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Organic : gas chromatography and Fenske eqn
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 Message 1 of 2 in Discussion 
From: victoriasarah  (Original Message)Sent: 11/12/2006 3:25 PM
These labs are killer! ;o)

This week we separated two components of a petroleum, cyclohexane and toluene, by fractional distillation. The point of this was to determine the efficiency of our packing material by assessing the composition of the fractions that distilled out.

So, the total number of theoretical plates can be obtained by the Fenske eqn, log (nA/nB) / log a. nA and nB are supposed to be the moles of cyclohexane divided by toluene. a is supposed to be the volatility factor of cyclohexane.

So my problem comes in trying to calculate the number of moles of cyclohexane and toluene. We did a gas chromatogram for one of our fractions, which we are supposed to use for the calculation. But I'm not sure how to get the number of moles from the chromatogram.

According to the chromatogram, the height of cyclohexane was 58.534 and the area was 1017.1175. The height of toluene was 6.271 and the area was 148.8420. Now I know I'm supposed to multiply these by a correction factor, but I'm not sure where to go from there, or how to get the number of moles of each component.

Thanks in advance!


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 11/13/2006 7:21 AM
You would normally run a reference chromatogram by injecting known amounts of the compounds you are measuring.  From the peak areas you can determine moles of compound per unit peak area.  Once you have this, you just have to multiply each peak area in your sample chromatograms by its corresponding mole/pk area factor to convert the peak areas to moles.

Or, another procedure, did you spike your samples with a reference compound?  1,4-dioxane has been mentioned here before in this regard.  By comparing the areas of the sample peaks with the dioxane peak, you can calculate the amounts of each compound that way.  You would need to know how the peak areas of the various compounds relate to the peak area of 1,4-dioxane (or other reference).
 

Steve