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.

1 Answer
May 13, 2018

Answer:

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...