The relative mass of aluminum chloride is 267 and its composition by mass is 20,3% Al and 79,7% chlorine. Determine the empirical and molecular formulas of Aluminum chloride.

Answer 1

Take a 100 g sample.
This will give you 20.3 g Al and 79.7 g Cl. Convert each of these grams into moles.
moles Al = 20.3/atomic mass Al
moles Cl = 79.7/atomic mass Cl.

Now find the ratio, in small whole numbers, to equate the two. The easiest way to do that is to divide the smaller number by itself (which assure you it will be 1.000), then divide the other number by the same small number. You should get AlxCly where x and y are small whole numbers. That will give you the empirical formula.
Then add up the atomic masses to find the mass of the empirical formula, divide into 267 to determine n, round n to a whole number, and multiply by the empirical formula to find the molecular formula. The molecular formula will be (AlxCly)n.