3.0 g of ethanoic acid reacts with excess ethanol and a little concentrated H2SO4. The yield of ethyl ethanoate is 60%. What mass of ester is obtained? (H = 1, C = 12, O = 16)
- CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O, a 1 : 1 ratio. Moles of acid = 3.0 ÷ 60 = 0.050 mol.
- Mr of ethyl ethanoate, C4H8O2 = 48 + 8 + 32 = 88, so the maximum mass = 0.050 × 88 = 4.4 g.
- Actual mass = 60% of 4.4 = 0.60 × 4.4 = 2.64 g.
Answer: 2.64 g of ethyl ethanoate (about 2.6 g)