[Kayla_N]

So i know that the equation for specific rotation is :   alpha/ (l *d), but i dont know how to use it with the questions below. Please help and explain. thanks.

1. Consider a solution that contains 64% R isomer and 36% S isomer. If the observed specific rotation of the mixture is -82.0 degrees, what is the specific rotation of the pure R isomer?
answer: ? deg

2. What observed rotation is expected when a 1.88 M solution of (R)-2-butanol is mixed with an equal volume of a 0.940 M solution of racemic 2-butanol, and the resulting solution is analyzed in a sample container that is 1 dm long? The specific rotation of (R)-2-butanol is -13.9 degrees mL g^(-1) dm^(-1).
answer: ? deg

[cpncoop]

If you know that a 50:50 mixture = 0, and that a 64:36 mixture = -82, you can set up two equations to solve for the answer:

Let x = R isomer, and y = S isomer

0.5x + 0.5y = 0 (eq 1)
0.64x + 0.36y = -82 (eq 2)

solving for x and y in eq 1, we get y = -x
plugging this into equiation 2 we get:

0.64x + -0.36(-x) = -82

Combining the x terms we get:

0.28x = -82

and solving for x gives -292.85, the specific rotation for the R isomer
since y = -x, the rotation for the S isomer is 292.85

 

[Kayla_N]

Thanks for explaining. BTW is number two using the same steps?

[cpncoop]

didn't realize there was a number 2.... For this one, you would have:

x = R isomer, y= S isomer

Let's assume we use 1 L of each solution.  For the racemic solution, we'd have:

.47 moles of R, and .47 moles of S (.94 moles total)

For the other solution, we'd only have

1.88 moles of the R isomer.

In total, we have 2.35 moles of R, and 0.47 moles of S.  From this point on, you should be able to find the answer using a similar approach to what we described before (i.e. you know the rotation for the pure material, you know the rotation for a 50:50 mixture, so you should be able to calculate the rotation for this specific percentage).

Let me know if this works,