Post on 22-Dec-2015
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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?
Rate Theory of Chromatography
Rate Theory of Chromatography
Rate Theory of Chromatography
Rate Theory of Chromatography
W = 4 where W =>width of peak at baseline
=>standard deviation expressed as time
Rate Theory of Chromatography
H = 2/Lwhere
=>standard deviation
expressed as length
Rate Theory of Chromatography
thus
L L L2
N = ---- = ------- = -------H 2/L 2
Rate Theory of Chromatography
and
= ------
L/tR
where tR =>retention timeL/tR => rate of travel of band
Rate Theory of Chromatography
therefore,LW = ---------
4tR
Rate Theory of Chromatography
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
Sources of Zone Broadening
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
Sources of Zone Broadening
Separations on Columns
ZRs = -------
Wwhere Rs => column resolution
W => av. width of peaksZ => time separation,
difference ofretention time
Separations on Columns
Z 2((tR )y - (tR )x )Rs = ------- = -------------------W Wx + Wy
where (tR )x => retention time, component x
(tR )y => retention time, component y
Separations on Columns
(tR )y - (tR )x (N)1/2
Rs = ------------------ * ---------(tR )y 4
Separations on Columns
Separations on Columns
Separations on Columns
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
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
N2 = ((Rs)2/(Rs)1)2 * N1 = (1.5/10.5)2 * 3.1 x 103 plates
N2 = 63 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: (d.) the length of column required to achieve a resolution of 1.5
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
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
(e)(tR )1 (Rs )1
2
-------- = ---------- = ((Rs )1 /(Rs )2 )2
(tR )2 (Rs )22
(tR )2 = ((Rs )1 /(Rs )2 )2 * (tR )1
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
(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