Answers to Post-Reading Questions

Chapter 1. Introduction to Chemical Analysis

Question 1

Question 2

Transduction does not produce a voltage or current that is easily related to the quantity of analyte. Calibration establishes an empirical, rather than a theoretical, relationship between the analytical response and the quantity of analyte that caused that response.

Question 3

Limit of detection (LoD): The smallest amount of analyte that can be readily detected.

Dynamic range: The range over which the instrument response changes predictably and significantly with changes in the quantity of analyte.

Sensitivity: The change in signal vs. change in quantity of analyte (e.g. the slope of a linear calibration curve).

Selectivity: The degree to which a method of analysis is sensitive to the analyte versus other species in the sample matrix.

Signal-to-noise (S/N) ratio: The ratio between the average value of the output signal and its variation between replicate measurements.

Signal-to-background (S/B) ratio: The ratio between the average value of the output signal in response to analyte and the output signal in the absence of analyte.

Chapter 2. Properties of Light

Question 1

The photons in the green laser beam have more energy than the photons in the red laser beam because they have a shorter wavelength (i.e. higher frequency).

Question 2

Because the refractive index increases, the wavelength of the red laser beam will be shorter within a BK7 glass lens than through air.

Question 3

No, this observation implies that leaves absorb red light (complimentary to green) most strongly.

Question 4

The angle should be 45°. The angle of incidence equals the angle of reflection, such that 2 × 45° = 90°.

Question 5

Refraction

Question 6

No, the width of the slits needs to be similar in size as the light passing through. Diffraction would not be anticipated until the slit width was on the order of microns.

Chapter 3. Light Sources

Question 1

Question 2

Chapter 4. Wavelength Selection

Question 1

Entrance slit, grating, and exit slit.

Question 2

The relative angle of the grating determines the center wavelength output by the monochromator.

The slit width and dispersion of the grating determine the range wavelength(s) output from the monochromator.

Question 3

Grating: diffraction

Prism: refraction

Question 4

Chapter 5. Photodetectors

Question 1

Question 2

Question 3

Chapter 6. UV-Visible Absorption

Question 1

Yes

Question 2

The energy difference between the S₀→S₂ electronic states is larger than for S₀→S₁. S₀→S₂ transitions therefore occur at shorter wavelengths (i.e. higher energy) than S₀→S₁ transitions.

Question 3

A(λ) = −log[P(λ)/P₀(λ)] = ε(λ)bc

A is absorbance, P is the intensity of light of wavelength λ transmitted through the sample, P₀ is the intensity of light of wavelength λ incident on the sample, ε is the molar absorption coefficient of the molecule at wavelength λ, b is the path length of light through the sample, and c is the concentration of the molecule.

Question 4

Concentration is given in units of M and path length is given in units of cm.

Question 5

A = –log(P/P₀) = –log(0.72/7.2) = –log(0.10) = 1.0

It can be handy to remember that the light intensity is attenuated 10-fold for every unit of absorbance. A = 1 is a ten-fold attenuation, A = 2 is a hundred-fold attenuation, and so on.

Question 6

c =[A(λ)]/[ε(λ)b] = 0.15/[(150 000 M^-1 cm^-1)(1.0 cm)] = 1.0 x 10^-6 M or 1.0 µM

Chapter 7. UV-Visible Spectrophotometers

Question 1

The cuvette must be transparent to the wavelength (or over the wavelength range) being measured.

Different cuvette materials have different costs and robustness.

Question 2

Emission from a xenon lamp is significant across both the UV and visible spectral regions.

A tungsten-halogen lamp emits mainly visible light and a deuterium lamp emits mainly UV, such that the two lamps must be combined to cover both spectral regions.

Question 3

Double-beam spectrophotometers more reliably measure higher and lower absorbance values than single-beam spectrophotometers.

Double-beam spectrophotometers have superior precision.

Question 4

Acquires full spectra rapidly

More robust design (no/fewer moving parts)

Enables more compact designs

Question 5

Wavelength selection.

Control of spectral bandwidth (i.e. range of wavelengths) and resolution.

Chapter 8. Molecular Fluorescence

Question 1

Φ = 0.50 = (photons emitted)/(photons absorbed)

Φ = 0.50 = (1000)/(photons absorbed)

Photons absorbed = 2000

Question 2

Stokes’ shift = emission peak wavelength – absorbance peak wavelength

22 nm = emission peak wavelength – 598 nm.

Emission peak wavelength = 620 nm

Question 3

Kasha’s rule states that fluorescence emission almost always occurs from S₁v₀ because of the rapid non-radiative conversion of higher-order electronic states (S₂) to the S₁ state.

Yes, the shape of the emission spectrum is the same regardless of excitation wavelength. Per Kasha’s rule, emission occurs from the same state (S₁v₀) regardless of the state to which a molecule is initially excited.

Question 4

Fluorescence occurs within hundreds of picoseconds to tens of nanoseconds after excitation.

Phosphorescence tends to occur within microseconds to minutes.

Since the timescale is on the order of nanoseconds, the likely mechanism of emission is fluorescence.

Note: As you might learn in lecture, the time to reach 37% (i.e. 1/e) of the initial intensity after pulsed excitation is a quantity called the fluorescence lifetime.

Chapter 9. Measurement of Fluorescence

Question 1

The fluorescence excitation spectrum will be analogous in that it will have the same shape and spectral position as the absorbance spectrum.

Question 2

A spectrofluorometer requires excitation and emission monochromators to select the excitation and emission wavelength(s) of interest. That is, the input wavelength is not the same as output wavelength, so two separate monochromators are needed.

A spectrophotometer requires just one monochromator because only the intensity of transmitted light is measured.

Question 3

The L-shaped path minimizes background due to stray excitation light because reflections and scattering are minimized at a right angle to the incident light.

Fluorescence emission is isotropic, such that an L-shape path does not affect how efficiently fluorescence is detected.

Question 4

The trend is linear at low sample concentrations, then plateaus (or starts to decrease) at high sample concentrations.

Question 5

The terms in Eqn. 9.1 are parameters that can be optimized to maximize fluorescence signal:

• Higher sample concentration
• Longer path length
• More intense excitation light
• Selection of the excitation wavelength with the highest molar absorption coefficient
• Selection of the emission wavelength with highest contribution to the total fluorescence emission
• Components or designs that increase the instrument efficiency (e.g. more sensitive detectors)

Chapter 10. Atomic Absorption and Emission

Question 1

Elements

Question 2

Unlike molecules, atoms do not have vibrational and rotations states. Transitions in atoms are therefore purely electronic.

Question 3

At 589.0 and 589.6 nm.

Question 4

For atomic spectroscopy, the sample is required to be in the gaseous atomic state. Samples are typically solid or liquid, with most of the matter in a non-atomic (e.g. molecular) state. During atomization, heat (i.e. energy) is applied to convert a molecular or other condensed-phase sample into a gaseous atomic population.

Question 5

The Boltzmann distribution.

Chapter 11. Atomic Spectroscopy

Question 1

Flames have enough energy to atomize the sample, but, for many elements, inadequate energy to excite an atomic population efficiently.

Question 2

0.004 nm/0.2 nm = 0.02 = 2%

Question 3

The light from the flame would cause an atomic absorbance reading to be incorrectly low (i.e. adds extra light), while soot would cause an atomic absorbance reading to be incorrectly high (i.e. absorbs or scatters light).

Question 4

The light measured comes from atoms that are thermally excited. By definition, fluorescence is emission of light following excitation via the absorption of light. Light absorption is not part of the process for atomic emission spectroscopy. (Note: atomic fluorescence spectroscopy does exist, but is relatively uncommon.)

Question 5

AAS is recommended because of its lower cost, and because the lesser versatility of AAS will not be problematic if the company needs to measure only three elements.

Chapter 12. Infrared Absorption

Question 1

The fundamental transition.

Question 2

6.06 µm

Question 3

5668 cm–1

Due to anharmonicity, the first overtone will be a little less than double the wavenumber of the fundamental.

Question 4

Using the group frequency region:

• Advantage: peaks are most easily linked to specific bond types and functional groups in an analyte structure.
• Disadvantage: lack of structural specificity (i.e. not enough detailed information about the analyte(s) overall structure).

Using the fingerprint region:

• Advantage: provides a unique pattern for a given analyte molecule.
• Disadvantage: the many absorption peaks are challenging to correlate to specific functional groups and interpretation often requires computer assistance.

Question 5

The symmetric stretching vibration of carbon dioxide produces no change in net dipole as the opposing motions of each C=O bond cancel out one another. In the asymmetric vibration, one oxygen atom moves away from the carbon atom while the carbon and other oxygen atoms move closer to one another, resulting in changes in the net dipole moment as the bonds vibrate.

Chapter 13. FTIR Spectroscopy

Question 1

0

An FTIR instrument uses a Michelson interferometer, which includes neither a grating nor slits.

Question 2

• FTIR spectrometers use a Michelson interferometer instead of a monochromator or polychromator.
• FTIR spectrometers uses an IR light source (e.g. globar), whereas UV-visible spectrophotometers use UV-visible light sources (e.g. xenon lamp).
• FTIR spectrometers use IR-sensitive detectors (e.g. MCT detector), whereas UV-visible spectrophotometers use UV-visible-sensitive detectors (e.g. PMT).
• FTIR spectrometers tend not to use sample cells/holders made from glass or quartz or plastic, whereas UV-visible-spectrophotometers mainly use these materials for sample cells.

Question 3

• Multiplex of Fellgett advantage (improve signal-to-noise ratio by averaging many rapidly-acquired scans).
• Throughput or Jacquinot advantage (no grating or slits, so light throughput is higher and yields superior signal-to-noise ratios).
• Higher spectral resolution is easily achieved by moving the interferometer mirror farther.
• Greater mechanical simplicity and robustness.

Question 4

Glass has strong IR absorption.