A compound was isolated and found to contain 72.2% magnesium and the rest nitrogen. What is the empirical formula of this compound?

Since the compound is 72.2% Mg then there would be 27.8% N. Then you can just change it to 72.2 g Mg and 27.8 g N.

then convert to moles by dividing by molecular mass.

72.2:24.31=2.96 moles Mg

27.8:14.01=1.98 moles N

then divided by the lowest number.

2.96:1.98=1.5

Since it is 1.5 you multiply by 2. so you have 3 Mg

1.98:1.98=1 N

so the empirical formula is Mg3N