Question

A compound of aluminium and chlorine is composed of 9.0 g Al for every 35.5 g of chlorine. The empirical formula of the compound is:

• A. $AlCl$
• B. $AlC{l}_2$
• C. $AlC{l}_4$
• D. $AlC{l}_3$

Answer 1

The correct option is D $AlC{l}_3$

To find moles of each element using the average atomic mass of an element to convert grams of each to moles as :

$Al$ : 9 g x (1 mole/27 g) = 0.33333 moles
$Cl$: 35.5 g x (1 mole/35.453 g) = 0.9987 moles

dividing both numbers by the smaller one as :
0.33333/0.33333 = 1
0.9987/0.33333 = 3

Therefore the formula is : $AlC{l}_3$

And its empirical formula is also $AlC{l}_3$ (a simple ratio of all elements of molecular formula)