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
