Question

Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 6.030×10−2 −mg sample of menthol is combusted, producing 0.1697 mg of CO2 and 6.954×10−2 mg of H2O. What is the empirical formula for menthol?

Answer 1

Answer:

The empirical formula is = $C_{10}H_{20}O$

Explanation:

Mass of water obtained = $6.954\times 10^{-2}\ mg$

Also, 1 mg = $10^{-3}$ g

So, mass of water = $6.954\times 10^{-5}\ g$

Molar mass of water = 18 g/mol

Moles of $H_2O$ = $6.954\times 10^{-5}\ g$ /18 g/mol =  3.8634×10⁻⁶ moles

2 moles of hydrogen atoms are present in 1 mole of water. So,

Moles of H = 2 x 3.8634×10⁻⁶ = 7.7267×10⁻⁶ moles

Molar mass of H atom = 1.008 g/mol

Mass of H in molecule = 7.7267×10⁻⁶ x 1.008 = 7.6653×10⁻⁶ g

Mass of carbon dioxide obtained = 0.1697 mg

Also, 1 mg = $10^{-3}$ g

So, mass of carbon dioxide = $1.697\times 10^{-4}\ g$

Molar mass of carbon dioxide = 44.01 g/mol

Moles of $CO_2$ = $1.697\times 10^{-4}\ g$  /44.01 g/mol = 3.856×10⁻⁶ moles

1 mole of carbon atoms are present in 1 mole of carbon dioxide. So,

Moles of C = 3.856×10⁻⁶ moles

Molar mass of C atom = 12.0107 g/mol

Mass of C in molecule = 3.856×10⁻⁶ x 12.0107 = 46.3132 ×10⁻⁶ g

Given that the Menthol only contains hydrogen, oxygen and carbon. So,

Mass of O in the sample = Total mass - Mass of C  - Mass of H

Mass of the sample = $6.030\times 10^{-2}\ mg$

Also, 1 mg = $10^{-3}$ g

So, mass of sample = $60.30\times 10^{-6}\ g$

Mass of O in sample = ( 60.30 - 7.6653 - 46.3132 ) ×10⁻⁶ g= 6.3215×10⁻⁶ g

Molar mass of O = 15.999 g/mol

Moles of O  = 6.3215×10⁻⁶  / 15.999  = 0.3951×10⁻⁶ moles

Taking the simplest ratio for H, O and C as:

7.7267×10⁻⁶ : 0.3951×10⁻⁶ : 3.856×10⁻⁶

= 20 : 1 : 10

The empirical formula is = $C_{10}H_{20}O$