Question

Pentyl Ethanoate, CH3COOC5H11, which smells like bananas, is produced from the esterification reaction:

CH3COOH(aq) +C5H11OH(aq) → CH3COOC5H11(aq) + H2O(l)

A reaction uses 3.58 g of CH3COOH and 4.75 g of C5H11OH and has a yield of 45.00%. Determine the mass of ester that forms.

Answer 1

Molar mass of CH3COOH,
MM = 2*MM(C) + 4*MM(H) + 2*MM(O)
= 2*12.01 + 4*1.008 + 2*16.0
= 60.052 g/mol


mass(CH3COOH)= 3.58 g

use:
number of mol of CH3COOH,
n = mass of CH3COOH/molar mass of CH3COOH
=(3.58 g)/(60.05 g/mol)
= 5.962*10^-2 mol

Molar mass of C5H11OH,
MM = 5*MM(C) + 12*MM(H) + 1*MM(O)
= 5*12.01 + 12*1.008 + 1*16.0
= 88.146 g/mol


mass(C5H11OH)= 4.75 g

use:
number of mol of C5H11OH,
n = mass of C5H11OH/molar mass of C5H11OH
=(4.75 g)/(88.15 g/mol)
= 5.389*10^-2 mol
Balanced chemical equation is:
CH3COOH + C5H11OH ---> CH3COOC5H11 + H2O


1 mol of CH3COOH reacts with 1 mol of C5H11OH
for 5.962*10^-2 mol of CH3COOH, 5.962*10^-2 mol of C5H11OH is required
But we have 5.389*10^-2 mol of C5H11OH

so, C5H11OH is limiting reagent
we will use C5H11OH in further calculation


Molar mass of CH3COOC5H11,
MM = 7*MM(C) + 14*MM(H) + 2*MM(O)
= 7*12.01 + 14*1.008 + 2*16.0
= 130.182 g/mol

According to balanced equation
mol of CH3COOC5H11 formed = (1/1)* moles of C5H11OH
= (1/1)*5.389*10^-2
= 5.389*10^-2 mol


use:
mass of CH3COOC5H11 = number of mol * molar mass
= 5.389*10^-2*1.302*10^2
= 7.015 g

% yield = actual mass*100/theoretical mass
45.0= actual mass*100/7.015
actual mass=3.157 g
Answer: 3.16 g