Answer:
Approximately [tex]43.9\%[/tex] by mass.
Explanation:
Look up relevant relative atomic mass data on a modern periodic table:
Calculate the formula mass of [tex]\rm ZnSO_4[/tex], [tex]\rm H_2O[/tex], and [tex]\rm ZnSO_4\cdot 7\, H_2O[/tex]:
[tex]M(\rm ZnSO_4) \approx 65.38 + 32.06 + 4 \times 15.999 = 161.436\; \rm g \cdot mol^{-1}[/tex].
[tex]M(\rm H_2O) \approx 2 \times 1.008 + 15.999 = 18.015\rm g \cdot mol^{-1}[/tex].
[tex]M(\rm ZnSO_4\cdot 7\, H_2O) \approx 161.436 + 7 \times 18.015 \approx 287.541\; \rm g\cdot mol^{-1}[/tex].
In other words, each mole of [tex]\rm ZnSO_4\cdot 7\, H_2O[/tex] formula units has a mass of approximately [tex]287.541\; \rm g[/tex].
However, that one mole of [tex]\rm ZnSO_4\cdot 7\, H_2O[/tex] also contains seven moles of [tex]\rm H_2O[/tex] molecules, which has a mass of approximately [tex]7 \times 18.015 = 126.105\; \rm g[/tex].
Calculate the mass ratio of water in this compound:
[tex]\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}[/tex].
