Chemistry • Year 12 • Module 8 • Lesson 2
Gravimetric Analysis
Apply stoichiometric reasoning to gravimetric data, interpret a multi-trial dataset, and sequence the analytical method in context.
1. Interpret multi-trial gravimetric data — Murray–Darling sulfate monitoring
Water quality officers from the Murray–Darling Basin Authority collected irrigation drainage water and analysed its sulfate content by gravimetric analysis. Each 50.0 mL sample was evaporated to a 0.500 g dried residue and treated with excess BaCl2(aq). The BaSO4(s) precipitate was filtered, dried to constant mass, and weighed. Results are given below. 10 marks
| Trial | Mass of dried residue (g) | Mass of BaSO4 precipitate (g) | Observation |
|---|---|---|---|
| 1 | 0.500 | 0.348 | White precipitate, dried to constant mass |
| 2 | 0.500 | 0.351 | White precipitate, dried to constant mass |
| 3 | 0.500 | 0.392 | Precipitate removed from oven early — slightly damp |
| 4 | 0.500 | 0.287 | Some precipitate adhered to beaker wall; not fully transferred |
M(BaSO4) = 233.39 g mol−1; M(SO42−) = 96.06 g mol−1
1.1 Identify which trials should be excluded from the calculation. For each excluded trial, state the procedural reason and explain the direction of the error on the measured precipitate mass. 4 marks
1.2 Calculate the average BaSO4 mass from the valid trials. 1 mark
1.3 Using the average valid mass, calculate the mass of sulfate ion (SO42−) in the 0.500 g dried residue. Show all working. 3 marks
1.4 Calculate the percentage by mass of sulfate ion in the dried residue. 1 mark
2. Graph interpretation — BaSO4 precipitate mass vs sulfate concentration
A chemist prepared six standard sulfate solutions of known concentration and carried out a gravimetric analysis on each 25.0 mL aliquot. The results are shown in the graph below. 7 marks
Figure 2. BaSO4 precipitate mass from 25.0 mL aliquots of standard SO42− solutions. Adapted from standard gravimetric practice.
2.1 Describe the relationship between sulfate concentration and precipitate mass shown in the graph. 2 marks
2.2 An unknown drainage sample yielded 263 mg of BaSO4. Use the graph to estimate the sulfate concentration of that sample in g L−1. Show your reading on the graph or in your working. 1 mark
2.3 Explain, using lesson content, why the graph passes through the origin. 2 marks
2.4 If incomplete drying had affected all six standards equally, predict how the graph line would change compared to the one shown. 2 marks
3. Sequence the steps — FSANZ chloride analysis of processed food
FSANZ (Food Standards Australia New Zealand) monitors sodium chloride in processed foods. The eight steps below describe a gravimetric determination of chloride in a processed cheese sample, but they have been shuffled. Write the correct order (1–8) in the “Order” column. 8 marks
| Order | Step description |
|---|---|
| Heat the crucible containing the precipitate in an oven at 105°C until the mass is constant between two successive readings. | |
| Weigh the dried AgCl precipitate in the crucible; record mass. | |
| Filter the suspension through ashless filter paper into a pre-weighed crucible, washing the precipitate with dilute HNO3(aq). | |
| Calculate moles of AgCl using n = m/M (M = 143.32 g mol−1). | |
| Dissolve a precisely weighed cheese sample in warm dilute HNO3(aq) to bring chloride ions into solution. | |
| Use stoichiometry (Ag+ + Cl− → AgCl, 1:1 ratio) to find n(Cl−) and hence m(Cl−). | |
| Add excess AgNO3(aq) to the sample solution; a white curdy precipitate of AgCl(s) immediately forms. | |
| Calculate the percentage by mass of chloride, and hence NaCl, in the original cheese sample. |
4. Cause-and-effect chains — gravimetric errors
For each procedural error, complete the effect chain. 6 marks
4.1 Cause: The precipitating agent is not added in excess (insufficient amount used).
So: some analyte remains instead of forming the precipitate.
So: the collected precipitate mass is too .
Overall outcome: the calculated analyte content is . (1 mark per effect, 3 total)
4.2 Cause: Impurity ions from the sample solution are trapped within the BaSO4 crystals as they form (co-precipitation).
So: the precipitate contains extra mass from .
So: the measured precipitate mass is too .
Overall outcome: the calculated analyte content is . (1 mark per effect, 3 total)
Q1 — Multi-trial data
1.1 Exclude Trial 3 (precipitate not fully dried; residual water inflates mass → measured mass too high, analyte overestimated) and Trial 4 (precipitate not fully transferred; lower mass collected → measured mass too low, analyte underestimated). 2 marks: 1 per trial with both reason and direction. Award 1 each.
1.2 Average valid mass = (0.348 + 0.351) / 2 = 0.3495 g.
1.3 n(BaSO4) = 0.3495 / 233.39 = 1.498 × 10−3 mol. By 1:1 ratio n(SO42−) = 1.498 × 10−3 mol. m(SO42−) = 1.498 × 10−3 × 96.06 = 0.1439 g (accept 0.144 g). Award 1 mark each step.
1.4 % SO42− = (0.144 / 0.500) × 100 = 28.8%.
Q2 — Graph interpretation
2.1 The relationship is directly proportional (linear through the origin): as sulfate concentration increases, the mass of BaSO4 precipitate increases at a constant rate. [1 mark for “linear/proportional”; 1 mark for through the origin or specifying constant rate.]
2.2 Reading from graph at 263 mg on y-axis → approximately 1.5 g L−1.
2.3 When there is zero sulfate, no precipitate forms and the mass of BaSO4 = 0; so the relationship logically passes through the origin. [1 mark for reasoning; 1 mark for no-analyte/no-precipitate link.]
2.4 The line would be shifted upward (higher y-intercept / steeper apparent gradient): every data point would show a larger measured mass than the true precipitate mass, because extra water mass is included. The slope would appear steeper than the true relationship. [1 mark for “higher/shifted up”; 1 mark for reason: water adds to every measured mass.]
Q3 — Sequence (correct order)
1 — Dissolve cheese in warm dilute HNO3(aq).
2 — Add excess AgNO3(aq); white AgCl precipitate forms.
3 — Filter through ashless filter paper, washing precipitate.
4 — Heat crucible at 105°C to constant mass.
5 — Weigh dried AgCl in crucible.
6 — Calculate n(AgCl) = m/M.
7 — Use stoichiometry to find n(Cl−) and m(Cl−).
8 — Calculate % Cl− and NaCl in sample.
Award 1 mark per correct position (accept ±0 errors in steps 6–8 if logical order maintained).
Q4 — Cause-and-effect chains
4.1 …remains dissolved. So measured mass too low. Overall: analyte content underestimated.
4.2 …extra mass from impurities/foreign ions. So measured mass too high. Overall: analyte content overestimated.