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