Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: wildgrl89 on April 14, 2005, 06:02:39 PM
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I have to calc the theoretical yield for the preparation of Ph3COH through Grignard synthesis....I though the Limiting reagent was Mg, however when I calcualted it My yield was .548...and my experimental yield was .84....I'm fairly sure my experiment went correctly so my calcualtions must be flawed?!
Mg- (0.053g) * (24.31g/mol)= 0.002139mol
0.002139mol * (256 g/mol)= 0.548g
Rxn: 1). 0.26 mL Bromo-Benzene plus 0.053g Mg ---->
Ph-MgBr
2) 0.125 mL PhCO2CH3/HCL ---> Ph3COH + CH3OH
Any idea where I'm going wrong!?
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i didnt bother to punch my calculator, but you can check if the answer given was mass yield and not molar yield, or vice versa
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No the product weight I have is from an experiment I ran...!?
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In reactions we always use molar ratios.
You should start from (balanced) reactions (at least for reagents on the left sider and the product on the right side), then calculate moles from masses and volumes of reagents.
In Grignard reactions we usually uses some excess of Grignard reagent, hence methyl benzoate is probably the limiting reagent. If you do not know a density of methyl benzoate, approximate its density to 1.0 g/mL (but not for C6H5Br - usually density is on the bottle od commercial reagent)
PhCOOCH3 + 2Mg + 2C6H5Br + H2O => Ph3COH
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Yes, I calculate a different limiting reagent!
This may help:
density of bromobenzene = 1.491 g/mL
MW of bromobenzene = 157.02 g/mol
density of methyl benzoate = 1.094 g/mL
MW of methyl benzoate = 136.15 g/mol
MW of triphenylmethanol = 260.34 g/mol