Question

According to the experimental procedure of Experiment F1, 135 microliters of acetophenone (120.15 g/mol, 1.03 g/mL) was reacted with 127 mg of 4-nitrobenzaldehyde (151.12 g/mol). What is the theoretical yield, in milligrams (mg), of trans-4-nitrochalcone (253.25 g/mol)? Enter your answer as digits only (no units), using the proper number of significant figures.

Answer 1

Answer: The theoretical yield of 4-nitrochalcone is, $2.13\times 10^2$

Explanation : Given,

Volume of acetophenone = 135  microliters = 135 × 10⁻⁶ L = 0.135 mL

conversion used : (1 microliter = 10⁻⁶ L) and (1 L = 1000 mL)

Density of acetophenone = 1.03 g/mL

Mass of acetophenone = Density × Volume = 1.03 g/mL × 0.135 mL = 0.139 g

Mass of 4-nitrobenzaldehyde = 127 mg  = 0.127 g

Conversion used : (1 mg = 0.001 g)

First we have to calculate the moles of acetophenone and 4-nitrobenzaldehyde

$\text{Moles of acetophenone}=\frac{\text{Given mass acetophenone}}{\text{Molar mass acetophenone}}$

$\text{Moles of acetophenone}=\frac{0.139g}{120.15g/mol}=0.00116mol$

and,

$\text{Moles of 4-nitrobenzaldehyde}=\frac{\text{Given mass 4-nitrobenzaldehyde}}{\text{Molar mass 4-nitrobenzaldehyde}}$

$\text{Moles of 4-nitrobenzaldehyde}=\frac{0.127g}{151.12g/mol}=0.000840mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$C_8H_8O+C_7H_5NO_3\rightarrow C_{15}H_{11}NO_3$

From the balanced reaction we conclude that

As, 1 mole of 4-nitrobenzaldehyde react with 1 mole of acetophenone

So, 0.000840 mole of 4-nitrobenzaldehyde react with 0.000840 mole of acetophenone

From this we conclude that, acetophenone is an excess reagent because the given moles are greater than the required moles and 4-nitrobenzaldehyde is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of 4-nitrochalcone

From the reaction, we conclude that

As, 1 mole of 4-nitrobenzaldehyde react to give 1 mole of 4-nitrochalcone

So, 0.000840 mole of 4-nitrobenzaldehyde react to give 0.000840 mole of 4-nitrochalcone

Now we have to calculate the mass of 4-nitrochalcone

$\text{ Mass of 4-nitrochalcone}=\text{ Moles of 4-nitrochalcone}\times \text{ Molar mass of 4-nitrochalcone}$

Molar mass of 4-nitrochalcone = 253.25 g/mole

$\text{ Mass of 4-nitrochalcone}=(0.000840moles)\times (253.25g/mole)=0.21273g=212.73mg=2.13\times 10^2mg$

(1 g = 1000 g)

Therefore, the theoretical yield of 4-nitrochalcone is, $2.13\times 10^2mg$