Organic Chemistry Problem- percent yield?

[studyguide12]

I forget how to calculate percent yield.We started with 1.5 ml of distilled h20, 1.0 ml of concentrated sulfuric acid and added 1.0 of 2-methyl-2-butanol. We ended with .386 g of 2-methyl-2-butene. Any help would be appreciated

[Johndude969]

Please note that I have had to make a few asumptions about this question. Firstly that the reaction was infact a DEhydration reaction and that water was removed. Also I have had to assu0me that the only product of the reaction was 2-methyl 2-butene and its isomer 2-methyl 1-butene was not created. Using that as a basis, I used the amounts provided to calculate the Moles of the compounds used. Knowing the substance in shortest supply is the limiting reagent in this question, the 2-methyl 2-butanol was taken as the limiting reagent at a concentration of 0.011moles (assuming the chemicals used were pure). Since the dehydration reaction from butanol to butene (shortening the name for handiness) is a 1:1 reaction the amount of then 0.011 moles of the 2-methyl 2-butene should have been created and that should be an amount of 0.77143g should be created. To calculate the percentage yield divde the amount created by the expected yield and multiply it by 100%. So, 0.386g/0.77143g *100 = 50.03%

 

So the answer to your question is that the yield in this reaction is approximatley 50%.