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A) The reaction between ethanol and ethanoic acid to produce ethyl ethanoate and water is an esterification reaction.

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2 parts
b

44.0 g44.0 \text{ g}

c

75%75\%

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Part a

The reaction between ethanol and ethanoic acid to produce ethyl ethanoate and water is an esterification reaction.

CH3CH2OH(l)+CH3COOH(l)⇌CH3COOCH2CH3(l)+H2O(l)\text{CH}_3\text{CH}_2\text{OH} (l) + \text{CH}_3\text{COOH} (l) \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 (l) + \text{H}_2\text{O} (l)

Part b

44.0 g44.0 \text{ g}

  1. Calculate the molar mass of ethanol (CH3CH2OH\text{CH}_3\text{CH}_2\text{OH}).

    Molar mass of C = 12.0 g/mol12.0 \text{ g/mol}
    Molar mass of H = 1.0 g/mol1.0 \text{ g/mol}
    Molar mass of O = 16.0 g/mol16.0 \text{ g/mol}

    M(CH3CH2OH)=(2×12.0)+(6×1.0)+(1×16.0)=24.0+6.0+16.0=46.0 g/molM(\text{CH}_3\text{CH}_2\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}

  2. Calculate the moles of ethanol.

    Moles of ethanol=Mass of ethanolMolar mass of ethanol=23 g46.0 g/mol=0.50 mol\text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{23 \text{ g}}{46.0 \text{ g/mol}} = 0.50 \text{ mol}

  3. Determine the moles of ethyl ethanoate produced.

    From the balanced chemical equation, 1 mole of ethanol reacts to produce 1 mole of ethyl ethanoate.

    Moles of ethyl ethanoate=0.50 mol\text{Moles of ethyl ethanoate} = 0.50 \text{ mol}

  4. Calculate the molar mass of ethyl ethanoate (CH3COOCH2CH3\text{CH}_3\text{COOCH}_2\text{CH}_3 or C4H8O2\text{C}_4\text{H}_8\text{O}_2).

    M(CH3COOCH2CH3)=(4×12.0)+(8×1.0)+(2×16.0)=48.0+8.0+32.0=88.0 g/molM(\text{CH}_3\text{COOCH}_2\text{CH}_3) = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 48.0 + 8.0 + 32.0 = 88.0 \text{ g/mol}

  5. Calculate the theoretical yield (mass) of ethyl ethanoate.

    Theoretical yield=Moles of ethyl ethanoate×Molar mass of ethyl ethanoate\text{Theoretical yield} = \text{Moles of ethyl ethanoate} \times \text{Molar mass of ethyl ethanoate} Theoretical yield=0.50 mol×88.0 g/mol=44.0 g\text{Theoretical yield} = 0.50 \text{ mol} \times 88.0 \text{ g/mol} = 44.0 \text{ g}

    The theoretical yield of ethyl ethanoate is 44.0 g\boxed{44.0 \text{ g}}.

Part c

75%75\%

  1. State the actual yield and theoretical yield.

    Actual yield = 33 g33 \text{ g}
    Theoretical yield = 44.0 g44.0 \text{ g} (from part b)

  2. Calculate the percentage yield.

    Percentage yield=Actual yieldTheoretical yield×100%\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% Percentage yield=33 g44.0 g×100%=0.75×100%=75%\text{Percentage yield} = \frac{33 \text{ g}}{44.0 \text{ g}} \times 100\% = 0.75 \times 100\% = 75\%

    The percentage yield is 75%\boxed{75\%}.

Part d

Ethyl ethanoate belongs to the homologous series of *esters

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