Problem: Nicotine, a component of tobacco, is composed of C, H, and N. A 5.775-mg sample of nicotine was combusted, producing 15.666 mg of CO2 and 4.491 mg of H2 O. What is the empirical formula for nicotine?

FREE Expert Solution

Recall that in combustion analysisa compound reacts with excess O2 to form products


mole-to-mole comparison:

  • 1 mole of C in 1 mole of CO(molar mass = 44.01 g/mol or mg/mmol)
  • 2 moles of H in 1 mole of H2(molar mass = 18.02 mg/mmol)


Finding the mass of C and H:


mass of C=15.666 mg CO2×1 mmol CO244.01 mg CO2×1 mmol C1 mmol CO2×12.01 mg C1 mmol C

mass of C = 4.275 mg C


mass of H=4.491 mg H2O×1 mmol H2O18.02 mg H2O×2 mmol H1 mmol H2O×1.01 mg H1 mmol H

mass of H = 0.503 mg H


mass of N = 5.775 mg – (4.275 mg C + 0.503 mg H) = 0.997 mg N


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Problem Details

Nicotine, a component of tobacco, is composed of C, H, and N. A 5.775-mg sample of nicotine was combusted, producing 15.666 mg of CO2 and 4.491 mg of H2 O. What is the empirical formula for nicotine?

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