Michele Chan Lisa James Chem 2212L Results Calculations 3.00g methybenzoate X (1 mol methybenzoate/136.15g )X(1 mol methy-nitrobenzene 1 mol methybenzoate)x (181.15g/1mol Methynitrobenzene) =3.99g methynitrobenzene 3.00mlHNO3 X( 1.42g/ml) (1 HNO3mol/63.01g) X( 1molHNO3/ 1 mol methynitrobenzene) X(181.15g/1mol Methynitrobenzene)=12.25g methynitrobenzene 7ml H2SO4 X( 1.84g/ml) X(1 mol H2SO4/ 98.09g) X (2 mol H2SO4/ mol methynitrobenzene) X(181.15g/1mol Methynitrobenzene)= 47.6g methynitrobenzene Limiting reactant is methybenzoate So percentage yield is (actual/theoretical) X100 =1.785g/3.99g X100 = 44.7% Discussion Looking at the mass of our final product, we ended up with a low percentage yield, which I think was due to the carelessness in lab, where we lost mass when trying to neutralize the product prior to recrystalization. It took us multiple tries before we were able to neutralize the product so we checked the PH a lot, and loss mass in the transporting of the funnel. After we ran
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