Question

The percent water in ZnSO4.7H20.

Answer 1

Approximately
`43.9\%` by mass.

Step-by-step 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 * 15.999 = 161.436\; \rm g \cdot mol^(-1)`.

`M(\rm H_2O) \approx 2 * 1.008 + 15.999 = 18.015\rm g \cdot mol^(-1)`.

`M(\rm ZnSO_4\cdot 7\, H_2O) \approx 161.436 + 7 * 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 * 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})} * 100\% \\ &\approx (126.105\; \rm g)/(287.541\; \rm g) * 100\%\approx 43.9\%\end{aligned}`.