[griffon87]

The mixture had an observed specific rotation of +15.9deg. It is known that the specific rotation of the R enantiomer is -38.9deg, determine the percentage of each isomer in the mixture? 

My attempt:
I calculated the difference btw the mix 15.9-(-38.9) = 54.8.
38.9/54.8= 71% R isomer.
1-x
1-.71=.29 or 29% for S isomer.

I clearly am lost. 

[cth]

You have:
100% of the R enantiomer gives -38.9deg.
100% of the S enantiomer gives +38.9deg.

If the mixture was 50% R and 50% S, the specific rotation would be zero.

For this example, the observed rotation is +15.9deg. It has a + sign, so it indicates the S enantiomer is more abundant than the R, because the S is "pushing" towards positive rotation.

consider the following table, it should help:

      S composition        0%      50%         x%       100%
      rotation             -38.9          0      +15.9      +38.9

[KritikalMass]

Try this:

1)
[(observed rotation)/(specific rotation)]x100 = %R (That will give you the percent of the R enantiomer in excess in the solution).

2)
(100 - %R)= Amount of racemic mixture.

3)
Divide that number in half since it will be half R and half S.

4)
Total concentration of R will be the amount from step 1 + amount from step 3.

Concentration of S will be the amount from step 3.

[KritikalMass]

Whoops need to make a correction! Where ever you see an R in my previous post replace it with an S, and where ever you see an S replace it with an R.

Sorry! 

[griffon87]

Thank you both. The "pushing" explanation was very clear, and thanks for the steps.