If a student starts with 0.387 g of p-anisaldehyde and ends up with 0.265 g of methyl-E-4-methoxycinnamate, what are their theoretical and percent yields?

Answer 1

1. Write the equation. It isn't necessary to balance everything but you must know how many mols of the prouct you obtain for 1 mol of the reactant.
2. Convert 0.387g to mols. mols = grams/molar mass.
3. Using the coefficients in the balanced equation, convert mols of the reactant to mols of the product.
4. Convert g to mols. g = mols x molar mass. This is the theoretical yield of the methyl E compound.
5. %yield = (actual grams yield/theoreticl yield)*100 = ?

Answer 2

Thanks! However, Do I convert the g I got by dividing by the molar mass of methyl E? The reaction is 1-to-1.

Answer 3

As I outlined it, you convert the para compd to mols. If the rxn is 1:1, you will get that many mols of the methyl E cmpd. Convert that to grams as I wrote; g = mols methyl E x molar mass. That is the theoretical yield of the methyl E.
Then %yield = (0.265 methyl E/theoretical yield of methyl E)*100 = ?

Answer 4

Thank you so much!!! :)

Answer 5

78