Esters — practice questions

(a) First, identify the structures of the reactants. Butan-1-ol is $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$ and ethanoic acid is $\text{CH}_3\text{COOH}$. The reaction removes the –OH from the carboxylic acid and the –H from the alcohol's hydroxyl group.

Displayed Formula:

      H   H   O    H   H   H   H
      |   |   ||   |   |   |   |
H--C--C--C--O--C--C--C--C--H
      |   |        |   |   |   |
      H   H        H   H   H   H

(b) The alcohol part is from butan-1-ol, so the first name is 'butyl'. The carboxylic acid part is from ethanoic acid, so the second name is 'ethanoate'.

Systematic Name: Butyl ethanoate

Step 1: Write the balanced equation. $\text{CH}_3\text{CH}_2\text{COOCH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{COONa} + \text{CH}_3\text{OH}$

Step 2: Calculate the molar mass ($M_r$) of the ester and the salt. $M_r(\text{methyl propanoate}) = (4 \times 12.0) + (8 \times 1.0) + (2 \times 16.0) = 88.0 \text{ g mol}^{-1}$ $M_r(\text{sodium propanoate}) = (3 \times 12.0) + (5 \times 1.0) + (2 \times 16.0) + 23.0 = 96.0 \text{ g mol}^{-1}$

Step 3: Calculate the moles of the ester. Moles = Mass / $M_r$ Moles of ester = 1.76 g / 88.0 g mol⁻¹ = 0.0200 mol

Step 4: Use the stoichiometry of the reaction. From the equation, the molar ratio of ester to salt is 1:1. Therefore, moles of sodium propanoate formed = 0.0200 mol.

Step 5: Calculate the mass of the salt. Mass = Moles × $M_r$ Mass of sodium propanoate = 0.0200 mol × 96.0 g mol⁻¹ = 1.92 g (to 3 s.f.)