Lecture 2 ChromatographyTheory HVCH V2

8/12/2019 Lecture 2 ChromatographyTheory HVCH V2

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Part II

Chromatographic

Theory

O Partition between mobile phase/stationary phase:

Sm Ss Sm: the solute in the mobile phase

Ss: the solute in the stationary phase.

Separation process Elution


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O Two approaches explain the separation process:

Plate theory Martin and Synge (1941), based on an analogy with distillation and countercurrent extraction.

Rate theory van Deemter (1956), accounts for the dynamics of a separation.

o Advantages o Limitations

Theories

O partition coefficient ( equilibrium constant), K

K is assumed to be independent of concentration.

Solute Retention Factor


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O Capacity ratio , k

t R - the adjusted retention t ime


Retention time

O retention time (tR ): time between sampleinjection and an analyte peak reaching adetector at the end of the column

O The time taken for the mobile phase to passthrough the column is called tM.


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O Solute fraction s

, 1

1

, 1

If constant mobile phase velocity:


Terms : retention factor, k, the capacity factor, the capacity ratio, andthe partition ratio, and is sometimes given the symbol k .

Example 1

In a chromatographic analysis of low molecular weight acids,butyric acid elutes with a retention time of 7.63 min. The

columns void time is 0.31 min. Calculate the retention factorfor butyric acid.

Solution

K butyric = ( t r tm) / t m= (7.63 min 0.31 min) / 0.31 min = 23.6


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Chromatogram for a two-component mixture. Determine theretention factor for each solute , sample injected time t = 0.

Relationship betweenelution time and distanceis proportional, we canmeasure tm , t r,1 , and t r,2 using a ruler.The measurements are7.8 mm, 40.2 mm, and51.5 mm, respectively.

Solution : The retention factors for solute A and solute B are:k 1 = ( t r tm) / tm = (40.2 mm 7.8 mm) / 7.8 mm = 4.15k 2 =( t r tm) / tm = (51.5 mm 7.8 mm) / 7.8 mm = 5.60

Example 2

Retention time and volume

O Retention volume , VR volume of mobilephase required to elute a solute to a maximum

from a column.O Retention time , t R time required to reach the

same maximum at constant flow.


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Retention time and volume

O For a constant column length:

O Retention time for mobile phase , tM

t 1

Component separation

O If solutes 1 and 2 have capacity factor of k 1 and k 2 thentheir retention times t R1 and t R2 :

O Peak separation:

2 1 2 1

tR2 > t R1


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O selectivity factor

the separation of two species (A and B) on thecolumn;

species A elutes faster than species B: >1.

Selectivity factor

Example 3

In the chromatographic analysis for low molecularweight acids, the columns void time is 0.31 min. Theretention time for isobutyric acid is 5.98 min. What is

the selectivity factor for isobutyric acid and butyric acid(retention time of 7.63 min)?

Solution

Calculate the retention factor for isobutyric acid and butyric acid.k iso = ( t r tm) / tm = (5.98 min 0.31 min) / 0.31 min = 18.3

The selectivity factor, therefore, is = k but / k iso =23.6 / 18.3 = 1.29


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Example 4

O Determine the selectivity factor

The selectivityfactor is = k 2 / k 1 = 5.60 / 4.15= 1.35

Column Efficiency

Band broadening

Sharp symmetrical chromatographic peaks

Efficiency of the column


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Plate theory of Chromatography

O A chromatographic column is mathematicallydivided into theoretical plates (N) .

O There is an equilibrium partitioning of thesolute between the stationary phase and themobile phase.

O The analyte moves down the column by transferof equilibrated mobile phase from one plate tothe next.

N = L / H

Column efficiency in terms of the number of theoreticalplates, N,

L - columns length; H plate height.

Note: more theoretical plates (the better) Column efficiencyimproves chromatographic peaks become narrower.

Height Equivalent to a Theoretical Plate

HETP = L / N

the smaller the better.

Determination of N


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The solutes average linear velocity is the distance it travels, L,divided by its retention time, t r.

If peak broadening: H is the variance per unit length of thecolumn (.

- standard deviation, sec or min,Time of elution, T ( /solutes average linear velocity, ).

H = 2 / L

= / = tr / L

Determination of N

w = 4

H = Lw 2 / 16 tr2

For a Gaussian peak shape, the width at thebaseline, w , is four times its standard deviation, .

Height of a theoretical plate in terms of t r and w :

N = 16( tr 2 / w 2 )

Number of theoretical plates:

Determination of N


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Since it is difficult to accurately measure the beginningand end of a peak, it is common to use the width at halfheight and assume the peak is Gaussian.

Determination of N

Example 5 A chromatographic analysis for the chlorinated pesticide Dieldringives a peak with a retention time of 8.68 min and a baselinewidth of 0.29 min. What is the number of theoretical plates?Given that the column is 2.0 m long, what is the height of a

theoretical plate in mm?Solution

The number of theoretical plates:N = 16 t r 2 / w 2 = N = 16(8.68 min) 2 / (0.29 min) 2 = 14300 plates

The average height of a theoretical plate:

H = L / N = (2.0 m / 14300 plates) (1000 mm / m) = 0.14mm/plate


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Resolution

O Resolution, R s , how well two neighboring peakscompletely separated from each other.

O The resolution of two peaks A and B:

O controlling the capacity factor, k , separationsgreatly improved:

O by changing the temperature (in GC) or thecomposition of the mobile phase (in LC).

Resolution


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Example 6

In a chromatographic analysis of lemon oil a peak forlimonene has a retention time of 8.36 min with a baselinewidth of 0.96 min. -Terpinene elutes at 9.54 min with abaseline width of 0.64 min. What is the resolution betweenthe two peaks? Please give your comments of the results.

Solution

The resolution isRAB = 2 t r / ( w B + w A) = 2(9.54 min 8.36 min) / (1.64 min +

0.96 min) = 1.48

Example 7

O This Figure shows theseparation of a twocomponent mixture. What isthe resolution between thetwo components? Use aruler to measure t r, w A, andw B in millimeters.

Measurements are 8.5 mm for t r,and 12.0 mm each for w A and w B

Using these values, the resolution isRAB = 2 t r / ( w A + w B) = 2(8.5 mm) / (12.0 mm + 12.0 mm) = 0.70


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Peak Capacity

n c = 1 + ( (N)/ 4)ln( V max / V min )

o A measure of the number of solutes that canbe separated, n c.

o Based on isothermal/isocratic conditions.

nc - columns peak capacity

V min and V max - the smallest and the largestvolumes of mobile phase in which we can eluteand detect a solute.

Example 8 O A column with 10 000 theoretical plates can resolve no

more than:

n c = 1 + ( (10000) / 4)ln(30 mL / 1 mL)

= 86 solutesassumed V min =1 mL and V max = 30 mL.

consideration a column enough theoretical plates toseparate a complex mixture.


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Asymmetric Peaks

O Poisson distributionO Gaussian distribution

Optimizing ChromatographicSeparations

RAB = ( t r,B t r,A) / (0.5( w B +w A)) (t r,B t r,A)/(0.5(2 w B)) = ( tr,B t r,A)/ w B

Define the effects ofsolute retention factorselectivity,column efficiencyresolution of two closely eluting peaks.

If the two peaks (A and B) have similar retention times, it isreasonable to assume that their peak widths are nearlyidentical:

RAB= ((N) / 4) ( t r,B t r,A) / t r,B


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RAB= ( (N) / 4) (( 1) / ) ( k B / (1 + k B))

Reretention times of solutes A and B.

tr,A = k AtM + tM tr,B = kBtM + tM

R AB = ( (N) / 4) (( kB - k A) / (1 + kB))

t r,B= (16 RAB2 H / u ) ( / ( 1)) 2 ((1 + k B)3 / k B2 )

u - mobile phases velocity

Optimizing ChromatographicSeparations

Using the Retention factor to Optimize Resolution


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Using Selectivity to Optimize Resolution

Using Column Efficiency to Optimize Resolution


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Rate Theory of Chromatography

O Plate theory neglects the concepts of solute diffusion andflow paths.

O Rate theory accounts for these and can be used to predictthe effect on column performance factors such as :

O Phase propertiesO Phase thicknessO Solute diffusivitiesO Support sizeO Partition coefficientsO Support porosityO Phase velocityO Flow rates

Rate Theory of Chromatography

O Rate theory (inside a column): time for the solute toequilibrate between the stationary and mobile phase.

O band shape of chromatographic peak is affected by:O the rate of elutionO the different paths available to solute molecules as they

travel between particles of stationary phase.O band broadening:

O variations in paths lengthO longitudinal diffusionO mass transfer in the stationary phase, andO mass transfer in the mobile phase.


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Equation of parameters

H = Hp + Hd + Hs + Hm

The height of a theoretical plate is a summationof the contributions:

An alternative form of this equation is the vanDeemter equation accounting for the dynamics ofthe separation process.

Van Deemter equation

factor characteristic of packingd P particle diameter factor for irregularity of interparticle spaces

Dg diffusion coefficient of compound in gasDl diffusion coefficient of compound in liquidu linear gas velocityk capacity ratiod f liquid phase effective film thicknessH height of theoretical plate


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The equation consists of three basic terms.

Packing related term

Gas (mobile phase) term

Liquid (stationary phase) term

O - Eddy diffusion/multipath

O B - Longitudinal diffusion/molecular diffusion

O C - Resistance to mass transfer

HETP = A + B / u + C u

u - average velocity of the mobile phase.



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Plot showing the relationship between the height of a theoretical plate, H,and the mobile phases velocity, u , based on the van Deemter equation.

Van Deemter plots

Other equations

H = ( B / u )+ ( C s + C m)u

C s and C m - mass transfer for the stationary phaseand the mobile phase

H = Au 1/3 + ( B / u ) + Cu

Lecture 2 ChromatographyTheory HVCH V2

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