Chapter 15.2, Problem 15LC

Interpretation Introduction

Interpretation: Whether the percentage of the mass of copper in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ is the same or not as in ${\text{CuSO}}_{\text{4}}$ needs to be determined.

Concept Introduction: The percent by mass of any species in other can be calculated by taking the ratio of their molar masses and multiplying by 100%. For example, the percentage of A in AB can be calculated as follow:

  $\%A=\frac{M_{\text{A}}}{M_{\text{AB}}}\times 100\%$

Explanation of Solution

The mass of copper in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ and ${\text{CuSO}}_{\text{4}}$ can be simply calculated by taking the ratio of the molar mass of Cu and given species. Since, the molar mass of ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ will be more than ${\text{CuSO}}_{\text{4}}$ , thus, the percentage of the mass of Cu should be more in case of ${\text{CuSO}}_{\text{4}}$ .

Here, the molar mass of Cu is 63.54 g/mol. Now, the molar mass of ${\text{CuSO}}_{\text{4}}$ and ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ can be calculated as follows:

  $\begin{array}{c}{\text{CuSO}}_{\text{4}}=63.54+32.065+4\left(16\text{ g/mol}\right)\\ =63.54+32.065+64\\ =159.605\text{ g/mol}\end{array}$

Or,

  $\begin{array}{c}{\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}=63.54+32.065+4\left(16\text{ g/mol}\right)+5\left(18\text{ g/mol}\right)\\ =63.54+32.065+154\\ =249.605\text{ g/mol}\end{array}$

Now, the percentage of Cu in ${\text{CuSO}}_{\text{4}}$ will be:

  $\begin{array}{c}\%\text{Cu}=\frac{63.54}{159.605}\times 100\%\\ =40.44\%\end{array}$

Similarly, the percentage of Cu in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ will be:

  $\begin{array}{c}\%\text{Cu}=\frac{63.54}{249.605}\times 100\%\\ =25.45\%\end{array}$

Therefore, the percentage of Cu in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ is less than that in ${\text{CuSO}}_{\text{4}}$ .