Question

Copper metal can be prepared by roasting copper ore, which can contain cuprite (Cu2S) and copper(II) sulfide.
Cu2S(s)+O2(g)-->2 Cu(s) +SO2(g)
CuS(g)+O2(g)-->Cu(s)+ SO2(g)
Suppose an ore sample contains 11.0% impurity in addition to a mixture of CuS and Cu2S . Heating 100.0 g of the mixture produces 76.3 g of copper metal with a purity of 89.3%. What is the weight percent of CuS in the ore? The weight percent of Cu2S ?
Weight Percent of CuS=____
Weight percent of Cu2S=_____

Answer 1

1.find the mass of impurities

$\frac{11}{100}\times100=11$

100-11=89 g

2.let Cu₂S - x g, CuS=89-x g

Cu from Cu₂S - 127x/159g

Total amount of Cu:

$\frac{127x}{159}+\frac{63.5(89.0-x)}{95.5}=\frac{76.3\times89.3}{100}$

$\frac{127x}{159}+\frac{63.5(89.0-x)}{95.5}=68.1359$

x=66.95 - m Cu₂S

$\omega (Cu2S)=\frac{66.95}{100}=0.6695$ or 66.95%

$m (CuS)=89-66.95=22.05$ or 22.05%