Acompound 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