How do I calculate theoretical yield here? - theoretical yield of synthesis of benzocaine
I begin with the p-toluidine -> p-methylacetanalide -> p-Acetamidobenzoic -> p-amino benzoic acid -> Benzocaine
I begin with 1.61g p-toluidine
Because it has more than reactants, there is a molar ratio of 1:1 in the synthesis of law? And if so, I'll just mole of p-toluidine, is multiplied by a ratio of 1:1 (which is the same, of course), then multiplying the molecular weight of benzocaine grams theoretical?
Theoretical Yield Of Synthesis Of Benzocaine How Do I Calculate Theoretical Yield Here?
4 comments:
First we set g mol
P-toluidine molar mass = 107.17 g / mol
1.61 g / 107.17 = 0.0150 mol of p-toluidine
the ratio of 1:1 for revealing 0.0150 mol benzocaine
Benzocaine molecular weight = 165,189 g / mol
0.0150 mol x 165,189 g / mol = 2.48 g formed
I do not think that this is multiplied by 1:1.
I think I will have the mass of the formula p-toluidine.
My chemistry is a little fuzzy, since the school closed. But lucky theoretical thought my performance really bad. Lol.
Theoretical
This is the amount of products, which is synthesized under ideal conditions.
To determine the theoretical yield, you multiply the number of moles of limiting reagent ratio between the limiting reactant and the product synthesis and the molecular weight of the product.
First, the calculation of the number of moles in 1.61g p-toluidine (p-toluidine 1.61g/MW say). The theoretical yield is expected that the% yield at each step of the sequence is 100%. If this is true, then the number of moles of final product (benzocaine) should be equal to the number of moles of p-toluidine. So to get the final grams of benzocaine, multiply the number of moles of p-toluidine and therefore moleculat weight benzocaine benzocaine. This gives the theoretical number of grams All the proposals as doing everything, the calculation is the same as I described. Good luck and I hope that helps.
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