What mass due to waters of crystallization is present in a 3.38 g sample of FeSO4 •7H2O?

First, it is necessary to have the molar mass of FeSO4 • 7H2O and molar mass of compound without water FeSO4.

M (FeSO4 • 7H2O) = 54+32 + (16x4) + 7 (2+16) = 276 g/mole

M (FeSO4) = 54+32 + (16x4) = 150 g/mole

When the molar mass value of iron (II) sulfate heptahydrate is subtracted from the molar mass value of iron (II) sulfate we get the total molar mass of water in crystal:

M (H20) = M (FeSO4 • 7H2O) - M (FeSO4) = 276 - 150 = 126 g mole

Now we can determine how much water there are in 3.2 g of the crystal:

M (FeSO4 • 7H2O) : M (H20) = m (FeSO4 • 7H2O) : m (H20)

276 : 126 = 3.38 : x

x = 425.88 / 276 = 1.54 g of H2O