Consider the above reaction where 6.1 g of benzoic acid is used to get 7.8 g of m-bromo benzoic acid. The percentage yield of the product is ___.
(Round off to the Nearest integer)
[Given : Atomic masses : C = 12.0 u, H : 1.0 u, O : 16.0 u, Br = 80.0 u]

To determine the percentage yield, first calculate the molar masses of the reactant and product.

For benzoic acid:

$$\mathrm{C_7H_6O_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (6\times1) + (2\times16)}$$

$$\mathrm{= 84 + 6 + 32 = 122\ g\ mol^{-1}}$$

For m-bromobenzoic acid:

$$\mathrm{C_7H_5BrO_2}$$

$$\mathrm{Molar\ Mass = (7\times12) + (5\times1) + 80 + (2\times16)}$$

$$\mathrm{= 84 + 5 + 80 + 32 = 201\ g\ mol^{-1}}$$

The balanced reaction shows a $$\mathrm{1:1}$$ molar ratio between benzoic acid and m-bromobenzoic acid.

Moles of benzoic acid used:

$$\mathrm{Moles = \frac{6.1}{122}}$$

$$\mathrm{= 0.05\ mol}$$

Therefore,

$$\mathrm{0.05\ mol\ Benzoic\ Acid \rightarrow 0.05\ mol\ Product}$$

Theoretical yield of product:

$$\mathrm{Mass = Moles \times Molar\ Mass}$$

$$\mathrm{= 0.05 \times 201}$$

$$\mathrm{= 10.05\ g}$$

Actual yield obtained:

$$\mathrm{7.8\ g}$$

Percentage yield:

$$\mathrm{\%\ Yield = \frac{Actual\ Yield}{Theoretical\ Yield}\times100}$$

$$\mathrm{= \frac{7.8}{10.05}\times100}$$

$$\mathrm{= 77.61\%}$$

Rounding to the nearest integer:

$$\boxed{\mathrm{78}}$$