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
Question 29
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-nit... 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 number of moles of methyl benzoate
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Answer
To find the number of moles of methyl benzoate, use the formula:
n = rac{m}{M_r}
where:
m = 1.70 g (mass of methyl benzoate)
M_r = 136.0 g/mol (molar mass of methyl benzoate)
Calculating:
n = rac{1.70}{136.0} = 0.0125 ext{ mol}
Step 2
Determine the theoretical yield of methyl 3-nitrobenzoate
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Answer
Since the reaction produces 1 mole of methyl 3-nitrobenzoate per mole of methyl benzoate, the theoretical moles of methyl 3-nitrobenzoate produced is also 0.0125 mol.
Now, calculate the mass of methyl 3-nitrobenzoate using:
m=nimesMr
where M_r = 181.0 g/mol.
Calculating:
m=0.0125imes181.0=2.2625extg
Step 3
Adjust for percentage yield
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Answer
Since the percentage yield is 65.0%, the actual yield can be calculated as:
extActualyield=extTheoreticalyieldimesextpercentageyield
Using the theoretical yield from the previous step:
ext{Actual yield} = 2.2625 imes rac{65.0}{100} = 1.471625 ext{ g}
Rounding to two decimal places, the mass of methyl 3-nitrobenzoate produced is approximately 1.47 g.
