Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250mg sample of nicotine was combusted, producing 14.242mg of CO2 and 4.083mg of H2O. What is the empirical formula for nicotine? If nicotine has a molar mass of 160±5 g/mol, what is its molecular formula?
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1 Answer

Molar mass of C = 12.0 g/mol
Molar mass of H = 1.0 g/mol
Molar mass of N = 14.0 g/mol
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol
In 5.250 mg of nicotine :
Mass of C = Mass of C in CO₂ = (14.242 mg) × (12.0/44.0) = 3.884 mg
Mass of H = Mass of H in H₂O = (4.083 mg) × (1.0×2/18.0) = 0.454 mg
Mass of N = (5.250 – 3.884 – 0.454) mg = 0.913 mg
Mole ratio C : H : N
= (3.884/12.0) : (0.454/1.0) : (0.913/14.0)
= 0.324 : 0.454 : 0.0652
= 4.97 : 6.96 : 1
≈ 5 : 7 : 1
Empirical formula = C₅H₇N
Let (C₅H₇N)n be the molecular formula.
Molar mass :
(12.0×5 + 1.0×7 + 14.0) × n ≈ 160
n = 2
Molecular formula = C₁₀H₁₄O₂