`0.1653 g` aluminium reacts completely with `0.652 g` chlorine to form chloride of aluminium.
a. What is the empirical formula of the compound?
b. If molecular mass of the compound is 267 amu, calculate the molecular formula of the compound.

Explanation

`0.1653 g of Al = 0.652 g` of chlorine
a. `27 g of Al = (0.652)/(0.1653) xx 27`
`106.95 g` of chlorine
Moles of Chlorine `= (106)/(35.5) = 2.998 ~~ 3`
Therefore formula `= AlCl_(3)`
b. `Mw = 267 g`
Empiricial formula weight of `AlCl_(3) = 27 + 3 xx 35.5`
`= 133.5`
`n = (Mw)/(EFw) = (267)/(133.5) = 2`
Molecular formula `= 2 xx AlCl_(2) = Al_(2) Cl_(6)`