Chromatography

Separation Methods

An Introduction to Chromatographic

Separations

ChromatographyChronology• column chromatography• paper chromatography• gas-liquid chromatography• thin layer chromatography (TLC)• high-pressure liquid chromatography

(HPLC - high-performance liquid chromatography)

Parts of Column

• column• support• stationary phase• mobile phase

Parts of Column

Column• copper tubing• stainless steel tubing• glass tubing

Parts of Column

Support• finely divided solids

– ground firebrick– alumina, specially treated

• walls of column for capillary columns

Parts of Column

Stationary Phase• stationary phase evenly dispersed on

surface of support– column chromatography

• non-volatile, viscous liquids dispersed evenly on surface of support

Parts of Column

Stationary Phase• stationary phase evenly dispersed on

surface of support– planar chromatography

• porous paper in paper chromatography• finely ground solid spread evenly on glass or plastic

plate for tlc (thin-layer chromatography

Parts of Column

Mobile Phase• sample mixture carried through

stationary phase by mobile phase• non-reactive gas in glc (gas-liquid

chromatography, gc)• non-reactive liquid in llc (liquid-liquid

chromatography, lc)

Linear Chromatography

CsK = -----

Cm

K is T dependentCs => conc. in stationary phaseCm => conc. in mobile phase

Linear Elution Chromatography

Theories of Elution Chromatography

some zone broadeningzone separation

Theories of Elution Chromatography

N = L/Hwhere n => number of plates

L => length of columnH =>height equivalent to a

theoretical plate (HETP)

Rate Theory of Chromatography

zone shapes (Gaussian curve)

Why?





W = 4 where W =>width of peak at baseline

=>standard deviation expressed as time


H = 2/Lwhere

=>standard deviation

expressed as length


thus

L L L2

N = ---- = ------- = -------H 2/L 2


and

= ------

L/tR

where tR =>retention timeL/tR => rate of travel of band


therefore,LW = ---------

4tR


andLW2

H = ----------16tR

2

N = 16 (tR /W)2

Sources of Zone Broadening

classical van Deemter EquationH = A + B/u + Cu

where A => eddy diffusionB => longitudinal diffusionu => flow rateC => mass transfer

Eddy Diffusion

A = 2dR

where

=>packing factordR => average diameter of

particle

• caused by many pathways• minimized by careful packing

Longitudinal Diffusion

B = 2DM

where

=> obstruction factorDM => diffusion coefficient of

solute in themobile phase

• minimized by lowering temperature of column oven and decreasing flow rate

Mass Transfer

qR (1 - R)df2 dR

2

C = -------------------- + -----------DS DM

where df => film thickness of stationary phase (most important factor)

dR =>diameter of support particle

Mass Transfer

qR (1 - R)df2 dR

2

C = -------------------- + -----------DS DM

where R =>retention ratio: tM /tR

DS =>diffusion coefficient for solutes in stationary phase

Mass Transfer

qR (1 - R)df2 dR

2

C = -------------------- + -----------DS DM

where DM =>diffusion coefficient for solutes in mobile phaseq &

=>constants for

column


modernized versionH = B/u + CSu + CMu

where CS => coefficient of mass transfer in the stationary phase

CM => coefficient of mass transfer in the mobile phase


Separations on Columns

ZRs = -------

Wwhere Rs => column resolution

W => av. width of peaksZ => time separation,

difference ofretention time


Z 2((tR )y - (tR )x )Rs = ------- = -------------------W Wx + Wy

where (tR )x => retention time, component x

(tR )y => retention time, component y


(tR )y - (tR )x (N)1/2

Rs = ------------------ * ---------(tR )y 4




EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (a.) column resolution

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (a.) column resolution

2((tR )y - (tR )x ) 2(14.4 - 6.4)Rs = ----------------- = ---------------- = 10.5

Wx + Wy (0.45 + 1.07)

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (b.) the av. no. of plates in the column

N = 16 * (tR /W)2

for component ANA = 16 * (6.4/0.45)2 = 3.2 x 103 plates

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (b.) the av. no. of plates in the column

for component BNB = 16 * (14.4/1.07)2 = 2.9 x 103 plates

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (c.) the plate height

H = L/Nfor component BH = L/NB = (22.6 cm)/(2.9 x 103 plates) = 7.8 x 10-3cm/plate

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An

unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the:

(d.) the length of column required to achieve a resolution of 1.5

N1 (Rs )12

--- = -------- = ((Rs )1 /(Rs )2 )2

N2 (Rs )22

EXAMPLE: Substances A and B were found to have retention times of 6.4 and 14.4 min, respectively, on a 22.6 cm column. An unretained sample of air passed through the column in 1.30 min. The widths of the peak bases were 0.45 and 1.07 min. Calculate the: (d.) the length of column required to achieve a resolution of 1.5

N1 (Rs )12

--- = -------- = ((Rs )1 /(Rs )2 )2

N2 (Rs )22

whereN1 = (NA + NB )/2 = (3.2 x 103 + 2.9 x 103)/2

= 3.1 x 103 platesR1 = 10.5


N2 = ((Rs)2/(Rs)1)2 * N1 = (1.5/10.5)2 * 3.1 x 103 plates

N2 = 63 plates


Hav = (HA + HB)/2 = ((7.8 + 7.0)x 10-3)/2= 7.4 x 10-3 cm/plate

L = Hav * N2 = (7.4 x 10-3cm/plate)(63 plates) = 0.46 cm


(e)(tR )1 (Rs )1

2

-------- = ---------- = ((Rs )1 /(Rs )2 )2

(tR )2 (Rs )22

(tR )2 = ((Rs )1 /(Rs )2 )2 * (tR )1


(e)(tR )2 = ((Rs )1 /(Rs )2 )2 * (tR )1

(tR )2 = (1.5/10.5)2 * (10.4 min) = 0.13 min = 7.8 sec

Applications of Chromatography

• Qualitative Analysis• Quantitative Analysis

– Analyses Based on Peak Height– Analyses Based on Peak Areas– Calibration and Standards– The Internal Standard Method– The Area Normalization Method

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