A sample of 2-butanol has a specific rotation of 3.25. Determine the % ee and molecular composition of this sample. The sepcific rotation of pure (+)-2-butanol is +13.0.

is it (3.25/13.0)x100%= 25%... i'm a little confused on the molecular composition though.

Yes, %ee is 25% and it can be used to check later calculations as shown.

If the + enantiomer is +13.0, then the - enantiomer is -13.0.
Let x = + enantiomer
1-x = - enantiomer.
Then x(13) + (1-x)(-13)=3.25
Solve for x and 1-x.
I get 0.625 for x and
0.375 for 1-x.
Therefore, % composition is 62.5% dextro and 37.5% levo.
As a way of checking the first number,
%ee = enan(1) - enan(2) = 62.5% - 37.5% = 25% which agrees with your first calculation done a different way (and called optical purity). Voila!
A third way of doing the same thing is
%ee = (R-S/R+S)*100 =
[(0.625-0.375)/(0.625+0.375)]*100 =
(0.25/1.0)*100 = 25% where R is one isomer and S is the other.
I hope this helps.

Well, you're on the right track with the calculation of the % ee (enantiomeric excess). The formula you mentioned is correct, so the % ee is indeed (3.25/13.0) x 100% = 25%.

Now, let's talk about the molecular composition. In this context, molecular composition refers to the proportion of each enantiomer present in the sample. Since we know that the specific rotation of pure (+)-2-butanol is +13.0, we can assume that the specific rotation of pure (-)-2-butanol is -13.0.

Since the specific rotation of your sample is 3.25, it means that it is slightly rotated towards the positive direction. This indicates that your sample is enriched with the (+)-enantiomer. However, we can't determine the exact proportion of each enantiomer based on this information alone. We need additional data or experimental results to make a quantitative determination of the molecular composition.

So, in summary, you correctly calculated the % ee as 25%, but without more information, we can't determine the exact molecular composition.

To determine the % ee (enantiomeric excess) and molecular composition of the sample, the following steps need to be carried out:

1. Calculate the observed specific rotation for the sample using the formula:
Observed specific rotation = (% ee x specific rotation of (+)-enantiomer) + ((100 - % ee) x specific rotation of (-)-enantiomer)

Plugging in the values, we have:
3.25 = (% ee x 13.0) + ((100 - % ee) x (-13.0))

2. Rearrange the equation to solve for % ee:
3.25 - ((100 - % ee) x (-13.0)) = % ee x 13.0
3.25 + (13.0 x % ee) = % ee x 13.0
3.25 = (% ee x 13.0) + (13.0 x % ee)
3.25 = 26.0 x % ee

Dividing both sides by 26.0:
% ee = 3.25 / 26.0
% ee = 0.125 or 12.5%

So, the % ee of the sample is 12.5%.

Regarding the molecular composition, it is not possible to determine it with the given information. The specific rotation alone does not provide enough information about the molecular composition. It only indicates the extent to which the sample is optically active and contains one enantiomer in excess. Additional analysis, such as chiral HPLC or spectroscopy, would be required to determine the exact molecular composition.

To determine the percent enantiomeric excess (% ee) and molecular composition of the sample, you need to understand a few concepts related to enantiomers and specific rotation.

Enantiomers are molecules that are mirror images of each other but are not superimposable. These molecules have the same physical and chemical properties but rotate plane-polarized light in opposite directions.

Specific rotation is a measure of the angle through which plane-polarized light is rotated by a chiral compound. The specific rotation value, denoted by [α], is dependent on the concentration and length of the sample cell used for the measurement. It is also influenced by temperature and the wavelength of light used.

To calculate percent enantiomeric excess (% ee), you need to compare the specific rotation of the sample to that of the pure enantiomer. Here's the correct calculation:

% ee = [(observed specific rotation)/(specific rotation of pure enantiomer)] × 100

So, in this case:

% ee = [(3.25)/(13.0)] × 100
% ee ≈ 25%

Your calculation of 25% is correct for the % ee.

Now, regarding the molecular composition, the percent enantiomeric excess (% ee) tells you the amount of excess of one enantiomer over the other in the sample. It does not directly provide information about the total molecular composition of the sample. To determine the molecular composition, you would generally need additional information, such as the concentration or amount of the sample used.

If you have additional information about the sample, such as its concentration (in moles or grams), you could use it along with the % ee to determine the composition. However, based on the information provided, the molecular composition cannot be determined solely from the given data.

To summarize, the calculation of % ee is correct, but without additional information, the molecular composition of the sample cannot be determined.