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Percent Yield After Dehydration Of 2-Methyl-2-Butanol

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#1 GlitterGirl



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Posted 23 October 2011 - 12:07 PM

Organic Chemistry Lab, we performed an experiment to dehydrate 2-methyl-2-butanol and formed 2-methyl-2-butene(85.888%) and 2-methyl-1-butene(14.112%), percentages derived from gas chromatoghraphy analysis after dehydration.

We were asked to calculate the percent yield but I'm not sure how to do this. The starting material was 18mL of H2O, 9mL of concentrated sulfuric acid, and 18mL (15grams) of 2-methyl-2-butanol. After dehydration was complete, the final product (which was the mixture of 2-methyl-2-butene and 2-methyl-1-butene, before GC analysis) was weighed at 3.055 grams.

I am confused, is the question of percent yield just asking after dehydration how much was left meaning 3.055g/15gx100 giving only a 20.4% yield?

Any explanation would be GREATLY appreciated. BTW, this calculation is required in a lab report that is due tomorrow. Thanks to anyone who is wiling to help!