Part B: Sulfate. A 1.109 g sample of a material known to contain sulfate is diss

Question

Part B: Sulfate. A 1.109 g sample of a material known to contain sulfate is dissolved and the sulfate is precipitated with Pb2 ion. After filtering, washing and drying, the precipitated PbS04 weighed 1.356 g. Calculate the percent of sulfate ion th Part C: Copper. A 0.275 M solution of Cu2' give an absorbance reading of 0.788 at 620 nm. Another solution gives an absorbance reading of 0.340 under the same conditions. What is the concentration of the second solution in moles of Cu"/L and grams of Cu2*/L Part D: Volatiles. A 1.232 gram sample of copper complex weighed 0.783 g after heating. What is the percent of volatile material in this sample? If the sample contains 17.4% ammonia, what is the percent water in the sample? Part E: Ammonia. A 0.467 g sample known to contain ammonia is titrated with 0.2791 M HCI. If 38.87 mL is required for the titration, calculate the percent ammonia in this sample.

Explanation / Answer

m = 1.109 g of sample

mass of PbSO4 = 1.356 g

mol = mass/MW = 1.356/303.26 = 0.00447mol of PbSO4

so

0.00447 mol of SO4-2 (sulfate

so..

MW of SO4-2 = 96.0626 g/mol

mass of SO4 = 96.0626 *0.00447 = 0.429399 g of SO4-2

% sulfate ion = 0.429399 / 1.109 * 100 = 38.719 %

m = 1.232 g of sample

m = 0.783 g after heating

% volatile material = (1.232 -0.783 ) / 1.232 *100 = 36.444 %

if sample is 17.4% ammonia...

find water..

100- 17.4 = 82.6 %

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