a 5.49g sample of iron(iii) chloride hydrate was found to contain 2.20g of water. what is the complete formula of the hydrate?

Question

anonymous · Answer

%(H2O)=2.2/5.49 x 100
           =40%
so m(FeCl3) = 60% m(total)

M(FeCl3)= 162.2g
M(FeCl3. xH2O)= 162.2 x100/60
                        =270.3g

so m(H2O) = 270.3-162.6
                 = 108.13g
     n(H2O) =m/M
                 =108.13/ (2x1.008+16)
                 6.002 mol (in one mol of FeCl3)

thus hydrate formula= FeCl3. 6H2O