a 1.32-g sample of aluminum reacted with 1.174 g oxygen to form aluminum oxide. Determine the empirical formula of the aluminum oxide.

Al2O3

Explanation:

Moles of Aluminium = 1.32 g/27 g/mole

= 0.0489 moles

Moles of oxygen = 1.174 g / 16 g/mole

= 0.0734 moles

Mole Ratio = 1 : 0.0734/0.0489

= 1 : 1.5

We multiply by 2 to make them whole numbers;

= (1 : 1.5) 2

= 2 : 3

Therefore;

The empirical formula is Al2O3