Question

In a 125-mL Erlenmeyer flask containing a magnetic stirring bar, mix 0.05 md benzaldehyde with the theoretical quantity of acetone, and add one-half the ture to a solution of 5 g of sodium hydroxide dissolved in 50 mL of water and 40 mL of ethanol at room temperature (<25°C). After 15 minutes of stirring ada remainder of the aldehyde-ketone mixture and rinse the container with a little ethanol to complete the transfer. Stir the mixture for 30 minutes; then collect the product by suction filtration on a Büchner funnel. Break the suction and carefully pour 100 mL of water on the product. Reapply the vacuum. Repeat this process

what is the theoretical yield and what is the percent yield?

Answer 1

The reaction is as shown below.

NaOH Acetone benzaldehyde dibenzalacetone 0.05 mol Molecular Weight: 234.30

As per the reaction stoichiometry, the conversion factor is $\frac{\text { 1 mol dibenzalacetone }}{\text { 1 mol benzaldehyde }}$

Molar mass of dibenzalacetone is 234.30 g/mol.

Theoretical yield of dibenzalacetone $=\text { 0.05 mol benzaldehyde } \times \frac{\text { 1 mol dibenzalacetone }}{ \text { 1 mol benzaldehyde }} \times \frac{\text { 234.30 g dibenzalacetone }}{ \text { 1 mol dibenzalacetone }}=\text { 11.72 g dibenzalacetone }$

Mass of Bowl + product

Mass of Bowl

Mass of product

Percentage yield $=\frac{3.1566 \ g}{11.72 \ g} \times 100 = 26.95 \%$