Let say formula of iron chlorides which have 56% and 65% chloride are fecl_{x} and fecl_{y} respectively.

then, molecular weight of fecl_{2} = (56 + 35.5x)

∴ Percentage of Cl^{-} = \(\frac{35.5x}{(56\ +\ 35.5x)}\times100=56\)

⇒ 3550x = 56(56 + 35.5x)

3550x = 3136 + 1988x

1562x = 3136

x = 2

molecular weight of fecl_{y} = (56 + 35.5y)

∴ Percentage of cl^{-} = \(\frac{35.5x}{(56\ +\ 35.5y)}\times100=65\)

⇒ 3550y = 3640 + 2307.5y

⇒ 1242.5y = 3640

y = 2.929

y = 3

Hence the formula of iron chloride containing 56% and 65% chlorine are fecl_{2} and fecl_{3} respectively.