How much chloride is in a sample of rock salt, or sulfur in a fertilizer? One of the oldest and most accurate answers is to weigh it, after turning it into a pure solid. Gravimetric analysis combines a precipitation reaction with Solution Stoichiometry.
Key idea
Dissolve the sample, add an excess of a reagent that precipitates the ion you want completely (chloride as AgCl, sulfate as $\mathrm{BaSO_4}$), then filter, dry and weigh the solid. $$\text{mass of ion} = \text{mass of precipitate} \times \frac{\text{mass of the ion in one formula unit}}{\text{molar mass of the precipitate}}$$ Then divide by the sample's mass for the percent.
Why it works
The excess reagent makes sure every ion of interest ends up in the precipitate, and a very insoluble precipitate with a definite formula carries a known fraction of that ion. AgCl, for example, is always $35.45/143.32 = 24.74\%$ chlorine by mass, whatever sample it came from (definite proportions again). A balance can weigh a solid to 0.0001 g, which makes this method very precise.
Worked example: Chloride in a salt sample
A $0.500\ \text{g}$ sample is dissolved and treated with excess $\mathrm{AgNO_3}$. The dried AgCl ($143.32\ \text{g/mol}$) weighs $1.025\ \text{g}$. What percent of the sample is chloride?
- Chloride's share of AgCl: $35.45/143.32 = 0.2474$.
- Mass of chloride: $1.025 \times 0.2474 = 0.2535\ \text{g}$.
- Percent: $0.2535/0.500 \times 100 = 50.7\%$.
Answer: $50.7\%$
Worked example: Sulfur in a fertilizer
A $0.800\ \text{g}$ fertilizer sample gives $1.167\ \text{g}$ of $\mathrm{BaSO_4}$ ($233.40\ \text{g/mol}$) when its sulfate is precipitated. Find the percent sulfur.
- Sulfur's share of $\mathrm{BaSO_4}$: $32.07/233.40 = 0.1374$.
- Mass of S: $1.167 \times 0.1374 = 0.1604\ \text{g}$.
- Percent: $0.1604/0.800 \times 100 = 20.0\%$.
Answer: $20.0\%$ sulfur.
| Ion sought | Precipitate | Fraction of its mass |
|---|---|---|
| chloride | AgCl | 35.45/143.32 = 0.2474 |
| sulfate (as S) | $\mathrm{BaSO_4}$ | 32.07/233.40 = 0.1374 |
| silver | AgCl | 107.87/143.32 = 0.7526 |
Common mistake
Taking the precipitate's mass as the mass of the ion: "1.025 g of AgCl means 1.025 g of chloride." Most of that mass is silver, which came from the reagent, not the sample. Check: the ion's mass must be smaller than the precipitate's, and the percent can never exceed 100%.
The precipitate must be truly insoluble and must contain nothing else; washing it removes spectator ions, and drying removes water, before it is weighed. The same arithmetic works in reverse: from the mass of an ion you expect, you can predict how much precipitate to look for, and a result far from it warns of an incomplete reaction or an impure solid.