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

The empirical formula is =
`C_(10)H_(20)O`

Step-by-step explanation:

Mass of water obtained =
`6.954* 10^(-2)\ mg`

Also, 1 mg =
`10^(-3)` g

So, mass of water =
`6.954* 10^(-5)\ g`

Molar mass of water = 18 g/mol

Moles of
`H_2O` =
`6.954* 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* 10^(-4)\ g`

Molar mass of carbon dioxide = 44.01 g/mol

Moles of
`CO_2` =
`1.697* 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* 10^(-2)\ mg`

Also, 1 mg =
`10^(-3)` g

So, mass of sample =
`60.30* 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`