[Lish]

A 0.110 M solution of an enantiomerically pure chiral compound D has an observed rotation of 0.16° in a 1-dm sample container. The molar mass of the compound is 161.0 g/mol.

What is the observed rotation of 100 mL of a solution that contains 0.01 mole of D and 0.005 mole of L? (Assume a 1-dm path length.)

So far, I've tried .054 degrees. I calculated the enantiomer excess of the mixture by the following method:

.01 moles of D  to .005 moles of L ~ so, .01+.005 = .015 total; using this total value, divide the portions of D and L.

so .01/.015 to .005/.015 ~ 67% D to 33% L.

And thus, the enantiomer excess will be 34%.

I calculated the specific rotation in another part of this question to be 9 degrees. The concentration is .01771 g/mL. The length is just 1 dm.

SO

a=[a]*c*l

a=9*.01771

now, since 34% is optically active and due to the portions, pure D - it would be .01771 (concentration) * 34% (I did write it as .34)

yet, I get the wrong answer apparently. What am I doing wrong?

[Borek]

I am not sure why do you calculate these percentages. Either calculate separately rotation of the D and L present and add angles (just watch the sign), or assume equimolar part of the solution is inactive, so only excess of D will rotate the light - and excess is just C_D-C_L.

[Lish]

Is there anything wrong in using the percentages? Not sure what you're suggesting as an alternative.

[Borek]

Perhaps it is not wrong (although rounding 1/3 to 34% is wrong and done way too early), just IMHO unusual.

PLease elaborate on what you don't understand.

Also - you are given all concentrations as molar, you can safely ignore molar mass of the compound and there is no need to express concentration in g/mL. Just use ratio of concentrations to find the rotation.

[Lish]

I am asking this question because when I entered in the answer I arrived at - the program stated that it was incorrect. Usually, if there is a small difference in the answer, like for instance: I enter .0898 and the answer is .09(computer's answer) or I enter .087 - it's usually accepted.

From what I understand - there is 34% optically active molecules (and they are D). I have the specific rotation, which is 9 degrees. I have the concentration, which is .01771 g/mL. So it's asking me what the observed rotation is in a .01 moles of D and the .005 moles of L mix.

I'm stuck basically. How would you guys do this problem? What other way is there to go about thinking of the moles ratio?

[Borek]

I already explained to you how I would approach the problem. If you don't understand what I wrote ask specific questions, otherwise all I can do is to repeat word for word what I posted earlier.

[Lish]

I couldn't figure it out. Apparently though the answer was to subtract .005 moles from .01 moles. I understand the logic behind it - that the .005 moles of L cancel the .005 moles of D, and .005 moles are left over. Then convert that to concentration and so on.

I just thought that I had to convert the mole ratio to a percentage, find the excess, then multiply the excess by the original concentration. Not sure if that is what you were suggesting or not.

[Borek]

I couldn't figure it out. Apparently though the answer was to subtract .005 moles from .01 moles. I understand the logic behind it - that the .005 moles of L cancel the .005 moles of D, and .005 moles are left over.

That's what I suggested - assuming C_D and C_L are concentrations of both forms, calculating the difference C_D-C_L is equivalent to subtracting number of moles of L from number of moles of D in a given volume.

Note that my other proposition is equivalent. I told you to calculate rotation of the solution assuming it contain only D (concentration given) and only L (concentration given), then to find overall rotation subtracting obtained values - which yields exactly the same result.

Then convert that to concentration and so on.

Molarity - mol per liter - IS a concentration. You don't have to convert it to anything, especially if the information in the question is given as angle per molarity (0.16° for 0.110M solution or 0.16°/0.110M=1.45°/M - you can use this value to calculate rotation for any given molar concentration; note that 1g/mL which is a concentration unit used when reporting specific rotation is directly proportional to molarity, which makes the exchangeable in all related equations, you just need some proportionality coefficient).

I just thought that I had to convert the mole ratio to a percentage, find the excess, then multiply the excess by the original concentration. Not sure if that is what you were suggesting or not.

It probably can be done through percentages, but I strongly advice against - as they are irrelevant to the problem. You can drive from New York to Washington through Los Angeles, but thera are much better way of getting from one city to another.