Question

Menthol is a flavoring agent extracted from peppermint oil. it contains c, h, and o. in one combustion analysis, 10.00 mg of the substance yields 11.53 mg h2o and 28.16 mg co2. what is the empirical formula of menthol? add subscripts to complete the empirical formula. cho

Answer 1

Combustion reaction for menthol is as follows;
CxHyOz + O₂ ---> xCO₂ +  H₂O
Mass of CO₂ formed - 28.16 mg
Therefore number of moles formed - 28.16/ 44 g/mol = 0.64 mmol
Mass of water formed  - 11.53 mg
number of water moles formed - 11.53 mg/18 g/mol = 0.64 mmol
From CO₂,
1 mol of CO₂ - 1 mol of C and 2 mol of O
therefore number of C moles - 0.64 mmol
                                 O moles - 1.28 mmol
from H₂O
1 mol of H₂O - 2 mol of H and 1 mol of O
number of H moles - 1.28 mmol
                 O moles - 0.64 mmol

Mass of menthol initially - 10 mg
in reactions, the masses of products are equal to the masses of reactants. The excess mass to the products formed is due to O₂ in air
Original mass of menthol - 10 mg
mass of water and CO₂ - 11.53 mg + 28.16 mg = 39.69
Difference in mass - 39.69 - 10 = 29.69 mg
This difference comes from O moles in air - 29.69 mg/ 16 g/mol = 1.8556 mmol
then O moles coming from menthol - (1.28 + 0.64) - 1.8556 = 0.064 mmol

In menthol 
C moles - 0.64 mmol
H moles - 1.28 mmol
O moles - 0.064 mmol
ratios of C:H:O
C          H         O
0.64    1.28      0.064
x1000  x1000    x1000 to get whole numbers
640       1280      64
10          20          1
Simplest ratio of C:H:O is 10:20:1
therefore empirical formula of menthol is C₁₀H₂₀O