Question

in the synthesis of propyl acetate 4ml of propanol and 4.5 ml of acetyl chloride was used what is the percent yield if a mass product of 3.04 grams ? was obtained

Answer 1

Given the reacting volume of propanol( HOCH2CH2CH3), V = 4 mL

Density of propanol, d = 0.803 g/mL

Hence mass of propanol( HOCH2CH2CH3) reacted = Vxd = 4 mL x 0.803 g/mL = 3.212 g

Molecular mass of propanol = 60.10 g/mol

Hence moles of propanol reacted = mass / molecular mass = 3.212 g / 60.10g/mol = 0.05344 mol

Given the reacting volume of acetyl chloride (CH3COCl), V = 4.5 mL

Density of acetyl chloride (CH3COCl), d = 1.104 g/mL

Hence mass of acetyl chloride (CH3COCl) reacted = Vxd = 4.5 mL x 1.104 g/mL = 4.968 g

Molecular mass of acetyl chloride (CH3COCl) = 78.49 g/mol

Hence moles of acetyl chloride (CH3COCl) reacted = mass / molecular mass

= 4.968g / 78.49g/mol = 0.06329 mol

The balanced reaction for the formation of propyl acetate from propanol and acetyl chloride is

CH3COCl + HOCH2CH2CH3 ------ > CH3COOCH2CH2CH3 + HCl

1 mol 1 mol 1 mol

Hence 1 mol of propanol reacts with 1 mol of acetyl chloride i.e the stoichiometric ratio of propanol to acetyl chloride is 1/1.

Hence 0.05344 mol of propanol that will react with the moles of acetyl chloride

= 0.05344 mol propanol x (1 mol acetyl chloride / 1 mol propanol) = 0.05344 mol acetyl chloride.

Hence propanol acts as a limiting reactant and is completely exhausted leaving the excess of acetyl chloride unreacted. Hence propanol will decide the amount of product, propyl acetate formed.

Hence moles of propyl acetate formed from 0.05344 mol propanol

= 0.05344 mol propanol x (1 mol propyl acetate / 1 mol propanol) = 0.05344 mol propyl acetate

Molecular mass of propyl acetate(CH3COOCH2CH2CH3) = 102.1 g/mol

Hence theoritical mass of propyl acetate formed = 0.05344 mol propyl acetate x (102.1 g/mol)

= 5.456 g  propyl acetate

Given that the actual mass of the product propyl acetate formed = 3.04 g

Hence Percent yield = (actual mass / theoritical mass)x100 = (3.04 / 5.456)x100 = 55.7 %