[JMaria]

So we had to do a lab for the fractional distillation of ethanol. The original lab stated that the lab was ethanol from sucrose but the ethanol was just given to us in a container.

I'm trying to calculate the percent yield of the ethanol. The density of my distilled ethanol solution had an average of .87 g/mL at 25C, which corresponds to under 75% ethanol by weight on the table given in the book. How would I even attempt to calculate the percent yield? I also know that with 95% ethanol by weight, the density is .8.

[curiouscat]

What was density of feed & hence conc.

[JMaria]

Feed as in the stock solution? If so, it wasn't given. The original volume obtained was 20mL and that's all the information I have about the stock.

[curiouscat]

Feed as in the stock solution? If so, it wasn't given. The original volume obtained was 20mL and that's all the information I have about the stock.

Yes, but you measured density of your distillate right?

Similarly why can't you measure it for your stock too?

[JMaria]

Ah, okay. That didn't occur to me at the time and the lab is done. My bad there. Is there any other way to calculate the percent yield or was that it?

[JMaria]

However, the book does say that it's 20mL of a 10% ethanol solution. Would that help for anything?

[curiouscat]

Ah, okay. That didn't occur to me at the time and the lab is done. My bad there. Is there any other way to calculate the percent yield or was that it?

Nope. Can't think how.

Pretty much you know essentially nothing about your feed.

[curiouscat]

However, the book does say that it's 20mL of a 10% ethanol solution. Would that help for anything?

Sure it helps. Now you know how much ethanol went in & how much was recovered. Divide & you get your yield.

[JMaria]

I know percent yield is actual/theoretical x 100 so are you saying to divide the mL of ethanol at the end by 2mL?

[curiouscat]

I know percent yield is actual/theoretical x 100 so are you saying to divide the mL of ethanol at the end by 2mL?

How many ml did you feed? How many did you recover?

[JMaria]

I put in 20 mL and I recovered 4mL.

[curiouscat]

I put in 20 mL and I recovered 4mL.

% yield = ( 4  * 0.75 * 0.87 ) / ( 20 * 0.1 * 0.98  ) * 100

Which in turn doesnt make sense because you'd be producing more ethanol than you put in.

Either your quantities or densities or conc. are wrong.

[JMaria]

Okay, I understand where I went wrong now. Thank you!

[curiouscat]

Okay, I understand where I went wrong now. Thank you!

Do tell. Where?

[zsinger]

Perhaps the differences in ethanol quantities are due to azeotrope formation with some other species other than water?