Question

A powder contains FeSO4·7H2O (molar mass = 278.01 g/mol) among other components. A 2.710-g sample of the powder was dissolved in HNO3 and heated to convert all iron to Fe3 . The addition of NH3 precipitated Fe2O3·xH2O which was subsequently ignited to produce 0.278 g of Fe2O3. What was the mass of FeSO4·7H2O in the 2.710-g sample?

Answer 1

solution :-

mass of Fe2O3 = 0.278 g

lets calculate the moles of Fe2O3

moles of Fe2O3 = 0.278 g / 159.7 g per mol = 0.001741 mol

now lets calculate the moles of the Fe

1 mol Fe2O3 = 2 mol Fe

So moles of Fe = 0.001741 mol * 2 = 0.003482 mol Fe

1 mole Fe = 1 mol FeSO4

So moles of FeSO4 = 0.003482 mol

Now lets convert the moles of the FeSO4 to its mass

Mass of FeSO4.7H2O = moles * molar mass

                           = 0.003482 mol * 278 g per mol

                          = 0.968 g

So the mass of the FeSO4.7H2O in the powder is 0.968 g