Question

Nicotine, a component of tobacco, is composed of c, h, and n. a 4.725 −mg sample of nicotine was combusted, producing 12.818 mg of co2 and 3.675 mg of h2o. what is the empirical formula for nicotine?

Answer 1

C = 12.818 mg CO2 (1 mmol CO2/44 mg CO2) (1 mmol C/1 mmol CO2) = 0.29 mmol

H = 3.675 mg of H2O (1 mmol H2O/18 mg H2O) (2 mmol H/1 mmol H2O) = 0.41 mmol

N = (4.725 mg – 0.29 * 12 – 0.41 * 1) * (1 mmol/14 mg) = 0.06 mmol

Divide everything by the smallest number:

C = 0.29/0.06 = 4.8 ~ 5

H = 0.41/0.06 = 6.8 ~ 7

N = 0.06/0.06 = 1

Empirical formula is:

C5H7N