Question

a) gravimetric analysis is very useful for routine monitoring of ions present in relatively high concentrations. why is gravimetric analysis not used for determining trace amounts of ions?

b) from .323g of an unknonw sample of M_xCL_y, .523 g AgCl was recovered using gravimetric analysis employed in this experiment. what is the percentage of Cl- in the unknown sample? what is the theoretical molat wieght of metal in the chlorite salt sample M_xCl_y for hte following given:

1. x=1, y= 1

2. x=1, y=2

c. x=2, y=4

Answer 1

a) Gravimetric anaylsis relies purely on the very low solubility product of the precipitate in the solvent used, calling for large ion concentrations to ensure that the precipitate dissolved in the media by the result of its equilibrium in solution is extremely small and thus negligible for calculation of the desired parameters. It involves using a precipitating agent, another soluble compound to produce an insoluble compound with the ion of interest, after reaction.

Solubility product is a measure of the equilibrium between the dissolved and insoluble compound in a solvent at a certain temperature and pressure. Therefore, if trace amounts of ions are present, then a large portion of the ion, even as the precipitate species will remain dissolved in the solution giving an inaccurate result as only a portion of the actual concentration of the ion of interest will be acquired by the observer. This will result in a much lower value for the desired parameter than actual. Hence, to avoid such imprecision, large ion concetrations are preferred for gravimetric analysis.

b) The molar mass of AgCl is 143.32g/mol and that of chloride ion is 35.453g/mol. In 0.523g of AgCl, there is 3.6492mmol of the silver salt. This gives the no.of moles of chloride in 0.323g of the unknown sample to also be 3.6492mmol as AgCl is comprised of equimolar quantities of silver and chloride ions.

So, for MCl, it shows 3.6492mmol of M in the 0.323g of sample.If 0.323g makes up 3.6492mmol, then molar mass of M is calculated by 0.323/0.0036492 = 88.5125g/mol. Now, the molar mass of the precipitate is 85.5125+35.453 = 120.9655g/mol. The percentage composition of chloride is now given by (35.453/120.9655)x100 = 29.31%

If the unknown sample's formula is MCl2, then it means that 3.6492mmol of chloride is there as two equivalents of M giving 3.6492/2 = 1.8246mmol of M which in turn gives it molar mass to be 0.323/0.0018246 = 177.0251g/mol. The molar mass of the chloride precipitate is 177.0251+(2x35.453) = 247.9311, giving a percentage composition of chloride to be {(2x35.453)/247.9311}x100 = 28.6%

If the unknown's formula is M2Cl4, it means that there is 3.6492mmol/4 = 0.9123x2 = 1.8246mmol of M in the sample. Since there are two equivalents of M to four equivalents of chloride, the no.of moles of M in one equivalence of M in 0.323g of sample is 0.9123mmol giving the molar mass of M in 0.323g to be 0.323/0.0009123 = 354.0502g/mol. The molar mass of the precipitate is (2x354.0502)+(4x35.453) = 849.9124g/mol having the percentage of chloride to be {(4x35.453)/849.9124}x100 = 16.6855%