A compound contains 46.7% nitrogen and 53.3% oxygen by mass. What is the empirical formula of the compound?

The empirical formula of a compound that contain 46.7 % nitrogen and 53.3% oxygen is calculated as below

find the moles of each element

that is, mole = % composition/molar mass

nitrogen (N) = 46.7 / 14=3.33 moles

oxygen (O) = 53.3 / 16 = 3.33 moles

divide each mole with the smallest number of mole to get the mole ratio that is 3.33 moles

nitrogen (N) = 3.33/3.33 = 1

oxygen (O) = 3.33/3.33 = 1

therefore the empirical formula = NO