evil_lurker - 11-8-2005 at 07:10
The did it screwy. I can't seem to grasp the mathmatical formula to convert the molar percentage to grams.
According to the publication, they are claiming a 40% yield based on 1 mol of product per 2 mol of metal ion promoter.
The metal ion promotor weighs in at 268g per mole, the product weighs in at 134g per mole.
Based on the numbers above, if 1 mole of metal ion promoter is used, how many grams of product are yielded at 40%?
Magpie - 11-8-2005 at 20:00
% yield = [moles product/(2 moles promoter)] x 100
rearranging:
moles product = (% yield)(2 moles promoter)/100
substituting:
moles product = (40)(0.5)/100 = 0.20
g product = (0.2)(134g) = 26.8 g
unionised - 14-8-2005 at 12:55
Is this "promoter" a reactant or a catalyst or even, a promoter?
Promoter is a technical term for something that enhances the effect of a catalyst without being a catalyst itself.
If they are using the term correctly the question is meaningless as there isn't any information about how much reactant there is.