An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula

for the oxide.

The answer is: Al2O3

Explanation:

The data they give us is:

0.545 gr Al 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

0.02 mol Al / 0.02 = 1 Al 0.03 mol O / 0.02 = 1.5 O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

1Al * 2 = 2Al 1.5O * 2 = 3O

Now the answer is given correctly:

Al2O3