Answer to Question #350738 in Organic Chemistry for tenyears

Question

In an experiment, 0.791 g of acetone reacts with 1.044 g of benzaldehyde to produce experimental 1.00 g of dibenzalacetone. Calculate the theoretical and percentage yield.

Answer 1

Given:

m(acetone)= 0.791 g

m(benzaldehyde) = 1.044 g

m(dibenzalacetone exp.) = 1.00 g

Find:

theoretical yield -?

percentage yield -?

Solution:

The mentioned reaction between acetone and benzaldehyde:

The same reaction just with molecular formulas:

C₃H₆O (acetone) + 2C₇H₆O (benzaldehyde) $\to$ C₁₇H₁₄O (dibenzalacetone)

M(C₃H₆O) = 58.08 g/mol

M(C₇H₆O) = 106.12 g/mol

M(C₁₇H₁₄O) = 234.29 g/mol

How many moles of reagents were used for the reaction?:

n(C₃H₆O) = m(C₃H₆O)/M(C₃H₆O) = 0.791 g/58.08 g/mol = 0.01362 mol

n(C₇H₆O) = m(C₇H₆O)/M(C₇H₆O) = 1.044g/106.12 g/mol = 0.009838 mol

According to the reaction for 1 mole of acetone, it is needed 2 moles of benzaldehyde.

In our case 0.01362 mol x 2 = 0.02724 mol that is much more than 0.009838. Therefore here there is the excess acetone and we have to use n(C₇H₆O) = 0.009838 mol for further calculations.

n(C₁₇H₁₄O theoretical) = 0.5 n(C₇H₆O) = 0.5 x 0.009838 mol = 0.004919 mol

theoretical yield = m(C₁₇H₁₄O theoretical) = n(C₁₇H₁₄O theoretical) x M(C₁₇H₁₄O) =

0.004919 mol x 234.29 g/mol = 1.152 g

percentage yield = $\eta$(C₁₇H₁₄O) = [m(dibenzalacetone exp.)/m(C₁₇H₁₄O theoretical)]x100% =

[1.00 g/1.152 g] x 100% = 86.8%

Answer: theoretical yield = 1.152 g, percentage yield = 86.8%.