Follow this example to do question VII-V 3.

For example, a compound is made up of:

Step 1

Convert to moles of atoms,

molar mass
grams <==> moles

aluminum(1.99 g / 26.982 g/mole) = 0.07375 mole Al
oxygen(1.76 g / 15.999 g/mole) = 0.1100 mole O

Step 2

Divide the number of moles of each element by the smallest of the mole ratios calculated.

Here, we will divide by 0.07375 moles.

Al:0.07375 mole / 0.07375 mole = 1.00
O:0.1100 mole / 0.07375 mole = 1.49

Change the mole ratios in Step 2 to the smallest set of whole numbers.

Al:1.00 x 2 =2
O:1.49 x 2 =3

The empirical formula is Al2O3