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sladkih [1.3K]
3 years ago
5

The percent water in ZnSO4.7H20.

Chemistry
1 answer:
inna [77]3 years ago
7 0

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}.

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