What is the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N?

1 Answer
Jun 25, 2016

C_5H_7NC5H7N

Explanation:

First, assume that there is 100 g of substance.

Next is to calculate the number of moles or each element:

74.0 g C * (1mol C )/ (12.01 g) ~~ 6.16 mol C74.0gC1molC12.01g6.16molC
8.65 g H * (1mol H )/ (1.008 g) ~~ 8.58 mol H8.65gH1molH1.008g8.58molH
17.4 g N * (1mol N )/ (14.01 g) ~~ 1.24 mol N17.4gN1molN14.01g1.24molN

After getting the mol of each element, you can calculate now the mole ratio. Do this by dividing each number of moles by the lowest number of moles obtained.

C= (6.16/1.24)~~ 5C=(6.161.24)5
H= (8.58/1.24)~~ 7H=(8.581.24)7
N= (1.24/1.24)~~ 1N=(1.241.24)1

Therefore, the empirical formula of a substance with 74.0% C , 8.65% H, and 17.4% N is C_5H_7NC5H7N