Problem 83 - Chemistry: The Central Science in SI Units, Global Edition

Question

When benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) reacts with bromine \(\left(\mathrm{Br}_{2}\right),\) bromobenzene \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\right)\) is obtained: $$ \mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}+\mathrm{HBr} $$ (a) When \(30.0 \mathrm{~g}\) of benzene reacts with \(65.0 \mathrm{~g}\) of bromine, what is the theoretical yield of bromobenzene? (b) If the actual yield of bromobenzene is \(42.3 \mathrm{~g}\), what is the percentage yield?

Step-by-Step Solution

Answer

The theoretical yield of bromobenzene is 60.2 g, and the percentage yield is approximately 70.3%.

Step 1: (a) Theoretical yield of bromobenzene

Step 1: Determine the moles of each reactant First, we need to find the number of moles of benzene and bromine. To do this, divide the mass by the molar mass of each substance. Number of moles of benzene = \(\frac{30.0 \ g}{78.11 \ g/mol} \approx0.384 \ mol\) Number of moles of bromine = \(\frac{65.0 \ g}{159.81 \ g/mol} \approx0.407 \ mol\) Step 2: Determine the limiting reactant For each reactant, calculate the mole ratio with respect to the balanced chemical equation: Mole ratio of benzene = \(\frac{0.384 \ mol}{1}\) Mole ratio of bromine = \(\frac{0.407 \ mol}{1}\) Since the mole ratio of benzene is smaller, it is the limiting reactant. Step 3: Calculate the theoretical yield of bromobenzene Use the given balanced chemical equation to calculate the number of moles of bromobenzene produced by 0.384 moles of benzene: \[\frac{\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br} \ moles}{\mathrm{C}_{6} \mathrm{H}_{6} \ moles} = \frac{1}{1}\] Number of moles of bromobenzene produced = \(0.384 \ mol\) Next, multiply the number of moles of bromobenzene produced by its molar mass to find the theoretical yield in grams. Theoretical Yield of bromobenzene = 0.384 mol × 157.01 g/mol = \(60.2 \ g\) So, the theoretical yield of bromobenzene is 60.2 g.

Step 2: (b) Percentage yield of bromobenzene

Step 1: Calculate the percentage yield Now, we'll use the actual yield and theoretical yield to calculate the percentage yield. Percentage Yield = \(\frac{\textrm{Actual Yield}}{\textrm{Theoretical Yield}}\times 100\%\) Percentage Yield = \(\frac{42.3 \ g}{60.2 \ g}\times 100\% \approx 70.3\%\) So, the percentage yield of bromobenzene is approximately 70.3%.

Key Concepts

Limiting Reactant

In any chemical reaction, reactants are converted into products. However, the amounts of products formed are often limited by one of the reactants, known as the limiting reactant. This concept is crucial as it determines the maximum output of a reaction.
To identify the limiting reactant, we first need to know the amounts of each reactant in moles. This is done by dividing the mass of each reactant by its molar mass. In the given exercise, benzene and bromine are the two reactants. By calculating their moles, we have approximately 0.384 moles for benzene and 0.407 moles for bromine.
Once we know the moles, we compare their ratios based on the balanced chemical equation. Here, the chemical equation shows a one-to-one ratio between benzene and bromine. Since benzene has fewer moles than bromine, it is the limiting reactant. This means that all of the benzene would react, but some bromine would remain unreacted. Understanding the limiting reactant helps predict the maximum amount of product that can be formed in the reaction.

Bromination Reaction

The bromination reaction is a process where bromine is added to an organic compound, which in this case, is benzene. This type of reaction is an example of halogenation, where a halogen is introduced into an organic molecule.
The specific reaction given is: \(\mathrm{C}_{6} \mathrm{H}_{6} + \mathrm{Br}_{2} \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br} + \mathrm{HBr} \). This reaction highlights the aromatic substitution where a hydrogen atom on the benzene ring is replaced by a bromine atom.
In such reactions, the aromaticity of benzene typically remains intact, leading to the production of bromobenzene. This mechanism involves the attack of the bromine molecule on the benzene ring, forming a temporary complex and eventually releasing hydrogen bromide (HBr) as a byproduct. The reaction is usually facilitated in the presence of a catalyst or under specific conditions to improve its efficiency. It illustrates how specific, well-understood concepts in organic chemistry help predict and control chemical transformations.

Percentage Yield

Calculating the percentage yield of a reaction is important because it compares the efficiency of an experimental process by showing how much of the theoretical yield is actually obtained.
Percentage yield is determined using the formula:

• \(\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100\%\)

In the exercise provided, the actual yield of bromobenzene obtained from the reaction is 42.3 g, while the theoretical yield was calculated to be 60.2 g.
Putting these values into the formula gives a percentage yield of approximately 70.3%.
This implies that only 70.3% of the product expected by theory was obtained in practice. Several factors could have caused this discrepancy, such as incomplete reactions, side reactions, or practical losses during the procedure. Understanding these factors and how to optimize conditions can help increase the yield in future reactions.