Analytical Chem istry - DePauw University

376 Analytical Chemistry 2.0Practice Exercise 8.2A 0.7336-g sample of an alloy containing copper and zinc is dissolved in 8M HCl and diluted to 100 mL in a volumetric flask. In one analysis, thezinc in a 25.00-mL portion of the solution is precipitated as ZnNH 4 PO 4 ,and subsequently isolated as Zn 2 P 2 O 7 , yielding 0.1163 g. The copperin a separate 25.00-mL portion of the solution is treated to precipitateCuSCN, yielding 0.2931 g. Calculate the %w/w Zn and the %w/w Cuin the sample.Click here to review your answer to this exercise.In Practice Exercise 8.2 the sample contains two analytes. Because wecan selectively precipitate each analyte, finding their respective concentrationsis a straightforward stoichiometric calculation. But what if we cannotprecipitate the two analytes separately? To find the concentrations of bothanalytes, we still need to generate two precipitates, at least one of whichmust contain both analytes. Although this complicates the calculations, wecan still use a conservation of mass to solve the problem.Example 8.2A 0.611-g sample of an alloy containing Al and Mg is dissolved and treatedto prevent interferences by the alloy’s other constituents. Aluminum andmagnesium are precipitated using 8-hydroxyquinoline, providing a mixedprecipitate of Al(C 9 H 6 NO) 3 and Mg(C 9 H 6 NO) 2 that weighs 7.815 g.Igniting the precipitate converts it to a mixture of Al 2 O 3 and MgO thatweighs 1.002 g. Calculate the %w/w Al and %w/w Mg in the alloy.So l u t i o nThe masses of the solids provide us with the following two equations.gAl(C HNO) + gMg(C HNO) = 7.815 g9 6 3 9 6 2gAlO + gMgO =1 . 002 g2 3With two equations and four unknowns, we need two additional equationsto solve the problem. A conservation of mass requires that all thealuminum in Al(C 9 H 6 NO) 3 is found in Al 2 O 3 ; thusgAlO1 molAl= gAl(C HNO) 3×459.45 gAl(C HNO)2 3 9 6gAlO9 6= 0.11096×gAl(C HNO)32 3 9 63101.96 gAlO×2 molAlO2 32 3Using the same approach, a conservation of mass for magnesium gives