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?

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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons