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 

The correct answer is c