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Topic: Empirical Formula  (Read 5067 times)

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Offline achibaby1974

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Empirical Formula
« on: February 08, 2008, 10:57:49 PM »
Menthol ( = 156.3 g/mol) is a compound of C, H, O. When 0.1595 g menthol was put into combustion, it produced 0.449 g CO2 and 0.184 g H2O. What is its molecular formula?

Well first I'd have to find the empirical formula right? How do I do that? They've given me molecular compounds. What do I do?

Maybe

.449 g CO2 x (1 mol CO2/ 44.01 g CO2) = 0.010202227 mol CO2

and

.184 g H2O x (1 mol H2O/ 18.016 g H2O) = 0.010213144 mol H2O

I have no clue what to do from here.
« Last Edit: February 08, 2008, 11:38:28 PM by achibaby1974 »

Offline Yggdrasil

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Re: Empirical Formula
« Reply #1 on: February 08, 2008, 11:07:59 PM »
1.3333 * 3 = 4

Offline achibaby1974

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Re: Empirical Formula
« Reply #2 on: February 08, 2008, 11:14:32 PM »
wow. i'm retarded.

Offline Yggdrasil

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Re: Empirical Formula
« Reply #3 on: February 09, 2008, 12:34:27 AM »
For your new question try thinking about the reactants in a combustion reaction and -- importantly -- where the atoms in the products come from.  For example, from where do the hydrogens in the water come? 

Offline achibaby1974

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Re: Empirical Formula
« Reply #4 on: February 09, 2008, 05:33:58 PM »
well i know that a complete combustion reaction starts with what's given this case CH/ CHO somethin, menthol, which goes through complete combustion with oxygen gas to make CO2 and water.  Basically, complete combustion always results in CO2 and H2O, at least according to my chem prof. 

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