Nitration of 1.70 g of methyl benzoate (M_r = 136.0) produces methyl 3-nitrobenzoate (M_r = 181.0) - AQA - A-Level Chemistry - Question 29 - 2021 - Paper 3

The reaction produces 1 mole of methyl 3-nitrobenzoate for every mole of methyl benzoate.

Thus, the theoretical yield in grams can be calculated as:

$\text{Theoretical yield} (g) = n \times M$

Where:

• n = number of moles of methyl benzoate = 0.0125 mol
• M = molar mass of methyl 3-nitrobenzoate = 181.0 g/mol

Calculating:

$\text{Theoretical yield} = 0.0125 \, \text{mol} \times 181.0 \, \text{g/mol} \approx 2.26 \, \text{g}$

The actual yield can be determined using the percentage yield formula:

$\text{Actual yield} = \left( \frac{\text{Percentage yield}}{100} \right) \times \text{Theoretical yield}$

Given that the percentage yield is 65.0%:

$\text{Actual yield} = \left( \frac{65.0}{100} \right) \times 2.26 \, \text{g} \approx 1.47 \, \text{g}$

Thus, the mass of methyl 3-nitrobenzoate produced is approximately 1.47 g.