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?

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:

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{=}\mathbf{15}\mathbf{.}\mathbf{666}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}{\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{C}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}}}\mathbf{\times }\frac{\mathbf{12}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{mg}\mathbf{ }\mathbf{C}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{C}}}$

mass of C = 4.275 mg C

$\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{H}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{491}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}{\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{ }\cancel{\mathbf{mg}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{2}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{H}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{mg}\mathbf{ }\mathbf{H}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mmol}\mathbf{ }\mathbf{H}}}$

mass of H = 0.503 mg H

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