The idea behind gravimetric analysis
Gravimetric analysis重量分析 turns the component being measured into a pure solid of known composition, weighs it, and converts using the formula. There are two common kinds:
- Precipitation: add a reagent that turns the ion being measured completely into an insoluble solid, such as Cl⁻ → AgCl or SO₄²⁻ → BaSO₄.
- Volatilization: heat to drive off a volatile component and calculate from the mass lost, such as the water in a hydrate.
The purpose of each step
- Weigh the sample accurately (analytical balance).
- Dissolve it and add a slight excess of the precipitating reagent, so the ion precipitates completely.
- Check that precipitation is complete: let it settle and add one more drop of reagent to the clear liquid; it’s complete when no more precipitate forms.
- Filter and wash: filter through a weighed crucible or filter paper and wash soluble impurities off the solid.
- Dry to constant mass: repeat drying, cooling and weighing until two weighings in a row agree within the allowed limit (such as 0.0005 g), which shows the water is gone.
- Calculate: m(precipitate) → n(precipitate) → n(component) → m(component) → mass percent.
A solid mixture containing NaCl weighs 0.5000 g. Click the steps in order; repeat the last one until the mass is constant.
- A white precipitate forms: Ag⁺(aq) + Cl⁻(aq) → AgCl(s).
- No new precipitate forms, so all the Cl⁻ has precipitated.
- This washes soluble impurities (such as NaNO₃ and excess AgNO₃) off the solid.
Two weighings in a row agree within 0.0005 g: the mass is constant.Trial Crucible + solid (g) Solid (g) Loss (g)
Example 1: percent chlorine in a soluble chloride
A 0.4513 g sample of a soluble chloride is dissolved in water and excess AgNO₃ is added. The AgCl, dried to constant mass, weighs 0.8826 g. Find the mass percent of Cl in the sample (AgCl 143.32 g/mol, Cl 35.45 g/mol).
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Moles of AgCl.
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Mass of Cl. Each AgCl contains one Cl:
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Mass percent.
Example 2: the formula of a hydrate
1.2223 g of BaCl₂·xH₂O is heated to constant mass, leaving 1.0420 g of anhydrous BaCl₂. Find x (BaCl₂ 208.23 g/mol, H₂O 18.02 g/mol).
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Water lost. 1.2223 − 1.0420 = 0.1803 g.
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Moles of each.
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The ratio.
The formula is .
- Measuring Cl⁻ by adding AgNO₃ solution
Insoluble AgCl forms.
- Measuring Cl⁻ by adding NaNO₃ solution
Sodium salts and nitrates are all soluble, so nothing precipitates.
- Measuring SO₄²⁻ by adding BaCl₂ solution
Insoluble BaSO₄ forms.
- Measuring SO₄²⁻ by adding KNO₃ solution
Potassium salts and nitrates are all soluble.
- Measuring Ag⁺ by adding NaCl solution
AgCl precipitates, so this works in reverse too.
- Measuring Na⁺ by adding K₂CO₃ solution
Common sodium salts are all soluble, so Na⁺ can’t be measured by precipitation.
- Measuring Ca²⁺ by adding Na₂CO₃ solution
Insoluble CaCO₃ forms.
- Measuring NO₃⁻ by adding BaCl₂ solution
All nitrates are soluble.
Common mistakes
Mistake 1: “One weighing after drying is enough.” One round of drying may not remove all the water. Dry and weigh repeatedly until the mass is constant; otherwise water remains in the solid and the result is too high.
Mistake 2: “The more precipitating reagent, the better.” A slight excess is enough. A large excess leaves reagent clinging to the precipitate that is hard to wash off, making the mass too large.
Mistake 3: “Percent Cl = mass of AgCl ÷ mass of sample.” Only part of the mass of AgCl is chlorine. Convert first, m(Cl) = m(AgCl) × 35.45 ÷ 143.32, then divide by the sample mass.
Quick check
Why is the precipitate dried to constant mass?
When two weighings in a row no longer change, the water is gone.
How do you check that precipitation is complete?
If no more precipitate forms, the ion has precipitated completely.
The main reason for a slight excess of precipitating reagent is to:
With too little reagent, some of the ion stays in solution and the result is too low.
How many grams of Cl are in 0.2866 g of AgCl?
0.2866 ÷ 143.32 = 2.000 × 10⁻³ mol, × 35.45 = 0.07089 g.
The purpose of washing the precipitate is to:
Impurities left on the solid make its mass too large.
Why is a heated crucible cooled in a desiccator before weighing?
Both effects make the mass reading wrong.
AP-style practice
To find the purity of an impure Na₂SO₄ sample, a student dissolves 0.7500 g of it in water and adds excess BaCl₂ solution, forming a white precipitate. The precipitate is filtered, washed and dried to a constant mass of 1.0815 g (BaSO₄ 233.39 g/mol, Na₂SO₄ 142.04 g/mol). Assume the impurities don’t react with BaCl₂.
(a) Show how the precipitate forms with a net ionic equation.
(b) Calculate the mass percent of Na₂SO₄ in the sample.
(c) How can the student confirm that all the SO₄²⁻ has precipitated?
(d) After the first drying the precipitate weighed 1.1102 g, after the second 1.0817 g, and after the third 1.0815 g. Why can’t the first mass be used? Would a result calculated from it be too high or too low?
(e) A little precipitate passes through the filter paper. Is the measured mass percent too high or too low? Explain.
Model answer and scoring
(a) Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s).
(b) n(BaSO₄) = 1.0815 ÷ 233.39 = 4.634 × 10⁻³ mol = n(Na₂SO₄); m(Na₂SO₄) = 4.634 × 10⁻³ × 142.04 = 0.6582 g; mass percent = 0.6582 ÷ 0.7500 × 100% = 87.76%.
(c) Let the mixture settle, then add more BaCl₂ solution to the clear liquid; if no more white precipitate forms, all the SO₄²⁻ has precipitated.
(d) After the first drying the solid still held water (the next two masses kept falling, so it wasn’t at constant mass). Using 1.1102 g makes the BaSO₄ mass too large, so the calculated Na₂SO₄ and its mass percent are too high.
(e) Too low. Some BaSO₄ was lost, so the measured mass is too small → n(BaSO₄) is too small → the calculated mass of Na₂SO₄ is too small.
Scoring: (a) a correct net ionic equation with states; (b) the 1 : 1 mole ratio and the correct result; (c) testing the clear liquid; (d) not at constant mass, water still present, result too high; (e) too low, with the chain of reasoning.
Summary
- Gravimetric analysis turns the component into a pure solid of known composition (or drives off a volatile part), weighs it, and converts using the formula.
- Key steps: a slight excess of reagent with a check that precipitation is complete, washing off impurities, and drying to constant mass.
- The calculation runs m(precipitate) → n → mole ratio from the formula → m(component) → mass percent.
AP tip: Gravimetric error questions almost always come down to whether the weighed mass is too large or too small: not fully dry or impurities not washed off → too high; incomplete precipitation or lost precipitate → too low.
Sources: atomic masses from the periodic table supplied with the AP Chemistry exam (giving AgCl 143.32, BaSO₄ 233.39 and BaCl₂ 208.23 g/mol); the experimental data are illustrative.