1. Identify the reaction stoichiometry

Benzoic acid undergoes bromination to form m-bromobenzoic acid.

$\text{Benzoic acid} \rightarrow \text{m-bromobenzoic acid}$

Since one hydrogen on the ring is replaced by one bromine atom, the mole ratio is:

$1 : 1$

2. Find molar masses

• Benzoic acid: $\mathrm{C_7H_6O_2}$

$M = 7(12) + 6(1) + 2(16) = 84 + 6 + 32 = 122\,\text{g mol}^{-1}$

• m-Bromobenzoic acid: $\mathrm{C_7H_5BrO_2}$

$M = 7(12) + 5(1) + 80 + 2(16) = 84 + 5 + 80 + 32 = 201\,\text{g mol}^{-1}$

3. Calculate moles of benzoic acid used

Given mass of benzoic acid = $6.1\,\text{g}$

$n = \frac{6.1}{122} = 0.05\,\text{mol}$

4. Calculate theoretical yield of m-bromobenzoic acid

Using $1:1$ mole ratio:

$0.05\,\text{mol benzoic acid} \Rightarrow 0.05\,\text{mol m-bromobenzoic acid}$

So theoretical mass of product is:

$m_{\text{theoretical}} = 0.05 \times 201 = 10.05\,\text{g}$

5. Calculate percentage yield

Actual mass obtained = $7.8\,\text{g}$

$\%\text{ yield} = \frac{7.8}{10.05} \times 100$

$\%\text{ yield} = 77.61\%$

Rounded to nearest integer:

$\boxed{78}$

6. Comparison with stored answer

Stored correct answer = $78$

Our derived answer matches the stored answer.