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Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Korokian on September 26, 2006, 05:05:53 PM

Title: combustion analysis
Post by: Korokian on September 26, 2006, 05:05:53 PM

Menthol, the substance we 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 CO2 and 0.1159 g of H2O.  What is the empirical formula for menthol?  If the compound has a molar mass of 156 g/mol, what is its molecular formula?

can someone check my calculations?

i got .07720 g C
.0065 g H
.0128 g O

.006428 mol C
.006436 mol H
.00105 mol O

emperical formula C₆H₆O

but it doesnt correspond with the molar mass

Title: Re: combustion analysis
Post by: Dan on September 26, 2006, 05:31:57 PM

Your calculations for H are not right, but you're almost there... Remember it is H₂O.

This error has carried forward and thrown off your O calculations.

Title: Re: combustion analysis
Post by: Korokian on September 26, 2006, 08:12:38 PM

okay i got it thx

Title: Re: combustion analysis
Post by: hobaoe on December 08, 2006, 10:22:59 PM

this is my solution:
m(C)=(0.2829/44)*12=0.077155(g)
m(H)=(0.1159/18)*2=0.01288(g)
m(O)=0.1005 - 0.077155 - 0.01288=0.010465(g)

C         :         H        :         O
10       :         20       :        1
emprical formula: C₁₀H₂₀O