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

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