# 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 = 4.275 mg C

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

95% (153 ratings) ###### 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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