When 4.50 g of ethanoic acid and 10.0 g of ethanol were allowed to reach equilibrium at60°C, the equilibrium mixture was found to contain 4.40 g of ethyl ethanoate. Calculate thepercentage yield of ethyl ethanoate, based on ethanoic acid. Show your working wherenecessary.
My method
Theoretical
CH3COOH + CH3CH3OH ===== CH3COOCH2CH3 + H2O
mol CH3COOH - 0.075
mol CH3CH2OH - 5/23
mol CH3COOCH2CH3 formed - 5/23
mass CH3COOCH2CH3 formed - 88 x 5/23
Therefore % yield is 4.4/mass above x 100 = 23%
What have I done wrong?