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

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