Question

In another experiment, you have 4.5000 g of a copper(ii) sulfate hydrate with an unknown number of attached water molecules. after heating the hydrate, you have 3.3608 g of the anhydrous compound (copper(ii) sulfate with no waters) left. using these data, calculate the number of water molecules that is present in the formula of this hydrate (obviously before heating).

Answer 1

Copper sulfate hydrate on heating gives out all the water of hydration to yield anhydrous copper sulfate.

`CuSO_(4).xH_(2)O(s)-->CuSO_(4)(s)+xH_(2)O(g)`

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 =
`1.1392 g * (1mol)/(18 g) =0.06329mol`

Moles of Copper sulfate =
`3.3608 g* (1mol)/(159.61g)=0.02106mol`

Mole ratio of water to copper sulfate =
`(0.06329mol)/(0.02106mol)=3`

Therefore, there are 3mol
`H_(2)O`per one mol
`CuSO_(4)`

Hence the formula of the hydrate will be
`CuSO_(4).3H_(2)O`