1. Exercises 5

2. 5.1) Describe the general elution problem
Exercise 5.1
5.1) Describe the general elution problem

3. Exercise 5.1 5.1) Describe the general elution problem
Elution patterns typically follow an exponential function
Too low eluent strength = too long time
“The general elution problem in chromatography”
Too high eluent strength = too low resolution

4. Exercise 5.1
The general elution problem is solved by applying a gradient of temperature in GC or of elution strength in LC
Solvent strength
Gradient elution:
The first compounds eluted with low eluent strength, the last compounds eluted with high eluent strength
Time
Gradient elution is
suitable for a large range of analyte properties
Detector response
Time

5. Exercise 5.2

6. Exercise 5.2
 A situation where k is not constant

7. Exercise 5.2
 A situation where k is not constant

8. Exercise 5.2
 A situation where k is not constant

9. Exercise 5.2
 A situation where k is not constant

10. Exercise 5.2  A situation where k is not constant
Valid, but not constant

11. Exercise 5.2  A situation where k is not constant
Valid, but not constant

12. Exercise 5.2
 A situation where k is not constant

13. Exercise 5.2
 A situation where k is not constant

14. Exercise 5.2  A situation where k is not constant
Part of the expression is valid but not the relationship involving k
Mathematically valid but not very useful

15. Exercise 5.2  A situation where k is not constant
Part of the expression is valid but not the relationship involving k
Mathematically valid but not very useful
Underlying factors depend on k
Underlying factors depend on k

16. Exercise 5.2  A situation where k is not constant
Part of the expression is valid but not the relationship involving k
Mathematically valid but not very useful
Underlying factors depend on k
Underlying factors depend on k

17. Exercise 5.2
 A situation where k is not constant 2

18. Exercise 5.2  A situation where k is not constant
The van Deemter eqation is not valid since it includes H, but the underlying relationships will be similar or identical as in isothermal chromatograpgy

19. Exercise 5.2  A situation where k is not constant
Strictly not valid since the van Deempter equation is not valid, but the same things that give low H will be good also in programmed chromatography

20. Exercise 5.2  A situation where k is not constant
The van Deemter eqation is not valid since it includes H, but the underlying relationships will be similar or identical as in isothermal chromatograpgy

21. Exercise 5.2
 A situation where k is not constant

22. Exercise 5.2
 A situation where k is not constant

23. Exercise 5.2
 A situation where k is not constant

24. Exercise 5.2
 A situation where k is not constant

25. Exercise 5.2
 A situation where k is not constant

26. Exercise 5.2
 A situation where k is not constant

27. Exercise 5.3
5.0 s
3.0 s
10%
3.4 s
5.6 s 5%

28. Exercise 5.3 wright,10% wleft,10% Af = w5% 2∙wleft,5% Tf = (Eq 24)

29. Exercise 5.3 wright,10% wleft,10% Af = w5% 2∙wleft,5% Tf =
(Eq 24)
Tf =
w5%
2∙wleft,5%
(Eq 25)
Af = 5.0 s / 3.0 s = 1.7
Tf = ( ) s / (2 ∙ 3.4) s = 1.3

30. Exercise 5.3 wright,10% wleft,10% Af = w5% 2∙wleft,5% Tf =
(Eq 24)
Tf =
w5%
2∙wleft,5%
(Eq 25)
5.0 s
3.0 s
10% 5%
5.6 s
3.4 s
Af = 3.0 s / 5.0 s = 0.60
Tf = ( ) s / (2 ∙ 5.6) s = 0.80

31. Exercise 5.3 wright,10% wleft,10% Af = w5% 2∙wleft,5% Tf =
(Eq 24)
Tf =
w5%
2∙wleft,5%
(Eq 25)
5.0 s
3.0 s
10% 5%
Both factors give value 1 for symmetric peaks
Fronting peaks have values below 1
Tailing peaks have values above 1
5.6 s
3.4 s
Af = 3.0 s / 5.0 s = 0.60
Tf = ( ) s / (2 ∙ 5.6) s = 0.80

32. Exercise 5.4

33. Exercise 5.4 Homologous series of alkanes.
Can be used to calculate retention indices:
n-pentane ≡ 500
n-hexane ≡ 600
n-heptane ≡ 700

34. Exercise 5.4 Homologous series of alkanes.
Can be used to calculate retention indices:
n-pentane ≡ 500
n-hexane ≡ 600
n-heptane ≡ 700
Differences between homologs z and z+1 are 100 retention index units

35. Exercise 5.4
Table of Kovats’ retention indices

36. Exercise 5.4
(Eq 22)
This is isothermal chromatography, so we have to apply the equation with log of adjusted retention times

37. Exercise 5.4 references 535 3.99 – 3.51 4.59 – 3.51
(Eq 22)
3.99 – 3.51
4.59 – 3.51
log 1.08 – log 0.48
log 4.83 – log 0.48
For compound A:
I = 100
+ 5 = 535
First reference has five carbons
8.34 – 3.51

38. Exercise 5.4 535 log 1.08 – log 0.48 log 4.83 – log 0.48
(Eq 22)
log 1.08 – log 0.48
log 4.83 – log 0.48
For compound A:
I = 100
+ 5 = 535

39. Exercise 5.4 references 535 635 log 10.82 – log 4.83
(Eq 22)
log – log 4.83
log – log 4.83
I = 100
+ 6 = 635
First reference has six carbons

40. Exercise 5.4 535 635 678 log 29.12 – log 4.83 log 48.33 – log 4.83
(Eq 22)
log – log 4.83
log – log 4.83
I = 100
+ 6 = 678

41. Exercise 5.4 691 535 635 678 log 39.29 – log 4.83 log 48.33 – log 4.83
(Eq 22)
log – log 4.83
log – log 4.83
I = 100
+ 6 = 691

42. Exercise 5.4
691
535
635
678
635
535
678
691

43. Exercise 5.5

44. Exercise 5.5

45. Exercise 5.5 A series of homologs
 we can calculate the separation number

46. The separation number:
Exercise 5.5
The separation number:
SN =
tR(z+1) – tR(z)
wh(z) + wh(z+1)
– 1
A series of homologs
 we can calculate the separation number

47. Exercise 5.5 SN = tR(z+1) – tR(z) wh(z) + wh(z+1) – 1 26.61 – 24.21
The separation number:
SN =
tR(z+1) – tR(z)
wh(z) + wh(z+1)
– 1
Calculated for the region between 18 and 19:
Program A
Program B
26.61 – 24.21
18.63 – 17.29
SN =
– 1 = 18.7
SN =
– 1 = 15.7
A has higher separation efficiency than than B
A series of homologs
 we can calculate the separation number

48. Exercise 5.5 A B SN = tR(z+1) – tR(z) wh(z) + wh(z+1) – 1
The separation number:
SN =
tR(z+1) – tR(z)
wh(z) + wh(z+1)
– 1
Calculated for the region between 18 and 19:
Program A
Program B
26.61 – 24.21
18.63 – 17.29
SN =
– 1 = 18.7
SN =
– 1 = 15.7
A has higher separation efficiency than than B 23 SN A B 18 13 8
Same pattern for the other chain lengths 3
Chain length 14 15 16 17 18 19 20 21

49. The separation number:
Exercise 5.5
The separation number:
SN =
tR(z+1) – tR(z)
wh(z) + wh(z+1)
– 1
This can also be solved using the peak capacity.
Use for instance the number of peaks between 14:0 and 20:0 (first and last peak)
(Eq. 21)

50. Exercise 5.5

51. Retention index for temperture programmed conditions
Exercise 5.5
Retention index for temperture programmed conditions
tR(i) – tR(z)
tR(z+1) – tR(z)
IT = B
+ z

52. Retention index for temperture programmed conditions
Exercise 5.5
Retention index for temperture programmed conditions
tR(i) – tR(z)
tR(z+1) – tR(z)
IT = B
+ z
For 18:1 n-9
Program A
Program B
25.14 – 24.21
26.61 – 24.21
17.85 – 17.30
18.63 – 17.30
IT = 1
+ 18
= 18.39
IT = 1
+ 18
= 18.41
The difference in retention indices between homologs is 1 in this case (ECL values)

53. Retention index for temperture programmed conditions
Exercise 5.5
Retention index for temperture programmed conditions
tR(i) – tR(z)
tR(z+1) – tR(z)
IT = B
+ z
For 18:1 n-9
Program A
Program B
25.14 – 24.21
26.61 – 24.21
17.85 – 17.30
18.63 – 17.30
IT = 1
+ 18
= 18.39
IT = 1
+ 18
= 18.41
Difference in retention index
18.41 – 18.39
18.40
∙ 100 % = 0.15 %
Difference in retention time
25.13 – 17.85
21.5
∙ 100 % = 33 %

54. Retention index for temperture programmed conditions
Exercise 5.5
Retention index for temperture programmed conditions
tR(i) – tR(z)
tR(z+1) – tR(z)
IT = B
+ z
For 18:1 n-9
Program A
Program B
25.14 – 24.21
26.61 – 24.21
17.85 – 17.30
18.63 – 17.30
IT = 1
+ 18
= 18.39
IT = 1
+ 18
= 18.41
Difference in retention index
Tells us that the selectivity is the same (which makes sense since the stationary phase is the same)
18.41 – 18.39
18.40
∙ 100 % = 0.15 %
Difference in retention time
25.13 – 17.85
21.5
∙ 100 % = 33 %
Tells nothing about selectivity

55. Retention index for temperture programmed conditions
Exercise 5.5
Retention index for temperture programmed conditions
tR(i) – tR(z)
tR(z+1) – tR(z)
IT = B
+ z
For 18:2 n-6
For 18:2 you do the same, but remember to change the references since it elutes between 19:0 and 20:0

57. 20 cm/s carrier gas velocity on a 25 m × 0
20 cm/s carrier gas velocity on a 25 m × 0.25 mm column requires an inlet pressure of 65 kPa (9.40 psi / bar)
The length may vary from to m
The diameter may vary from to mm.

58. Apply 65 kPa with these dimensions:
0.265
0.260
0.255
Diameter (mm)
0.250
0.245
0.240
0.235 23 24 25 26 27
Length (m)

59. Apply 65 kPa with these dimensions:
0.265
0.260
0.255
Diameter (mm)
20 cm/s
0.250
0.245
0.240
0.235 23 24 25 26 27
Length (m)

60. Apply 65 kPa with these dimensions:
0.265
0.260
23.3
21.1
0.255
Diameter (mm)
20.0 cm/s
0.250
0.245
19.1
17.3
0.240
0.235 23 24 25 26 27
Length (m)

61. Apply 65 kPa with these dimensions:
The carrier gas velocity is within 17.3 to 23.3 cm/s
0.265
0.260
23.3
21.1
0.255
Diameter (mm)
20.0 cm/s
0.250
0.245
19.1
17.3
0.240
0.235 23 24 25 26 27
Length (m)