Determine the empirical formula of a compound containing 72.22% of C, 7.07% of H, 4.68% of N ,16.03% of O?

Question

Determine the empirical formula of a compound containing 72.22% of C, 7.07% of H, 4.68% of N ,16.03% of O?

Answer 1

C: 72.22% / 12  = C6

H: 7.07% / 1 = H7

N: 4.68% / 14 = 0.33N

O: 16.03% / 16 = O1

With N = 0.33 then, we need to multiply everything by 3 to give...

C18H21NO3 which is Codeine (A pain killer).

Answer 2

Empirical Formula is just basically the ratio of the molecular formula. For example H202 is the molecular formula of hydrogen perox. and the ratio of H:0 is 2:2 or 1:1. so the empirical formula is HO.

to solve this problem:

C= 72.22 ( 1M / 12)=6.02

H= 7.07 (1m/ 1) = 7.07

N = 4.68 (1/14) = 0.334

O = 16.03 (1/ 16)= 1.001

until this point, divide by by the smallest number which in this case is 0.334.

C= 6.02/ .334 = 18 C

H= 7.07/ .334  = 21 H

N 0.334/ 0.334 = 1 N

O= 1.01/ 0.334 =3 0

conclusion: the empirical formula is C18H21N03 (codeine) alleviates pain

hope this helped!