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

Question

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

Asked on March 29, 2026Chemistry

This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.

Accepted Answer · 2026-03-29T15:19:35.895Z

a) The reaction between ethanol and ethanoic acid to produce ethyl ethanoate and water is an esterification reaction. $$\text{CH}_3\text{CH}_2\text{OH} (l) + \text{CH}_3\text{COOH} (l) \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 (l) + \text{H}_2\text{O} (l)$$ b) Step 1: Calculate the molar mass of ethanol ($\text{CH}_3\text{CH}_2\text{OH}$). Molar mass of C = $12.0 \text{ g/mol}$ Molar mass of H = $1.0 \text{ g/mol}$ Molar mass of O = $16.0 \text{ g/mol}$ $$M(\text{CH}_3\text{CH}_2\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}$$ Step 2: Calculate the moles of ethanol. $$\text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{23 \text{ g}}{46.0 \text{ g/mol}} = 0.50 \text{ mol}$$ Step 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. $$\text{Moles of ethyl ethanoate} = 0.50 \text{ mol}$$ Step 4: Calculate the molar mass of ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$ or $\text{C}_4\text{H}_8\text{O}_2$). $$M(\text{CH}_3\text{COOCH}_2\text{CH}_3) = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 48.0 + 8.0 + 32.0 = 88.0 \text{ g/mol}$$ Step 5: Calculate the theoretical yield (mass) of ethyl ethanoate. $$\text{Theoretical yield} = \text{Moles of ethyl ethanoate} \times \text{Molar mass of ethyl ethanoate}$$ $$\text{Theoretical yield} = 0.50 \text{ mol} \times 88.0 \text{ g/mol} = 44.0 \text{ g}$$ The theoretical yield of ethyl ethanoate is $\boxed{44.0 \text{ g}}$. c) Step 1: State the actual yield and theoretical yield. Actual yield = $33 \text{ g}$ Theoretical yield = $44.0 \text{ g}$ (from part b) Step 2: Calculate the percentage yield. $$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$$ $$\text{Percentage yield} = \frac{33 \text{ g}}{44.0 \text{ g}} \times 100\% = 0.75 \times 100\% = 75\%$$ The percentage yield is $\boxed{75\%}$. d) Ethyl ethanoate belongs to the homologous series of esters.

Accepted Answer · 2026-03-29T15:19:35.895Z

a) The reaction between ethanol and ethanoic acid to produce ethyl ethanoate and water is an esterification reaction. $$\text{CH}_3\text{CH}_2\text{OH} (l) + \text{CH}_3\text{COOH} (l) \rightleftharpoons \text{CH}_3\text{COOCH}_2\text{CH}_3 (l) + \text{H}_2\text{O} (l)$$ b) Step 1: Calculate the molar mass of ethanol ($\text{CH}_3\text{CH}_2\text{OH}$). Molar mass of C = $12.0 \text{ g/mol}$ Molar mass of H = $1.0 \text{ g/mol}$ Molar mass of O = $16.0 \text{ g/mol}$ $$M(\text{CH}_3\text{CH}_2\text{OH}) = (2 \times 12.0) + (6 \times 1.0) + (1 \times 16.0) = 24.0 + 6.0 + 16.0 = 46.0 \text{ g/mol}$$ Step 2: Calculate the moles of ethanol. $$\text{Moles of ethanol} = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}} = \frac{23 \text{ g}}{46.0 \text{ g/mol}} = 0.50 \text{ mol}$$ Step 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. $$\text{Moles of ethyl ethanoate} = 0.50 \text{ mol}$$ Step 4: Calculate the molar mass of ethyl ethanoate ($\text{CH}_3\text{COOCH}_2\text{CH}_3$ or $\text{C}_4\text{H}_8\text{O}_2$). $$M(\text{CH}_3\text{COOCH}_2\text{CH}_3) = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 48.0 + 8.0 + 32.0 = 88.0 \text{ g/mol}$$ Step 5: Calculate the theoretical yield (mass) of ethyl ethanoate. $$\text{Theoretical yield} = \text{Moles of ethyl ethanoate} \times \text{Molar mass of ethyl ethanoate}$$ $$\text{Theoretical yield} = 0.50 \text{ mol} \times 88.0 \text{ g/mol} = 44.0 \text{ g}$$ The theoretical yield of ethyl ethanoate is $\boxed{44.0 \text{ g}}$. c) Step 1: State the actual yield and theoretical yield. Actual yield = $33 \text{ g}$ Theoretical yield = $44.0 \text{ g}$ (from part b) Step 2: Calculate the percentage yield. $$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$$ $$\text{Percentage yield} = \frac{33 \text{ g}}{44.0 \text{ g}} \times 100\% = 0.75 \times 100\% = 75\%$$ The percentage yield is $\boxed{75\%}$. d) Ethyl ethanoate belongs to the homologous series of esters.