Ask Question
16 May, 03:34

Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic acid. After the reaction and workup, the student ended up with 0.1866 g of brominated product. Calculate the student/'s theoretical and percent yields.

+1
Answers (1)
  1. 16 May, 06:01
    0
    0.2889 g brominated product

    64.6 %

    Step-by-step explanation:

    This is a bromination chemical reaction of an alkene and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.

    The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.

    trans-cinnamic acid + pyridinium tribromide ⇒ 2,3-dibromo-3-

    phenylpropanoic acid

    Molar weight trans-cinnamic acid = 148.16 g/mol

    mass trans-cinnamic acid = 139.0 mg x 1g/1000 mg = 0.139 g

    # mol trans-cinnamic acid = 0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol

    Since our reaction is 1 mol trans-cinnamic acid produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:

    1 mol 2,3-dibromo-3-phenylpropanoic acid / trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid

    = 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid

    In grams the the theoretical yield is:

    molar mass 2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol

    The theoretical mass 2,3-dibromo-3-phenylpropanoic acid:

    = 9.38 x 10⁻⁴ mol 2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol

    = 0.2889 g

    % yield = mass experimental/mass theoretical

    = 0.1866 g / 0.2889 g x 100 = 64.6 %
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic acid. After ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers