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Topic: Observed Rotation of a Sample  (Read 5407 times)

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Offline tiny101

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Observed Rotation of a Sample
« on: May 26, 2009, 12:49:08 AM »
Hi there, I am finding it difficult to get through this question and hope that someone can shed some light upon it.

Problem: Bob wanted to weigh out 2 grams of chiral molecule X ([alpha]D = -30.1 degrees) but instead weighed out 0.2 grams of its enantiomer Y. Upon realizing his error, Bob decided to add 2.2 grams of enantiomer X, thinking this would balance off Y, leaving him with 2 grams of X and a mixture of 0.2 grams. What would be the observed rotation of this sample if 0.2 grams were dissolved in 10 mL of water and placed in a sample tube with a length of 0.7 dm?

Attempt at thinking this through: Utilizing the formula (specific rotation = observed rotation / (c*l)), I should be able to find the observed rotation of the sample. Except, besides that unknown, I'm also uncertain of the specific rotation. I was thinking I could use the specific rotation for the pure sample of A to find one of the unknowns but I am not sure of how.

Any help would be great,
thank you so much in advance.

Offline sjb

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Re: Observed Rotation of a Sample
« Reply #1 on: May 27, 2009, 03:49:39 PM »
I shall call (+)-X the X with alpha-D of -30.1deg, just for convenience here, it makes no odds whether it is the (+) or (-) enantiomer.

The specific rotation is usually based on a 10 cm length, if I recall correctly. So how would that affect things if you only had pure (+)-X?

What is the excess of (+)-X in the mixture, and what is the concentration, assuming that the bit that cancels with Y is inert.

Offline tiny101

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Re: Observed Rotation of a Sample
« Reply #2 on: May 28, 2009, 11:08:26 PM »
Thank you, I have figured out the answer!  ;D

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