Question

Menthol, the substance we can smell in mentholated cough drops, is composed of c, h, and o. a 0.1206 −mg sample of menthol is combusted, producing 0.3395 mg of co2 and 0.1391 mg of h2o. what is the empirical formula for menthol?

Answer 1

The solution would be like this for this specific problem:

(0.3395 mg CO2) / (44.00964 g CO2/mol) × (1 mol C / 1 mol CO2) × (12.01078 g C/mol) = 0.0926538 mg C

(0.1391 mg H2O) / (18.01532 g H2O/mol) × (2 mol H / 1 mol H2O) × (1.007947 g H/mol) = 0.0155651 mg H

(0.1206 mg total) - (0.0926538 mg C) - (0.0155651 mg H) = 0.0123811 mg O

(0.0926538 mg C) / (12.01078 g C/mol) = 7.71422 × 10^-3 mmol C

(0.0155651 mg H) / (1.007947 g H/mol) = 1.54424 × 10^-2 mmol H

(0.0123811 mg O) / (15.99943 g O/mol) = 7.73846 × 10^-4 mmol O

Now, we divide by the smallest number of moles:

(7.71422 × 10^-3 mmol C) / (7.73846 × 10^-4 mmol) = 9.969
(1.54424 × 10^-2 mmol H) / (7.73846 × 10^-4 mmol) = 19.955
(7.73846 × 10^-4 mmol O) / (7.73846 × 10^-4 mmol) = 1.000

After rounding to the nearest whole numbers, we can find the empirical formula which is:

C10H20O