What mass of ethanoyl chloride (Mr 78.5) reacts exactly with 3.70 g of butan-1-ol (Mr 74.0), and which ester forms?
- Moles of butan-1-ol = 3.70 ÷ 74.0 = 0.0500 mol.
- One –OH group reacts with one molecule of ethanoyl chloride, so 0.0500 mol is needed.
- Mass of ethanoyl chloride = 0.0500 × 78.5 = 3.925 g.
- Ester: butyl (from butan-1-ol) ethanoate (from ethanoyl chloride), CH3COOCH2CH2CH2CH3.
Answer: 3.93 g of ethanoyl chloride (3 significant figures); the ester is butyl ethanoate.