evil_lurker
National Hazard
Posts: 767
Registered: 12-3-2005
Location: United States of Elbonia
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Mood: On the wagon again.
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Need help calculating yield from publication...
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%?
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Magpie
lab constructor
Posts: 5939
Registered: 1-11-2003
Location: USA
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Mood: Chemistry: the subtle science.
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% 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
The single most important condition for a successful synthesis is good mixing - Nicodem
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unionised
International Hazard
Posts: 5126
Registered: 1-11-2003
Location: UK
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Mood: No Mood
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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.
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