Question

Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg of CO2 and 4.083 mg of H20. What is the empirical formula of nicotine?

Answer 1

n - number of mole

M(CO2) = 12.0 +16.0*2 = 44.0 mg/mmol

M(H2O) = 2*1.0 +16.0= 18.0 mg/mmol

1) 14.242 mg of CO2*1 mmol/44.0g =0.32368 mmol CO2

0.32368 mmol CO2 has 0.32368 mmol C

2) 0.32368 mmol C*12 mg/1 mmol= 3.88418 mg C

3) 4.083 mg of H2O * 1 mmol/18 mg = 0.226833 mmol H2O

0.226833 mmol H2O has 2*0.226833 mmol H = 0.453666 mmol H

4) 0.453666 mmol H*1 mg/1mmol = 0.453666 mg H

5) Mass of N = 5.250 mg - 3.88418 mg (C) - 0.453666 mg (H)≈ 0.912 mg N

0.912 mg N * 1 mmol/14.0 mg = 0.0652 mmol N

6) n(C) : n(H) : n(N) = 0.32368 mmol C : 0.453666 mmol H : 0.0652 mmol N

0.0652 mmol N is a smallest value.

n(C) : n(H) : n(N) = (0.32368 mmol/0.0652 mmol) C : (0.453666 mmol H/0.0652 mmol) : (0.0652mmol/0.0652 mmol)N = 5 : 7 : 1

n(C) : n(H) : n(N) = 5 : 7 : 1

Empirical formula : C5H7N.