Problem: 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 H2O. What is the empirical formula for nicotine?If nicotine has a molar mass of 160.5 g/mol, what is its molecular formula?

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=14.242 mg CO2×1 mmol CO244.01 mg CO2×1 mmol C1 mmol CO2×12.01 mg C1 mmol C

mass of C = 3.887 mg C


mass of H=4.083 mg H2O×1 mmol H2O18.02 mg H2O×2 mmol H1 mmol H2O×1.01 mg H1 mmol H

mass of H = 0.458 mg H


mass of N = 5.250 mg – (3.887 mg C + 0.458 mg H) = 0.905 mg N


The moles of C, H, and N are:


mmoles of C=3.887 mg C×1 mmol C12.01 mg C0.324 mmol C


mmoles of H=0.458 mg H×1 mmol H1.01 mg H0.453 mmol H


mmoles of N=0.905 mg N×1 mmol N14.00 mg N0.0646 mmol N


Divide by the smallest value, 0.0646 mmol, to get the smallest ratio of C, H, and N.


0.324 mmol C0.0646=5


0.453 mmol H0.0646= 7


 0.0646 mmol N0.0646=1


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Problem Details

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 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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