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?

Short answer

(a) The theoretical yield of bromobenzene is 60.3 g. (b) The percentage yield of bromobenzene is 70.2%.

Step-by-step solution

1. Write the balanced chemical equation

The balanced chemical equation is: \[ C_6H_6 + Br_2 \longrightarrow C_6H_5Br + HBr \]

2. Calculate the number of moles of reactants

We need the molar masses of benzene and bromine to convert the given mass to moles. Molar mass of benzene, \(C_6H_6 = 6 \times 12.01 + 6 \times 1.01 = 78.12\,\text{g/mol}\) Molar mass of bromine, \(Br_2 = 2 \times 79.90 = 159.80\,\text{g/mol}\) Now, find the moles of benzene and bromine. Moles of benzene = \(\frac{30.0\,\text{g}}{78.12\,\text{g/mol}} = 0.384\,\text{mol}\) Moles of bromine = \(\frac{65.0\,\text{g}}{159.80\,\text{g/mol}} = 0.407\,\text{mol}\)

3. Determine the limiting reactant

To determine the limiting reactant, we will compare the mole ratios of the reactants to the balanced equation. Mole ratio of benzene to bromine: \(\frac{0.384\,\text{mol}}{0.407\,\text{mol}} = 0.943\) From the balanced equation, the mole ratio of benzene to bromine should be 1:1. Since the ratio is less than 1 in our case, benzene is the limiting reactant.

4. Calculate the theoretical yield of bromobenzene

The balanced equation tells us that one mole of benzene reacts to produce one mole of bromobenzene. Thus, the theoretical yield of bromobenzene is equal to the moles of the limiting reactant (benzene). With the molar mass of bromobenzene, we can calculate its theoretical yield: Molar mass of bromobenzene, \(C_6H_5Br = 12.01\times 6+ 1.01\times 5 + 79.90 = 157.03\,\text{g/mol}\) Theoretical yield = moles of benzene \(\times\) molar mass of bromobenzene Theoretical yield = \(0.384\,\text{mol} \times 157.03\,\text{g/mol} = 60.3\,\text{g}\)

5. Calculate the percentage yield

Now we can calculate the percentage yield using the theoretical yield (60.3 g) and the actual yield (42.3 g): Percentage yield = \(\frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\) Percentage yield = \(\frac{42.3\,\text{g}}{60.3\,\text{g}} \times 100 = 70.2\%\) The percentage yield of bromobenzene is 70.2%.

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that determines the amount of product that can be formed. It is completely consumed first, stopping the reaction from continuing because no more product can be made once this reactant is gone. To find the limiting reactant, look at the mole ratio of reactants given in the balanced chemical equation. Compare these to the ratios of the actual moles you have in your reaction.
In our example, the reaction between benzene (\(C_6H_6\)) and bromine (\(Br_2\)) gives one mole of product, bromobenzene (\(C_6H_5Br\)), for each mole of benzene and bromine used. If you calculate the moles from the given mass of each reactant, you can determine which is the limiting reactant by seeing which has the smaller ratio compared to the product produced. In this case, benzene is the limiting reactant since its mole ratio is less than one when compared to bromine.

Mole Ratio

The mole ratio in a balanced chemical reaction allows us to predict the amount of product formed from a given amount of reactant or to determine the limiting reactant. This ratio is derived from the coefficients of the reactants and products in the balanced equation. Each coefficient tells you how many moles of a substance participate in the reaction.
For example, consider the balanced equation: \[ C_6H_6 + Br_2 \longrightarrow C_6H_5Br + HBr \] This equation shows a 1:1 ratio for benzene to bromine. For every mole of benzene, one mole of bromine is required to react fully. In practice, by knowing the moles of each reactant, you compare their composition to the ideal mole ratio in the equation. If the moles of benzene are less than the moles of bromine, then benzene is the limiting reactant and will determine the maximum amount of the product formed.

Percentage Yield

Percentage yield is a measure of the efficiency of a reaction. It compares the actual yield (the amount of product actually obtained from a reaction) to the theoretical yield (the amount predicted based on stoichiometry if everything reacted perfectly). This calculation helps gauge how close an experiment comes to the ideal reaction.
The formula to find the percentage yield is: \[ \text{Percentage yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100 \] In our scenario, the theoretical yield of bromobenzene was 60.3 g, but the actual yield was 42.3 g. By plugging these values into our formula, we find the percentage yield to be 70.2%. This indicates that 70.2% of the maximum possible bromobenzene was actually formed, showing some loss in the process likely due to incomplete reactions or side reactions.

Bromobenzene

Bromobenzene is a chemical compound formed when benzene reacts with bromine. It is an aromatic compound, characterized by a bromine atom substituting one of the hydrogen atoms in benzene. The presence of the bromine atom modifies the molecule's properties and makes it useful in organic synthesis and as an intermediate in the production of other chemicals.
Bromobenzene appears as a colorless liquid but carrying out the reaction with benzene and bromine requires careful handling because these reactants are volatile and can be hazardous. The calculation of the theoretical and percentage yield, as in our case, helps determine how much bromobenzene can be expected versus what was actually obtained. This way, chemists can optimize their processes to get as close as possible to theoretical yields.