RESULT AND CALCULATION
Mass of solids
= 0.5 kg
Solvent flow rate
= 0.102 L/min
Solvent feed temperature =
30 °C
20
min
30
min
Filling
Effluent
Amount of
Volume
Volume
Density
Solute
(mL)
(mL)
(g/mL)
composition
solute (g)
Batch
Contact time
Solvent
time
1
37
3000
5.5
0.764
0.80
3.178
2
37
3000
7.5
0.772
0.86
3.604
(min)
During the experiment, we taking out some of the effluent volume to determine its density and
amount of solute at different contact time. By applying density formula as below:
Let take the volume of effluent at 20min is 5.5ml and the weight is 4.2g
Density P = Mass /Volume
P= 4.2g/5.5ml
=0.764g/ml
By referring calibration curve, the solute composition of density 0.764g/ml is approximately
0.80. Thus, we are able to calculate the amount of solute generate.
Since, 5.2ml x 0.764 = 3.973g of total mixture.
Amount Solute = Total Mixture X Solute Composition
= 3.973g x 0.80
= 3.178g
TABLE FOR CALIBRATION OF SOLUTE COMPOSITION
Solute composition
Mass of PO
Mass off n-Hex
(g)
(g)
0.00
1.00
0.000
0.653
4.86
48.54
0.100
0.668
6.13
30.66
0.200
0.682
16.86
56.77
0.297
0.696
23.41
58.67
0.399
0.715
32.17
64.33
0.500
0.725
46.53
77.55
0.600
0.744
67.18
96.11
0.699
0.754
54.19
67.99
0.797
0.762
79.9
88.65
0.900
0.783
100.23
100.82
1.00
0.795
(g oil / g
(g oil / g
solvent)
solution)
Density (g/mL)
Density Against Solute Compostion
0,9
DENSITY g/ml
0,85
0,783
0,8
0,744
0,75
0,715
0,7
0,653
0,668
0,682
0,754
0,795
0,762
0,725
0,696
0,65
0,6
0
0,2
0,4
0,6
0,8
1
1,2
Solute Compostion
Figure 1: Calibration Curve of Peanut Oil with Isopropanol
DATA ANALYSIS
The experiment conducted is to determine the properties of peanut oil (lipids) extracted using
0.5 kg of peanut as a soluble material with concentrated solution of isopropanol as solvent
using batch mode. For this experiment, 500 g of crushed peanut is measured, packaged and
placed in an extraction vessel which then is fed with 2500 ml of solvent isopropanol. Purpose
of isopropanol is to dissolve non-polar compounds which satisfies the criteria of dissolving oil
from solid compounds.
From the calibration curve of solute composition, we can analyze that the density is
increase approximate linearly when the solute composition is getting higher. The solute
composition is zero when only 100ml of isopropanol (solvent) is added and the density is
0.653g/ml. When the amount of solute which is peanut oil is added accordingly, the solute
composition started. When 100g of solute is added to 100ml of isopropanol, the density becomes
0.668g/ml and the solute composition is 1.00.
The density at 20min is less dense compared to 30min. The differences density between
20 and 30 min is 0.008g/ml. This probably due to the longer time of solvent-solute extraction
will result higher density of effluent being produced. This probably due to the longer time of
solvent-solute extraction will result higher density of effluent being produced. As we know, the
effluent is the peanut oil which extracted from the peanut. Besides, the amount of solute
produced will become higher when the time taken for extraction is longer. For example, the
amount of solute at 30 min is 3.604g while 20min would only generate 3.178g. Furthermore,
the volume at 20min is 5.5ml while 30min is 7.5ml which the difference is 2ml. Other than
volume, the solute composition is also increase from 0.80 to 0.86.