# Menthol is composed of C, H, and O. A 0.1005g sample of menthol is combusted, producing 0.2829g of CO2 and 0.1159g H2O. What is the empirical and molecular formula for menthol?

## The compound has a molar mass of 156.30g/mol.

May 13, 2018

Well, let us see.....I make it ${C}_{10} {H}_{20} O$

#### Explanation:

$\text{Moles of carbon dioxide} = \frac{0.2829 \cdot g}{44.01 \cdot g \cdot m o {l}^{-} 1} = 6.43 \cdot m m o l$...there were thus...................

$6.43 \times {10}^{-} 3 \times 12.011 \cdot g \cdot m o {l}^{-} 1 = 0.07721 \cdot g$ with respect to CARBON in the original $0.1005 \cdot g$ mass.

$\text{Moles of water} = \frac{0.1159 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1} = 6.43 \cdot m m o l$...there were thus...................

$6.44 \times {10}^{-} 3 \times 2 \times 1.00794 \cdot g \cdot m o {l}^{-} 1 = 0.01927 \cdot g$ with respect to HYDROGEN in the original $0.1005 \cdot g$ mass.

And so we have accounted for $90.18 \cdot m g$...the balance was oxygen...$0.01032 \cdot g \equiv 6.45 \times {10}^{-} 4 \cdot m o l$...

And so empirical formula....${C}_{\frac{0.00643 \cdot m o l}{6.45 \times {10}^{-} 4 \cdot m o l}} {H}_{\frac{0.01286 \cdot m o l}{6.45 \times {10}^{-} 4 \cdot m o l}} {O}_{\frac{6.45 \times {10}^{-} 4 \cdot m o l}{6.45 \times {10}^{-} 4 \cdot m o l}}$

Here we have divided thru by the LEAST molar quantity, i.e. that of oxygen...

$\equiv {C}_{10} {H}_{20} {O}_{1} \equiv {C}_{10} {H}_{20} O$

But we know that the $\text{molecular formula}$ is a whole number of the multiple of the $\text{empirical formula}$....therefore....

$156.3 \cdot g \cdot m o {l}^{-} 1 \equiv n \times \left(12.011 \times 10 + 1.00794 \times 20 + 16.00\right) \cdot g \cdot m o {l}^{-} 1$...

Clearly, $n = 1$, and here the $\text{molecular formula}$ is the same as the $\text{empirical formula}$. I hope you can see the formatting in this answer. I am having trouble. Menthol is truly a lovely organic chemical with a beautiful smell....and it is very commonly used as a flavouring...