Chapter 10, Problem 174A

Interpretation Introduction

Interpretation:

Mass percentage of Cu needs to be determined in ${\text{Cu}}_{\text{3}}{\left(\text{OH}\right)}_{\text{4}}{\text{SO}}_{\text{4}}$ and ${\text{Cu}}_{\text{4}}{\left(\text{OH}\right)}_{\text{6}}{\text{SO}}_{\text{4}}$.

Concept introduction:

The 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 174A

Percentage of Cu in Cu₃ (OH) ₄SO₄ and Cu₄ (OH) ₆SO₄ are 53.74 and 56.19 respectively.

Explanation of Solution

Molar mass of Cu₃ (OH) ₄SO₄ and Cu₄ (OH) ₆SO₄ are 354.5 g/mol and 452 g/mol respectively.

Atomic mass of Cu is 63.5 amu.

Now, there are 3 Cu atoms in Cu₃ (OH) ₄SO₄and 4 Cu atoms in Cu₄ (OH) ₆SO₄thus, mass percent of Cu in both compounds will be:

% of Cu in Cu₃ (OH) ₄SO₄$\text{=}\frac{\text{3}\times 63.5\text{×100\%}}{354.5}=\frac{19050}{354.5}\%=53.74\%$

% of Cu in Cu₄ (OH) ₆SO₄$\text{=}\frac{4\times 63.5\text{×100\%}}{452}=\frac{25400}{452}\%=56.19\%$

Conclusion

Thus, the percentage of Cu in Cu₃ (OH) ₄SO₄ and Cu₄ (OH) ₆SO₄ are 53.74% and 56.19% respectively.