An oxide of aluminium is formed by the reaction of 4.151g Al with 3.692 g O. How to calculate its empirical formula ?
FIrst, you calculate the number of moles that reacted for both elements - it's a little easier to use monoatomic oxygen, not diatomic oxygen
Now just divide these two numbers to get the mole ratio of aluminium to oxygen in the oxide
This means that the empirical formula for your oxide will be
SIDE NOTE Here's why I've used monoatomic oxygen instead of
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.