Suppose a student started with 129 mg of trans-cinnamic acid and 0.50 mL of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.216 g of brominated product. Calculate the student\'s theoretical and percent yields.

Answer 1

1. Write the equation and balance it.
2. Convert 129 mg to mols. mols = grams/molar mass
3. Convert 0.50 mL of 10% bromine to mols. I believe this is 0.05 mL bromine. You may need to look up the density of liquid bromine to and use that to obtain grams, then convert to mols.
4. Using the coefficients in the balanced equation to convert mols cinnamic acid to mols of the product.
5. Do the same for bromine.
6. The two answers probably will be different; the correct answer in limiting reagent problems (such as this) is ALWAYS the smaller value tna the reagent producing that number is the limiting reagent.
7. Convert mols of the product to grams. This is the theoretical yield (TE).
8. Then %yield = (actual yield/TE)*100 = ?
Note: If you KNOW which is the limiting reagent you can use it for step 1, 2or 3, 4or 5, then 6 and 7.

Answer 2

What does the equation look like? What is the grams?

Answer 3

The lim reg is trans-cinn.

so convert 129 mg trans-cinn to moles using its molar mass (148.16g/mol), then use 1 to 1 mole ratio to get to moles of product, then convert to grams product using molar mass 307.97g/mol. this gives you theo yield.

note: u don't need to use the ml bromine in the calculations, it was only to find which is the lim reagent--which i've given you.