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Section III - Spectroscopy

Chp 18Sections - allProblems - all as needed

Chp 20Sections - 20.1-3, 6Problems – 20.1-14, 26, 27

Chp 19Sections 19.1,2 5,6Problems 19.1-12, 16-21 (the problems in this chp are less than ideal)

Chp 21Pages 494-498, 502 (Boltz Dist), 505 (HCL), 508 (LOD), 509, 510Problems 21.1, 7, 9, 12, 15, 16, 17

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The Nature of Electromagnetic Radiation

Is EM radiation a Wave?

Two ways to transfer energy – waves & particles

v = v λ or c = v λv = velocity c = speed of light in vacuum

v = c/n (n is refractive index)v = frequency λ = wavelength

Is EM radiation a particle?

PhotonsE = h v

H is Planks Constant6.6 x 10-34 Joule sec

2-D profiles ofA cone as ananalogy

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The EM Spectrum

The EM Spectrum

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λcv =frequency

speed of light

wavelength λhchvE ==

energy

Planck’s constant

Recall that light acts as both a wave and a particle

•Frequency and wavelength are inversely proportional

•Higher frequency light has more energy per photon

•Longer wavelength light has less energy per photon.

Modes of Interaction Between Analytes & EM Radiation

Molecular & Atomic- Electronic transitions- Vibrational transition-Rotational transitions

Molecular fluorescenceand phosphoresenceand scattering

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Electronic TransitionsFormaldehyde as a Case Study

Electronic TransitionsIn the UV and Visible Spectrum

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Vibrational TransitionsIn the infrared Spectrum

3N – 6 modes

Some Chapter 18 Problems

2x

1/22x

E = h ν = h c/λ = E = 6.6x10-34 Js (3.0x1017nm/s / 650 nm)E = 3.05x10-19 J (x Av # yields)E = 1.84x105 J or 184 kJ

More than in (a) by factor of 650/400

ν = c / λ = 3.0x1017 nm/s / 562 nmν = 5.33x1014 s-1 (Hz)

Rot. Vibr. Electronic

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Spectrophotometry

Light In Measure Light Out

• The amount of light absorbed by a chemical species in the sample is used to determine the concentration of that species.

Beer’s Law…. A = εbcAbsorbance Concentration

Sample

Spectrophotometry

Light In Measure Light Out

Sample

Absorbance to determine concentration is typically measured using light at a single wavelength (λ) (monochromatic light).

Absorbance measurements can be made made at a series of different wavelengths. A plot of absorbance versus λ for a compound is the absorbance spectrum for that compound.

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Measuring AbsorbanceSample

LightSource

LightDetector

WavelengthSelector(Monochromator)

P0

b (often 1.00 cm)

P

Absorbance, PP

PPTA 0

0

logloglog =−=−=

Absorbance, (Beer’s Law) bcA ε=A – no units; ε M-1cm-1; b pathlength in cm; C –concentration in M

Beer’s Law

bcA ε=Absorbance is linear with concentration (very convenient for calibration)

Absorbances are additive: AT = A1 + A2 + A3

Beer’s Law can fail (become non-linear) for a variety of reasons

Chemical Deviations (shifts in equlibria)Fundamental (reflectivity function of refractive index) Polychromatic lightMultiple pathlengthsStray light A = log (P0 + Ps) / (P + Ps)

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Some Chapter 18 Problems

A simultaneous determination for cobalt and nickel can be based upon absorption by their respective 8-hydroxyquinolinol complexes. Molar absorptivities corresponding to their absorption maxima are as follows:

Molar Absorptivity (M-1 cm-1)365 nm 700 nm

Co 3529 428.9

Ni 3228 10.2

Calculate the molar concentration of nickel and cobalt in each of the following solutions based upon the following data:

Absorbance (1.00-cm cells)365 nm 700 nm

Solution A 0.598 0.039

Solution B 0.902 0.072

Some Chapter 18 Problems

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(Problem From Other text)

(Problem From Other text)

IBP ~530nm

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(Problem From Other text)

Suppose that absorbance had been givenBut Kin had not. Solve for Kin!

Instead of starting withA = εHIn CHIn + εIn Cin

WriteKin = [H+][In-] / [Hin]

Luminescence If = K C

Advantages:• low background• both excitation & emission spectra for selectivity• temporal properties can be exploited

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Luminescence If = K C

Why is Laser Induced Fluorescence so popular?“The P0Term”

(Problems From Other text)

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Double-Beam Instruments

Tungstun lampsLasersGlobar (IR)Etc.

Beamsplitter is replaced by a segmented mirror For double beam in time spectrometer

λ selectors

Double-Beam InstrumentsWill pass a simple grating monochromator around !

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Monochromator bandwidth shouldBe smaller than spectral features

For Beer’s Law based quanti-tative analysis, pick the rightwavelength

Double-Beam Instruments

This point will becritical when dealingwith AA technique

Photomultiplier tube as example

Double-Beam Instruments

PhotodetectorsPMT

50 ohmresistor

ground

Million e-, 10 nsec pulse

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Scatchard Plots: How are [PX] and [X] Determined?

P + X ⇔ PX

X0 is known, [PX] can be measured(This treatment assumes that X does not absorb at the measurement λ, and P and PX do absorb at this λ)

][][0 PXXX +=

][][ PbPXbA PPX εε +=

Scatchard Plots: How are [PX] and [X] Determined?

][][ PPXA PPX εε +=

][][ 0 XPP −=and

][][ 0 PXPPXA PPPX εεε −+=so

A0

0])[( APXA PPX +−= εεand

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Scatchard Plots: How are [PX] and [X] Determined?

0])[( APXA PPX +−= εε

PPX εεε −=∆0AAA −=∆

ε∆∆

=APX ][

])[(][][][

0 PXPKPKX

PX−==

AKPKXA

∆−∆=∆

0][ε

Plot ∆A/[X] vs ∆A

Chapter 19 Problem AKPKXA

∆−∆=∆

0][ε

Answer – K = 88

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Atomic techniques from Chp 21

AA – our focus here

Basic Facts• Atomic Absorption looks at ground state, unexcited, neutral

bare atoms.• Atomic Emission looks at excited state atoms or ions.• Gas phase atomic spectra are extremely sharp, with

bandwidths of ~0.003 nm.• 50-5000 sharp lines per element.• Applicable to all metals on periodic table and a few others.

Basic Calibration Curve for Absorption(could draw similar curve for emission, or for internal standard calibration)

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Method of Standard Additions is also widely used in Atomic Spectroscopy

Apparatus Used in AA

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Spectral Bandwidth Considerations

Absorption line is wider thanlamp emission due to greaterpressure broadening and Doppler broadening.

The emitted light has the spectrum of the element(s) which make up the cathode.

Flame AA Problems• Background Absorbance

– (broad, from molecules, and from light scatter off particles)

– Measure it, and subtract it.• Spectral Interferences

– Choose analytical wavelength carefully.• Chemical Interferences

– Use higher temperature– Add protective or releasing agents

• Ionization Interference– Use ionization “buffer”

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Solving Ionization Interferences• Elements which partially ionize in the flame are sensitive to the [electrons] in the

flame.M ↔ M+ + e-

Only M contributes to atomic absorption.• Other sample components that ionize in the flame alter the [electrons], making the

analyzed element’s signal CHANGE.• Solve by adding HIGH concentration of easily ionized metal (Cs, Li, K) to all

standards and samples – an electron “buffer”, or ionization suppressor.

Solving Chemical InterferencesProblem: Phosphate interferes with calcium measurements

Can use higher T, to break down molecule.Can add lanthanum, which binds phosphate even tighter than calcium, freeing calcium.Can add EDTA, which complexes Ca, preventing formation of the calcium phosphate.

Exercise 21-B- A mixture containing 2.00 µg Mn/mL (as int. std.) and 2.50 µg Fe/mL

gave a signal ratio (Fe/Mn) of 1.05/1.00. - A mixture of 5.00 mL of Fe unknown and 1.00 mL of 13.5 µg Mn/mL

gave a signal ratio (Fe/Mn) of 0.185/0.128.- What is the Fe concentration in the unknown, in µg/mL and in mole/L?

Use equation 19 from p. 91:F = (AFe/AMn) x ([Mn]/[Fe]) for the calibration experiment. F is a “response factor”F = (1.05/1.00) x (2.00/2.50) = 0.840Now for the unknown experiment: [Mn] = 13.5 x (1.00 / 6.00) = 2.25 µg/mL0.840 = (0.185/0.128) x (2.25 µg/mL / [Fe]). Solve for [Fe]: [Fe] = 3.871 µg/mL (not yet rounded)Finally, correct for the dilution of the unknown:Original unknown conc. = 3.871 µg/mL x (6.00 mL / 5.00 mL) = 4.65 µg/mLFor molarity: 4.65 µg/mL x (1 g / 106 µg) x (1 mole/55.845 g) x (1000 mL/L)[Fe] = 8.33 x 10-5 M

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