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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: hsucellmolstudent on September 27, 2009, 07:46:59 PM

Title: % recovery of a mixture
Post by: hsucellmolstudent on September 27, 2009, 07:46:59 PM

Hi everyone,

My lab partner and I are at an impasse, can any of you answer our questions?

we recently did an extraction to separate benzoic acid and p-dimethoxybenzene from a mixture. We're working on our lab report and need to record % recovery of each crude solid we obtained. We began with 0.800g of a benzoic acid and p-dimethoxybenzene mixture that is1:1 by weight.

I think that we should use half of this (0.400g) as the original weight to find the % recovery of each solid separately since ideally benzoic acid and p-dimethoxybenzene were each half of that 0.800g.

My partner thinks that 0.800 was the original weight we started with and therefore is the amount we should use for the equation. 

I have one other question-- when I calculated % recovery (using the assumption that i started with 0.400 g of each compound), I came up with over 116.5% recovery of benzoic acid and 91.175% p-dimethoxybenzene. I feel like I should have lost some of the benzoic acid in the process. Could this have been caused by impurities? Can anyone think of a reason I gained benzoic acid?

Thank you all for your time!

Title: Re: % recovery of a mixture
Post by: Borek on September 28, 2009, 03:27:27 AM

I would go for 0.4 g - after all, when all substance was recovered, it should be 100%, not 50%.

Not knowing the procedure it is hard to advise on the second question. Sometimes it is solvent that is difficult to remove.

Note, that sum of percentages should be 200.