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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nickname on April 13, 2011, 11:10:45 AM

Title: Percent yield
Post by: nickname on April 13, 2011, 11:10:45 AM

I have to determine the limiting reactant and the percent yield of dibenzalacetone

2 benzaldehyde + 1 acetone---> 1 dibenzalacetone

0.116g acetone
mw acetone= 58.08g/mol
moles acetone= .116g/58.08 g/mol= 1.997245179x10^-3 moles

0.424g benzaldhyde
mw benzaldehyde= 106.12g/mole
moles of benzaldehyde? 0.424g/106.12g/mole= 3.99546819x10^-3 but there is 2 benzaldehyde for each dibenzalacetone= moles of benzaldehyde/ 2= 1.997738409x10^-3 moles

so acetone is the limiting reagent?

use acetone as thereotical moles:

moles of dibenzalacetone= 0.085g/224g/mole= 3.79x10^-3 moles

percent yield using acetone= (3.79x10^-3/1.997245179x10^-3)x100= 180%

Are these answers correct? Thank you!

Title: Re: Percent yield
Post by: DevaDevil on April 13, 2011, 11:17:08 AM

I did not check your calculations, but according to your numbers, yes, acetone is limiting (just). I would almost say they are present in stoechiometric ratio.


180% yield is not possible of course, you cannot get more than expected...

%yield = 100% * actual yield / theoretical yield

so calculate what you would get based on the moles of benzaldehyde (if acetone wasn't limiting), and calculate the actual yield. Then take the fraction to give %yield.

Title: Re: Percent yield
Post by: rabolisk on April 13, 2011, 11:19:19 AM

It's a simple exponent error. Acetone is the limiting reactant (barely). Where you messed up is the second part.

Title: Re: Percent yield
Post by: Dan on April 14, 2011, 03:37:59 AM

Quote from: rabolisk on April 13, 2011, 11:19:19 AM

It's a simple exponent error.

Yup. Check your "moles of dibenzalacetone". Also, dibenzalacetone has a molecular weight of 234 (not 224).