete (some glucose remains unreacted); side reactions occur producing by-products such as glycerol; yeast may also use some glucose for cellular respiration, not fermentation; product inhibition stops the reaction before glucose is fully consumed.
Q3 — Compare and contrast table
What is measured? Gravimetric: mass of a dried precipitate formed from the analyte. Titration: volume of a standard solution required to reach the equivalence point. Type of reaction: Gravimetric: precipitation reaction producing an insoluble product. Titration: acid–base, redox, or complexation reaction with a sharp endpoint. Key equipment: Gravimetric: analytical balance, filter paper, drying oven. Titration: burette, pipette, conical flask, indicator. Primary standard needed? Gravimetric: No; accuracy depends only on the balance and molar mass. Titration: Yes; a primary standard is needed to prepare the standard solution with a precisely known concentration. One advantage: Gravimetric: does not require a primary standard; concentration accuracy is independent of another solution’s concentration. Titration: faster; requires smaller sample quantities; suitable when no easily filterable precipitate exists.
Q4.1 — Theoretical yield of Cu(OH)2
m(pure CuSO4•5H2O) = 25.0 × 0.80 = 20.0 g [1]. n(CuSO4•5H2O) = 20.0 ÷ 249.68 = 0.08010 mol [1]. Ratio 1:1; n(Cu(OH)2) = 0.08010 mol; m(Cu(OH)2) = 0.08010 × 97.56 = 7.814 g ≈ 7.81 g [1].
Q4.2 — Percentage yield and reason
% yield = (5.20 ÷ 7.81) × 100 = 66.6% [1]. One reason: some precipitate was lost during filtering and washing (some Cu(OH)2 may have dissolved slightly or been left on the filter paper); incomplete precipitation (solution not sufficiently concentrated); loss during transfer to/from filter [1]. Accept any valid reason.
Q4.3 — Back-calculate actual purity
If yield = 100%, then n(Cu(OH)2) = 5.20 ÷ 97.56 = 0.05330 mol = n(CuSO4•5H2O). m(pure) = 0.05330 × 249.68 = 13.30 g. % purity = (13.30 ÷ 25.0) × 100 = 53.2%. [1] The sample is approximately 53% pure, not 80% as labelled.