Question

A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. what is the empirical formula

Answer 1

MgThe  empirical  formula   is  calculated as follows
fin the  moles of each  element
moles =  %  composition/ molar mass

magnesium(Mg) =  72.2/24 = 3moles
Nitrogen(N) = 27.8/14  = 1.986 moles

find the  moles  ration by  dividing each mole  by the smallest mole ( 1.986 moles)

that is 

magnesium)Mg)  = 3/1.986 =  1.5
Nitrogen (N)=  1.986/1.986 =1
multiply  both  the mole  ratio to remove the decimal

magnesium (Mg) = 1.5 x2 = 3
nitrogen (N)= 1 x2 =2

therefore  the empirical formula = Mg3N2