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Trapped in aPr?lude

Registered: 02/15/05
Posts: 24
Last seen: 11 years, 2 months
Stoichiometry Problem
    #3795952 - 02/18/05 12:06 AM (12 years, 11 months ago)

I received all my college chemistry credit out of high school, and haven't had to take a chemistry course in two years. I decided to go into Organic next semester, and I'm trying to refresh my memory. I picked up a textbook, and I was working on the end-of-chapter problems in the Stoichiometry chapter and only ran into trouble with the very last problem.

"A sample of cocaine, C17H21O4N, is diluted with sugar, C12H22O11. When a 1.00 mg sample of this mixture is burned, 1.00 mL of carbon dioxide (rho = 1.80 g/L) is formed. What is the percentage of cocaine in the mixture?"

The general procedure in all of the other similar problems was to find the number of moles (or mass) of one of the elements of the product. I did this and found that in the 1.00 mL of CO2, there are:

0.00180 g CO2
4.09 x 10-5 mol CO2
4.91 x 10-4 g C
4.09 x 10-5 mol C
.00131 g O
8.18 x 10-5 mol O

This is as far as I got. Because both carbon and oxygen are common to cocaine and sugar, I don't see any way to relate the calculated values to the mixture without a balanced equation. In my head, I was fudging around with a method that would have two equations and two unknowns (percent of coke in the mix and percent of sugar), but I didn't try any of it out of paper because I don't remember ever needing to use two equations to solve a basic chemistry problem.

Can someone point out what I'm missing, or if I approached the problem wrong? Thanks for the help!

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Registered: 07/20/02
Posts: 784
Loc: Northern Europe
Re: Stoichiometry Problem [Re: Boccherini]
    #3798477 - 02/18/05 02:55 PM (12 years, 11 months ago)

The number of oxygens in the molecules are irrelevant, you can assume that the oxygen comes from the air. If you set the percent of sugar in the mixture to x, the percent of cocaine is 100-x, so there are not two unknown. :wink:

Let me know if you want a solution to the problem.

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