document1

Chem 241 - Summer 2010

Section 1 1

Chemistry 241Professor: Gary L. Glish

Office: Caudill 320Phone: 962-2303

email: [email protected]: [email protected] Hours:

» Monday 2:00-3:00; Thursday 11:00–12:00 » Other afternoons - drop-in or by appointment

About Me

B.A. 1976 Wabash College» majors: Chemistry and Economics

Ph.D. 1980 Purdue University1980-1992 Oak Ridge National

Laboratory1992 UNC

» Teach: Chem 070, 101, 241, 241L, 395, 396, 448, 742, 742L, 743, 941, 992, 993, 994

Chemistry 241 Prerequisites

Chem 102 or 102H

It is an Honor Code violation to enroll in a course for which you have not taken prerequisites.

Class logistics

blackboard.unc.edu»Syllabus»Lecture Notes»Exam keys»Old exams»Equation sheet

Class logistics

Book: Quantitative Chemical Analysis, 7th Edition, by Daniel C. Harris (UNC Custom text is subset of chapters from this)chapters from this)» Also recommended: Solutions Manual

Grading -» 3 in class exams - 20% each» final exam - 40% » Ask the Class (extra credit)

Class logistics – Schedule

Week of Monday Tuesday Wednesday Thursday Friday

May 10 Chpts. 3, 4,5 Chpt. 23,24 Chpt. 24,25 No Class

May 17 Chpt. 25,26 Exam 1 Chpt. 9 Chpt. 10 No Class

May 24 No Class No Class No Class No Class No Class

Final Exam 3-6 PM, Monday, June 14th

May 31 No Class Chpt. 11 Exam 2 Chpt. 18,20 No Class

June 7 Chpt. 20, 21 Chpt. 21, 22 Exam 3 Chpt. 14,15 No Class


Section 1 2

Class logistics - ExamsFinal is comprehensiveThe final will be divided into sections, a

better score on a section of the final than the in-class exam will replace that grade

Missed exams will use the corresponding section of the final as grade

Equation sheet will be providedLeft handers – email me by Feb. 1 if you

want a left-handed desk for exams

Honor System (http://honor.unc.edu) General Responsibilities.

It shall be the responsibility of every student at the University of North Carolina at Chapel Hill to:

1. Obey and support the enforcement of the Honor Code;

2 Refrain from lying cheating or stealing;2. Refrain from lying, cheating, or stealing;

3. Conduct themselves so as not to impair significantly the welfare or the educational opportunities of others in the University community; and

4. Refrain from conduct that impairs or may impair the capacity of University and associated personnel to perform their duties, manage resources, protect the safety and welfare of members of the University community, and maintain the integrity of the University.

Honor System Procedures

Professor reports possible violationStudent attorney general decides if

charges should be filedgHonor system notifies accused

student – defense counsel providedHonor Court TrialIf convicted - Sanctions

Exam Policies

Cheating on an exam - automatic 0 (can NOT be replaced)

NO CELL PHONES or IPODS – automatic 0 (can NOT be replaced)( p )

Exam leaves room other than with me –automatic 0 (can NOT be replaced)

Exam not turned in within 60 seconds after I say STOP – automatic 0 (can be replaced)

Exam Policies

Replacement of grade from Final Exam not valid for grades resulting from disciplinary actionIf l k h Fi l E hIf you only take the Final Exam, that grade is your grade

If you miss more than 1 exam AND the Final Exam, no replacement of grades

Class logistics - Grading scale

Gx x

sx

your scoreA 1.25B 0.45

Grade

x

s

class average

standard deviation

C -0.75D -1.4

You must earn at least 50% of the points to pass


Section 1 3

Class logistics - Alternative scale

92 -100% A84 - 91 % B76 - 83% C68 - 75% D

The scale that gives you the highest grade will be used

Class logistics - Equations

2H A

+ 2

+ 2 +a1 a1 a2

= [ H ]

[ H ] + [ H ] K + K K

N*

N =

g*

ge

o o

- E

kT

= c

n E = h

[ H ] =K K F + K K

K + F+

1 / 2a1 a2 a1 w

a1

[ ]

' '

" "

' 'x

A b

A b

b b

y

y

x y

[ ]

' '

" "

' 'y

b A

b A

b b

x

x

x y

=

1a1

" "b bx y

x y " "b bx y

x y

pH = pK + [Base]

[Acid]a log x =-b (b - 4ac )

2a

2 1/ 2HA

+

+ =

[ H ]

Ka+ [ H ]

A= -logT = bc

solute

standard

solute

standard

C

C = F area

area % % % ...e e et x y 2 2

1

2 e e et x y 2 21

2... a b wK K = K

kresponse to y

response to xx y, E = E -.05916

nQ log

q nFG = - nFE

Class logistics - Ask the Class

Participation optional – must have a “clicker”

Questions asked during classF i f l i hFraction of total points over the semester multiplied by 10 and added to final exam grade to calculate final grade

Class logistics - Ask the ClassMost questions will be worth 3 pointsMusic questions – 1 point for answering It is an Honor Code violation to answer

with someone else’s clicker

Class logistics - Homework

Suggested homework problems on Blackboard

Homework will not be graded or collectedcollected

Work on homework problems with classmates

At least one question on each exam will be a homework problem

Course Goals

Learn some basics about Analytical Chemistry

Understand principles of p ptechniques

Apply principles to problem solving (THINKING)


Section 1 4

What is Analytical Chemistry?

Analytical chemistry is the science of chemical measurementsmeasurements

What is an “Analyte”

object of the chemical measurements, sometimes the “sample”sometimes the sample sometimes a component in

the “sample”.

What are we trying to measure?

Qualitative Analysis - what is it?»Newcastle or Coors?»What are the components that make

C ?up Coors?

Quantitative Analysis - how much is there?»six pack or case?»percent ethanol?

Define what is being measured

surface compositionbulk compositionaverage compositionspatially resolvedtemporally resolved

Main Topics for Chem 241

separationschemical equilibria - acid/basespectroscopy the use of radiationspectroscopy - the use of radiation

to probe chemical propertieselectrochemistry - measurement of

electrical properties of solutions

Let’s get started

Chapter 0 – nice general overviewChapter 1 – review from 101/102Chapter 2 – not necessary for this

l b h l f l f 241Lclass, but helpful for 241L


Section 1 5

Significant Figures

You should know sig figsPoints will be deducted on examsRounding off is done only on the

FINALFINAL answer

Significant Figures - Logs

log 1034 = log 1.034 x 103 = 3.0145

log 1.034 x 10-3 = -2.9855

antilog -4.756 = 1.75x 10-5

ERRORS

Accuracy and Precision

Why do measurements disagree?

ERRORS

Accuracy and Precision

Accuracy - how close to “true” valuePrecision - how reproducibleAccuracy and precision not correlated

i b i di.e. a measurement can be precise and not accurate or accurate and not precise

Types of Errors

Systematic (determinate) - something involved with the measurement system that can be detected and correctedcorrected

Random (indeterminate) - natural limitation in ability to make measurements

Uncertainty

absolute (ex)» estimated uncertainty associated with

measurement e.g. ±0.2mm » note that uncertainty has same units as» note that uncertainty has same units as

measurement


Section 1 6

Uncertainty

relative (ex/x) percent relative (%ex)

» %ex = (ex/x) (100)» for constant absolute uncertainty, the

relative uncertainty decreases with increasing magnitude of measurement

√

√

Propagation of Uncertainty

√

√

Gaussian Distribution

normal error, or bell shaped, curveA curve which predicts the

distribution of data - only random errorerror

MUST define limits of data set


curve characterized by two parameters» arithmetic mean

t d d d i ti x s


» standard deviation s xi


Arithmetic mean

xn

x

n

i

i

value of measurement i

number of measurements


Section 1 7

x xi

21

2


Standard deviation

s

x x

n

n

i

1

1 degrees of freedom

s variance2


standard deviation is a measure of the width of the distribution - how close the values are clustered about the meanmean

ss


standard deviation of the mean

reduce standard deviation of the mean by making more measurements

sn

x 1 2


Gaussian equation

y e

1

21

2

2 2


for x = - to + , probability must equal 1

probability that a value falls in some range is the area under curve in thatrange is the area under curve in that range


95.5% of all values fall within ± 2s of the mean

99.7% of all values fall within ± 3s of the meanthe mean

Table 4-1, page 56z = (x-x)/s


Section 1 8

Gaussian Distribution Gaussian Distribution

y

xs

xxz

1

1% > x1

From Table Find area

x

y

x1

From Table, Find area for z1 (=Az1)

%= (.5-Az1) x 100


y

xs

xxz

s

xxz

2

21

1 % between x1 and x2

From Table, find area f ( A ) d

x

y

x1

for z1 (=Az1) and z2

(=Az2)

%= (Az1+Az2) x 100

x2

Confidence Intervals

Tells us what fraction of time the “true” value will be within the range of our error bars

What is the likelihood that the “true” value (µ) lies in some range about the mean ( )?x


xts

n


“t” is called “Student’s t” and is a statistically derived value

the value of “t” is dependent upon the number of measurements and thenumber of measurements and the level of confidence desired


Section 1 9


p. 58


Increasing the level of confidence increases the range of values

Increasing the number of measurements decreases the range ofmeasurements decreases the range of values


Data set (12.6, 11.9, 13.0, 12.7, 12.5)mean = 12.5, std. dev. = 0.4There is a 50% likelihood that the

l i i h ?true value is in what range?

12 50 741 0 4. .


t50% = 0.741

12 55

.

12.4 < < 12.6 at 90% confidence (t90% = 2.132)

12.1 < < 12.9

Comparison of Means

Are two sets of data the same or different?

Calculate the % confidence that they are different or is difference onlyare different - or is difference only due to random chance

Comparison of Means

Comparison of measurement to known value - same as confidence interval

calctns

x

If tcalc > t in Table, data sets are DIFFERENT at that confidence level


Section 1 10

Comparison of Means

Comparison of two sets of measurements

calculate “t”, compare to Table to determine level of confidencedetermine level of confidence

21

21

2121

nn

nn

s

xxt

pooled

Comparison of Means

21 pooled

s

x x x x

n npooled

i jsetset

1

2

2

2

21

1 2

12

2

Comparison of Means

Comparison of individual differencescalculate “t” compare to Table to

determine level of confidence

xxd

ns

dt

BiAi

d

deviationaverage

21

Comparison of Means

ng

21

2

1

n

dds i

d

22

21

s

sF

Comparison of Standard Deviations

21 ss

Fcalculated > Ftable means significant difference

Q test for bad data

Sometimes, for unknown reasons, you get a bad data point

Determine whether it is statistically significant with Q testsignificant with Q test


Section 1 11

Q test for bad data

Need at least 4 measurementsThe magnitude of Q to reject data is

dependent upon # of measurements and degree of likelihood (Table 4 5)and degree of likelihood (Table 4-5)

QGap

Range

Q test for bad data

Gap = difference between suspect data point and next closest data point

Range = difference between suspect data point and furthest data point

Quantification

Desire linear response with amount of sample

plot a parameter of signal (e.g. area, height) vs sample amount (weightheight) vs. sample amount (weight, concentration, etc.)

Quantification

can use signal height if peak is symmetrical - better to use area

IF detector responds EQUALLY to all analytes relative areas = relativeall analytes, relative areas = relative amounts

Quantification

Three common approaches to quantification» Calibration (Standard, Working)

CurveCurve» Standard Addition» Internal Standard

Calibration Curve

Generate a plot of response versus known amount (e.g. concentration) of the analyte of interest

Response

Amount

xx

O xx

x


Section 1 12

Calibration Curve

Need to minimize matrix differences to reduce matrix effects

Need stable instrumentation

Standard Addition

Add known amounts of analyte to aliquots of unknown amount

[ ]x A[ ]

[ ] [ ]

x

x s

A

Ax

x s

Know [s], measure Ax and Ax+s

solve for [x]

Standard Addition

x

x

[s]0

xx

[-x]

Standard Addition

Constant matrix - no matrix effectsInstrument stability still important

Internal Standard

Add a known amount of an analyte (standard) similar to the analyte of interest to the sample

Measure response of analyte andMeasure response of analyte and standard

Need to know response factor

Internal Standard

[ ]

[ ]

x

sF

response of x

response of s

F is response factorideally =1


Section 1 13

Internal Standard

x

xR

Rx

xx

][

][

s

x

Rs

Internal Standard

Corrects for» sample losses during work-up» matrix effects» instrument instability

Calibration

Sensitivity (m) = slope of Calibration Curve

signalm

amount

Limit of Detection (LOD) - minimum detectable amount

Calibration

sLOD

3

Limit of Quantitation

mLOD

m

sLOD

10

Calibration Least Squares

many measurements use calibration curve of response vs. quantity ([ ], weight, etc.)

get calibration curve collecting dataget calibration curve collecting data at several different quantities


Section 1 14

Least Squares

assume response is linear and uncertainty in response (ey) is >> than uncertainty in quantity (ex) (standard)(standard)

y = mx + b

Least Squares

find value of m and b that minimize deviations of y

deviation of yi for xi

ddi = yi - ydi = yi - (mxi + b)

Least Squares

di

[x]

y = mx + b

Least Squares

must square to make all deviations, positive or negative, weighted the same

d 2 = (y mx b)2di2 = (yi - mxi - b)2

Least Squares

Use matrix algebra to determine m and b

D

ny

xyxm

i

iii

D

yx

yxxb iii

2

Least Squares

yx ii

nx

xxD

i

ii

2


Section 1 15

Least Sq. - Uncertainty

Because there is an uncertainty in y, there must be an uncertainty associated with m and b

21

221

2

22

n

d

n

dds ii

y

21

2

D

nss y

m


21

22

D

xss

D

iyb

m


the first decimal place of the standard deviation is last sig. fig for slope or intercept

to express uncertainty in terms ofto express uncertainty in terms of confidence interval, multiply sm or sbby appropriate “t”

b


Uncertainty in quantity of unknown is:

x

y s b s

m s

y b

m

Analytical Separations

Why separate compounds?» to isolate or concentrate component(s)

from a mixture» to separate a component(s) from other» to separate a component(s) from other

species that would interfere in the analysis

Methods of Separation

Extraction » washing clothes

Crystallization» drugs

Distillation» moonshine

Chromatography


Section 1 16

Solvent Extraction

Extraction: transfer of a solute from one phase to another.

Can use most any combination of phases ( lid li id iti l fl id)(solid, liquid, gas, supercritical fluid)

Solvent extractions use two immiscible liquids.» Typically aqueous/organic solvent combos

Solvent Extraction

Organic solvents less dense than water » diethyl ether, toluene, hexane

Organic solvents more dense than water» chloroform, CCl4, dichloromethane

Like dissolves like so ideally, the extracting solvent should be similar to the solute (analyte)

Separatoryfunnel

Solvent Extraction

shake

add second immiscible

solvent

Solute partitions between the two phases

Solvent Extraction

[S]1

[S]2

Phase 1

Phase 2

Solvent Extraction

Equilibrium constant for this partitioning is K (partition coefficient)

K=[S]2

[S]1

Solvent Extraction

Determination of solute concentration in each phase

Define some variables:» V1 & V2 are volumes of solvents 1&2» m = total # of moles of solute (S)

present» q = fraction of solute remaining in

phase 1 at equilibrium


Section 1 17

Solvent Extraction

[S]1 = qm/V1

[S]2 = (1-q)m/V2 K=[S]2

[S]1qm/V1

(1-q)m/V2= =q/V1

(1-q) /V2

Solvent Extraction

q =KV2 + V1

V1 (1-q) =KV2 + V1

KV2

fraction of S in: phase 1 phase 2

Rearrange:

Solvent Extraction

If remove V2 and extract V1 with fresh layer of V2, what fraction remains in V1?» Initial moles = m» Initial moles = m» after first extraction = qm» after second extraction = q(qm)=q2m

q(2) =KV2 + V1

V12

Solvent Extraction

Fraction in V1 after n extractions:

q(n) =KV2 + V1

V1n

Solvent Extraction

Example: Solute A has a partition coefficient of 4.000 between hexane and water. (K = [S]hexane/[S]water = 4) ( [ ]hexane [ ]water )If 150.0 ml of 0.03000 M aqueous A is extracted with hexane, what fraction of A remains if:

Solvent Extraction

a) one 600.0 ml aliquot of hexane is used?

150mlq =

4(600ml) + 150ml

150ml= 0.05882 = 5.882%

# moles remaining0.05882 (0.03M•0.150L) = 2.647x10-4 moles


Section 1 18

Solvent Extraction

b) 6 successive 100.0 ml aliquots of hexane are used?

q =4(100ml) + 150ml

150ml= 0.0004115

6

# moles remaining4.115 x 10-4 (0.03M•0.150L) = 1.852x10-6moles

Solvent Extraction

Although same volume of hexane is used, it is more efficient to do several small extractions than one big one! » 1 600 ml extraction extracts 94 12%» 1 - 600 ml extraction extracts 94.12% » 6 - 100 ml extractions extract 99.96%

HA H+ + A-Ka

Solvent Extraction (pH effects)

with organic acids/bases:

B + H2O BH+ + OH-Kb

Generally, neutral species are more soluble in an organic solvent and charged species are more soluble in aqueous solution

very little here, ions have poor solubility


Partitioning of organic acids between two phases:

organic

HA H+ + A-Kaaqueous

HA H+ + A-have poor solubility


When the solute (acid/base) can exist in different forms, D (di ib i ffi i ) i d(distribution coefficient) is used instead of K (partition coefficient)


D =total conc. in phase 2total conc. in phase 1

D =[HA]org

[HA]aq + [A-]aq

HA

HA H+ + A-Ka

K


Section 1 19

[H+][A ]


Substitute for [A-] in D eq. and rearrange

Ka =[H+][A-]

[HA][A-] =

Ka [HA]

[H+]

D =[HA]

[HA]

2

1 K HA

Ha[ ]

[ ]1


D =[HA]

[HA]

[HA]

[HA]

2

1

2

1

K HA

HK

Ha a[ ]

[ ] [ ]1

1 H H[ ] [ ]

D =[HA]

[HA]2

1

1

K

Ha

[ ]= K


D = K1

K

Ha

[ ]

D =K

1

K

H

K H

H Ka a

[ ]

[ ]

[ ]

[H+]=Ka

pH=pK[H+]>>Ka


pH effect on D for organic acids

HK

][D

D

pH

pH pKaK

mainlyHA

[H+]<<Ka

mainlyA-

aKH ][

D


Example problem: Want to separate two organic acids using a scheme based on pH. Acid 1 (pKa = 4), Acid 2 (pKa = 8)

K1

4 8

K2

D

pH

2 (pKa 8)Acid 2 stays in organic phase, acid 1 is extracted into aqueous phase

[H+]=Ka[H+] K


Analogous treatment for organic bases (proton acceptors, not KOH)

[H+] + Ka

D =K Ka

D

pH

K

[ ] a

pH=pKa

[H+]>>Ka

mainlyBH+

[H+]<<Ka

mainlyB


Section 1 20

Kacid base


In general:

D

pH

K

Initial Aq. phase

pH=1, extract with ether

Separate organic acid, base and neutral

Aq. PhaseOrg. acid

Aq. PhaseOrg. base

Ether PhaseOrg. acid, Org. neutral

Ether PhaseOrg. neutral

extract with pH=12 Aq. Sol’n

What is Chromatography

Method to separate components in a mixture based on different Distribution coefficients between the two phases.

Same principle as solvent extraction, but one phase is stationary and one phase is “mobile”

Stationary phases are most commonly coated or packed in a column


Chromatography categorized on the basis of interaction between solute and stationary phase.

Mobile phase either gas or liquid Mobile phase either gas or liquid Stationary phase either liquid or

solid


Liq/Liq (Partition)Liq/Solid (Adsorption)Gas/Liq (Partition) Gas

Liquid Chromatography

Gas/Solid (Adsorption) Chromatography


Other modes of chromatography (Fig. 23-6)» Ion-exchange: separates charged

speciesspecies» Size exclusion or Gel Permeation:

separates molecular size» Affinity: separates on the basis of

antibody-anitgen, enzyme-substrate interactions


Section 1 21

Chromatography Basics

Typical chromatogram:

Det

ecto

rR

espo

nse

time or volume


tect

orpo

nse

tm

tr1

tr2

tr1

Det

Res

p

time or volumeinjection time

tm = time for mobile phase to travel length of column (dead time)

t r1

tr = retention time

tr = adjusted retention time = tr - tm

= tr2/tr1 = relative retention


Mobile phase flow rate:» volumetric flow rate (F): ml/min » linear flow rate (v): cm/min (mm/min)T d ib lTwo ways to describe solute “retention”» retention time, tr

» retention volume, Vr

Vr = Ftr


Partition coefficient K = Cs/Cm» C = Concentration of analyte» s = stationary phase

bil h» m = mobile phase Vs = volume of stationary phaseVm = volume of mobile phase


Capacity factor, k: a measure of retention» higher k = greater solute retention

k = K =VsVm

molessmolesm

k = =ts

tm

tr - tm

tm


What fraction of time does the solute spend in mobile phase?

q = fraction of solute in mobile phasephase


Section 1 22

CmVmCmVm + CsVs

q = =molesm

moless + molesm

1 1


q =1

1 +CsVsCmVm

q =1

1 + KVs

Vm

KVsVm

= capacity factor = k

1 + kq = 1


Fraction of time solute spends in mobile phase

1 + k

1 + k(1-q) = k

larger k means greater retention times Fraction of time solute spends in stationary phase = (1-q)

linear flow rate (cm/min)


Rate of travel of solute molecule through column:

rate = v1

1 + KVsVm

rate = v (fraction of time in mp)

linear flow rate (cm/min)

1

1 + k= v

L column length


Retention time, tr: time it takes solute to go from beginning to end of column.

tr = rate of solute travelL column length

v1

1 + KVsVm

tr =L

tm =Lv

tr = tm ( 1 + )K = tm (1+k’)VsVm


Retention time, tr:

m

Vr = Vm ( 1 + )K = Vm + KVsVsVm

Retention volume, Vr: multiply retention time by volumetric flow rate, F

Efficiency of Separation

Typical chromatogram:

Det

ecto

rR

espo

nse

time or volume


Section 1 23


Two factors affect how well two components are separated» difference in retention time

k idth» peak widths Solutes in a column spread into a

Gaussian profile:


Gaussian peak shape:

trtime

t0

w1/2=2.35 h

1/2hw=4

The resolution (separation) of two solutes:» resolution = = wavg wavg

tr Vr


tr = difference between retention times of two peaks tr = (tr2 - tr1)

wavg = average of the peak widths at baseline ( 4)

R=0.50 R=0.75


Resolution: higher R, better separation

timet0 timet0

timet0

R=1.00

timet0

R=1.50R1 is good


Plate theory:» treats separation in discrete stages,

more stages = more plates.Th ti l l t (N) bTheoretical plates (N): a number

indicating how good a column is for a separation


i ifi f h l

221

2

2

2

2

2 55.516

w

t

w

ttN rrr

N is specific for each solute on a

given columnIncreasing retention time increases N


Section 1 24

1N

R


relation of N to Resolution (R):

14

NR

=tr2tr1

2261

N R


N required to obtain a certain resolution:

21

timet0

N1

timet0

N2N2>N1


N depends upon the length of the column Independent of the column length is the

Height Equivalent of a Theoretical Plate

N

LHETP 2

2

16 rt

Lw

As HETP , resolution increases (N )

Causes of band broadening:

Why Bands Spread

Band broadening

Causes of band broadening:» Longitudinal diffusion» Resistance to mass transfer (RMT)» Eddy diffusion

Why Bands Spread

Longitudinal diffusion: solute [ ] is lower at the edges of a band; solute diffuses to the edges.

HB = B/v : B = constant, v = flow rate decrease HB by increasing v

Same Conc.at Equil.Low Conc.

Low Conc.

High Conc.

Resistance to mass transfer (RMT):Mobilephase

St ti

slow equil.

Why Bands Spread

bandwidth

Stationaryphase

bandwidth

HC = Cv : C = constant, v = flow rate decrease HC by decreasing v


Section 1 25

Why Bands Spread

Eddy diffusion (not simple diffusion):

time

HA = A : A = constant, depends on size of particles

Why Bands Spread

Van Deemter Equation:HETP = HA + HB + Hc

HETP = A + (B/v) + CvHETP = A + (B/v) + Cv

Why Bands Spread

van Deemter Plot:

H

v

Bv Cv

A

vopt

Hmin

Why Bands Spread

Asymmetric peak shapes: K depends on [ ] at high [ ] (solute becomes solvent)

Cs

Cm

Linear ideal peak shape

(-) tailed

(+) overloaded

(+) deviation:

slow slow

Observedchromatogram

Why Bands Spread

Asymmetric peak shapes

(+) deviation:

fast fast fast

(-) deviation:

slow

fastfast

slow slow

time

Chromatography

Gas Liquid

General Chromatography

GSC GLC LSC LLC IEC GPC AC

GC: volatile solutesLC: any mobile phase soluble solutes


Section 1 26

Gas Chromatography

Gas chromatography:» Analytes (volatile) are vaporized and

transported through the column» Gaseous mobile phase: (He N or» Gaseous mobile phase: (He, N2, or

H2) as long as mobile phase is inert, choice

is not critical» Stationary phase: non-volatile liquid

or solid particles

Gas Chromatography

Gas Chromatography

Schematic diagram of GC:

exit

detector

septum

carrier gas

injectionport

Tinj,det Toven + 50°Ccolumn oven

column

Gas Chromatograph

Gas Chromatograph Gas Chromatograph


Section 1 27

Gas Chromatograph Gas Chromatography

Solid stationary phase:» fine solid particles packed into

stainless steel tubing b ( b bl k) carbon (carbon black) SiO2 (silica gel), Al2O3 (alumina)

» analyte adsorbs directly on solid particles

» gives strong retention of solutes (large surface area)

Gas Chromatography

Liquid stationary phase:» non-volatile liquid coated on inside of

column or on fine solid support » ideal characteristics of solid support:» ideal characteristics of solid support: strong, porous, high surface area and

inert (non-adsorptive)» real example of solid support: diatomaceous earth (algae skeletons)

Gas Chromatography

Liquid stationary phase cont’d:» ideal characteristics - wide liquid range

(min/max temp limits), stable, low volatility, low viscosity (D and RMT)

» examples of liquid stationary phases : silicones: non to strong polarity depending on R groups

R

O Si O Si

R R

R n

Gas Chromatography

Liquid stationary phase cont’d:» examples of liquid stationary phases carbowax: strong polarity

CH CH O)—CH2CH2O)n— other examples in Table 24-1

» chose stationary phase to match polarity of solute (like dissolves like) solute polarities listed in Table 24-2

Gas Chromatography

Columns: packed» ~ 3-6mm inner diameter

tubing, 1-5 m long » used for preparative

iseparations or to separate gases that are poorly retained

» lower resolution» small, uniform particle

size decreases Eddy diffusion (requiring higher pressures)


Section 1 28

Gas Chromatography

Columns: open tubular » 0.1-0.5 mm inner dia.,

10-100 m long10-100 m long» 0.1-5 m thick sp

coated on inner walls » higher resolution,

shorter analysis times, greater sensitivity compared to packed columns

Gas Chromatography

Detectors

Thermal Conductivity (most common):

e-

He has high thermal conductivity, is main component entering detector

gas

Detectors

Detectors

Thermal Conductivitye-

analyte appears, T.C., Temp , Wire Resistance , detector current “read-out”

gas

Detectors

Thermal Conductivity» universal» non destructive» non-destructive» linear range (>105) » Detection limit is 400 pg


Section 1 29

Flame Ionization Detector (FID):

cathode (collectsCHO+ ions)

Detectors

column effluent

H2

air

CHO ions)

anode

Detectors

Flame Ionization Detector (FID):» organic solutes are burned in flame

producing CH radicals and eventually CHO+CHO

» CH.

+ O. CHO+ + e-

» CHO+ ions are collected by cathode, produces current as the response

Detectors

Flame Ionization Detector (FID):» organics (reduced carbon only)» Destructive

li ( 7)» linear range (>107)» detection limit is 2 pg

Radioactive-emitter

Ni63 +insulator

Detectors

Electron Capture (ECD):

electrodes

Ni63 +-

+-

Detectors

Electron Capture (ECD):» - + gas - + gas+ + e- (standing

current)» e- + solute solute- (e- capture)

time

Cur

rent

(I)

standing current

» e + solute solute (e capture)» solute- + gas+ solute + gas

Detectors

Electron Capture (ECD):» Non-destructive» non-linear

l i i l» selective to e- capturing solutes» detection limit 5 fg


Section 1 30

Detectors

Variations of FID:» Flame Photometric» Alkali Flame

S lf Ch il i» Sulfur Chemiluminescence Characteristics in Table 24-5

Qualitative Analysis

All previous detectors non-specific retention times alone can’t identify a

compound» Mass spectrometer» Fourier Transform Infrared Spectrometer» compare spectra obtained by these two

detectors with known sample spectra (fingerprint)


Problems with comparing retention times for qualitative analysis:» tr dependent upon: LLvK f(temp)Vs, Vm f(packing)

» Can compare tr to retention of standard solutes, typically alkanes

rnrxtltltlogt log

n)-(Nn 100I


Kovats retention index (I):

rnrN tlogt log

)(

n = # of C atoms in smaller alkaneN = # of C atoms in larger alkanetrx = adjusted retention time of unknowntrn = adjusted retention time of smaller alkanetrN = adjusted retention time of larger alkane


Kovats index for linear alkanes equals 100 times the number of carbon atoms

Time (min)

3.50.25

Det

ecto

rR

espo

nse C7

C8 C9

xair

5.6 7.46.3


Using Kovats retention index:

rnrx tlogt logn)(Nn100I

0.25)-6.5log(0.25)-4.7( log0.25)-6.5log(0.25)-(6.3 log

8)-(9 8100 I

I = 842

rnrN tlogt log

n)-(Nn 100I


Section 1 31

Low T

Res

pons

e

C6 C7C9

air C8

Temperature Programing

RR

espo

nse

C6 C7 C9air

C8 C10 C11 C12

High T

Res

pons

e

C6 C7 C9air

C8 C10 C12C11 C13 Ttime


Preparative LC- separates milligrams to grams of analyte

Analytical - separate micrograms to picograms HPLCpicograms - HPLC

High Performance LC» Liquid/Solid (LSC)» Liquid/Liquid (LLC)


Small diameter packing (stationary phase):» provides more uniform flow (A )» less distance for solutes to diffuse in» less distance for solutes to diffuse in

mobile phase to interact with stationary phase (C )

» sacrifice: much higher pressure is required to “push” mobile phase through column (~3000psi = 200 atmospheres)

10m) 40

60

Packing


0 2 4 6 8

H (m

)

Flow rate (ml/min)

20

40

10

5m

3m

Packingdiameter

Liquid Chromatography Liquid Chromatography

Liquid-Solid chromatography (LSC): (adsorption)» Stationary phases: very porous =

high surface area for interaction withhigh surface area for interaction with mobile phase silica gel: SiO2 xH2Oalumina: Al2O3 xH2O


Section 1 32

OH OH


Silica - OH groups very polar

O Si O Si

O nO

O Si O Si

O

O Si OH

OH

5SiO2 2H2O


Liquid-Solid chromatography (LSC):» Mobile phases: solvent displaces solute

from stationary phase, rather than solute partitioningpartitioning

» Elutropic series: relative ability of solvent to displace solute (Table 25-2)Methanol>acetonitrile>chloroform>hexane

» The greater the eluent strength, , the more rapidly solutes will be eluted


Toluene °=0.22


Elutropic elution:

Start w/ 100% Benzene then add Acetonitrile

“gradientelution”

Acetonitrile °=0.52

“isocraticelution”


Gradient elution - composition of the mobile phase changes with timesolvent A = benzene, solvent B =

methanolmethanol

% A

time


LSC:» can use thin-layer plates (TLC):

cheap, easy to do and learn, simultaneous separations, but poor quantification, reproducibility and resolution


Section 1 33


Liquid-liquid chromatography (LLC):» Stationary phase: viscous liquid

coated or chemically bonded tocoated or chemically bonded to exposed silanol groups of silica (solid support)

Si OH + Cl Si R

CH3

CH3

Si O Si R

CH3

CH3


Common liquid stationary phases:» Polar phases: Amino [R = (CH2)3NH2 ] Cyano [R = (CH2)3CN ]Cyano [R (CH2)3CN ] Diol [R = (CH2)2OCH2CH(OH)CH2OH ]

» Non-Polar phases: Octadecyl, C18 [R = (CH2)17CH3 ] Octyl [R = (CH2)7CH3 ] Phenyl [R = (CH2)3C6H5 ]


“Normal” phase LC:» Polar stationary phase» Non-polar mobile phase» Order of elution: non polar polar

Like dissolves Like!

» Order of elution: non-polar polar» Solvent strength: non-polar (weak), polar

(strong)


“Reversed” phase LC:» Non-polar stationary phase» Polar mobile phasep» Order of elution: polar non-polar» Solvent strength: non-polar (strong),

polar (weak)

Like dissolves Like!


“Classical” vs HPLC:Packing Pressure N

classical: 50 m gravity 500HPLC 3 5 3000 i 15 000HPLC: 3-5 m 3000psi 15,000

» HPLC: higher resolution, faster and good quantification, but expensive and complex

Pump I

Injector

LC Instrumentation

Schematic diagram of HPLC:

Pump

Pump

Solvent B

isocratic: 1 solvent

Column Detector

I

Solvent A

To waste or fraction collector

gradient: 2 or more solvents


Section 1 34

LC Instrumentation

Injection valvefrom pump

to column

Injection port

Sample loopwaste

LC Instrumentation

Primarily use packed columns although capillary columns are gaining in popularity

» 5-30 cm long» 1-5 mm i.d. » Often use a guard column to protect

main column

Eluate in

LC Instrumentation

UV spectrophotometric Detectors:

Lightsource

Eluate out

Photo Detector

LC Instrumentation

UV Spectrophotometric Detectors:» linear (105 range in solute [ ] )» 0.1-1 ng detection limit

f i l i l ( l li i d)» fairly universal (solvent limited)» can handle gradients

Fluorescence Detectors are 100x more sensitive, but less universal

LC Instrumentation

Mass Spectrometer Detectors:» linear (105 range in solute [ ])» sensitive (0.1-1 fg detection limit)

h dl di» can handle gradients» universal » response is fairly specific for each

compound» often used in conjunction with UV det.

Commercial LC Instrument


Section 1 35

Ion Exchange

Partition solute between solvent and charged sites on stationary phase» solute must be ionic

O it tt tOpposites attract» anion exchange - positively charged

stationary phase» cation exchange - negatively charged

stationary phase

Ion Exchange

Stationary phase is a cross-linked copolymer » vinyl benzene/divinylbenzene

i l b l t» vinyl benzene polystyrene» divinylbenzene is cross linker - links

polystyrene chains together

Ion Exchange

Stationary phase

CH CH2CH CH2

Vinyl benzeneCH CH2

divinyl benzene

Use 1 - 16% divinyl benzene, amount changes pore size, rigidity - cross linking, selectivity

Ion Exchange

CH CH2 CHCH CH2

Stationary phase - cross linked polymer

CH CH2 CHCH CH2 CH CH2

R

Ion Exchange Ion Exchange


Section 1 36

Ion Exchange

Binding of analyte depends upon» Magnitude of chargeCoulombs Law - F q1q2/r2

3+ >2+ > 1+3+ >2+ > 1+

» size of hydrated ion (related to r in Coulomb eq.)

Ion Exchange

Selectivity coefficient

R-Na+ + Li+ R-Li+ + Na+

KR Li Na

R Na Li

[ ][ ]

[ ][ ]

Selectivity coefficients relative to Li = 1.00

Ion Exchange Ion Exchange

Gradient elution with increasing ionic strength or pH change

KAg > KNaI i l [Li+] l N +» Increasing solvent [Li+] elutes Na+

before Ag+

» Note that Li+ and H+ are the most weakly bound ions, but can still use them to elute other ions if their [ ] is sufficiently high

Ion Exchange Molecular Exclusion

More commonly called Gel Permeation Chromatography (GPC)

Also called Size Exclusion Chromatography (SEC)Chromatography (SEC)


Section 1 37

Molecular Exclusion

Stationary phase contains small pores that analytes can diffuse into (depending upon size)

Larger molecules cannot fit intoLarger molecules cannot fit into pores so they elute faster than smaller molecules

Molecular Exclusion

Molecular Exclusion

Stationary Phase» pore size determines range of MW

which can be separated (see Table 26-4)4)

» exclusion limit: smallest molecule which can not fit into the pores; any larger molecule will have same Vr

Molecular Exclusion

Stationary phase typically a cross-linked polymer

Common stationary phases» Sephadex (glucose/glycerol polymer Fig» Sephadex (glucose/glycerol polymer, Fig.

26-2)» Bio-Gel (polyacrylamide gel, Fig. 26-12)

Molecular Exclusion

Stationary Phase - solvent trapped in pores (Vs)

Mobile Phase - solvent outside of porespores

Mobile Phase Volume (Vm) = Void Volume (V0)

Molecular Exclusion

smr KVVV

sr KVVV 0

s

r

V

VVK 0

If solute is larger than pores,

Vr = V0 and K=0


Section 1 38

Molecular Exclusion

sT VVV 0

0Ts VVV

0

0

VV

VVK

T

rav

If solute is small and penetrates all pores,

Vr = VT and K=1

Molecular Exclusion

For intermediate sized molecules, 0 < K < 1If K > 1, analyte has adsorped to sites

lon columnVT = volume of column

Molecular Exclusion

Applications of GPC» Molecular weight determination: separation based on hydrodynamic

radii hich is ro ghl correlated to MWradii which is roughly correlated to MW » Desalting - remove salt impurities

from biological samples

Molecular Exclusion

Molecular weight determination:» compare Vr of unknown to that of

standards with known MW and similar str ct re

Vr

Log MW

2

4

6 Different classesof stationary phase

similar structure

Affinity Chromatography

Makes use of tight, specific complex (highly selective):» antigen/antibody

/ b t t» enzyme/substrate Very powerful method of

purification in biology

support


Process

supportlinkerarm affinity

ligand


Section 1 39


Can release bound species by changing pH, temp, ionic strength, etc.

Example chromatogram:

change conditions

“garbage”

desired solute

Example chromatogram:

Electrophoresis

Separation is based on differences in migration of charged ions in an electric field in solution

typically carried out on buffer soaked

(+) anode (-) cathode

-

+

++n

typically carried out on buffer soaked paper or gel

Electrophoresis

Rate of migration dependent upon: size of solute

» larger size, more friction, slower tmovement

charge of solute» greater charge, increased force

(Coulomb’s law), faster migration

Electrophoresis

Rate of migration: » electrophoretic velocity, Vep = epEep: electrophoretic mobility = q/f

h i C l bq = charge in Coulombs f = friction coefficient (f = 6r, Stokes

Eq.)E = electric field strength (volts/cm)

fused silica

flow

Capillary Electrophoresis

Schematic:

fused silica capillary

50cm

HV(30kV)

pressurized

Commercial CE instrument


Section 1 40

fused silicapolyimide

coating


Fused silica capillary:

330m dia.g

15m thick

opening25-75m dia.


electroosmotic flow:» fused silica has exposed silanol

groups (Si-OH); deprotonated at pHs 2

- - - - - - - - - - - - -

- - - - - - - - - - - - -

+ + + + + + + + + + + + +

+ + + + + + + + + + + + + (+) anode

(-) cathode

flow

2 » Causes a layer of + charge to build up

at surface

O- O- O- O- O- O- O- O- O- O- O- O- O- O- O- O-

O-O-O-O-O-O-O-O-O-O-O-O-O-O-O-O-


electroosmotic flow:» surface charges move to cathode,

dragging solution» surface charge “removes” friction

electroosmotic flow profile hydrodynamic flow profile

laminar

» surface charge removes friction between solvent and walls of capillary


electroosmotic flow velocity Veo = eoE:

» eo: electroosmotic mobility ( surf. eo y (charge dens. 1/(ionic strength)1/2

» E = electric field strength (volts/cm) apparent mobility: app = ep + eo

apparent velocity: Vapp = appE


Vt

LL

LV

tL

Etd

t

dappapp

V

Ld = length of column from injector to detectort = migration timeLt = total length of columnV = voltage applied to column


Veo Vep Vapp

(+) anode (-) cathode

-

+

n


Section 1 41


Efficiency:» no particles so no multiple paths (A = 0)» no stationary phase so no RMT (C = 0)» HETP = B/v» HETP B/v» can increase velocity by applied voltage,

but due to resistance, this generates heat which longitudinal diffusion (B)

» still get high efficiencies (N=105-106)


LLtLV

21

2

2

2 DtL

N

Vt

LL

LV

tL

Etd

t

dappapp

V

2D

VN app

Capillary Electrophoresis Capillary Electrophoresis

Sample injection» Very small sample sizes - 10-9 liters» Hydrodynamic injection, uses pressure

to force sample onto columnto force sample onto column» electrokinetic - based on ep thus

injected sample has different relative composition of analytes


Detection:» UV/Fluorescence» Mass SpectrometryCh i i f CECharacteristics of CE» very fast» high efficiencies» especially useful for biopolymers

document1

Documents

class final exam

class chpt

class chem

class exams

class logistics exams

final exam grade

exam automatic

eachfinal exam