Question

Calculate the percentage of water of crystallization in ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$$\left[\mathrm{H}=1,\mathrm{O}=16,\mathrm{S}=32,\mathrm{Cu}=64\right]$

Answer 1

Step 1. Given information

Copper sulfate pentahydrate $\left({\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}\right)$ is given. It is required to calculate the percentage of water of crystallization.

Step 2. Calculation of Gram Molecular mass of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$

Gram Molecular mass of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is calculated as follows

$$\begin{gathered} \mathrm{Gram}\mathrm{molecular}{\mathrm{mass}}_{{\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}}=\left(1\text{Cu}+1\text{S+4O}\right)\text{+}\left(5(2\text{H}+1\text{O}\right)=\left(1\times 64+1\times 32\text{+4×16}\right)\text{g+}\left(5(2\times 1+1\times 16\right)\text{g} \\ \text{=}\left(64+32+64\right)\text{g+}\left(90\right)\text{g}=250\text{g} \end{gathered}$$

Step 3. Formula for the determination of mass percentage
$\text{Percentage composition by mass}=\frac{\text{Mass of the element in 1 molecule of a compound}}{\text{Gram molecular mass of the compound}}\times 100$

Step4. Calculation of percentage of water in crystallization in ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$

• There are $5$ molecules of water present as water of crystallization in one molecule of ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$.
• The mass of one molecule of water is $18\mathrm{g}$.
• The mass of water present in ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ as water of crystallization is $5\times 18\mathrm{g}=90\mathrm{g}$

The percentage of water in crystallization in ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is calculated as follows

$$\begin{gathered} \text{\% water of crystallization=}\frac{90\text{g}}{250\text{g}}\text{×100} \\ =36\% \end{gathered}$$

Hence, the percentage of water of crystallization in ${\mathrm{CuSO}}_4.5{\mathrm{H}}_2\mathrm{O}$ is $36\%$.