Question

Based on the use of 1.150 mol of isopentyl alcohol and 0.300 mol of acetic acid and an equilibrium constant value of 4.2. Estimate the mass of isopentyl acetate lost 1) due to incomplete reaction, 2) during the washing step, and 3) during the distillation. (Assume the solubility of the ester in aqueous sodium bicarbonate is equal to that in water, 0.25 g/100 mL).

Answer 1

1) Initial moles of isopentyl acetate = 1.150 mmol

Moles of acetic acid = 0.300 mol

equilibrium constant = 4.2

The ICE table will be:

	alcohol	acid	ester
initial	1.150	0.300	0
change	-x	-x	+x
Equilibrium	1.150-x	0.300-x	x

K = 4.2 = [ester]/[alc][acid]

4.2 = x/(1.150-x)(0.300-x)

4.2(1.150-x)(0.300-x) = x

4.2(0.345-1.450x+x²) = x

1.449 - 6.09x + 4.2 x² = x

4.2x² -7.09x + 1.449 = 0

x = 0.238 mmol

So, ester formed = x = 0.238 mmol

Isopentyl formed should be 0.300 mmol

So, moles of isopentyl acetate lost = 0.300-0.238 = 0.062 mmol =

Molar mass of isopentyl acetate = 130 g/mol = 0.130 g/mmol

mass of isopentyl acetate lost = mol*molar mass = 0.062*0.130 = 0.008 g

mass of isopentyl acetate lost = 0.008 g

2) during washing, solubility of ester is 0.25 g/100 ml

Mass of ester formed = 0.238*0.130 = 0.031 g

As the aqueous sodium bicarbonate is added initially 10 ml,

so, solubility = 0.25g/100 ml

in 10 ml, solubility = 0.025 g

So, in washing ester lost = 0.025 g

3) we are left with ester = 0.0031-0.025 = 0.006 g

So, in distillation, we will get 0.0006 g of ester.