Correct. If more water was present, then all butanol would distill as an azeotrope before water is removed to distill any remaining and higher boiling butanol. If you start with 5% water, the water will be removed as the lower boiling azeotrope with butanol. After the water is exhausted, the pot will now only contain the higher boiling 1-butanol. Continued distillation should see a jump in the boiling point to the pure butanol.
Exactly. Thanks.
Even better, since the H2O-Butanol azeotropic composition falls within the miscibility gap it ought to seperate into a Butanol-rich and a butanol-poor layer. The butanol-rich layer being ~80% Butanol will be above the azeotropic-point and thus amenable to the same distillation procedure as before. Right?
The Butanol-poor layer can be distilled into pure-H2O and azeotrope and there you go again.
Is my analysis correct? In theory you'd get a pure seperation, in spite of the azeotrope. My point being, the azeotrope won't really impair your seperation ability here. Not a 100% sure if I am right.
Okay, that is the theory, but it will also depend upon the efficiency of the distillation. Water and butanol can co-distill. If excess heat is applied, the temperature of the distillate will increase before all of the water has exited the distillation. Decreasing amounts of water will be present depending on the liquid-vapor equilibrium being established during the distillation. Too much heat will co-distill fractions with less than the azeotropic ratio, depending on the amount of heat, column length, and rate of distillation.
Agreed. But distillation efficiency is a general matter and not specific to azeotropes. So you could even screw up a simple non-azeotropic distillation if you didn't do it right (say pumped in too much heat as you mention.)