Suatu senyawa organik tersusun dari 40 % karbon, 6,6 % hidrogen, dan sisanya oksigen. Jika Ar C= 12, H=1, O= 16, Mr = 90. Maka rumus empiris dan rumus molekul s

C = 40% = 40 gram = 40/12 = 3,3
H = 6,6% =6,6 gram = 6,6/1 = 6,6
O = 100% - 40% - 6,6% = 53,4% = 53,4 gram = 53,4/16 = 3,34
Perbandingan C:H:O = 3,3:6,6:3,34 = 1:2:1,012 = 1:2:1
RE = CH2O
Mr = 90
Mr(CH2O)*n = Mr Senyawa
(12+2+16)*n = 90
30*n = 90
n = 90/30
n = 3

Jadi RMnya (CH2O)n => (CH2O)3 => C3H6O3