Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene. CH₃COCl + C₆H₆ → C₆H₅C(O)CH₃ + HCl In a preparation, with an excess of benzene, the mass of ethanoyl chloride used (Mₑ = 78.5) was 5.7 × 10⁻² kg. The percentage yield of phenylethanone was 62%. What mass, in grams, of phenylethanone was produced? [1 mark]

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Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene - AQA - A-Level Chemistry - Question 33 - 2017 - Paper 3

Question 33

Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene. CH₃COCl + C₆H₆ → C₆H₅C(O)CH₃ + HCl In a preparation, with an excess of benzen... show full transcript

Worked Solution & Example Answer:Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene - AQA - A-Level Chemistry - Question 33 - 2017 - Paper 3

Step 1

Calculate the moles of ethanoyl chloride used

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Answer

To find the moles of ethanoyl chloride, use the formula:

$\text{moles} = \frac{\text{mass}}{\text{molar mass}}$

Given the mass = 5.7 × 10⁻² kg = 57 g and the molar mass (Mₑ) = 78.5 g/mol, we find:

$\text{moles of C₂H₅COCl} = \frac{57 \text{ g}}{78.5 \text{ g/mol}} \approx 0.726 \text{ moles}$

Step 2

Calculate the theoretical yield of phenylethanone

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Answer

The theoretical yield of phenylethanone is equal to the moles of ethanoyl chloride, since the reaction is a 1:1 conversion.

So, the theoretical yield in grams is:

$\text{mass} = \text{moles} \times \text{molar mass}$

The molar mass of phenylethanone (C₈H₈O) is approximately 120 g/mol, hence:

$\text{mass}_{\text{theoretical}} = 0.726 \text{ moles} \times 120 \text{ g/mol} = 87.1 \text{ g}$

Step 3

Calculate the actual yield based on percentage yield

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Answer

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

$\text{actual yield} = \left(\frac{\text{percentage yield}}{100}\right) \times \text{theoretical yield}$

Given the percentage yield is 62%, we have:

$\text{actual yield} = \left(\frac{62}{100}\right) \times 87.1 \text{ g} \approx 54.05 \text{ g}$

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Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene - AQA - A-Level Chemistry - Question 33 - 2017 - Paper 3

Worked Solution & Example Answer:Phenylethanone can be prepared by the reaction between ethanoyl chloride and benzene - AQA - A-Level Chemistry - Question 33 - 2017 - Paper 3

Calculate the moles of ethanoyl chloride used

Calculate the theoretical yield of phenylethanone

Calculate the actual yield based on percentage yield

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