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classicfan
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PostPosted: Mon Nov 02, 2009 7:15 pm    Post subject: Molar equivalent Reply with quoteFind all posts by classicfan

If I have 0.5 g of sodium borohydride, how do I figure out the molar equivalent of this?
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expert
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PostPosted: Mon Nov 02, 2009 8:27 pm    Post subject: Reply with quoteFind all posts by expert

Divide it by molecular weight of NaBH4
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classicfan
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PostPosted: Tue Nov 03, 2009 8:27 am    Post subject: Reply with quoteFind all posts by classicfan

expert wrote:
Divide it by molecular weight of NaBH4


But isn't that just moles of NaBH4? In my lab, we have to state the weight in grams, moles of substance and molar equivalent (eq)... so they consider it as different from moles.
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PostPosted: Tue Nov 03, 2009 11:58 am    Post subject: Reply with quoteFind all posts by expert

"Molar equivalent" can be only clear with a relevance to a particular reaction. Without telling us details about your reaction I can only guess that this is a reduction consuming hydride 100% (this is not always the case for NaBH4). The question is “how many molar equivalents of NaBH4 required for your reaction?”
For example you reduce aldehyde by the following process:
RCHO + NaBH4 --> RCH2OH + NaH2BO3

To understand better how NaBH4 works simply present it as a “hydrogen preserve” that can be “uncanned” by the following reaction:
NaBH4 = 3H2O = 4H2 + NaH2BO3
It is obvious from this reaction that one equivalent of NaBH4 will produce four equivalents of hydrogen. On the next step write formal reduction of your substrate by hydrogen, for example:
RCHO + H2= RCH2OH, so it’s clear that 1 mol of aldehyde will require 1 mol of H2, so 4 mols released by 1 mol of NaBH4 can reduce as many as 4 mols of aldehyde:

4RCHO + NaBH4 + 3H2O = 4RCH2OH + NaH2BO2

In this particular case molar equivalent of NaBH4 should be calculated as the amount of NaBH4 in grams divided by molecular weight of NaBH4 and everything yet divided by four.

Since NaBH4 reacts with water, alcohols and acids with a release of hydrogen, it usually taken in excess for the reduction. Practically I would always recommend using it with at least 2-fold excess.
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