A pure sample of (R)-limonene has a specific rotation of +125.6. If a sample had a specific rotation of +35.6, how much of each enantiomer is present in the sample? Show your work.

I don't even know where to begin....please help :(

I suggest X = + enantiomer

and 1-X = - enantiomer.
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125.6X + (1-X)*-125.6 = 35.6 and solve for X and 1-X. These will be fractions and you can convert to percent by multiplying by 100.
Note: If the + enantiomer has +125.6 then the - enantiomer will have -125.6.
I have seen this worked a quickie way but I've forgotten how to do it the short way. This should give you the correct answers.

To determine the amount of each enantiomer present in the sample, we can use the relationship between specific rotation and enantiomeric purity. The specific rotation, denoted by [α], is defined as the observed rotation divided by the concentration (in g/mL) and the path length (in dm). It is expressed in degrees per decimeter per gram per milliliter (°/dm·g/mL).

The specific rotation of (R)-limonene is given as +125.6. This means that a pure sample of (R)-limonene with a concentration of 1 g/mL and a path length of 1 dm will exhibit a rotation of +125.6°.

If a sample has a specific rotation of +35.6°, we can calculate the enantiomeric excess (ee) using the formula:

ee = ([α]observed - [α]pure) / ([α]observed + [α]pure)

where [α]observed is the observed specific rotation (+35.6°) and [α]pure is the specific rotation of the pure enantiomer (+125.6°).

Using the given values, we can calculate the enantiomeric excess:

ee = (+35.6 - +125.6) / (+35.6 + +125.6)
= -90 / 161.2
≈ -0.558

The enantiomeric excess represents the ratio of the excess of one enantiomer to the total amount of both enantiomers. In this case, the negative sign indicates that the excess is primarily of the (-)-enantiomer.

To determine the percentage of each enantiomer in the sample, we can use the formula:

% of (+)-enantiomer = (1 + ee) / 2 * 100
% of (-)-enantiomer = (1 - ee) / 2 * 100

% of (+)-enantiomer = (1 + (-0.558)) / 2 * 100
≈ 22.1%

% of (-)-enantiomer = (1 - (-0.558)) / 2 * 100
≈ 77.9%

Therefore, in the given sample, approximately 22.1% is the (+)-limonene and 77.9% is the (-)-limonene.

To determine the amount of each enantiomer present in the sample, we need to use a formula known as the optical purity equation. This equation relates the observed specific rotation to the enantiomeric excess (EE) and the specific rotation of each enantiomer.

The optical purity equation can be written as follows:

Observed Specific Rotation (SRobs) = (SR1 * %ee1) + (SR2 * %ee2)

where:
- SRobs is the observed specific rotation of the sample (+35.6 in this case)
- SR1 and SR2 are the specific rotations of the enantiomers (+125.6 and -125.6 for (R)-limonene)
- %ee1 and %ee2 are the enantiomeric excesses of the respective enantiomers (to be determined)

To solve this equation, we can rearrange it to isolate %ee2:

%ee2 = (SRobs - (SR1 * %ee1)) / SR2

Now, let's substitute the given values into the equation and solve for %ee2.

SRobs = +35.6
SR1 = +125.6
SR2 = -125.6

%ee2 = (35.6 - (125.6 * %ee1)) / -125.6

To proceed, we need to know the enantiomeric excess (%ee1) of (R)-limonene, which is not provided in the question. The enantiomeric excess represents the percentage of the preferred enantiomer in the mixture. Let's assume a value for %ee1 (e.g., 100%) and calculate %ee2 accordingly:

%ee1 = 100%

%ee2 = (35.6 - (125.6 * 1)) / -125.6
%ee2 = (35.6 - 125.6) / -125.6
%ee2 = -90 / -125.6
%ee2 ≈ 0.7155

This result of %ee2 ≈ 0.7155 means that the enantiomer with a specific rotation of -125.6 is approximately 71.55% of the mixture. To calculate the %ee1, we can use the equation:

%ee1 = 100% - %ee2
%ee1 = 100% - 0.7155
%ee1 ≈ 99.2845

Therefore, the (R)-limonene enantiomer (%ee1) is approximately 99.28% present in the sample, while the (S)-limonene enantiomer (%ee2) is approximately 0.72% present in the sample.