Question

Determine the empirical formula of a compound containing 22.6% nitrogen and 77.4% oxygen

Answer 1

Empirical formula is the simplest ratio of whole numbers of components in a compound 
for 100 g of the compound 
                                          N                                  O
mass                                 22.6 g                         77.4 g
number of moles               22.6 g / 14 g/mol        77.4 g / 16 g/mol 
                                          = 1.61 mol                  = 4.84 mol
divide by the least number of moles 
                                          1.61 / 1.61  = 1.00      4.84 / 1.61 = 3.01
rounded off to nearest whole numbers 
N - 1
O - 3

therefore empirical formula - NO₃

Answer 2

The empirical  formula  of a compound that  contain 22.6%  nitrogen and 77.4% oxygen  is calculated as below

calculate the moles of each element
= % composition/molar mass
that is
         N(nitrogen) = 22.6/14= 1.61 moles
         O(oxygen)  = 77.4/16 = 4.84  mole

find the mole ratio by dividing each mole  by the  smallest  mole ( 1.61 moles)

that is 
        N = 1.61/1.61 = 1 
        O = 4.84/1.61 = 3

therefore the empirical  formula = NO3