Optical rotation of mixture The specific rotation of optically pure adrenaline is $-53$. A mixture of (R)- and (S)-adrenaline was found to have a specific rotation of $-45$. Calculate the % ee of the mixture (enter numeric value only, without % sign, rounded to the nearest whole percent).


This should be $\%~\text{ee}= \text{specific rotation sample}/\text{specific rotation enantiomer}$

But $100\times (-53/{-45})=117.78~\%$

The fraction the other way is $84.9~\%$ but doesn't stick to the formula.
Enantiomeric excess (ee) is defined as:

$$\text{ee} = \frac{\alpha(\text{sample})}{\alpha(\text{pure enantiomer})}\tag{1}$$

Here, we are given the specific rotation of our sample ($-45^\circ$) and that of a pure enantiomer ($-53^\circ$). To calculate the ee, all you need to do is to plug these two values into equation $(1)$ and work out how to turn a fractional number into a value in $\%$.


   The answer is $85~\%~\text{ee}$.