Calculating the Optical Rotation Value

[Interception]

Homework Statement

So the problem states that we have a (+) enantiomer of a coumpund with an optical rotation of 50*. If a pure sample containes 42% of the (+) enantiomer and 58% of the (-) enantioimer, what is the optical rotation value. (By * i mean degrees)

Homework Equations

I'm not sure which equation would apply to this situation. It's a small school, and our proffesor hardly went over this subject. We'd done a few problems, but none where like this.

The Attempt at a Solution

I know that it will be a negative value since the the (-) enantiomer outnumbers the (+). I didn't know how to approach the problem, so I took 58%-the 42% to see how much of the (-) will contribute to the rotations since the 42% of the (+) enantiomer will counteract 42% of the (-) enantiomer. I got -16. I took this mean that 16% of the (-) enantiomer would be optically active, and 16% of 50 is 8, or in this case -8* of optical rotation. The negative value makes sense, and the fact that it's not a large number. Does this look right?

 

[Borek]

Looks OK to me.