Question

Suppose a student started with 155.0 mg of trans-cinnamic acid, 458 mg of pyridinium tribromide, and 2.35 mL of glacial acetic acid. After the reaction and workup, the student ended up with 0.2170 g of brominated product. Calculate the student\'s theoretical and percent yields.

Answer 1

Answer: The theoretical yield of the product is 0.320 grams and its percent yield is 67.81 %

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For trans-cinnamic acid:

Given mass of trans-cinnamic acid = 155 mg = 0.155 g   (Conversion factor: 1 g = 1000 mg)

Molar mass of trans-cinnamic acid = 148.2 g/mol

Putting values in equation 1, we get:

$\text{Moles of trans-cinnamic acid}=\frac{0.155g}{148.2g/mol}=0.00104mol$

• For pyridinium tribromide:

Given mass of pyridinium tribromide = 458 mg = 0.458 g

Molar mass of pyridinium tribromide = 319.82 g/mol

Putting values in equation 1, we get:

$\text{Moles of pyridinium tribromide}=\frac{0.458g}{319.82g/mol}=0.00143mol$

• The chemical equation for the reaction of trans-cinnamic acid and pyridinium tribromide follows:

$\text{trans-cinnamic acid}+\text{pyridinium tribromide}\xrightarrow[]{CH_3COOH}\text{2,3-dibromo-3-phenyl propanoic acid}$

Here, glacial acetic acid is considered as a buffer reagent so as to maintain the pH of the reaction.

By Stoichiometry of the reaction:

1 mole of trans-cinnamic acid reacts with 1 mole of pyridinium tribromide

So, 0.00104 moles of trans-cinnamic acid will react with = $\frac{1}{1}\times 0.00104=0.00104mol$ of pyridinium tribromide

As, given amount of pyridinium tribromide is more than the required amount. So, it is considered as an excess reagent.

Thus, trans-cinnamic acid is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of trans-cinnamic acid produces 1 mole of 2,3-dibromo-3-phenyl propanoic acid

So, 0.00104 moles of trans-cinnamic acid will produce = $\frac{1}{1}\times 0.00104=0.00104moles$ of 2,3-dibromo-3-phenyl propanoic acid

Now, calculating the mass of 2,3-dibromo-3-phenyl propanoic acid from equation 1, we get:

Molar mass of 2,3-dibromo-3-phenyl propanoic acid = 308 g/mol

Moles of 2,3-dibromo-3-phenyl propanoic acid = 0.00104 moles

Putting values in equation 1, we get:

$0.00104mol=\frac{\text{Mass of 2,3-dibromo-3-phenyl propanoic acid}}{308g/mol}\\\\\text{Mass of 2,3-dibromo-3-phenyl propanoic acid}=0.320g$

• To calculate the percentage yield of 2,3-dibromo-3-phenyl propanoic acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of 2,3-dibromo-3-phenyl propanoic acid = 0.217 g

Theoretical yield of 2,3-dibromo-3-phenyl propanoic acid = 0.320 g

Putting values in above equation, we get:

$\%\text{ yield of 2,3-dibromo-3-phenyl propanoic acid}=\frac{0.217g}{0.320g}\times 100\\\\\% \text{yield of 2,3-dibromo-3-phenyl propanoic acid}=67.81\%$

Hence, the theoretical yield of the product is 0.320 grams and its percent yield is 67.81 %