Question

Iron(II) Sulfate forms a blue-green hydrate with the formula FeSO4 ·n H2O(s). If this hydrate is heated to a high enough temperature, H2O(g) can be driven off, leaving the pale yellow anhydrous salt FeSO4(s). A 10.140-g sample of the hydrate was heated to 300 °C. The resulting FeSO4(s) had a mass of 5.5406 g. Calculate the value of n in FeSO4 ·n H2O(s).

Answer 1

Solution :-

mass of the hydrate = 10.140 g

mass of the anhydrous salt = 5.5406 g

mass of water loss = mass of hydrate - mass of anhydrous salt

                           = 10.140 g - 5.5406 g

                           = 4.5994 g

now lets calculate the moles of anhydrous salt and water

moles = mass / molar mass

moles of FeSO4 = 5.5406 g / 151.9076 g per mol

                          = 0.036473 mol

moles of H2O = 4.5994 g / 18.016 g per mol

                      = 0.255295 mol water

now lets calculate the mole ratio of the water to anhydrous salt

moles of water / moles of FeSO4 = 0.255295 mol / 0.036473 mol = 7

So the value of the n = 7

formula of the hydrate = FeSO4.7H2O