There are two weights of halides of the same trivalent metal. Samples have the same mass

There are two weights of halides of the same trivalent metal. Samples have the same mass, but in one of them there are 2 times more metal atoms than in the other. Establish the formulas of halides.

Decision:
1) Write the formula for halides:
Halides – MeA3 and MeB3;

2) Find the molar mass of the halides МеА3 and МеВ3:
Take Ar (Me) for x, Ar (A) for y, Ar (B) for z;
Mr (MeA3) = Ar (Me) + 3 * Ar (A) = x + 3y;
Mr (MeB3) = Ar (Me) + 3 * Ar (B) = x + 3z;

3) Since the weighed portions have the same mass, the amount of the substance of the halides will be equal, but the molar masses are different, because different halogens are found in the molecules. Then
Mr (MeA3) / Mr (MeB3) = 2;
(x + 3y) / (x + 3z) = 2;
x + 3y = 2 * (x + 3z);
x + 3y = 2x + 6z;
x = 3y – 6z;

4) Using the table D.I. Mendeleev, by the selection method, substituting the atomic masses of halogens (F, Cl, Br, I, At, Ts) into the equation, find the atomic mass Me. In this case, the atomic mass of the element A (y) must be at least 2 times greater than the atomic mass of the element B (z) for the solution to the equation to be correct.
Thus, element A is bromine (Br), element B is chlorine (Cl). Then
x = 3 * 80 – 6 * 35.5 = 27;
Ar (Me) = x = 27 – this is aluminum (valence III).

5) Formulas of halides:
AlBr3 and AlCl3.

Answer: Formulas of halides are AlBr3 and AlCl3.