CH3COOH + HO-CH2-CH3 ⇒ CH3COOCH2-CH3 + H2O
M ester = 88g/mol ⇒ cant. practica de ester = 0,1 moli ⇒
⇒ cant.teoret. = 0,1·100/75 = 2/15 moli ester
au fost necesari 2/15 moli acid (2/15 ·60 = 8g) si 2/15 moli alcool (2/15 ·46 = 92/15 g)
initial : ms = 100g ( 92/15 ≈ 6,13g etanol + 8g acid acetic + 85,87 g HOH)