Katelyn Algas
CHE 133-202
2/26/13
Experiment 5: Gas Chromatography and Raoult’s Law
Purpose
The purpose of this experiment was to determine whether or not a solution follows Raoult’s
law by determining the composition of both the condensed and vapor phases of the solution. A
gas chromatograph was used to determine the composition of the phases. Example data involving
acetone and DCM were used to show the liquid data and the vapor data and to show how to
perform the calculations needed to create a vapor pressure plot of the data. Then in the lab,
hexane and pentane were used, and a vapor pressure plot was created from the calculations of the
experimental data. The calculations that were performed were the moles per unit area of each
pure liquid component, the moles of each component in each mixture, the mole fraction of
component A for each mixture, the pure vapor pressure of components A and B, the pressure per
unit area of each pure vapor component, and the partial pressures of compounds A and B.
Procedure
CHE 133 Experiment 5, General Chemistry Lab II, Winter Quarter 2012-2013, DePaul
University. [Online] https://www.d2l.depaul.edu. February 18, 2013.

There were no changes made to the experiment.
Data and Results
Below are the data and results for the example data, as well as the data and results for the
experimental data. With these, vapor pressure plots were created to see whether or not the
solutions follow Raoult’s law.
2
Example Data
Table 1: Given Data for Components A and B
Liquid
Density
Molar
Integrated
Vapor
injection
(g/mL)
mass
peak area of pressure
volume
(g/mol)
pure liquid (mmHg)
(mL)
2.00×10-4
0.791
58.08004
108428.8
229.5
Integrated
peak area
of pure
vapor
1617.8
Pure
Acetone
(A)
2.00×10-4
1.325
84.93288
59457.2
429.5
3190.7
Pure
DCM
(B)
Table 1 shows the given data needed to perform the calculations. The calculations include the
number of moles of acetone and DCM, the total number of moles, the mole fractions of acetone
and DCM, the partial pressures of acetone and DCM, and the total pressure.
Table 2: Liquid Phase Data
Area A
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Area B
108428.8
99164.6
85637.7
75500.9
65854.4
54991.4
47531.6
36200.0
24431.0
12842.8
5727.8
10156.4
15521.4
20819.9
24797.3
32165.1
37672.5
42940.0
49725.5
59457.2
Moles of
A (mol)
Moles of
B (mol)
2.72E-06
2.49E-06
2.15E-06
1.90E-06
1.65E-06
1.38E-06
1.19E-06
9.09E-07
6.14E-07
3.23E-07
0.00E+00
0.00E+00
3.01E-07
5.33E-07
8.15E-07
1.09E-06
1.30E-06
1.69E-06
1.98E-06
2.25E-06
2.61E-06
3.12E-06
Total
number of
moles
(mol)
2.72E-06
2.79E-06
2.68E-06
2.71E-06
2.75E-06
2.68E-06
2.88E-06
2.89E-06
2.87E-06
2.93E-06
3.12E-06
Mole
fraction of
A
Mole
fraction of
B
1.00E+00
8.92E-01
8.01E-01
7.00E-01
6.02E-01
5.15E-01
4.14E-01
3.15E-01
2.14E-01
1.10E-01
0.00E+00
0.00E+00
1.08E-01
1.99E-01
3.00E-01
3.98E-01
4.85E-01
5.86E-01
6.85E-01
7.86E-01
8.90E-01
1.00E+00
Table 2 shows the data and calculations of liquid acetone and DCM. The colored boxes are the
integrated peak areas of acetone (A) and DCM (B).
Table 3: Vapor Phase Data
Area A
1.0
0.9
0.8
0.7
0.6
1617.8
1622.4
1148.6
1187.4
851.5
Area B
158.3
293.2
569.7
838.8
Partial
pressure of A
(mmHg)
229.5
230.2
163.0
168.5
120.8
Partial
pressure of
B (mmHg)
0.0
21.3
39.5
76.7
112.9
Total pressure
(mmHg)
229.5
251.5
202.4
245.2
233.7
3
0.5
0.4
0.3
0.2
0.1
0.0
645.1
408
369.3
150
69.1
1184.3
1403.3
2134.5
2177.5
2728.8
3190.7
91.5
57.9
52.4
21.3
9.8
0.0
159.4
188.9
287.3
293.1
367.4
429.5
251.0
246.8
339.7
314.4
377.2
429.5
Table 3 shows the data and calculations of acetone and DCM in the vapor phase. The colored
boxes are the integrated peak areas of acetone (A) and DCM (B).
Figure 1: Vapor Pressure Plot for Acetone and DCM
500
Pressure (mmHg)
400
300
Acetone
200
DCMM
Total
100
0
0
-100
0.2
0.4
0.6
0.8
1
1.2
Mole Fraction Acetone
Figure 1 shows the partial pressures of acetone and DCM and the total vapor pressure. Together,
they are a function of the mole fraction of acetone in the solution. The legend of the graph
distinguishes the partial pressures from the total vapor pressure.
Gas Chromatography Data
Pure
Pentane
(A)
Pure
Hexane
(B)
Table 4: Given Data for Components A and B
Liquid
Density
Molar
Integrated
Vapor
injection
(g/mL)
mass
peak area of pressure
volume
(g/mol)
pure liquid (mmHg)
(mL)
2.00×10-4
0.683
72.15028
417171.0
514.0
2.00×10-4
0.659
86.11116
115263.0
150.0
Integrated
peak area
of pure
vapor
6520.4
1834.0
4
Table 4 shows the given data needed to perform the calculations. The calculations include the
number of moles of pentane and hexane, the total number of moles, the mole fractions of pentane
and hexane, the partial pressures of pentane and hexane, and the total pressure.
Table 5: Liquid Phase Data
Moles of A
(mol)
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Area A
Area B
417171.0
153405.0
23583.0
134633.0
29861.0
110752.0
39669.0
90360.0
49156.0
76276.0
61153.0
50217.0
53190.0
51839.0
75584.0
33247.0
88345.0
1094.0
3029.0
115263.0
1.89E-06
6.96E-07
6.11E-07
5.03E-07
4.10E-07
3.46E-07
2.28E-07
2.35E-07
1.51E-07
4.96E-09
0.00E+00
Moles of
B (mol)
0.00E+00
3.13E-07
3.97E-07
5.27E-07
6.53E-07
8.12E-07
7.06E-07
1.00E-06
1.17E-06
4.02E-08
1.53E-06
Total
number of
moles (mol)
1.89E-06
1.01E-06
1.01E-06
1.03E-06
1.06E-06
1.16E-06
9.34E-07
1.24E-06
1.32E-06
4.52E-08
1.53E-06
Mole
fraction of
A
1.00E+00
6.90E-01
6.06E-01
4.88E-01
3.86E-01
2.99E-01
2.44E-01
1.90E-01
1.14E-01
1.10E-01
0.00E+00
Mole
fraction of
B
0.00E+00
3.10E-01
3.94E-01
5.12E-01
6.14E-01
7.01E-01
7.56E-01
8.10E-01
8.86E-01
8.90E-01
1.00E+00
Table 5 shows the data and calculations of liquid pentane (A) and hexane (B). The colored boxes
are the integrated peak areas of pentane (A) and hexane (B).
Table 6: Vapor Phase Data
Area A
1.0
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0.0
6520.4
4893.3
5961.7
5719.1
5120.0
3769.4
4031.7
2821.0
1618.2
103.2
Area B
215.9
335.3
501.1
711.4
785.9
1093.3
1068.6
1206.1
1837.0
1834.0
Partial
pressure of A
(mmHg)
514.0
385.7
470.0
450.8
403.6
297.1
317.8
222.4
127.6
8.1
0.0
Partial
pressure of B
(mmHg)
0.0
17.7
27.4
41.0
58.2
64.3
89.4
87.4
98.6
150.2
150.0
Total pressure
(mmHg)
514.0
403.4
497.4
491.8
461.8
361.4
407.2
309.8
226.2
158.4
150.0
Table 6 shows the data and calculations of pentane and hexane in the vapor phase. The colored
boxes are the integrated peak areas of pentane (A) and hexane (B).
5
Figure 2: Vapor Pressure Plot for Pentane and Hexane
700
600
Pressure (mmHg)
500
400
Pentane
300
Hexane
200
Total
100
0
0
-100
0.2
0.4
0.6
0.8
1
1.2
Mole Fraction Pentane
Figure 2 shows the partial pressures of pentane and hexane and the total vapor pressure, and
together, they are a function to the mole fraction of pentane in the solution. The legend of the
graph distinguishes the partial pressures from the total vapor pressure.
Sample Calculations
Moles per unit area for each pure liquid component
The moles per unit area for each pure liquid component was determined by first calculating the
moles of species i per unit area (C*i,l):
𝑉𝜌
𝑙 𝑖
C*i,l = 𝑀 𝐴∗
𝑖
𝑖,𝑙
The Vl is the volume of the liquid sample injected into the gas chromatograph (mL), ρi is the
density of the liquid (g/mL), Mi is the molar mass, and A*i,l is the integrated peak area of the pure
compound reported by the gas chromatograph.
𝑔
(2.00 × 10−4 𝑚𝐿)(0.683
)
𝑚𝐿 = 4.54 × 10−12 𝑚𝑜𝑙 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎
𝑃𝑒𝑛𝑡𝑎𝑛𝑒:
𝑔
(72.15028
) (417171.0)
𝑚𝑜𝑙
Moles of each component in each mixture
The number of moles in each mixture component was calculated by multiplying the integrated
peak area for each component of the mixture, Ai,l, by the correct value of C*i,l.
𝑛𝑖,𝑙 = 𝐴𝑖,𝑙 𝐶 ∗𝑖,𝑙
6
𝑃𝑒𝑛𝑡𝑎𝑛𝑒: (417171.0)(4.54 × 10−12 𝑚𝑜𝑙) = 1.89 × 10−6 𝑚𝑜𝑙
Mole fraction of component A for each mixture
Once the number of moles was calculated, the mole fraction of component A for each mixture
was determined by using this equation:
𝑋 𝑙𝐴 =
𝑃𝑒𝑛𝑡𝑎𝑛𝑒:
𝑛𝐴,𝑙
𝑛𝐴,𝑙 + 𝑛𝐵,𝑙
1.89 × 10−6 𝑚𝑜𝑙
= 1.00
(1.89 × 10−6 𝑚𝑜𝑙) + (0.00 𝑚𝑜𝑙)
Pure vapor pressure of compounds A and B
The vapor pressures of pure compounds A and B were given. These were assigned to P*A and
P*B.
P*A = 514.0 mmHg
P*B = 0.00 mmHg
Pressure per unit area for each vapor component
The pressure per unit area for each vapor component was calculated by dividing P*i by A*i,v.
𝐶 ∗𝑖,𝑣 =
𝑃𝑒𝑛𝑡𝑎𝑛𝑒 =
𝑃 ∗𝑖
𝐴 ∗𝑖,𝑣
514.0 𝑚𝑚𝐻𝑔
= 7.883 × 10−2 𝑚𝑚𝐻𝑔 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑎𝑟𝑒𝑎
6520.4
Partial pressures of compounds A and B
The partial pressures of compounds A and B were calculated by multiplying the area under each
vapor peak by the pressure per unit area.
𝑃𝐴 = 𝐴𝐴,𝑣 𝐶 ∗𝐴𝑣
𝑃𝑒𝑛𝑡𝑎𝑛𝑒 = (6520.4)(7.883 × 10−2 𝑚𝑚𝐻𝑔) = 514.0 𝑚𝑚𝐻𝑔
7
Discussion
Based on the experimental results, the system made up of pentane and hexane did not
follow Raoult’s Law. One possible explanation would be the sources of error involved. One
source of error could be that each vial did not have the same amount of pentane and hexane.
Perhaps when one group was mixing their vial, some of the solution came out, or when the
pentane and hexane were being put into the vial, some of each compound dissolved in the
process, since both compounds are volatile.
Intermolecular forces affect solutions. There are intermolecular forces between the solute
and the solvent when they are mixed together. The solute and the solvent by themselves also
have intermolecular forces between each other. The best or ideal way intermolecular forces
affect solutions is when the forces between the solute and solvent are about the same as the
forces between the solute particles and the solvent particles. The vapor pressure plot of acetone
and DCM is a perfect example. This showed how Raoult’s Law is true. However, not all
solutions follow it. Raoult’s Law occurs when there is a nonvolatile solute present. In this
experiment with pentane and hexane, it was mentioned that pentane and hexane were volatile,
which was why the experimental results did not agree with Raoult’s law and the plot does not
look like the plot for the example data.