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Topic: combustion analysis  (Read 10458 times)

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Offline Korokian

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combustion analysis
« 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

Offline Dan

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Re: combustion analysis
« Reply #1 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.
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Offline Korokian

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Re: combustion analysis
« Reply #2 on: September 26, 2006, 08:12:38 PM »
okay i got it thx
« Last Edit: September 26, 2006, 10:22:41 PM by Korokian »

Offline hobaoe

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Re: combustion analysis
« Reply #3 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

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