Gravimetric Analysis

Key Concepts

Gravimetric analysis is the quantitative isolation of a substance by precipitation and weighing of the precipitate.
An analyte is the substance to be analysed.
A precipitating reagent is the reactant used to precipitate the analyte.
Procedure:

Weigh the sample to be analysed
Dissolve the sample in a suitable solvent, eg, water
Add an excess of the precipitating reagent to precipitate the analyte
Filter the mixture to separate the precipitate from the solution
Wash the precipitate to remove any impurities
Dry the precipitate by heating to remove water
Cool the precipitate in a dessicator to prevent the precipitate absorbing moisture from the air
Weigh the cooled precipitate
Repeat the drying and weighing process until a constant mass for the precipitate is achieved
Calculate the percent by mass of analyte in the sample

Wite the balanced chemical equation for the precipitation reaction
Calculate the moles of precipitate: moles = mass ÷ molecular mass
Calculate moles of analyte from the balanced chemical equation using the mole ratio of analyte : precipitate
Calculate mass of analyte: mass = moles x molecular mass
Calculate percent by mass of analyte in sample: (mass analyte ÷ mass sample) x 100

Incomplete precipitation results in a value for the percentage of analyte in the sample that is too low.
Incomplete drying of the sample results in a value for the percentage of analyte in the sample that is too high
Other ions in the sample may also be precipitated resulting in a value for the percentage of analyte in the sample that is too high

A 2.00g sample of limestone was dissolved in hydrochloric acid and all the calcium present in the sample was converted to Ca²⁺_(aq).

Excess ammonium oxalate solution, (NH₄)₂C₂O_4(aq), was added to the solution to precipitate the calcium ions as calcium oxalate, CaC₂O_4(s).

The precipitate was filtered, dried and weighed to a constant mass of 2.43g.

Determine the percentage by mass of calcium in the limestone sample.

Wite the balanced chemical equation for the precipitation reaction:
Ca²⁺_(aq) + C₂O₄^2-_(aq) -----> CaC₂O_4(s)
Calculate the moles of calcium oxalate precipitated.
n(CaC₂O_4(s)) = mass ÷ MM
n(CaC₂O_4(s)) = 2.43 ÷ (40.08 + 2 x 12.01 + 4 x 16.00)
n(CaC₂O_4(s)) = 2.43 ÷ 128.10
n(CaC₂O_4(s)) = 0.019 mol
Find the moles of Ca²⁺_(aq).
From the balanced chemical equation, the mole ratio of Ca²⁺ : CaC₂O_4(s) is 1 : 1
So, n(Ca²⁺_(aq)) = n(CaC₂O_4(s)) = 0.019mol
Calculate the mass of calcium in grams
mass (Ca) = n x MM
mass (Ca) = 0.019 x 40.08 = 0.76g
Calculate the percentage by mass of calcium in the original sample:
%Ca = (mass Ca ÷ mass sample) x 100
%Ca = (0.76 ÷ 2.00) x 100 = 38%