The question is: A student performs a crystallization on an impure sample of biphenyl. The sample weighs 0.5g and contains about 5% impurity. Using his knowledge of solubility, he decides to use benzene as a solvent. The final weight is only 0.02g. Why is recovery so low?
My answer: The recovery is low because there will be some experimental loss, the original sample was not 100% biphenyl, and some biphenyl is soluble in the solvent even at 0 degrees C.
Is this sufficient?
Thanks from Sheryl
I would hit it little harder in spots. With a known impurity of 5%, we might expect to recover about 0.45 g. Some additional loss could be expected due to solubility in the solvent regardless of the solvent used. However, a recovery of only 0.02 g far exceeds looses due to solubility if the proper solvent was used for recrystallization and in the right amount. The loss in this case is due to the high solubility in benzene. Perhaps too much solvent was used. (The Merck Index says biphenyl is insoluble in water--which we would expect--and soluble in alcohol and ether. It doesn't mention solubility in benzene' however, solubility in ether suggests solubility in benzene,too My guess, then is that the student simply used too much benzene and his/her problem is loss in the solvent.)
2 answers
2 answers