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?
1 answer:
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
<span>H = 0.41/0.06 = 6.8 ~
7</span>
N = 0.06/0.06 = 1
Empirical formula is:
<span>C5H7N</span>
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