# 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 = 3.887 mg C

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:

0.324 mmol C

0.453 mmol H

0.0646 mmol N

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

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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?