I want to calculate the yield percent of ZIF-67.
In the attached image there are a experimental route for synthesis of ZIF-67. I want to know how it calculate the yield percent is 73%?
Hello F. Sh.
Thank you for sharing your question with the RG community. Although MOFs are coordination networks, they are still electrically neutral salt compounds in terms of their molecular formula.
n Co2+ (58.9) + 2n (C4H5N2)- (81.1) => [Co(C4H5N2)2]n (221.1)
For this reason, a maximum of 4.9 mmol Co-ZIF-67 can be formed from 4.9 mmol Co salt. The yield is then calculated as follows:
n = 100% * 0.7957 g / (0.0049 mol * 221.1 gmol-1) = 73%.
I would like to add that in many cases the metallic nodes are not single metal cations but consist of e.g. oxocentric metal clusters. In these cases, of course, the molecular formula is a bit more difficult to formulate.
@Florian Morsbach
Thanks a lot. Your answer really help me.
Now I have another question, if I repeat the experiment with half times of all reagnatnts and solvent (I mean: cobalt nitrate hexahydrate =1.436g÷2; 2-mitylimidazole 3.244g÷2 and methanol 100ml÷2),it must results in approximately half yield (0.7957÷2)?
You are right, the absolute yield is reduced by half. In percentage terms, of course, the yield remains similar.
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