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.
m(acetone)= 0.791 g
m(benzaldehyde) = 1.044 g
m(dibenzalacetone exp.) = 1.00 g
theoretical yield -?
percentage yield -?
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%.