Chapter 4, Problem 103GQ

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(s) + 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 75.4 g of copper metal with a purity of 89.5%. What is the weight percent of CuS in the ore? The weight percent of Cu₂S?

Interpretation Introduction

Interpretation:

The weight percentage of $CuSandC{u}_{2}S$ in the given sample of ore has to be determined.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.
• Weight percent of elements of a compound is the ratio of weight of element to the weight of whole compound and multiplied with hundred.

Answer to Problem 103GQ

The weight percentage of $CuSandC{u}_{2}S$ in the given ore is $\text{26}\text{.55}\text{\%}$ and $\text{62}\text{.45}\text{\%}$ respectively.

Explanation of Solution

Balanced chemical equation for the reaction occurred in the preparation of copper from cuprite $(C{u}_{2}S)$ and copper (II) sulfide is,

  $\begin{array}{l}C{u}_2{S}_{(s)}+O_{2(g)}\stackrel{}{\to }2C{u}_{(s)}+S{O}_{2(g)}\\ Cu{S}_{(s)}+O_{2(g)}\stackrel{}{\to }C{u}_{(s)}+S{O}_{2(g)}\end{array}$

$\begin{array}{l}Massofimpurityintheore= \frac{11}{100}\times 100\\ =11g\end{array}$

Then the mass of $CuSandC{u}_{2}S$ is $100-11=89g$

Mass of $Cu$ from $CuS$ =  $\frac{massofCuS}{MolarmassofCuS}\times \frac{1molCuinCuS}{1molCuS}\times molarmassofCuS$

Considering A as the mass of $CuS$ and B as the mass of $C{u}_{2}S$,

Mass of $Cu$ from $CuS$ is,

  $\frac{\text{A}}{\text{95}\text{.61}}\text{×}\frac{\text{1}}{\text{1}}\text{×63}\text{.55}\text{g/mol}\text{=}\text{0}\text{.6626}\text{A}$

Similarly mass of $Cu$ from $C{u}_{2}S$ is,

    $\frac{massofC{u}_{2}S}{MolarmassofC{u}_{2}S}\times \frac{2molCuinCuS}{1molC{u}_{2}S}\times molarmassofC{u}_{2}S$

Mass of $Cu$ from $C{u}_{2}S$ is,

  $\frac{\text{B}}{\text{159}\text{.1}}\text{×}\frac{\text{2}}{\text{1}}\text{×63}\text{.55}\text{=}\text{0}\text{.7989}\text{B}$

Total mass of copper metal produced,

  $Cu$ from $CuS$+$Cu$ from $C{u}_{2}S$+ impurities

Thus,

  $TotalmassofCumetalproduced=\text{0}\text{.6626}\text{A+}\text{0}\text{.7989}\text{B+}\text{impurities}$

But here total Cu metal produced with impurity = $\text{75}\text{.4}\text{g}$

Therefore mass of impurity is,

  $\frac{\text{10}\text{.5}}{\text{100}}\text{×75}\text{.4}\text{=}\text{7}\text{.917}\text{g}$

${\text{Total mass of Cu metal produced = mass of Cu from CuS + mass of Cu from Cu}}_{\text{2}}\text{S+ impurities}$

That is,

  $\text{75}\text{.4}\text{=}\text{0}\text{.6626}\text{A}\text{+}\text{0}\text{.7989}\text{B}\text{+}\text{7}\text{.917}$

Calculated mass of $CuS$ and $C{u}_{2}S$ is $89g$.  By using this value the mass of $CuS$ and the mass of $C{u}_{2}S$ can be calculated as follows,

  $A+B=89,A=89-B$

Substituting this in the above equation,

  $\text{75}\text{.4}\text{=}\text{0}\text{.6626}\text{(89-B)}\text{+}\text{0}\text{.7989}\text{B}\text{+}\text{7}\text{.917}$

Then the value of B is $\text{62}\text{.45}\text{g}$

  $\text{A=}\text{89-62}\text{.45}\text{=}\text{26}\text{.55}\text{g}$

$\begin{array}{l}WeightpercentofCuSintheore\text{=}\frac{\text{26}\text{.55}}{\text{100}}\text{×100}\text{\%}\\ \text{= 26}\text{.55}\text{\%}\end{array}$

$\begin{array}{l}WeightpercentofC{u}_{2}Sintheore=\frac{\text{62}\text{.45}}{\text{100}}\text{×100}\text{\%}\\ \text{=}\text{62}\text{.45}\text{\%}\end{array}$

Conclusion

The weight percentage of $CuSandC{u}_{2}S$ in the given sample of ore was determined.