Copper sulfate hydrate on heating gives out all the water of hydration to yield anhydrous copper sulfate.
[tex]CuSO_{4}.xH_{2}O(s)-->CuSO_{4}(s)+xH_{2}O(g)[/tex]
Mass of copper sulfate hydrate = 4.5000 g
Mass of anhydrous compound = 3.3608 g
So, the mass of water lost = 4.5000 g - 3.3608 g = 1.1392 g
Moles of water = [tex]1.1392 g * \frac{1mol}{18 g} =0.06329mol[/tex]
Moles of Copper sulfate = [tex]3.3608 g* \frac{1mol}{159.61g}=0.02106mol[/tex]
Mole ratio of water to copper sulfate = [tex]\frac{0.06329mol}{0.02106mol}=3[/tex]
Therefore, there are 3mol [tex]H_{2}O[/tex]per one mol [tex]CuSO_{4}[/tex]
Hence the formula of the hydrate will be [tex]CuSO_{4}.3H_{2}O[/tex]
