CHAPTER 5a: PRINCIPLES AND BASIC THEORY OF CHROMATOGRAPHY
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Transcript of CHAPTER 5a: PRINCIPLES AND BASIC THEORY OF CHROMATOGRAPHY
CHAPTER 5a:PRINCIPLES AND BASIC THEORY OF CHROMATOGRAPHY
Department of ChemistryFaculty of Science
Universiti Teknologi Malaysia
ANALYTICAL CHEMISTRYSSC 1293
The word “chromatography” originated from two Greek words, chroma which means “color” and graphy which means “writing”.
Chromatography was founded in 1906 by a Russian botanist, Micheal Tswett, when he separated plant extracts on a column packed with finely divided calcium carbonate.
Chromatography
Chromatography is a method to separate two or more compounds in a mixture based on the differences in the property of each individual substance.
The properties include polarity, solubility, ionic strength, and size.
Chromatography is today one of the most useful analytical methods for separation, identification, and quantitation of chemical compounds.
Purpose of Chromatography
In chromatography, the compounds are physically separated by distributing themselves between two phases: (a) a stationary phase(b) a mobile phase which flows continuously across the stationary phase.
Principles
Injector
Detector was located after the column or stationary phase
Mobile phase flow
Injected or applied sample
Packing material
The stationary phase can be (a) a solid packed into a column(b) a solid coating the surface of a flat, plane material (c) a liquid supported on a solid (d) a liquid supported on the inside wall of an open tube
The mobile phase can be a gas, a liquid or a supercritical fluid.
Principles
The mixture to be analysed is introduced onto the stationary phase and mobile phase carries the components through it.
Each individual analyte interacts with the two phases in a different manner.
Principles (cont.)
Injector
Detector
Mobile phase flow
t1 t2 t3Time
Interaction with stationary phase
Most
Least
Because analytes differ in their affinity for the stationary phase vs the mobile phase, each analyte exhibits different migration and elution patterns and thus a mixture of analytes can be separated and quantified.
The tracing of the output signal vs time or mobile phase volume is called a chromatogram.
Principles (cont.)
Time
Signal
Classification of elution chromatographic techniques
Adsorption Chromatography
Eluent flow
Legend: Active site
Analyte
Partition Chromatography
Solute dissolved in liquid phase coated on surface of solid support
Affinity Chromatography
+
+ +
+
Matrix Ligand Immobilized ligand
Ligand Sample Complex Impurities
Complex Purified sample
Size Exclusion Chromatography
3 2 1Signal
Volume
1 2 3
Pore size
Stationary phase particle
Chromatographic Techniques
Chromatography
Gas LiquidSupercritical Fluid
SolidLiquid Liquid Liquid
IEC
Solid Solid
GLC GSC SFC BPCLSCLLC TLC SEC
GLC Gas-Liquid Chromatography GSC Gas-Solid ChromatographySFC Supercritical Fluid Chromatography LLC Liquid-Liquid ChromatographyTLC Thin Layer chromatography LSC Liquid-solid ChromatographyBPC Bonded Phase Chromatography IEC Ion Exchange ChromatographySEC Size Exclusion Chromatography
Legend
Mobile phase
Stationary phase
General Theory of Chromatography
Properties of a Gaussian peak. is standard deviation, wi is width at the inflection point, wh is the width at half height, and wb is the width at the baseline intercept.
The plate theory
Tangents to points of inflection
Inflection point
Fraction of peak height
wh = 2.354
wi = 2
wb = 4
1.213
1.000
0.882
0.607
0.500
0.134
0
Schematic diagram showing important parameters in chromatography. to is retention time of unretained compound, tR1 and tR2 are the retention times for component 1 and component 2, respectively.
Chromatogram
w w
a b10% ketinggian
puncak
to
t
t R2
R1
Suntikan
w1/2
1 2
Injection10% of peak height
Chromatographic definitions Retention time, tr - time elapsed between injection of
sample and emergence of peak maximum Mobile phase time, tm - time required for an injected
solvent molecule or any other unretained compound to traverse the column - also known as to (column dead volume or column void volume)
Adjusted retention time, t’r - difference between analyte retention time and column void volume, given by t’r = tr - to
Chromatographic definitions
Capacity factor, k’ - A measure of how long a sample is retained with respect to the dead volume. Given by
k’ = (tr- to)/to
Column efficiency, N - A measure of the broadening of the sample peak as it passes through the column. Given by
N = 16 (tr’ /wb)2 or N = 5.54 (tr’ /wh)2
where wb is the width of peak at base peak
wh is the width of peak at half height
Chromatographic definitions
Height equivalent to a theoretical plate (HETP), H - used to compare column efficiency of different lengths, L.
H = L /N
Resolution, R s = degree of separation of two components leaving the column.
Rt t
w wsR1 R2( )
2
1 2
Continued …
Resolution
d
w1 w2
Data:d = 1.8 cmw1 = 0.8 cmw2 = 0.9 cm
Rs = 2 --------------- = ------------- = 2.1 d 3.6 cm (w1 + w2 ) 1.7 cm
Separation factor or selectivity factor
= ( t r (B) - t o )
( t r (A) - t o )=
t’r (B)
t’r (A)
=k’(B)
k’(A)
or
Separation factor or selectivity factor
t’r (B) 2.8 min = ________ = _________ = 1.4
t’r (A) 2.0 min
1 2 30 4 time (min)
InjectionInjection
Air peakAir peak
t’r (B) t’r (A)
Band broadening
It is due to finite rate at which several mass transfer process occur during migration of solute down a column.
Non-equilibrium theory - movement of solute through the column treated as a random walk.
Random walk - progress of a molecule through column is a succession of random stops and starts about a mean equilibrium concentration or band centre.
Band broadening
When solute desorbed from stationary phase and transferred to mobile phase, it moves more rapidly than the band centre.
Three factors contributing:-Eddy diffusion (A)Longitudinal molecular diffusion (B)Rate of mass transfer (C)
Typical pathways for two solute molecules during elution. The distance travelled by molecule 2 is greater than that travelled by molecule 1. Thus, molecule 2 arrieves at B later than molecule 1.
Eddy diffusion
(1)
(2)
A B
Eddy diffusion
Effect of different path lengths due to irregular flow of molecules through packed particles in a column
Each molecule see different paths causing the solute molecules to arrive at the column outlet at different times
Independent of mobile phase velocity
Band broadening due to longitudinal diffusion in mobile phase (A) initial band and (B) Diffusion of band with time.
Longitudinal diffusionA B
Profil kepekatan jalurA B
Arah laluan fasa bergerak
Lebar jalur
Profile of band conc.
Band width
Direction of mobile phase
Longitudinal diffusion
Results from tendency of solutes to diffuse from concentration centre of a band to the more dilute regions on either sides.
Occur primarily in the mobile phase
Longitudinal diffusion inversely proportional to mobile phase velocity
Higher velocity provides less time for diffusion to occur: Means no band broadening
Illustration of the influence of local nonequilibrium on band broadening.
Non-equilibrium
Mobile phase
Stationary phase
Interface
Flow
Equilibrium concentrationActual concentration
Legend:
Concentration profile of analyte at the interface between the stationary phase and the mobile phase: (a) ideal condition before longitudinal movement of mobile phase takes place. (b) Actual condition after longitudinal movement of the mobile phase.
Influence of local nonequilibrium on band broadening
(a)
(b)Analyte
concentration
Mobile phase
Stationary phase
Mobile phase
Stationary phase
New equilibrium
Analyte concentration
Resistance to mass transfer
Finite time required for the solutes to transfer in and out of the stationary phase
Kinetic lag in attaining equilibrium between two phases
Broadening occur when either of these rates is slow
Broadening depend on diffusion rate of analyte (time dependent)
Broadening worsen with increasing mobile phase flow rate
Plot of HETP against flow velocity, showing the contributions of A (eddy diffusion), B (longitudinal diffusion) and C (interface mass transfer).
Van Deemter plot
Flow velocity
C
B
(Hmin)
C
HETP
Van Deemter curve
H A Bu
Cu
Van Deemter’s Equation
H A Bu
Cu
Where;(A) Eddy diffusion(B) Longitudinal molecular (C) Diffusion rate of mass transfer
Effect of N and on resolution
to
to
to
to
Poor resolutionGood resolution
dueto column efficiencyGood resolution
due to column selectivityPoor resolution despite adequate column efficiency and selectivity (low k’)
Resolution, selectivity, and effeciency
(a)
(b)
(c)
Initial profile
Increased selectivity, N unchanged
Selectivity unchanged, N increased
Good resolution
Good resolution
Poor resolution
Qualitative Analysis
Retention time comparison By using identical column and operating conditions, the identity of
the analyte can be determined by comparing the retention time with the standard compound.
The the absence of a peak at the same retention time as that of a standard under identical conditions suggests that either the respective compound is absent in the sample or the compound is present at a concentration level below the detection limit of the method.
Kovats’ retention index (RI)
I zt t
t t
100 100
log log
log logR(x) R(z)
R(z+1) R(z)
' '
' '
I zt t
t t
100 100
R(x) R(z)
R(z+1) R(z)
' '
' '
For linear temperature-programmed GC
For isothermal GC Plots of log tR’ vs C-no. gives a straight line
Plots of tR’ vs C-no. gives a straight line
Kovats RI
The Kovats retention index. The log tR' values of a series of n-alkanes area plotted against (100 x carbon number). Compound x (log tR' = 0.6) has a retention index value of 680.
200 400 600 800 1000
0.2
0.4
0.6
0.8
1.0
0
log t'R
Indeks penahananRetention Index
Quantitative Analysis
Quantitative Analysis: Peak area
Measurement of peak area: (a) By triangulation. Peak area area of the triangle = (h' x w)/2. (b) Peak area area of the rectangle = h x w1/2.
Gerak
Garis tangen
w
h'
Masa atau isipadu
w
h
Masa atau isipadu
1/2
(a) (b)
balas
Response factor
XAmount X AreaPeak
x F
Response factor of compound X = FX
Det
ecto
r re
spon
se
X
0 5 10 min
S
Response factor
Response factor of compound X relative to that of compound S:
x
s Xof X/Amountof areaPeak S of S/Amount of areaPeak
FFRF
100 x peaks all of area TotalX of areaPeak %mixturein X%
A
If the response factors of all compounds in the sample are the same,
External standard
Mx Molarity of compound X, Ms Molarity of standard S, Fx Response factor of compound X, Fs Response factor of standard S, V(x) Volume of injected sample X, V(s) Volume of injected standard S, Ax Peak area of sample X,As Peak area of standard S.
MF A VF AV
Mxs x (s)
x s (x)s
External standardExternal standard poses a problem with reproducibility.
•Injection – the volume of injection may vary from one injection to another injection.
•Detection – the sensitivity of the detector may vary from time to time.
Internal standard
MF A NF AVx
s x s
x s x
Mx Molarity of compound X, Ms Molarity of standard S, Fx Response factor of compound X, Fs Response factor of standard S, V(x) Volume of injected sample X, Ax Peak area of sample X,As Peak area of standard S. N(s) Amount of internal standard S,
Criteria for internal standardsCharacteristics of ideal internal standards
(a) It is well separated from the components in the analyzed mixture.
(b) Its response factor is similar to that of the analyte.
(c) Its retention time is close to the that of the analyte to minimize errors in peak area measurement.
(d) Its concentration should be similar to that of the analyte.
Example 1. External standard
Consider the following data of a typical quantitative gas chromatographic analysis where a compound, X, is used as the external standard.
• An injection (1 L) of a mixture containing 10, 12, and 13 ppm of X, Y, and Z, respectively, gave respective peak areas of 515, 748 and 939 arbitrary units.
• An injection (2 L) of the sample containing compounds X, Y, and Z gave peak areas of 232, 657, and 984 arbitrary units, respectively.
• Calculate the concentrations of compounds X, Y and Z.
Example 1. Solution
Example 2. Internal standard
In a chromatographic analysis, a mixture containing 0.0567 M of a compound, X, and 0.0402 mM of a standard, S, gave peak areas of Ax = 498 and AS = 526 arbitrary units.
An aliquot (10.0 mL) of the unknown was added with 10.00 mL of 0.102 M S and the mixture was diluted to 25 mL in a volumetric flask. The mixture gave a chromatogram with Ax = 678 and AS = 588 arbitrary units.
• Calculate the concentration of X in the unknown.
Example 2. (Cont.)
Det
ecto
r re
spo
nse
X
0 5 10 min
S Determination of the relative detector response for the standard Add a known amount of the standard compound into a known quantity (volume or weight) of sample, Vx, Measure the peak area, and calculate the concentration of the analyte using the proper equation.
Example 2. Solution