# Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and O. A 0.1005 g sample of menthol is combusted, producing 0.2829 g of carbon dioxide and 0.1159 g of water. What is the empirical formula for menthol?

Aug 31, 2016

The empirical formula is $\text{C"_10"H"_20"O}$.

#### Explanation:

We can calculate the masses of $\text{C}$ and $\text{H}$ from the masses of their oxides (${\text{CO}}_{2}$ and $\text{H"_2"O}$).

$\text{Mass of C" = 0.2829 color(red)(cancel(color(black)("g CO"_2))) × "12.01 g C"/(44.01 color(red)(cancel(color(black)("g CO"_2)))) = "0.077 20 g C}$

$\text{Mass of H" = 0.1159 color(red)(cancel(color(black)("g H"_2"O"))) × "2.016 g H"/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "0.012 97 g H}$

$\text{Mass of O" = "Mass of menthol - mass of C - mass of O" = "0.1005 g - 0.077 20 g - 0.012 97 g" = "0.01 033 g}$

Now, we must convert these masses to moles and find their ratios.

From here on, I like to summarize the calculations in a table.

$\text{Element"color(white)(Xll) "Mass/g"color(white)(Xmll) "Moles"color(white)(mml) "Ratio" color(white)(m)"Integers}$
stackrel(—————————————————-——)(color(white)(ll)"C" color(white)(XXXml)"0.077 20" color(white)(mll)"0.006 428" color(white)(Xlll)9.956color(white)(Xm)10
$\textcolor{w h i t e}{l l} \text{H" color(white)(XXXXl)"0.012 97" color(white)(mll)"0.012 87} \textcolor{w h i t e}{m l l l} 19.93 \textcolor{w h i t e}{X X l l} 20$
$\textcolor{w h i t e}{l l} \text{O" color(white)(mmmml)"0.010 33"color(white)(mll)"0.000 6456} \textcolor{w h i t e}{m l} 1 \textcolor{w h i t e}{m m m m l l} 1$

The empirical formula is $\text{C"_10"H"_20"O}$.