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Example of gravimetric analysis
An example of a gravimetric analysis is the determination of chloride in a compound. In order to do a gravimetric analysis, a cation must be found that forms an insoluble compound with chloride. This compound must also be pure and easily filtered. The solubility rules indicate that Ag+, Pb2+, and Hg22+ form insoluble chlorides.
Therefore silver chloride could be used to determine % Cl–, because it is insoluble (that is, about 99.9% of the silver is converted to AgCl) and it can be formed pure and is easily filtered.
- Put enough unknown into a weighing bottle with the lid on sideways and dry in the oven. Cool in a desiccator.
- Indirectly weighsome mass, determined to 0.1 mg, of unknown into beaker.
- Dissolve the unknown.
- Add a precipiating agent to the solution.
- Optional – heat the solution on a hot plate to increase the particle size for easier filtering .This is usually referred to as digestion.
- Test for complete precipitation by adding a drop of the precipitating agent and looking for any sign of precipitate.
- Filter the solution using vacuum filtration. Use a rubber policemanto make sure all the precipitate has been transferred from the beaker to the filter. It is important that the precipitate is quantitatively transferred to the filter. If any remains in the beaker, the mass obtained will be inaccurate.
- Dry and weigh the precipitate.
- Use stoichiometry to determine the mass of the ion being analyzed.
- Find percent by mass of analyte by dividing the mass of the anayte by the mass of the unknown.
The following calculations would be done for this gravimetric determination of chloride:
Mass of sample of unknown chloride after drying =0.0984 g
Mass of AgCl precipitate = 0.2290 g
One mole of AgCl contains one mole of Cl–.
Therefore:
(0.2290 g AgCl)
(143.323g/mol) = 1.598×10-3 molAgCl
(1.598 x 10-3 mol AgCl) x (35.453 g/mol Cl) = 0.0566 g Cl
(0.0566 g Cl) x 100%
(0.0984 g sample)
= 57.57% Cl in unknown chloride sample
Notice that even though the mass of sample (0.0984) only contains three significant figures, the number is known to one part in a thousand (0.0001/0.0984 = 1/1000). The number 0.0984 therefore actually is “good” to four significant figures and the answer can be expressed to four significant figures.
If Pb2+ had been used to precipitate the chloride, the calculuation would need to be modified to account for the fact that each mole off PbCl2 contains two moles of chloride. Lead would not be a good precipitating reagent, however, because PbCl2 is moderately soluble and therefore a small amount of chloride would remain in solution, rather than in the precipitate.