Chem 241 - Summer 2010 Section 1 1 Chemistry 241 Professor: Gary L. Glish Office: Caudill 320 Phone: 962-2303 email: [email protected]email: [email protected]Office 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 University 1980-1992 Oak Ridge National Laboratory 1992 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 14 th 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
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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
Chem 241 - Summer 2010
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
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
Chem 241 - Summer 2010
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)
Chem 241 - Summer 2010
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?
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Gaussian Distribution
curve characterized by two parameters» arithmetic mean
t d d d i ti x s
Gaussian Distribution
» standard deviation s xi
Gaussian Distribution
Arithmetic mean
xn
x
n
i
i
value of measurement i
number of measurements
Chem 241 - Summer 2010
Section 1 7
x xi
21
2
Gaussian Distribution
Standard deviation
s
x x
n
n
i
1
1 degrees of freedom
s variance2
Gaussian Distribution
standard deviation is a measure of the width of the distribution - how close the values are clustered about the meanmean
ss
Gaussian Distribution
standard deviation of the mean
reduce standard deviation of the mean by making more measurements
sn
x 1 2
Gaussian Distribution
Gaussian equation
y e
1
21
2
2 2
Gaussian Distribution
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
Gaussian Distribution
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
Chem 241 - Summer 2010
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
Gaussian Distribution
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
Confidence Intervals
xts
n
Confidence Intervals
“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
Chem 241 - Summer 2010
Section 1 9
Confidence Intervals
p. 58
Confidence Intervals
Increasing the level of confidence increases the range of values
Increasing the number of measurements decreases the range ofmeasurements decreases the range of values
Confidence Intervals
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. .
Confidence Intervals
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Least Sq. - Uncertainty
21
22
D
xss
D
iyb
m
Least Sq. - Uncertainty
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
Least Sq. - Uncertainty
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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:
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
Solvent Extraction (pH effects)
Partitioning of organic acids between two phases:
organic
HA H+ + A-Kaaqueous
HA H+ + A-have poor solubility
Solvent Extraction (pH effects)
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)
Solvent Extraction (pH effects)
D =total conc. in phase 2total conc. in phase 1
D =[HA]org
[HA]aq + [A-]aq
HA
HA H+ + A-Ka
K
Chem 241 - Summer 2010
Section 1 19
[H+][A ]
Solvent Extraction (pH effects)
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
Solvent Extraction (pH effects)
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
Solvent Extraction (pH effects)
D = K1
K
Ha
[ ]
D =K
1
K
H
K H
H Ka a
[ ]
[ ]
[ ]
[H+]=Ka
pH=pK[H+]>>Ka
Solvent Extraction (pH effects)
pH effect on D for organic acids
HK
][D
D
pH
pH pKaK
mainlyHA
[H+]<<Ka
mainlyA-
aKH ][
D
Solvent Extraction (pH effects)
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
Solvent Extraction (pH effects)
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
Chem 241 - Summer 2010
Section 1 20
Kacid base
Solvent Extraction (pH effects)
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
What is Chromatography
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
What is Chromatography
Liq/Liq (Partition)Liq/Solid (Adsorption)Gas/Liq (Partition) Gas
Liquid Chromatography
Gas/Solid (Adsorption) Chromatography
What is 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
Chem 241 - Summer 2010
Section 1 21
Chromatography Basics
Typical chromatogram:
Det
ecto
rR
espo
nse
time or volume
Chromatography Basics
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
Chromatography Basics
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
Chromatography Basics
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
Chromatography Basics
Capacity factor, k: a measure of retention» higher k = greater solute retention
k = K =VsVm
molessmolesm
k = =ts
tm
tr - tm
tm
Chromatography Basics
What fraction of time does the solute spend in mobile phase?
q = fraction of solute in mobile phasephase
Chem 241 - Summer 2010
Section 1 22
CmVmCmVm + CsVs
q = =molesm
moless + molesm
1 1
Chromatography Basics
q =1
1 +CsVsCmVm
q =1
1 + KVs
Vm
KVsVm
= capacity factor = k
1 + kq = 1
Chromatography Basics
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)
Chromatography Basics
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
Chromatography Basics
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
Chromatography Basics
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
Chem 241 - Summer 2010
Section 1 23
Efficiency of Separation
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:
Efficiency of Separation
Gaussian peak shape:
trtime
t0
w1/2=2.35 h
1/2hw=4
The resolution (separation) of two solutes:» resolution = = wavg wavg
tr Vr
Efficiency of Separation
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
Efficiency of Separation
Resolution: higher R, better separation
timet0 timet0
timet0
R=1.00
timet0
R=1.50R1 is good
Efficiency of Separation
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
Efficiency of 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
Chem 241 - Summer 2010
Section 1 24
1N
R
Efficiency of Separation
relation of N to Resolution (R):
14
NR
=tr2tr1
2261
N R
Efficiency of Separation
N required to obtain a certain resolution:
21
timet0
N1
timet0
N2N2>N1
Efficiency of Separation
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
Chem 241 - Summer 2010
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
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
Chem 241 - Summer 2010
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)
Qualitative Analysis
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
Qualitative Analysis
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
Qualitative Analysis
Kovats index for linear alkanes equals 100 times the number of carbon atoms
Preparative LC- separates milligrams to grams of analyte
Analytical - separate micrograms to picograms HPLCpicograms - HPLC
High Performance LC» Liquid/Solid (LSC)» Liquid/Liquid (LLC)
Liquid Chromatography
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
Liquid Chromatography
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
Chem 241 - Summer 2010
Section 1 32
OH OH
Liquid Chromatography
Silica - OH groups very polar
O Si O Si
O nO
O Si O Si
O
O Si OH
OH
5SiO2 2H2O
Liquid Chromatography
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
Liquid Chromatography
Toluene °=0.22
Liquid Chromatography
Elutropic elution:
Start w/ 100% Benzene then add Acetonitrile
“gradientelution”
Acetonitrile °=0.52
“isocraticelution”
Liquid Chromatography
Gradient elution - composition of the mobile phase changes with timesolvent A = benzene, solvent B =
methanolmethanol
% A
time
Liquid Chromatography
LSC:» can use thin-layer plates (TLC):
cheap, easy to do and learn, simultaneous separations, but poor quantification, reproducibility and resolution
» surface charge removes friction between solvent and walls of capillary
Capillary Electrophoresis
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
Capillary Electrophoresis
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
Capillary Electrophoresis
Veo Vep Vapp
(+) anode (-) cathode
-
+
n
Chem 241 - Summer 2010
Section 1 41
Capillary Electrophoresis
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)