2. V. Chromatography
What is it?
• Chromatography = Color Writing
• Used to separate mixtures of organic and
inorganic compounds
• Discovered in 1903 by M. S. Tswett, a
Russian botanist who was able to
separate plant pigments
2
4. Uses of Chromatography
• Separation in prep. chemical synthesis
• Separation in drug analysis
• Separation of naturally occurring
compounds
• Results/outcomes:
Identification AND quantification of
compounds
4
5. How it works?
• All chromatographic methods have a
stationary and a mobile phase
• Stationary phase doesn’t move
• Mobile phase moves through the
stationary phase
• Separation depends on affinity of each
component to stationary or mobile phases
5
7. Types of Chromatography
7
1. Adsorption Chromatography:
A. According to Separation Mechanism:
Difference in adsorption
coefficients of solute
components and the
Stationary Phase.
Examples:
i. Liquid-Solid Chrom.
ii. Gas-Solid Chrom.
iii. Gas-Liquid Chrom.
8. 8
Types of Chromatography
8
A. According to Separation Mechanism:
2. Partition Chromatography:
Differences in Partition
Coefficients of solute
components between
Mobile and Stationary
phases.
Examples:
i. Liquid-Liquid Chrom.
ii. Gas-Liquid Chrom.
9. 9
Types of Chromatography
A. According to Separation Mechanism:
3. Ion-exchange Chromatography:
Differences in charges
of solute components,
e.g. electrolytes and
colloids.
Examples:
i. Electrophoresis
Chrom.
ii. Ion-exchange
Chrom.
10. 10
10
Types of Chromatography
A. According to Separation Mechanism:
4. Molecular exclusion Chromatography:
Difference in size of Ions
or Molecules of the
solute which diffuse in
sieve structure.
Examples:
i. Size Exclusion
Chrom.
ii. Gel-Filteration.
11. 11
Types of Chromatography
A. According to Separation Mechanism:
5. Affinity Chromatography:
Specific interactions between one kind of solute
molecule and a second molecule that is covalently
attached to the stationary phase.
Examples:
i. Protein
Separation
12. 12
Types of Chromatography
B. According to Mobile Phase:
1. Gas
Chromatography
i. Gas-liquid
Chromatography
(GLC)
ii. Gas-solid
Chromatography
(GSC)
13. 13
13
Types of Chromatography
B. According to Mobile Phase:
1. Gas
Chromatography
2. Liquid
Chromatography
3. Supercritical-
fluid
Chromatography
i. Column
Chromatography
ii. Planar
Chromatography
a. Liquid-liquid (Partition)
b. Liquid-solid (Adsorption)
c. Ion-exchange
d. Molecular exclusion
e. Affinity
f. Capillary electrophoresis
a. Thin layer
b. Paper
c. Slab electrophoresis
15. • Retention time, tr: time from sample
injection to center of the peak
• Adjusted retention time, tr’: retention time
(tr) – mobile phase retention time (tm);
tr’ = tr – tm
• Relative retention, , for two peaks:
• Unadjusted relative retention, :
=
tr2’
tr1’
15
=
tr2
tr1
16. 16
• Retention factor, k:
k =
tr - tm
tm
=
tr’
tm
=
CsVs
CmVm
Where:
Cs and Cm are the concentrations of
solute component in the stationary and
mobile phases
Vs and Vm are the volumes of the
stationary and mobile phases
K =
Cs
Cm
• Partition coefficient, K:
17. 17
Scaling up
Industrial scale
preparative column
Large mass
Small mass
(Large column radius)2
(Small column radius)2
=
Mass2
Mass1
(Radius2)2
(Radius1)2
=
k =
tr’
tm
=
Vs
Vm
K and =
tr2’
tr1’
=
k2
k1
=
K2
K1
23. ● Example 1:
Solvent passes through a column in 3
minutes, while solute requires 9 minutes.
(a) Calculate the adjusted retention time of
solute, tr’.
(b) Calculate the solute retention factor, k.
(c) If the volume of stationary phase is one-
tenth of the volume of the mobile phase in
the column (Vs = 0.1 Vm), find the partition
coefficient, K, for this system.
23
24. ● Solution:
(a) tr’ = tr - tm tr’ = 9 – 3 = 6 min.
(b) k =
tr’
tm
k =
6
3
= 2
(c) k = K
Vs
Vm
2 = K x 0.1
K = 20
24
25. ● Example 2:
Consider a chromatography experiment in
which 2 components with retention factors
=4 and =5 are injected into a column with
N = 1x103 theoretical plates. The retention
time for the less-retained component is 10
min.
(a) Calculate tm and find w1/2 and w for each
peak.
(b) Calculate the resolution for the two
peaks.
25
28. ; N =
5.55 tr
2
w1/2
2
103 =
5.55 x tr2
2
(w1/2)2
2
=
5.55 x 122
(w1/2)2
2
(w1/2)2
2
= 0.799 (w1/2)
2
= 0.894
; N =
16 tr
2
w 2
103 =
16 x tr1
2
w 2
1
=
16 x 102
w 2
1
w 2 = 1.6
1
w1 = 1.26 cm
28
29. ; N =
16 tr
2
w 2
103 =
16 x tr2
2
w 2
2
=
16 x 122
w 2
2
w 2 = 2.304
2
w = 1.52 cm
2
(b) ; resolution =
Dtr
wav
resolution =
12 - 10
(1.52 + 1.26)/2
= 1.44
29
30. ● Example 3:
Chromatograms of a mixture of A and B were obtained at the
same flow rate with two columns of equal length (see below):
i. Which column has more theoretical plates?
ii. Which column has a larger plate height?
iii. Which column gives higher resolution?
iv. Which column gives a greater relative retention?
v. Which compound has a higher capacity factor?
Column 1
Column 2
Column 1
Same
Cpd B
30
31. A mixture of benzene, toluene, and methane was injected into a gas chromatograph.
Methane gave a sharp spike in 42 s, whereas benzene required 251 s and toluene
was eluted in 333 s. Find the adjusted retention time and retention factor for each
solute, the relative retention, and the unadjusted relative retention.
The adjusted retention times are
The retention factors are
31
● Example 4:
33. 33
Ethylbenzene was eluted at 350 s. Find its retention factor
and the relative retention and unadjusted relative retention
for ethylbenzene and toluene.
(Answer: 7.33, 1.058, 1.051)
34. 34
What difference in retention times is required for an adequate
resolution of 1.5?
(Answer: 21.6 s)
