Chemistry Forum

If I have 0.5 g of sodium borohydride, how do I figure out the molar equivalent of this?

Divide it by molecular weight of NaBH₄

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.

"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.