Recall that in combustion analysis, a compound reacts with excess O2 to form products.
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.
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?
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