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Author Topic: coenzyme synthesis of benzoin  (Read 8834 times)

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webguy54

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coenzyme synthesis of benzoin
« on: February 25, 2006, 06:07:01 PM »

coenzyme synthesis of benzoin

2 C6H5CHO + C12H17ClN4OS (?) ---> C14H12O2
(benzaldehyde + thiamine hydrochloride ---> benzoin)

Is C12H17ClN4OS the correct formula for thiamine hydrochloride?  I found so many online...not sure which is correct.

Anyhow, the real question...which one is the limiting reagent? I got a ridiculus percent yield.

Data:
weight of thiamine hydrochloride used = 0.3005 g
weight of benzaldehyde used = 0.938 g
weight of benzoin produced = 0.677 g
MW of thiamine hydrochloride = 337.27 g/mol (? - I found so many different MWs along with chem formulas for this compound)
MW of benzaldehyde = 106.13 g/mol
MW of benzoin = 212.25 g/mol

molarity of thiamine hydrochloride = (weight of thiamine hydrochloride) / (MW of thiamine hydrochloride) = (0.3005 g) / (337.27 g/mol) = 0.0008909775 mol

molarity of benzaldehyde = (weight of benzaldehyde) / (MW of benzaldehyde) = (0.938 g) / (106.13 g/mol) = 0.008838217 mol

calculating limiting reagent:
thiamine hydrochloride = (0.0008909775 mol) / (1 mol) = 0.0008909775 mol
benzaldehyde = (0.008838217 mol) / (2 mol) = 0.0044191085 mol
thiamine hydrochloride is the limiting reagent, therefore moles of thiamine hydrochloride = moles of benzoin = theoretical yield

actual yield:
molarity of benzoin = (weight of benzoin) / (MW of benzoin) = (0.677 g) / (212.25 g/mol) = 0.0031896 mol

percent yield of benzoin = [(actual yield) / (theoretical yield)] x 100% = [(0.0031896 mol) / (0.0008909775 mol)] x 100% = 358%???!!!

F.Y.I. MP = 125C - 133C

Any help would be greatly appreciated. Thank you!
« Last Edit: February 25, 2006, 06:13:48 PM by webguy54 »
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