Question

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.

Answer 1

Well, let us see.....I make it `C_10H_20O`

Explanation:

`"Moles of carbon dioxide"=(0.2829*g)/(44.01*g*mol^-1)=6.43*mmol`...there were thus...................

`6.43xx10^-3xx12.011*g*mol^-1=0.07721*g` with respect to CARBON in the original `0.1005*g` mass.

`"Moles of water"=(0.1159*g)/(18.01*g*mol^-1)=6.43*mmol`...there were thus...................

`6.44xx10^-3xx2xx1.00794*g*mol^-1=0.01927*g` with respect to HYDROGEN in the original `0.1005*g` mass.

And so we have accounted for `90.18*mg`...the balance was oxygen...`0.01032*g-=6.45xx10^-4*mol`...

And so empirical formula....`C_((0.00643*mol)/(6.45xx10^-4*mol))H_((0.01286*mol)/(6.45xx10^-4*mol))O_((6.45xx10^-4*mol)/(6.45xx10^-4*mol))`

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

`-=C_10H_20O_1-=C_10H_20O`

But we know that the `"molecular formula"` is a whole number of the multiple of the `"empirical formula"`....therefore....

`156.3*g*mol^-1-=nxx(12.011xx10+1.00794xx20+16.00)*g*mol^-1`...

Clearly, `n=1`, and here the `"molecular formula"` is the same as the `"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...