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

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.