A gold mining company needs to know how much gold is in a tonne of ore — not approximately, exactly. A water treatment plant needs to measure sulfate contamination down to 0.1 mg. Both use gravimetric analysis: the oldest quantitative technique in chemistry, and still one of the most accurate. The principle is beautifully simple — cause precipitation, filter, dry, weigh.
📚 Core Content
Gravimetric analysis is a quantitative technique that determines the amount of an analyte (the substance being measured) by converting it into a pure, insoluble precipitate, then measuring the mass of that precipitate.
The key insight: if you know the mass of precipitate and the chemical formula of the precipitate, you can calculate exactly how many moles were formed — and from the balanced equation, work out how many moles of the original analyte were present.
If the precipitate dissolves even slightly, some of the analyte stays in solution and is never collected on the filter. This means the mass you weigh underestimates the true amount — your result will be too low. The more insoluble the precipitate, the more complete the reaction and the more accurate the result.
Adding a slight excess of the precipitating reagent (e.g. excess BaCl₂ when precipitating SO₄²⁻) drives the reaction to completion — ensuring all of the analyte has been converted to precipitate. The excess reagent remains dissolved in solution and is washed away during filtration.
Once you have the mass of precipitate, the calculation follows a chain of three conversions:
| Feature | Gravimetric | Volumetric (Titration) |
|---|---|---|
| What is measured | Mass of precipitate | Volume of solution |
| Key equipment | Analytical balance, oven, filter | Burette, pipette, conical flask |
| Typical precision | Very high (±0.0001 g balance) | High (±0.05 mL burette) |
| Speed | Slow (hours — drying time) | Fast (minutes per titration) |
| Best for | Insoluble or sparingly soluble analytes | Acid-base, redox reactions |
🧮 Worked Examples
🧪 Activities
1 A 0.2332 g precipitate of BaSO₄ is obtained from a water sample. Calculate the moles of SO₄²⁻ in the sample. (MM of BaSO₄ = 233.39 g mol⁻¹)
2 A 100 mL seawater sample produces 0.5735 g of AgCl precipitate. Calculate the concentration of Cl⁻ in mol L⁻¹. (Ag = 107.87, Cl = 35.453)
3 The filter paper used in a gravimetric experiment had a mass of 1.2455 g. After drying, the filter paper plus precipitate weighed 1.6823 g. What is the mass of the precipitate?
4 A 200 mL sample of industrial wastewater is analysed and found to contain 0.6798 g of BaSO₄ precipitate. Calculate the concentration of SO₄²⁻ in the sample in g L⁻¹. (MM of BaSO₄ = 233.39, S = 32.06, O = 15.999)
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❓ Multiple Choice
1. Why must excess precipitating reagent be added in gravimetric analysis?
2. A sample produces 0.4660 g of BaSO₄. Using MM(BaSO₄) = 233.39 g mol⁻¹, which is the correct first step?
3. Why is the precipitate cooled in a desiccator before weighing?
4. In an experiment, a 0.9576 g precipitate of AgCl is obtained. How many moles of Cl⁻ were in the sample? (Ag = 107.87, Cl = 35.453)
5. Which of the following best explains why BaSO₄ is ideal for gravimetric determination of sulfate?
✍️ Short Answer
6. A student performs a gravimetric analysis of a 250 mL sample of water to determine its sulfate content. After adding excess BaCl₂ and collecting the precipitate, the filter paper weighed 1.1234 g before and 1.5566 g after drying. Calculate: (a) the mass of BaSO₄ precipitate, (b) the moles of SO₄²⁻ in the sample, and (c) the concentration of SO₄²⁻ in mol L⁻¹. (MM of BaSO₄ = 233.39 g mol⁻¹) 5 MARKS
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7. Describe the gravimetric procedure a student would follow to determine the mass of chloride ions (Cl⁻) in a 500 mL sample of tap water, using silver nitrate (AgNO₃) as the precipitating reagent. In your answer, identify the precipitate formed, justify why excess AgNO₃ is used, and explain why the precipitate must be dried before weighing. 5 MARKS
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1. C — Excess reagent drives the precipitation reaction to completion, ensuring all analyte forms precipitate.
2. B — The first step always uses the MM of the precipitate (BaSO₄ = 233.39) to convert mass to moles.
3. A — After oven drying, the hot precipitate would absorb water from air on the bench. The desiccator contains drying agent (e.g. silica gel) that removes atmospheric moisture.
4. D — MM(AgCl) = 143.32. n = 0.9576 ÷ 143.32 = 6.68 × 10⁻³ mol. 1:1 ratio → n(Cl⁻) = 6.68 × 10⁻³ mol.
5. B — The extremely low solubility (Ksp ≈ 1.1 × 10⁻¹⁰) ensures virtually all SO₄²⁻ precipitates — the reaction goes to completion, giving an accurate result.
Q6 (5 marks):
(a) m(BaSO₄) = 1.5566 − 1.1234 = 0.4332 g (b) n(BaSO₄) = 0.4332 ÷ 233.39 = 1.857 × 10⁻³ mol; n(SO₄²⁻) = 1.857 × 10⁻³ mol (1:1 ratio) (c) c(SO₄²⁻) = 1.857 × 10⁻³ ÷ 0.250 = 7.43 × 10⁻³ mol L⁻¹Q7 (5 marks): Add excess AgNO₃(aq) to the 500 mL tap water sample and stir. The precipitate formed is AgCl(s) — a white, insoluble solid — via: Ag⁺(aq) + Cl⁻(aq) → AgCl(s). Excess AgNO₃ is used to ensure all Cl⁻ ions react and are converted to precipitate, driving the reaction to completion. Filter the precipitate through a pre-weighed filter paper, washing with distilled water to remove soluble impurities. Dry the precipitate in an oven at ~100°C and cool in a desiccator. The precipitate must be completely dry before weighing because any remaining water would add mass to the measurement, causing the calculated Cl⁻ content to be overestimated. Weigh the dry precipitate and subtract the filter paper mass to get m(AgCl), then calculate: n(AgCl) = m ÷ 143.32; n(Cl⁻) = n(AgCl); m(Cl⁻) = n × 35.453.
Tick when you've finished all activities and checked your answers.