Question

The percent water in ZnSO4.7H20.

Answer 1

Answer:

Approximately $43.9\%$ by mass.

Explanation:

Look up relevant relative atomic mass data on a modern periodic table:

• $\rm Zn$: $65.38$.
• $\rm S$: $32.06$.
• $\rm O$: $15.999$.
• $\rm H$: $1.008$

Calculate the formula mass of $\rm ZnSO_4$, $\rm H_2O$, and $\rm ZnSO_4\cdot 7\, H_2O$:

$M(\rm ZnSO_4) \approx 65.38 + 32.06 + 4 \times 15.999 = 161.436\; \rm g \cdot mol^{-1}$.

$M(\rm H_2O) \approx 2 \times 1.008 + 15.999 = 18.015\rm g \cdot mol^{-1}$.

$M(\rm ZnSO_4\cdot 7\, H_2O) \approx 161.436 + 7 \times 18.015 \approx 287.541\; \rm g\cdot mol^{-1}$.

In other words, each mole of $\rm ZnSO_4\cdot 7\, H_2O$ formula units has a mass of approximately $287.541\; \rm g$.

However, that one mole of $\rm ZnSO_4\cdot 7\, H_2O$ also contains seven moles of $\rm H_2O$ molecules, which has a mass of approximately $7 \times 18.015 = 126.105\; \rm g$.

Calculate the mass ratio of water in this compound:

$\begin{aligned}& \%m(\text{water}) \\ &= \frac{m(\text{water})}{m(\text{compound})} \times 100\% \\ &\approx \frac{126.105\; \rm g}{287.541\; \rm g} \times 100\%\approx 43.9\%\end{aligned}$.