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 is assumed to yield 1 mole of methyl 3-nitrobenzoate for each mole of methyl benzoate used.

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

$\text{Theoretical yield} = \text{moles of methyl benzoate} \times M_r \text{ of methyl 3-nitrobenzoate}$

which gives:

$\text{Theoretical yield} = 0.0125 \text{ moles} \times 181.0 \text{ g/mol} = 2.28125 \text{ g}$

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

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

Thus:

$\text{Actual yield} = \left( \frac{65.0}{100} \right) \times 2.28125 \text{ g} \approx 1.47981 \text{ g}$

This rounds to 1.48 g, and the closest option provided is 1.47 g.