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April 16, 2014

April 16, 2014

Posted by **Susie** on Wednesday, November 2, 2011 at 7:35pm.

What is the %ee?

______%ee

What is the molecular composition of this same sample?

S isomer = ______%

R isomer = ______%

This is how I solved for the R and S isomer:

Let x = % of R

Let 100-x = % of S

125.6x + [(-122.1)(100-x)/100] = 81.5

125.6x - 12210 + 122.1x = 8150

247.7x = 20360

x = 82% = R isomer

100-82 = 18% = S isomer

That's how I solved for the % of R and S isomer. I'm not even sure if I solved correctly.

And for the %ee, I have no idea how to do it! I know that you divide measured specific rotation of mixture by specific rotation of pure enantiomer and then multiply by 100, but in this case, the the values for the R and S are different, so I'm not sure where to go from there.

Please help!

- OChem -
**Anonymous**, Tuesday, November 8, 2011 at 3:45pmR-S= %ee I think

- OChem -
**Anonymous**, Wednesday, November 9, 2011 at 1:13am%ee = (+81.5/+125.5) x 100%

it was in the techniques book! :)

- OChem -
**Anonymous**, Wednesday, November 9, 2011 at 1:15amor +125.6. idk which one is the one in your problem.

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