Question

A 2.815-g sample of CaSO4 . XH20 was heated until all of the water was removed. Calculate the percentage water of hydration and the formula of the hydrate if the residue after heating weighed 2.485g.

Answer 1

Ans: CuSO4.XH2O(s) + Heat ----------------> CuSO4(s) + XH2O(g)

          2.815 g          2.485g

First, find the difference between the mass of hydrate before heating and the mass of the anhydrate after heating. This will give the difference that how much mass of water last.

So, 2.815 g – 2.485 g = 0.33 g of water loss

Now, dividing the mass of the water lost by the mass of hydrate used is equal to the fraction of water in the compound. Then multiplying this fraction by 100 gives the percent water in the hydrate.

(0.33 g / 2.815 g)(100) = 11.72 or 12%

To derive the formula: Use molar masses of CuSO4 (136.14g/mol) and H₂O (18.01g/mol) and find moles of both and compare:
2.485 g of CaSO4 = 0.0183 moles of CuSO4
0.33 g of H2O = 0.0183 moles of H2O

Now, molar ratio is 1:1 hence formula = CuSO4^.H2O