Question

multiple choice 2 points
a substance has a specific rotation of -40.00°. in a polarimeter tube that is 5.000 cm long, a solution of the compound (0.8100 g/100.0 ml) has an observed rotation of -7.520°. how much of each enantiomer is in the solution?
% of the (+)-enantiomer = 18.5%
% of the (-)-enantiomer = 40.0%
% of the (+)-enantiomer = 26.8%
% of the (-)-enantiomer = 73.2%
% of the (+)-enantiomer = 46.4%
% of the (-)-enantiomer = 73.2%
% of the (+)-enantiomer = 73.2%
% of the (-)-enantiomer = 26.8%

Explanation:

Step1: Calculate the specific rotation of the sample

The formula for specific rotation $[\alpha]=\frac{\alpha}{l\times c}$, where $\alpha$ is the observed rotation, $l$ is the path - length in decimeters, and $c$ is the concentration in g/mL. First, convert the path - length $l = 5.000\ cm=0.5\ dm$ and $c = 0.8100\ g/100.0\ mL = 0.0081\ g/mL$. Given $\alpha=-7.520^{\circ}$, then $[\alpha]_{obs}=\frac{-7.520^{\circ}}{0.5\ dm\times0.0081\ g/mL}\approx - 185.6^{\circ}$.

Step2: Calculate the enantiomeric excess (ee)

The enantiomeric excess $ee=\frac{[\alpha]_{obs}}{[\alpha]_{pure}}\times100\%$. Given $[\alpha]_{pure}=- 40.00^{\circ}$, then $ee=\frac{-185.6^{\circ}}{-40.00^{\circ}}\times100\% = 46.4\%$. Since the observed rotation is negative, the (-)-enantiomer is in excess.

Step3: Calculate the percentage of each enantiomer

Let $x$ be the percentage of the (-)-enantiomer and $y$ be the percentage of the (+)-enantiomer. We know that $x + y=100\%$ and $x - y=ee$. Solving the system of equations

$$\begin{cases}x + y = 100\\x - y=46.4\end{cases}$$

, add the two equations: $2x=146.4$, so $x = 73.2\%$ and $y = 26.8\%$.

Answer:

B. % of the (+)-enantiomer = 26.8%, % of the (-)-enantiomer = 73.2%