Benzoic Acid Lab Report

[Chemical Engineering Lab Practice 4]

EXTRACTION OF BENZOIC ACID
Candidate’s Name: Raajeswari A/P Radahakrishnan
Student ID: SCM-023562
Group Member’s Name: Angela A/P Gunasegaran Shubashinee A/P Murugan

Lecturer/ Supervisor: Ms Nazlina & Dr Ong Chi Siang
Date of Submission: 13 October 2015

1.0 ABSTRACT

This experiment is about extraction of benzoic acid from kerosene by water droplets that was conducted on 6 October 2015 with few group of students. The main purpose of this experiment is to determine the overall mass transfer coefficient of benzoic acid between kerosene and water. In the introduction section, it is included about the information of benzoic acid and its uses as well as the extraction of benzoic acid. …show more content…
Benzoic acid is slightly soluble in water and it is widely used in food preservative. The main purpose of conducting this experiment is to determine the overall mass transfer coefficient of benzoic acid between kerosene and water.
Kerosene
(Saturated with Benzoic acid)

C*B

NB Figure 1.1

Figure 1.1 shows a schematic diagram of a drop of water moving through kerosene saturated with benzoic acid. Because of the concentration difference between the benzoic acid in the kerosene and a water droplet, benzoic acid is transferred across the droplet interface into the water. The molar flux of benzoic acid through the interface, NB (mol/m2•s) is given by: NB = KC ( )

Where C*B is the concentration of benzoic acid in kerosene at saturation, is the average concentration of benzoic acid in the aqueous phase and KC is the mass transfer coefficient. To determine KC, we need to know NB, C*B and . The experiment is designed to determine …show more content…
Sample Calculation Average time travelled for each droplet:

= (4.00+4.59+4.72+5.63+4.28+4.97+4.85+4.54+4.76+4.73)/10 = 4.71 s

Average volume for 50 droplets: = (2.2+1.6+1.8)/3 = 1.87 ml

Average volume for a droplet = 1.87ml/50 = 0.037ml = 0.037ml x (( 1x〖10〗^(-3) l ))/1ml x 1ml/1000l = 3.7x〖10〗^(-8) m^3

Average surface area of the droplet
V= 4/3 πr^3
3.7x〖10〗^(-8) m^3=4/3 πr^3 r^3=((3.7x〖10〗^(-8) m^3 x3))/4π r= ∛(9.549x〖10〗^(-9) m^3 ) r=2.067x〖10〗^(-3) m

Area=4πr^2
=4π(2.067x〖10〗^(-3) m)^2
=5.39x〖10〗^(-5) m^2

Concentration of benzoic acid
NaOH + C6H5COOH C6H5COONa + H2O
C6H5COOH = A NaOH = B
MB = 0.03mol 1dm3 dm3 0.001m3 = 30 mol/m3

MAVA = MBVB

MA = MBVB VA = 30 (7.1 × 10-6) 2.0 × 10-5 = 10.65 mol/m3
The concentration of benzoic acid in aqueous, CB = 10.65 mol/m3

Concentration of benzoic acid in kerosene
MA = MBVB VA = 30 (3.13 × 10-5) 1.0 × 10-5 = 93.9 mol/m3
The concentration of benzoic acid in aqueous, CB* = 93.9 mol/m3

Mole of benzoic acid transferred
= mV
= 10.65 mol/m3 (3.7x〖10〗^(-8) m^3)
= 3.94 ×