Chapter 10, Problem 198A

Interpretation Introduction

Interpretation:

The ore of copper that yield the greater quantity of copper must be calculated.

Concept introduction:

Percentage of an element in a compound can be calculated as follows:

Percentage of an element in a compound $\text{=}\frac{\text{Number of atom}\times \text{Atomic mass of the element×100}}{\text{Molar mass of the compound}}\text{.}$

Answer to Problem 198A

Cu₂S would yield greater quantity of copper.

Explanation of Solution

Molar mass of Cu₂S is 159 g/mol

Molar mass of CuFeS₂ is 183.5 g/mol

Atomic mass of Cu is 63.5 g/mol

The percentage of Cu in given compound can be calculated as follows:

% of Cu in Cu₂S $\text{=}\frac{\text{2}\times \text{63}\text{.5×100\%}}{159}=\frac{12700}{159}\%=79.87\%$

% of Cu in CuFeS₂$\text{=}\frac{1\times \text{63}\text{.5×100\%}}{183.5}=\frac{6350}{183.5}\%=76.05\%$

Conclusion

Thus, Cu₂S would yield greater quantity of copper as it contains higher % of Cu.