Three elements – A, B, and C form three binary compounds. Each element has the same valence in these — Physical Chemistry Chemistry Question

Let the valences of elements A, B, and C be a, b, c, respectively. (Do not confuse valences and oxidation numbers!) The formulas of three compounds are: AbBa, BcCb, AcCa. From the mass fractions we can determine the ratios of atomic masses to valences:
w (A in AbBa) = b M(A) / (b M(A⁺ a M(B)) = 0.75
=> M(A)/a = 3 M(B)/b

w (B in BcCb) = c M(B) / (c M(B⁺ b M(C)) = 0.078
=> M(C)/c = 11.8 M(B)/b

We see that the ratio M(B) / b is the smallest one. Considering several elements with small ratios of atomic mass to valence, we easily find, that B is carbon: M = 12, b = 4, then A is aluminium: M = 27, a = 3, and C is chlorine: M = 35.5, c = 1. The compounds are: Al₄C₃, CCl₄, and AlCl₃.
The mass fraction of chlorine in aluminum chloride is:
w(Cl in AlCl₃) = (3 × 35.5) / (3 × 35.5 + 27) = 0.798 = 79.8%.

This result can be obtained without determining the exact formula of AcCa. Indeed, from the above relations, we find that (M(C)/c) / (M(A)/a) = 3.93. Therefore,
w(C in AcCa) = (a c M(C)) / (c a M(A) + a c M(C)) = (M(C)/c) / (M(A)/a + M(C)/c) = 3.93 / (1 + 3.93) = 0.798