# An oxide of aluminium is formed by the reaction of 4.151g Al with 3.692 g O. How to calculate its empirical formula ?

Feb 28, 2015

You can determine the empirical formula for this compound (or any compound, for that matter) by finding the mole ratio between oxygen and aluminium.

FIrst, you calculate the number of moles that reacted for both elements - it's a little easier to use monoatomic oxygen, not diatomic oxygen

$\text{4.151 g" * "1 mole aluminium"/"26.98 g" = "0.1539 moles }$ $A l$, and

$\text{3.692 g" * ("1 mole monoatomic oxygen")/"16.00 g" = "0.2308 moles }$ $O$

Now just divide these two numbers to get the mole ratio of aluminium to oxygen in the oxide

$\text{aluminium"/"oxygen" = "0.1539"/"0.2308" = "0.667} = \frac{2}{3}$

This means that the empirical formula for your oxide will be

$A {l}_{2} {O}_{3}$

SIDE NOTE Here's why I've used monoatomic oxygen instead of ${O}_{2}$. If you do the calculations with diatomic oxygen, the ratio will come out to be 1.33, or 4/3; however, this is the ratio of aluminium to ${O}_{2}$, not to O, so your empirical formula would have been

$A {l}_{4} {\left({O}_{2}\right)}_{3} = A {l}_{4} {O}_{6} = A {l}_{2} {O}_{3}$

The result would have been the same, of course, but I think it's easier to work with individual atoms - also, I think this is the recommended method as well.