6g of metal M react completely with 23.66g of chlorine to form 29.66g of the metallic chloride. Find the empirical formula of the metallic chloride. (M=27,Cl=35.5)

mols M = 6/27 = 0.222

molsl Cl = 23.66/35.5 = 0.666

Find the ratio. The easy way to do that is to divide both numbers by the smaller of the two.
0.222/0.222 = 1
0.666/0.222 = 3
Empirical formula is MCl3.
M must be Al.

Pls can you make the solvings clear to my understanding thanks

Thanks love the answer to the question.

The workings are correct .thanks

To find the empirical formula of the metallic chloride, we need to determine the ratio of atoms in the compound.

1. Start by finding the number of moles of metal M, chlorine, and the metallic chloride.

Moles of metal M = mass of metal M / molar mass of metal M
Moles of chlorine = mass of chlorine / molar mass of chlorine
Moles of metallic chloride = mass of metallic chloride / molar mass of chloride

Moles of metal M = 6g / 27 g/mol = 0.222 mol
Moles of chlorine = 23.66g / 35.5 g/mol = 0.666 mol
Moles of metallic chloride = 29.66g / (27 g/mol + 35.5 g/mol) = 0.428 mol

2. Divide the moles of each element by the smallest value among them (in this case, the moles of metal M, 0.222 mol) to get the simplest whole-number ratio.

Moles of metal M = 0.222 mol / 0.222 mol = 1
Moles of chlorine = 0.666 mol / 0.222 mol ≈ 3
Moles of metallic chloride = 0.428 mol / 0.222 mol ≈ 2

The simplest whole-number ratio of the atoms is 1:3:2.

3. Write the empirical formula of the metallic chloride using the subscripts obtained in the previous step.

The empirical formula of the metallic chloride is MCl2, where M represents the metal.

Therefore, the empirical formula of the metallic chloride is MCl2.

6 g of M is 6/27 mol

23.66 g of Cl is 2/3 mol

2/3 = 14/27

so there are 14 Cl atoms for every 6 atoms of M
ratio is 3/7 M/Cl

M3Cl7