# OChem

posted by .

A sample of limonene has a specific rotation of +81.5. (Given that R-(+)-limonene = +125.5 and, S-(-)-limonene = -122.1).

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.

• OChem -

R-S= %ee I think

• OChem -

%ee = (+81.5/+125.5) x 100%

it was in the techniques book! :)

• OChem -

or +125.6. idk which one is the one in your problem.