Caffeine, a stimulant found in coffee, tea, and certain soft drinks, contains C, H, O, and N. Combustion of 1.000 mg of caffeine produces 1.813 mg CO2, 0.4639 mg H2O, and 0.2885 mg N2. Estimate the molar mass of caffeine, which lies between 150 and 200 g/mol.

Solution: What we must determine is the molecular formula and the only way to get it with the data given is to determine the empirical formula. One we have that, we can get the "empirical formula weight" and multiply it by the proper scaling factor to get the molar mass.

Step One: determine the mass of each element present.

Carbon: 1.813 mg x (12.011 / 44.0098) = 0.4948 mg
Hydrogen: 0.4639 mg x (2.016 / 18.0152) = 0.0519 mg
Nitrogen: 0.2885 mg is given in problem
Oxygen: 1.000 mg minus (0.4948 + 0.519 + 0.2885) = 0.1648 mg

If the mass of N was given in terms of NH3, we would use the factor (14.007 / 17.031) to get mass of N. If the mass of N was given as N2O5, we would use the factor (28.014 / 108.009) to get the mass of N.

Step Two: Convert mass of each element to moles.

Carbon: 0.4948 mg ÷ 12.011 mg/mmol = 0.0412 mmol
Hydrogen: 0.0519 mg ÷ 1.008 mg/mmol = 0.0515 mmol
Nitrogen: 0.2885 mg ÷ 14.007 mg/mmol = 0.0206 mmol
Oxygen: 0.1648 mg ÷ 15.999 mg/mmol = 0.0103 mmol

Comment #1: notice that the mass for nitrogen is is given as mass of N2, but 14.007 (the atomic weight) is used. Why? Answer: we are interested in moles of N atoms involved, NOT moles of N2 molecules.

Comment #2: notice that I used the unit mg/mmol (milligrams per millimole) for the molar mass. When milligram amounts of substance are used, using mg/mmol as the unit on the molar mass (rather than g/mol) allows us to use the mass unit of mg rather than g. Note also that the numerical value is not changed in using mg/mmol. That is because both the numerator (grams) and the denominator (mol) of the original unit have been divided by 1000 to obtain the "milli" prefix.

Step Three: find the ratio of molar amounts, expressed in smallest, whole numbers.

Carbon: 0.0412 mmol ÷ 0.0103 mmol = 4
Hydrogen: 0.0515 mmol ÷ 0.0103 mmol = 5
Nitrogen: 0.0206 mmol ÷ 0.0103 mmol = 2
Oxygen: 0.0103 mmol ÷ 0.0103 mmol = 1

The empirical formula is C4H5N2O. Since the "EFW" of this is about 97, let's multiply by two. This gives us a molar mass for caffeine of about 194. The value given in reference books is 194.19.