Chem. 31 – 4/6 Lecture
Announcements I
• Exam 2 – Next Monday
– Covering Ch. 6 (topics since exam 1), 7, 8-1, 17,
and parts of 22 (parts in today’s lecture – even if
we don’t get to them today)
– Will review topics on Wednesday
• Lab Reports
– Water Hardness lab report resubmissions due
Wednesday
– AA report due Wednesday 4/15 (delay possible)
Announcements II
• Homework Set 2
– Set 2.3 set shortened and posted
• Today’s Lecture
– Chapter 17 Spectroscopy
• Beer’s Law/Basics on Instrumentation
– Chapter 22: Chromatography
• Overview
• Partitioning Between Two Phases
• Partitioning and Retention in Chromatography
Spectroscopy
Beer’s Law
Transmittance = T = P/Po
Absorbance = A = -logT
sample in cuvette
Light source
Absorbance used because it is
proportional to concentration
A = εbC
Where ε = molar absorptivity
and b = path length (usually in
cm) and C = concentration (M)
ε = constant for given
compound at specific λ value
Light
intensity
in = Po
b
Light
intensity
out = P
Spectroscopy
Beer’s Law Question
• Half of the 284 nm light is absorbed when
benzoic acid at a concentration of 0.0080
M is in a cuvette with a path length of 0.5
cm. What is the molar absorptivity of
benzoic acid at this wavelength?
Spectroscopy
More on Beer’s Law
– Law not valid for high
concentrations
– Deviations to law appear to
occur when multiple
wavelengths of light used
or when multiple species
exist but absorb light
differently
– Uncertainties are lowest
when 0.1 < A < 1
Example of deviations to Beer’s Law:
Unbuffered Indicator with ε(In-) = 300 M1 cm-1, ε(HIn) = 20 M-1 cm-1; pKa = 4.0
HIn ↔ H+ + In-
Absorbance
• Useful for determination
of analyte concentrations
• Some limitations
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.005
0.01
Total HIn Conc.
0.015
Spectroscopy
Spectrometers
sample in
cuvette
light detector –
measures light
intensity by converting
it to an electrical signal
Data
processor
light source
light discriminator:
monochromator
(passes only a
small range of
wavelengths)
Components can look very different in different types of
spectrometers, but spectrometers will have all of the major
components (except other methods of wavelength discrimination
may replace monochromators)
Spectroscopy
Example Measurement: Ozone
• Can also make measurements
remotely (e.g. absorbance between
two skyscrapers)
light source
(l = 254 nm)
reference cell
– compares absorbance through sample
cell vs.
– absorbance through reference cell
sample cell
• Ozone (O3) is a pollutant (lower
chopper
atmosphere) and in stratosphere
air in
provides UV protection
• Instrument is used for measurement
at station or in airplane
O3 scrubber
light detector
Chromatography – Ch. 22
Introduction
• Purpose of Chromatography
– To separate and detect components of a mixture
– Analytical chemists are more interested in the detection part
• Advantages of Chromatography
– Can handle more complex samples than typical spectroscopic
methods
– Also results in purification of mixtures (if desired)
• Disadvantage of Chromatography
– Separation takes time (so generally not as fast as pure
spectroscopic methods)
• Basis for Separation:
– differential partitioning between a stationary and a mobile phase
Chromatography
Partitioning – Ch. 22 Sect. 1
• Covering to understand
partitioning in
chromatography
• Partitioning can occur
between any two phases
(as long as one phase is
a fluid)
• Liquid-liquid is chosen as
an example
• Partitioning governed by
equilibrium equation
X(org)
X(aq)
K
[ X ]org
[ X ]aq
K = partition
coefficient (a
constant)
note: technically, upper conc. is for “raffinate” phase
while lower is for “extractant” phase
Chromatography
Partitioning
• Partitioning coefficient
depends on stability in solvents
(related to solubility in
solvents)
• Most common rule is likes
dissolve likes
• Example of water – hexane
partitioning
HO
larger K
OH
Although both have K > 1,
the OH group makes right
molecule more polar (favors
water more vs left)
• Other Effects on
Partitioning
– K gives distribution if
compound does not react
further in either phase
– However, compounds may
react further (e.g. acid HA
→ H+ + A- in aqueous phase)
– Ions (e.g. A-) will be found
almost exclusively in
aqueous phase
– Distribution coefficient (D)
gives ratio of total species
concentration (only
covering qualitatively)
Chromatography
Partitioning
• Example of Effect of
Aqueous Reactions on
Compound Distribution
• Compound A is nearly as
polar as B
• However, acidity affects
distribution between
water and organic layer
• Compound B will undergo
dissociation in water:
HA ↔ H+ + A-
• Distribution of B given by:
D = [HA]org/{[HA] + [A-]}aq
O
H3C
H3C
O
Compound A
K = 7.59
Not very
acidic
OH
Compound B
K = 6.17
pKa = 4.62
If aqueous phase is buffered at pH >
pKa (e.g. pH = 6), most of B will be in
anion form and very little of B will be
in organic phase
With a low pH buffer, D ~ K
Chromatography
Partitioning - Questions
1.
2.
3.
A compound with an octanol water partition coefficient of 52 is
placed in a separatory funnel with water and octanol and shaken.
The concentration of it in octanol is found to be 0.150 M. What is
its concentration in water?
It is desired to separate the following two compounds:
CH3(CH2)3OH and CH3(CH2)3NH2.
The two compounds have similar KOW values (around 11) but the
second compound is basic. What can be done to separate the
two?
It is desired to transfer butanol (left compound in #2) from water
to an organic phase. Would it be transferred most efficiently
using 1-octanol, a less polar solvent (e.g. octane), or a more
polar solvent (e.g. 1-hexanol) as the organic solvent?
Chromatography
Partitioning in Chromatography
• Separation Occurs in Column
• Partitioning Requires Two Phases:
– Mobile phase
•
•
•
•
fluid flowing through the column
type of fluid determines type of chromatography
fluid = gas means gas chromatography (GC)
liquid chromatography (high performance liquid
chromatography or HPLC)
• supercritical fluid (SFC) [supercritical fluid = fluid at high
temperature and pressure with properties intermediate
between liquid and gas]
– Stationary phase (solid or liquid within column)
• most commonly liquid-like substance on solid support
Chromatography
More on Stationary Phases
Open Tubular – in GC
(end on, cross section view)
Packed column (side view) (e.g.
Silica in normal phase HPLC)
Column Wall
Mobile phase
Stationary phase
(wall coating)
Expanded View
Stationary Phase
Chemically bonded to
packing material
Packing Material
Packing Material (solid)
Stationary phase is surface (larger
area than shown because its
porous)
Bonded phase (liquid-like)
Chromatography
Flow – Volume Relation
• Relationship between volume (used with gravity
columns) and time (most common with more
advanced instruments):
V = t·uV
V = volume passing through column part in time t at
flow rate uV
Also, VR = tR·uV where R refers to retention
time/volume (time it takes component to go through
column or volume of solvent needed to elute
compound)
Chromatography
More on Volume
• Hold-up volume = VM = volume occupied
by mobile phase in column
• Stationary phase volume = VS
• Calculation of VM:
VM = tM·uV, where tM = time needed for
unretained compounds to elute from column
Chromatography
Partition and Retention
• Partition Coefficient = K = [X]S/[X]M
• K is constant for X if T and/or solvent remain
constant
• K is not used that frequently in chromatography
• Retention Factor = k = main measure of
partioning/retention in column
• k = (moles X)S/(moles X)M = K(VS/VM)
• Retention Factor is more commonly used
because of ease in measuring, and since VM/VS
= constant, k = constant·K (for a given column)
• Note: kColumn1 ≠ k Column2 (if VM/VS changes)
Chromatography
Definition Section – Partition and Retention
• Since the fraction of time a solute
molecule spends in a given phase is
proportional to the fraction of moles in
that phase,
k = (time in stationary phase)/(time in
mobile phase)
• Experimentally, k = (tR – tM)/tM
• Note: t’R = tR – tM = adjusted retention
time
Chromatography
•
•
•
•
Definition Section – Relative Retention
NOT ON EXAM 2
For a separation to occur, two compounds, A
and B must have different k values
The greater the difference in k values, the easier
the separation
Relative Retention = a = kB/kA (where B elutes
after A) = measure of separation ease =
“selectivity coefficient”
a value close to 1 means difficult separation
Chromatography
Reading Chromatograms
• Determination of parameters from reading chromatogram (HPLC
example)
• tM = 2.374 min. (normally determined by finding 1st peak for
unretained compounds – contaminant below)
• VM = uV·tM = (1.0 mL/min)(2.37 min) = 2.37 mL
• 1st peak, tR = 4.958 min.; k = (4.958 - 2.374)/2.374 = 1.088
• a (for 1st 2 peaks) = kB/ kA = tRB’/ tRA’ = (5.757 – 2.374)/(4.958 –
2.374) = 1.31 [NOT ON EXAM 2]