Question

Calculate the mass of 10.0 mmol (SS) or 1.00 mmol of trans-cinnamic acid and the theoretical yield of 2,3-dibromo-3-phenylpropanoic acid?

m.w of trans-cinnamic acid: 148.2

m.w of 2,3-dibromo-3-phenylpropanoic acid: 308

Answer 1

Amount given initially for trans-cinnamic Acid is 1.00 g. I know that theoretical yield is a 1:1 ratio, The theoretical yeild is the maximum amount of product with possible given the amounts of reagent So .0068 mols Mass of 2,3-dibromo-3-phenylpropanoic acid obtained is .80g, .0026

Theorectial yield = moles limiting reagent x (mol ratio product / limiting reagent) x (molar mass product/ 1 mol)

% yield = actual yield / theoretical yield x 100/1

So .0068 mols Mass of 2,3-dibromo-3-phenylpropanoic acid obtained is .80g, .0026
= 0.80 g / 2.09 g x 100/1
= 38%

the whole 0.0068 mol of reagent is convert to product in a 1:1 ratio.

Now convert to mass (mass = molar mass x moles) to give the theoretical yield in grams