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?

Recall that in combustion analysis, a compound reacts with excess O₂ to form products. 

mole-to-mole comparison:

• 1 mole of C in 1 mole of CO₂ (molar mass = 44.01 g/mol or mg/mmol)
• 2 moles of H in 1 mole of H₂O (molar mass = 18.02 mg/mmol)

Finding the mass of C and H:

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{14}\mathbf{.}\mathbf{242}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{C}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{mg}\mathbf{ }\mathbf{C}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{C}}}$

mass of C = 3.887 mg C

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{083}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{H}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{mg}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{H}}}$

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:

$\mathbf{mmoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{887}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{C}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmol}\mathbf{ }\mathbf{C}}{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{C}}}$= 0.324 mmol C

$\mathbf{mmoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{458}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{H}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmol}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{H}}}$= 0.453 mmol H

$\mathbf{mmoles}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{N}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{905}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{N}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmol}\mathbf{ }\mathbf{N}}{\mathbf{14}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }\mathbf{N}}}$= 0.0646 mmol N

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

$\frac{0.324 \mathrm{mmol} \mathrm{C}}{0.0646}=$5

$\frac{0.453 \mathrm{mmol} \mathrm{H}}{0.0646}=$ 7

 $\frac{0.0646 \mathrm{mmol} \mathrm{N}}{0.0646}=$1