A 15.79 g sample of a hydrate of an iron(II) sulfate compound was heated, without decomposing the sulfate to drive off the water. The mass was reduced to 8.63 g . What is the formula of the hydrate?

Answer 1

First, find out how much water was present:

15.79 g - 8.63 g = 7.16 g of H₂O

How many moles of water is this?

7.16 g H₂O x 1 mol H₂O/18 g = 0.398 moles H₂O

Find moles of Iron(II) sulfate remaining:

8.63 g FeSO₄ x 1 mol FeSO₄/151.9 g = 0.0568 moles FeSO₄

Find the mole ratio of H₂O to FeSO₄

0.398 mol H₂O / 0.0568 mol FeSO₄ = 7.00

Formula of the hydrate = FeSO₄•7H₂O