Question

1. A 0.7336-g sample of an alloy containing copper and zinc is dissolved in 8 M HCl and diluted to 100 mL in a volumetric flask. In one analysis, the zinc in a 25.00-mL portion of the solution is precipitated as ZnNH4PO4, and subsequently isolated as Zn2P2O7, yielding 0.1163 g. The copper in a separate 25.00-mL portion of the solution is treated to precipitate CuSCN, yielding 0.1931 g. Calculate the %w/w Zn and the %w/w Cu in the sample.

Answer 1

From the recovered Zn2P2O7 we calculate the grams of Zn recovered:

g Zn = 0.1163 g Zn2P2O7 * (1 mol Zn2P2O7 / 304.71 g) * (2 mol Zn / 1 mol Zn2P2O7) * (65.38 g Zn / 1 mol) * (100 mL / 25 mL) = 0.2 g Zn recovered

From the precipitate of CuSCN we calculate the grams of Cu recovered:

g Cu = 0.1931 g CuSCN * (1 mole CuSCN / 121.63 g) * (1 mole Cu / 1 mole CuSCN) * (63.55 g Cu / 1 mole) * (100 mL / 25 mL) = 0.4 g Cu recovered

We calculate the percentages by weight of each metal:

% Zn = g Zn * 100 / g total = 0.2 * 100 / 0.7336 = 27.26%

% Cu = 0.4 * 100 / 0.7336 = 54.53%

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