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Gravimetric analysis finds the amount of a substance by weighing: either a pure precipitate formed from it, or the mass lost when it’s heated. The chemistry is straightforward, but the calculations need care with molar masses, ratios and significant figures. These twelve problems cover the main question types. For the method itself, see gravimetric analysis.
Molar masses used (g/mol): H 1.008; C 12.01; O 16.00; Na 22.99; Mg 24.31; S 32.06; Cl 35.45; K 39.10; Ca 40.08; Fe 55.85; Cu 63.55; Ag 107.87; Ba 137.33.
Derived: BaSO₄ 233.39; AgCl 143.32; H₂O 18.02; CaCO₃ 100.09; CaO 56.08; CO₂ 44.01; MgSO₄ 120.37; CuSO₄ 159.61; Fe₂O₃ 159.70.
The general method
- Find moles of the weighed product (mass ÷ molar mass).
- Use the formula or equation to convert to moles of the analyte.
- Convert to mass, percentage or concentration.
A shortcut: the gravimetric factor = (molar mass of analyte × ratio) ÷ molar mass of weighed product. Mass of analyte = mass of product × gravimetric factor.
Part A: Precipitation methods
1. Excess barium chloride is added to a solution of a sulfate fertiliser. The dried BaSO₄ precipitate weighs 0.4668 g. Find the mass of sulfate ions.
Answer: n(BaSO₄) = 0.4668 ÷ 233.39 = 2.000 × 10⁻³ mol n(SO₄²⁻) = 2.000 × 10⁻³ mol (1 : 1) M(SO₄²⁻) = 32.06 + 4 × 16.00 = 96.06 Mass = 2.000 × 10⁻³ × 96.06 = 0.1921 g
2. A 0.5000 g sample of an impure chloride salt gives 1.0749 g of AgCl with excess silver nitrate. Calculate the percentage of chlorine by mass in the sample.
Answer: Gravimetric factor for Cl in AgCl = 35.45 ÷ 143.32 = 0.24735 Mass of Cl = 1.0749 × 0.24735 = 0.2659 g % Cl = 0.2659 ÷ 0.5000 × 100 = 53.18%
3. The sample in question 2 is thought to be potassium chloride. Is this consistent with the result? (KCl: 74.55 g/mol)
Answer: Pure KCl contains 35.45 ÷ 74.55 × 100 = 47.55% Cl. The sample contains 53.18%, more chlorine than pure KCl could. So it can’t be KCl with an inert impurity. It might contain a chloride with a higher chlorine percentage, such as NaCl (60.66% Cl), perhaps mixed with KCl. See question 11 for how to find such a mixture’s composition.
4. 25.00 cm³ of a solution of magnesium sulfate gives 0.5835 g of BaSO₄. Find the concentration of MgSO₄ in mol/dm³ and g/dm³.
Answer: n(BaSO₄) = 0.5835 ÷ 233.39 = 2.500 × 10⁻³ mol = n(MgSO₄) c = 2.500 × 10⁻³ ÷ 0.02500 = 0.1000 mol/dm³ In g/dm³: 0.1000 × 120.37 = 12.04 g/dm³
Part B: Heating methods and water of crystallisation
5. 2.495 g of hydrated copper(II) sulfate, CuSO₄·xH₂O, is heated to constant mass. The anhydrous CuSO₄ left weighs 1.596 g. Find x.
Answer: Mass of water = 2.495 − 1.596 = 0.899 g n(CuSO₄) = 1.596 ÷ 159.61 = 0.01000 mol n(H₂O) = 0.899 ÷ 18.02 = 0.04989 mol Ratio = 0.04989 ÷ 0.01000 = 4.99 ≈ 5, so CuSO₄·5H₂O See hydrates and water of crystallisation.
6. Why must a sample be heated “to constant mass”?
Answer: To be sure all the water (or other volatile product) has been driven off. The sample is heated, cooled and weighed repeatedly until two consecutive masses agree within the balance’s precision. If heating stops too soon, the mass loss is too small and x comes out too low.
7. A student heats a hydrate but lets it cool on the open bench before weighing. How will this affect the value of x, and how should it be avoided?
Answer: The anhydrous solid absorbs water vapour from the air as it cools, so its mass appears too high, the calculated mass of water lost is too small, and x comes out too low. Cool the sample in a desiccator before weighing.
8. 5.00 g of impure limestone is heated strongly until the mass is constant. The final mass is 3.02 g. Assuming only CaCO₃ decomposes (CaCO₃ → CaO + CO₂), find the percentage purity.
Answer: Mass of CO₂ lost = 5.00 − 3.02 = 1.98 g n(CO₂) = 1.98 ÷ 44.01 = 0.04499 mol = n(CaCO₃) Mass of CaCO₃ = 0.04499 × 100.09 = 4.503 g Purity = 4.503 ÷ 5.00 × 100 = 90.1%
Part C: Harder problems
9. Iron in an ore is converted to Fe₂O₃, which is weighed. A 1.250 g sample of ore gives 0.8520 g of Fe₂O₃. Find the percentage of iron in the ore.
Answer: Gravimetric factor = (2 × 55.85) ÷ 159.70 = 0.69944 Mass of Fe = 0.8520 × 0.69944 = 0.5959 g % Fe = 0.5959 ÷ 1.250 × 100 = 47.67% The factor of 2 matters: each Fe₂O₃ contains two iron atoms.
10. Magnesium sulfate heptahydrate (Epsom salt), MgSO₄·7H₂O, is heated to constant mass. What percentage of the original mass should be lost? (MgSO₄·7H₂O = 246.48 g/mol)
Answer: 7 × 18.02 = 126.14 g of water per 246.48 g of hydrate. % lost = 126.14 ÷ 246.48 × 100 = 51.18% This kind of “expected value” calculation is useful to check an experiment is working.
11. A 1.000 g mixture of NaCl and KCl is dissolved and treated with excess AgNO₃, giving 2.1450 g of AgCl. Find the mass of each salt. (NaCl 58.44, KCl 74.55)
Answer: Let the mass of NaCl = m, so KCl = 1.000 − m. Total moles of Cl⁻ = n(AgCl) = 2.1450 ÷ 143.32 = 0.014966 mol m ÷ 58.44 + (1.000 − m) ÷ 74.55 = 0.014966 0.017112m + 0.013414 − 0.013414m = 0.014966 0.003698m = 0.001552 m = 0.420 g NaCl, so KCl = 0.580 g Check: 0.420 ÷ 58.44 + 0.580 ÷ 74.55 = 0.007187 + 0.007780 = 0.014967 mol ✓
12. Give three sources of error in a BaSO₄ precipitation analysis, and say whether each makes the result too high or too low.
Answer: Any three, for example:
- Precipitate passes through the filter (particles too small) → too low. Fix: digest the precipitate (keep it hot) so crystals grow; use fine filter paper.
- Precipitate not fully washed, so soluble salts remain → too high.
- Precipitate not dried to constant mass (still damp) → too high.
- Incomplete precipitation (not enough BaCl₂) → too low. Fix: add excess and test the filtrate with a further drop of BaCl₂.
- Co-precipitation of other ions trapped in the crystals → usually too high.
Tips for gravimetric questions
- Keep four significant figures in intermediate steps when masses are given to four decimal places.
- Always check the ratio of analyte atoms per formula unit of product (e.g. 2 Fe per Fe₂O₃).
- Sense-check: the analyte mass can’t exceed the sample mass, and percentages should be ≤ 100%.
- For mixture problems, set up one equation per unknown.
- Write the balanced equation or formula first; most wrong answers come from a wrong ratio, not wrong arithmetic.
- Link errors to the direction of the effect: “too high” or “too low”, and why.
For related calculations, see percent composition, empirical vs molecular formula and percent yield.
Key takeaways
- Gravimetric calculations run: mass of product → moles → ratio → moles of analyte → mass or percentage.
- The gravimetric factor combines these steps in one multiplication.
- Heating to constant mass and cooling in a desiccator are essential for accurate hydrate and decomposition results.
- Mixture problems need simultaneous equations.
- Every error should be linked to whether it makes the result too high or too low.
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