A compound is found to contain 36.84% nitrogen and 63.16% oxygen by mass. What is the empirical formula?

Question

Find the empirical formula please please

Answer 1

Assuming 100 g sample of the compound, there is  36.84 g N and 63.16 g O

Now convert the grams of each element into moles as follows—

Moles of N—     36.84 g x. 1 mole N       =   2.63 moles of N

                                         14.01 g

[ 14.01 g is the molar mass of N]

Moles of O —    63.16 g  x   1 mole O     =.   3.95 moles of O

                                         16.00 g

[ 16.00 g is the molar mass of O]

Now to get the smallest ratio between the two moles divide both by the smaller of the two

N —  2.63/2.63. = 1

O —  3.95/2.63  = 1.5

N : O ratio =. 1 : 1.5

For Empirical formula, we need the whole # ratio of the moles, so to get the whole #s , multiply both by 2  so

N : O   =  2 (1 :  1.5)

So whole # ratio   N : O.  = 2 : 3

Hence the Empirical formula is N₂O₃