Calculating the percentage of (+)-2- butanol in the sample

Question

Specific rotation of (+)-2-butanol is 12.5$^{\circ}$ . A sample of 2-butanol containing both the enantiomers was found to have specific rotation value of –2.8$^{\circ}$ . What is the percentage of (+)-2- butanol in the sample?

I attempted it by letting the percentage of (+)-2-butanol be $x \%$ and therefore percentage of enantiomers in the mixture would be $2x \%$ due to equal contribution by (-)-2-butanol. Since the net specific rotation of the mixture is given to be negative, we know that percentage of excess of (-)-2-butanol would be $(100-2x)\%$. It is also given that specific rotation of pure (+)-2-butanol is 12.5$^{\circ}$, but I am unable to understand how is that useful here.

In it's solution, they have calculated the percentage of (-)-2-butanol as $\dfrac{2.8}{12.5} \times 100=22.4 \%$ and thus percentage of (+)-2-butanol came out to be $\dfrac{100-22.4}{2}=38.8\%$.

I didn't understand this last step, why is the division of 2.8 and 12.5 giving the percentage of (-)-2-butanol?

Answer 1

A different, more logical solution gives the correct answer as well.

Since the optical rotation of (+)-2-butanol is 12.5 degrees, the optical rotation of (-)-2-butanol is -12.5 degrees. If the mole fraction of the (+) isomer is $x$ and the mole fraction of the (-) isomer is $1-x$, then $$-2.8 = 12.5x -12.5(1-x)$$

Solving this simple linear equation gives $x = 0.388$, which when converted to mole percentage, is $38.8\%$.