Nicotine, a compound that contains carbon, hydrogen, and nirtogen is found to contain 74.02% carbon, 8.700% hyrdogen and 17.28% nitrogen

Question

A) Determine the empirical formula of nicotine

B)The molecular mass of nicotine is 161 emu. Determine the molecular formula.

Answer 1

a) Empirical formula

C = 74.02%

H = 8.700%

N = 17.28%

Assume that we have a 100 g sample of nicotine. In this 100 g sample, mass of each element equals

C = 74.02% x 100 g = 74.02 g

H = 8.700% x 100 g = 8.700 g

N = 17.28% x 100 g = 17.28 g

Convert the grams of each element to their respective # of moles.

C = 74.02 g x (1 mol / 12.01 g) = 6.163 mol C

H = 8.700 g x (1 mol / 1.01 g) = 8.614 mol H

N = 17.28 g x (1 mol / 14.01 g) = 1.233 mol N

To determine the mole ratio of each element in the nicotine compound, divide the moles of each element by the lowest # of moles (= 1.233).

C = 6.163 / 1.233 = 4.988 mol C ≈ 5 mol C

H = 8.614 / 1.233 = 6.986 mol H ≈ 7 mol H

N = 1.233 / 1.233 = 1.000 mol N ≈ 1 mol N

The empirical formula is C₅H₇N.

b) Molecular Formula

Multiple = molar mass / empirical formula mass

Molar mass = 161 g/mol

Empirical formula mass = (5 x 12.01 g/mol C) + (7 x 1.01 g/mol H) + (1 x 14.01 g/mol N) = 81.07 g/mol

Multiple = 161 / 81.07 ≅ 2

Molecular formula = 2 x C₅H₇N = C₁₀H₁₄N₂