Answer:
56.44 %
Explanation:
The formula of copper(II) sulfate pentahydrate is [tex]CuSO_4\cdot 5H_2O[/tex].
In order to calculate the mass percentage of water in it, we may assume that we have 1 mol of copper(II) sulfate pentahydrate. In general, the mass percentage is a ratio between the mass of a component and the total mass of the compound expressed in percent:
[tex]\omega = \frac{m_c}{m_t}\cdot 100 \%[/tex]
Since we're taking 1 mole of a substance here, we may state that the mass percentage of water will be calculating using molar masses instead:
[tex]\omega_{H_2O} = \frac{5M_{H_2O}}{M_{CuSO_4\cdot 5 H_2O}}\cdot 100 \%[/tex]
Notice that we take 5 molar masses of water, as 1 mole of copper(II) sulfate pentahydrate contains 5 moles of water molecules, then:
[tex]\omega_{H_2O} = \frac{5\cdot 18.016 g/mol}{159.609 g/mol}\cdot 100 \% = 56.44 \%[/tex]
