Thats my point aswell the 96% ethanol azeotrope is not required for this calculation and if enahs took the time to read my first post they'll see i spelt out how to do the % yield.

No, what you said is just wrong. Unless you distill every bit of the solution, and then calculate the amount of ethanol in it. But that is not the point of distillation, that is just a slow method of transferring one solution into another container.

If you start with 50mL of ethanol and 50mL of water. You then distill it and you only collect the distillate while the B.P. is appropriate. When you stop collecting the distillate, you at most have 96% ethanol in it. Your method (original/result) assumes you can collect 100% of the ethanol. Once you pass that critical point (in this example) of the ~50 mL of distillate, the B.P. will start shifting towards water, and you get more and more water in each drop then you do ethanol.

The only possible way to get 100% of the ethanol back is to distill the entire solution; but t

*he point of distillation is to separate compounds*.

You collect a volume of distillate, you then recognize that at the maximum possible efficiency, only 96% of that is going to be ethanol. That is your theoretical yield for that possible volume of distillate. You then calculate how much ethanol you actual have in your distillate (say, using densities). This is then what you use to find your % yield, and determine the efficiency of your distillation apparati and usage.

Your method is a % yield, but it is pointless when applied to distillation like this; as your maximum possible separated compound yield is not what you started with. The only way to get it what you started with is to distill all 100mL of the 50/50mL solution; which is not distillation!