# 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.

5

7

1

98% (298 ratings) ###### 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?

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Combustion Analysis concept. You can view video lessons to learn Combustion Analysis. Or if you need more Combustion Analysis practice, you can also practice Combustion Analysis practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Kowach's class at CCNY.