Question

An organic compound, containing only C, H, N, has a molar mass of 162 g/mol. The combustion of 5.64 mg of the compound released 15.30 mg CO2 and 4.39 mg H2O. What is the molecular formula of this organic compound?

Answer 1

From the given data % of C , % of H can be calculated as follows:

% of C = [weight of CO2/Weight of organic compound][12/44]x100

% of H =  [weight of H2O/Weight of organic compound][2/18]x100

Thus % of C = [15.30mg/5.64mg]x[12/44] x100 = 73.98

% of H = [4.39/5.64]x[2/18]x 100 = 8.64

Hence % of N = 100-( 73.98 + 8.64) = 17.37

Element C H N

%by weight 73.98 8.64 17.37

atomic weight 12 1 14

number of moles 6.165 8.64 1.24

mole ratio 6.165/1.24 8.64/1.24 1.24/1.24

= 4.99 =6.96 = 1

Thus theratio of number of atoms of C = 5, H =7 and N =1 in the empirical formula.

The empirical formula is C₅H₇N

Given molecular weight = 162g/mol.

From the empirical formula calculated, we obtain the molecular formula as

Molecular formula = [empirical formula]_n  where n = molecular weight/empirical formula weight

The empirical formula weight = (12x5) + (1x7) + (14x1) =81

Hence n = 162/81 = 2

Thus the molcular formula = [C₅H₇N]₂ = C₁₀H₁₄N₂