Acertain substance is analyzed and found to contain the following weight percentages 36.84% nitrogen (n) and 63.16% oxygen (o). determine the empirical formula of this compound. (atomic weight n=14,o=16)

Empirical formula is the simplest ratio of whole numbers of components in a compound 
in the given compound lets calculate for 100 g of the compound 
                                           N                                              O
mass                                36.84 g                                     63.16 g
number of moles             36.84 g / 14 g/mol                     63.16 g / 16 g/mol 
                                        = 2.63 mol                                  = 3.94 mol
divide by least number of moles 
                                         2.63/2.63 = 1.00                        3.94 / 2.63 = 1.50
multiply by 2 to get numbers that can be rounded off to whole numbers
N  - 1.00 x 2 = 2.00
O - 1.50 x 2 = 3.00
therefore empirical formula is N₂O₃

Is a logical series of steps

Do you have any options?