Nicotine, a component of tobacco, is composed of C, H, and N. A 6.825-mg sample of nicotine was combusted, producing 18.515 mg of CO2 and 5.308 mg of H2O. What is the empirical formula for nicotine

C₅H₇N

Explanation:

Empirical formula is defined as the simplest positive integer ratio of atoms present in a compound

The reaction of combustion of nicotine, CₐHₓNₙ produce:

CₐHₓNₙ + O₂ → a CO₂ + x/2 H₂O + n NO₂

As you can see, moles of carbon are the same of CO₂, that is:

18.515mg CO₂ ₓ (1mmol / 44mg) = 0.4208mmol Carbon

In mass:

0.4208mmol Carbon ₓ (12mg / 1mmol) = 5.0496mg C

Moles of water are 1/2 moles of hydrogen, that are:

5.308mg H₂O ₓ (1mmol / 18gm) = 0.2959mmol water ₓ (2mol H / 1mol H₂O) =0.5898mmol H

In mass:

0.5898mmol H ₓ (1mg / 1mmol) = 0.5898mg

mg of nitrogen are:

6.825mg - 5.0496 - 0.5898 = 1.1856mg N

Moles are:

1.1856 mg N ₓ (1mmol / 14mg) = 0.0847mmol N

The ratio between moles of nitrogen, hydrogen and carbon are:

C: 0.4208mmol / 0.0847mmol = 5

H: 0.5898mmol / 0.0847mmol = 7

N: 0.0847mmol / 0.0847mmol = 1

That means empirical formula is:

C₅H₇N

I hope it helps!

the answer is 303.15 or 303

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