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

SpectroscopyBeer’s Law

Light intensity in = Po

Light intensity out = P

Transmittance = T = P/Po

Absorbance = A = -logTLight 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)

b

ε = constant for given compound at specific λ value

sample in cuvette

SpectroscopyBeer’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?

SpectroscopyMore on Beer’s Law

• Useful for determination of analyte concentrations

• Some limitations– 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

0

0.05

0.10.15

0.2

0.25

0.3

0.350.4

0.45

0.5

0 0.005 0.01 0.015

Total HIn Conc.

Ab

sorb

ance

Example of deviations to Beer’s Law: Unbuffered Indicator with ε(In-) = 300 M-

1 cm-1, ε(HIn) = 20 M-1 cm-1; pKa = 4.0

HIn ↔ H+ + In-

SpectroscopySpectrometers

light source

sample in cuvette

light discriminator: monochromator (passes only a small range of wavelengths)

light detector – measures light intensity by converting it to an electrical signal

Data processor

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)

SpectroscopyExample Measurement: Ozone

• Ozone (O3) is a pollutant (lower atmosphere) and in stratosphere provides UV protection

• Instrument is used for measurement at station or in airplane– compares absorbance through

sample cell vs.– absorbance through reference cell

• Can also make measurements remotely (e.g. absorbance between two skyscrapers)

light source (l = 254 nm)

light detector

air in

O3 scrubbersa

mpl

e ce

ll

refe

renc

e ce

ll

chopper

Chromatography – Ch. 22Introduction

• 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

ChromatographyPartitioning – 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)

aq

org

X

XK

][

][ K = partition

coefficient (a constant)

note: technically, upper conc. is for “raffinate” phase while lower is for “extractant” phase

ChromatographyPartitioning

• Partitioning coefficient depends on stability in solvents (related to solubility in solvents)

• Most common rule is likes dissolve likes• Example of water – hexane partitioning

• 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)

OHOH

larger K Although both have K > 1, the OH group makes right molecule more polar (favors water more vs left)

ChromatographyPartitioning

• 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

Compound A Compound B

K = 7.59 K = 6.17

pKa = 4.62Not very acidic

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

OCH3 CH3

O

OH

ChromatographyPartitioning - Questions

1. 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?

2. 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?

3. 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?

ChromatographyPartitioning 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)

Column Wall

Mobile phase

Stationary phase (wall coating)

Packed column (side view) (e.g. Silica in normal phase HPLC)

Packing Material (solid)

Stationary phase is surface (larger area than shown because its porous)

Bonded phase (liquid-like)Expanded View

Stationary Phase

Chemically bonded to packing material

Packing Material

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]