Chemistry • Year 11 • Module 2 • Lesson 9
Gravimetric Analysis
Apply the gravimetric calculation pathway and procedure knowledge to real experimental data, tables, and scenarios.
1. Interpret experimental data — sulfate in irrigation water
A water quality officer collected samples from four irrigation channels in the Murray–Darling Basin and tested each for sulfate (SO42−) by gravimetric analysis. The table records the raw data from each trial. 10 marks
| Sample | Volume of water (mL) | Mass of filter paper (g) | Mass of filter paper + BaSO4 (g) | Mass of BaSO4 (g) | n(BaSO4) (mol) | c(SO42−) (mol L−1) |
|---|---|---|---|---|---|---|
| A | 250 | 1.1024 | 1.5698 | |||
| B | 500 | 1.0873 | 1.6785 | |||
| C | 200 | 1.0951 | 1.3284 | |||
| D | 100 | 1.1102 | 1.2549 |
MM(BaSO4) = 233.39 g mol−1. Use 4 significant figures throughout.
1.1 Complete all empty cells in the table. Show your working for Sample A below. 6 marks (1.5 per sample)
1.2 The NSW EPA guideline for irrigation water is a maximum sulfate concentration of 3.0 × 10−3 mol L−1. Identify which sample(s), if any, exceed this guideline. 2 marks
1.3 Explain why using a larger sample volume (e.g. 500 mL vs 100 mL) improves the accuracy of the gravimetric result for a dilute sample. 2 marks
2. Interpret graph — repeated weighings to constant mass
A student dried a BaSO4 precipitate in an oven, cooled it in a desiccator, and weighed it multiple times. The graph records the mass after each drying cycle. 7 marks
Figure 2. Measured mass of BaSO4 precipitate after each 30-minute oven-drying cycle (120 °C), cooled in desiccator. Masses recorded in grams. Illustrative data.
2.1 Describe the trend in mass from cycle 1 to cycle 5. 2 marks
2.2 Explain, in terms of water content, why the mass decreases between cycles 1 and 3 and then levels off. 2 marks
2.3 Identify the correct mass to use in all subsequent calculations, from the graph. Explain what would happen to the calculated SO42− concentration if cycle 1 mass (0.5241 g) were used instead of the constant mass. 3 marks
3. Compare gravimetric and volumetric analysis across five features
Complete the two-column table below. For each feature, write a concise description contrasting the two techniques. 10 marks (1 per cell)
| Feature | Gravimetric analysis | Volumetric analysis (titration) |
|---|---|---|
| What is measured? | ||
| Key equipment | ||
| Typical speed | ||
| Main source of error | ||
| Best suited to |
4. Predict and justify — error analysis scenario
A student performed a gravimetric analysis for chloride (Cl−) ions in a 250 mL bore water sample using AgNO3 as the precipitating reagent. After filtering and oven-drying the AgCl precipitate, she realised she had forgotten to subtract the mass of the filter paper before calculating. The filter paper weighed 1.1055 g; the total dried mass was 1.6729 g.
5 marks
4.1 Calculate the correct mass of AgCl precipitate. Use this to find the correct concentration of Cl− in the sample in mol L−1. (MM(AgCl) = 143.32 g mol−1, MM(Cl) = 35.453 g mol−1) 3 marks
4.2 Predict whether the student’s uncorrected result (using the total mass 1.6729 g) would give a concentration that is too high or too low. Justify your answer quantitatively. 2 marks
Q1.1 — Completed data table
Sample A: m(BaSO4) = 1.5698 − 1.1024 = 0.4674 g. n(BaSO4) = 0.4674 ÷ 233.39 = 2.003 × 10−3 mol. c = 2.003 × 10−3 ÷ 0.250 = 8.01 × 10−3 mol L−1.
Sample B: m = 1.6785 − 1.0873 = 0.5912 g. n = 0.5912 ÷ 233.39 = 2.533 × 10−3 mol. c = 2.533 × 10−3 ÷ 0.500 = 5.07 × 10−3 mol L−1.
Sample C: m = 1.3284 − 1.0951 = 0.2333 g. n = 0.2333 ÷ 233.39 = 9.996 × 10−4 mol. c = 9.996 × 10−4 ÷ 0.200 = 5.00 × 10−3 mol L−1.
Sample D: m = 1.2549 − 1.1102 = 0.1447 g. n = 0.1447 ÷ 233.39 = 6.201 × 10−4 mol. c = 6.201 × 10−4 ÷ 0.100 = 6.20 × 10−3 mol L−1.
Q1.2 — Samples exceeding guideline
All four samples exceed 3.0 × 10−3 mol L−1: A (8.01), B (5.07), C (5.00), D (6.20) × 10−3 mol L−1.
Q1.3 — Why larger volume improves accuracy
For a dilute sample, a larger volume produces a larger mass of precipitate. A larger precipitate mass reduces the relative impact of the analytical balance uncertainty (±0.0001 g) and any filter paper mass error, improving the precision and accuracy of the calculated concentration.
Q2.1 — Trend in mass (2 marks)
The mass decreases sharply from 0.5241 g at cycle 1 to approximately 0.5105 g at cycle 3 [1]. From cycle 3 onwards the mass levels off, reaching a constant value of approximately 0.5102 g — indicating all residual water has been removed [1].
Q2.2 — Why mass decreases then levels off (2 marks)
Each drying cycle removes more residual water (adsorbed or trapped moisture) from the precipitate, reducing its measured mass [1]. Once all the water has been driven off, there is no further mass loss and the mass reaches a constant minimum value, indicating the precipitate is completely dry [1].
Q2.3 — Constant mass to use; effect of using cycle 1 mass (3 marks)
Correct mass to use: 0.5102 g (the constant value reached at cycles 4–5) [1]. If the cycle 1 mass (0.5241 g) were used, the calculated moles of BaSO4 would be too high (0.5241 ÷ 233.39 = 2.246 × 10−3 vs correct 2.186 × 10−3 mol), giving a calculated SO42− concentration that is too high [1]. This occurs because the extra mass recorded at cycle 1 was residual water, not BaSO4 — using it would overestimate the analyte content of the sample [1].
Q3 — Gravimetric vs volumetric comparison
What is measured: Gravimetric: mass of precipitate. Volumetric: volume of standard solution at endpoint. Key equipment: Gravimetric: analytical balance, oven, desiccator, filter paper. Volumetric: burette, pipette, conical flask, indicator. Typical speed: Gravimetric: slow (hours, due to drying time). Volumetric: fast (minutes per titration). Main source of error: Gravimetric: incomplete precipitation, coprecipitation, incomplete drying. Volumetric: reading the burette meniscus, endpoint detection. Best suited to: Gravimetric: very insoluble analytes (Cl−, SO42−); situations requiring high precision by mass. Volumetric: acid–base or redox analytes; situations requiring rapid results.
Q4.1 — Correct Cl− concentration (3 marks)
m(AgCl) = 1.6729 − 1.1055 = 0.5674 g [1]. n(AgCl) = 0.5674 ÷ 143.32 = 3.959 × 10−3 mol = n(Cl−) [1]. c(Cl−) = 3.959 × 10−3 ÷ 0.250 = 1.58 × 10−2 mol L−1 [1].
Q4.2 — Uncorrected result (2 marks)
The uncorrected result (using 1.6729 g) would give a concentration that is too high [1]. The uncorrected n(AgCl) = 1.6729 ÷ 143.32 = 1.167 × 10−2 mol, giving c = 4.67 × 10−2 mol L−1 — nearly three times the correct value, because the filter paper mass (1.1055 g) inflated the apparent precipitate mass [1].