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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

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Nitration of 1.70 g of methyl benzoate (M_r = 136.0) produces methyl 3-nitrobenzoate (M_r = 181.0). The percentage yield is 65.0%. What mass, in g, of methyl 3-nitr... show full transcript

Worked Solution & Example Answer: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

Step 1

Calculate the moles of methyl benzoate

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Answer

To find the moles of methyl benzoate, we use the formula:

moles=massMr\text{moles} = \frac{\text{mass}}{M_r}

Given that the mass is 1.70 g and the molar mass (M_r) is 136.0 g/mol:

moles of methyl benzoate=1.70 g136.0 g/mol=0.0125 moles\text{moles of methyl benzoate} = \frac{1.70 \text{ g}}{136.0 \text{ g/mol}} = 0.0125 \text{ moles}

Step 2

Determine the theoretical yield of methyl 3-nitrobenzoate

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Answer

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:

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

which gives:

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

Step 3

Calculate the actual yield of methyl 3-nitrobenzoate

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Answer

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

Actual yield=(Percentage yield100)×Theoretical yield\text{Actual yield} = \left( \frac{\text{Percentage yield}}{100} \right) \times \text{Theoretical yield}

Thus:

Actual yield=(65.0100)×2.28125 g≈1.47981 g\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.

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