Chapter 15.2, Problem 8SP

Interpretation Introduction

Interpretation: To determine the percent by mass of water in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$

Concept Introduction: The mass percentage used to describe the solution composition indicates the mass of solute contained in a given mass of solution.

Answer to Problem 8SP

The percent by mass of water in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ will be 36 %.

Explanation of Solution

It is known that,

Mass of $\text{Cu}$ = 63.54 u

Mass of S = 32 u

Mass of oxygen = 16 u

Mass of hydrogen = 1.008 u

  $\begin{array}{c}\text{Mass of copper sulfate = mass of copper + mass of sulfur+ mass of oxygen}\\ \text{Mass}\text{of}\text{copper}\text{sulfate}\text{=}\text{63}\text{.54 u}+\text{32 u}+\text{16 u}\\ \text{Mass}\text{of}\text{copper}\text{sulfate=}\text{159u}\end{array}$

Now, add the amount of water in ${\text{CuSO}}_{\text{4}}$ as shown:

  $\begin{array}{l}{\text{The mass of CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}\text{= mass of copper sulfate + mass of water molecule}\\ {\text{The mass of CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}\text{=}\text{159}+\text{5×18}\\ {\text{The mass of CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}\text{=}\text{249}\end{array}$

The percentage of water molecules can be determined by using the formula:

  $\text{\%}\text{of}{\text{H}}_{\text{2}}\text{O=}\frac{\text{Mass}\text{of}\text{water}}{\begin{array}{l}\text{Mass}\text{of}\text{copper}\text{sulphate}\\ =\frac{\text{90}}{\text{249}}\times 100\%\\ =36\%\end{array}}\times 100\%$

Therefore, the percent by mass of water in ${\text{CuSO}}_{\text{4}}{\text{.5H}}_{\text{2}}\text{O}$ will be 36 %.