Chapter 11 – Infrared Spectroscopy

Problem 11.1: Convert each of the following frequencies into wavenumbers. a. 4.98 X 1013 Hz b. 6.24 X 1013 Hz

c. 1.2 X 1013 Hz d. 1.2 X 1014 Hz

e. 9.0 X 1013 Hz f. 2.4 X 1013 Hz

(a) c = λν 2.998 x 108 m/s = (λ)(4.98 x 1013 s-1)

λ = 6.020 x 10-6 m = 6.020 x 10-4 cm

�̅� = 1/6.020 x 10-4 cm = 1661 cm-1 = 1660 cm-1

(d) c = λν 2.998 x 108 m/s = (λ)(1.2 x 1013 s-1)

λ = 2.498 x 10-5 m = 2.498 x 10-3 cm

�̅� = 1/2.498 x 10-3 cm = 400.266 cm-1 = 4.0 x 102 cm-1

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

Problem 11.2: Determine the reduced mass for each pair of atoms a. HCl b. DCl

c. HBr d. CO

e. CN f. CH

(d) CO

𝑢 = 𝑚1𝑚2

(𝑚1+𝑚2)=

(12.011)(16.00)

(12.011+16.00) = 6.861 g/mol

in kg/molecule: 6.861 𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 1.139 x 10-26 kg

(d) CH

𝑢 = 𝑚1𝑚2

(𝑚1+𝑚2)=

(12.011)(1.0079)

(12.011+1.0079) = 0.9248 g/mol

in kg/molecule: 0.9248 𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 1.536 x 10-27 kg

The remaining parts are solved in an analogous way by making appropriate substitutions of the atomic masses of

the atoms involved.

Problem 11.3: Convert each of the following wavenumbers into (i) wavelength (in μm); (ii) frequency (in Hertz, s-1), and (iii) energy (J). a. 3000 cm-1 b. 4000 cm-1

c. 1710 cm-1 d. 2100 cm-1

e. 800 cm-1 f. 1500 cm-1

(d) 2100 cm-1

(i) μm = (2100 cm-1)-1 x (1 m/100 cm) x (106 μm / m) = 4.76 μm ≈ 4.8 μm

(ii) c=λν (2.998x108 m/s) = (4.76 x 10-6 m)ν ν = 6.295 x 1013 = 6.3 x 1013 Hz

(iii) E = hν = (6.626 x 10-34 Js)(6.295 x 1013 s-1) = 4.172 x 10-20 = 4.2 x 10-20 J

(d) 800 cm-1

(i) μm = (800 cm-1)-1 x (1 m/100 cm) x (106 μm / m) = 12.5 μm ≈ 13 μm

(ii) c=λν (2.998x108 m/s) = (12.5 x 10-6 m)ν ν = 2.398 x 1013 = 2.4 x 1013 Hz

(iii) E = hν = (6.626 x 10-34 Js)(2.398 x 1013 s-1) = 1.589 x 10-20 = 1.6 x 10-20 J

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

Problem 11.4: Convert each of the following wavelengths into wavenumbers.

a. 3.3 m

b. 2.5 m

c. 5.85 m

d. 4.76 m

e. 12.5 m

f. 6.7 m

(a) �̅� = 1/(3.3 x 10-6 m x 100 cm/m) = 3030 cm-1 = 3.0 x 103 cm-1

(e) �̅� = 1/12.5 x 10-4 cm = 800 cm-1 = 800. cm-1

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

Problem 11.5: Using Table 11.1 and Example 11.1 as a guide, analyze the following spectrum:

Problem 11.6: (c&d) Determine the number of normal modes for each of the following molecules.

(c) nonlinear, so we use (3N-6) = (3)(14)-6 = 36 normal modes

(d) linear, so we use (3N-5) = (3)(3)-5 = 4 normal modes

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

Problem 11.7: In Example 11.3, we saw that the symmetric vibrational mode resulting from the trans-orientation of the two C=O bonds in CO2 results in the symmetric vibrational mode being IR-inactive. Based only on symmetry arguments, determine which of each pair of molecules below will have the greatest number of IR active vibrations and therefore the most complicated IR spectrum.

Depending upon the prerequisites for this course, instructors may wish to incorporate elements of group theory in the derivation of this answer. Assuming the instructors require less than a rigorous determination of irreducible representation of all normal modes and the use of character tables we can make the following deductions based solely on symmetry arguments.

In paring (A), the CO2 molecule has a C axis of rotation and NO2 has a C2 axis of rotation; CO2 is more symmetric than is NO2. Because NO2 has a lower symmetry, we would expect more of its vibrational modes to be IR active. In paring (B), the cis version of Pt(NH3)2Cl2 has a single C2 axis of rotation and two vertical planes of reflection. The trans version of Pt(NH3)2Cl2 has three C2 axes of rotation two vertical plane of reflection and a horizontal plane of reflection. Because cis version has a lower symmetry, we would expect more of its vibrational modes to be IR active.

In paring (C) we can make the same arguments that we made for paring (B). The cis version has less symmetry than the trans version and therefore will have a more complicated IR spectrum. A look at the point group assignments reinforces this idea. The trans version has D4h symmetry and the cis version has C2v symmetry. Problem 11.8: A Michelson interferometer generates a beat pattern for a monochromatic radiation source. The frequency of the resulting interferogram is recorded below. Determine the frequency of incident radiation that created that interferogram given the fact that the mirror was pulsing over a distance of 3 cm at a frequency of 1 Hz. (a) 2.40 × 104 Hz (c) 1.38 × 104 Hz (e) 7.20 × 103 Hz (b) 3.60 × 103 Hz (d) 1.20 × 104 Hz (f) 1.08 × 104 Hz The relevant equation is

𝒇 = 𝝂𝟐𝒖

𝒄or =

𝒇𝒄

𝟐𝒖 Eq. 11.7

= frequency of the incident radiation f = frequency of the interferogram reaching the detector c = speed of light u = the velocity of the pulsing mirror. First we need to convert the velocity of the pulsing mirror into units of meters/second.

𝑢 = 3 𝑐𝑚𝑠⁄ 𝑋

1𝑚

100 𝑐𝑚= 0.03 𝑚

𝑠⁄

Making appropriate substitutions,

(a) =𝑓𝑐

2𝑢=

(𝟐.𝟒𝟎 𝑿 𝟏𝟎𝟒 𝑯𝒛)(2.99 𝑋 108𝑚𝑠⁄ )

2(0.03𝑚𝑠⁄ )

= 1.2 𝑋 1014𝐻𝑧

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

Problem 11.9: A He-Ne laser in an FTIR instrument emits light having a wavelength of 632.8 nm. What is the expected frequency of the interferogram of the laser, assuming the moving mirror is oscillating over a distance of 3 cm at a frequency of 1 Hz? The relevant equation is

𝒇 = 𝝂𝟐𝒖

𝒄or =

𝒇𝒄

𝟐𝒖 Eq. 11.7

= frequency of the incident radiation f = frequency of the interferogram reaching the detector c = speed of light u = the velocity of the pulsing mirror. First we need to convert the velocity of the pulsing mirror into units of meters/second.

𝑢 = 3 𝑐𝑚𝑠⁄ 𝑋

1𝑚

100 𝑐𝑚= 0.03 𝑚

𝑠⁄

Next we need to convert 632.8 nm into units of hertz.

𝜈 = 𝑐

𝜆=

2.99 𝑋 108 𝑚𝑠⁄

6.328 𝑋 10−7𝑚= 4.551 𝑋 1014𝐻𝑧

Making appropriate substitutions

𝑓 = 𝜈2𝑢

𝑐=

(4.551 𝑋 1014𝐻𝑧)(2)(0.03 𝑚𝑠⁄ )

2.99 X 108 ms⁄

= 𝟗. 𝟏𝟑𝟐 𝑿 𝟏𝟎𝟒𝑯𝒛

Problem 11.10: What is the maximum distance the moving mirror in a Michelson interferometer would need to travel to be able to distinguish between (a) 1,000 and 1,004 cm-1 in an FTIR spectrum? (b) Between 1,000 and 999.5 cm-1 ? The relevant equation is,

∆𝜈 = 1

𝛿𝑚𝑎𝑥=

1

2𝑑𝑚𝑎𝑥 Eq. 11.8

Δ�̃� = spectral resolution, cm-1

δmax = maximum retardation, cm

dmax = maximum mirror movement, cm

Making appropriate substitutions

(a) ∆𝜈 = (1004 𝑐𝑚−1 − 1000 𝑐𝑚−1) =1

2𝑑𝑚𝑎𝑥

Solving for dmax we obtain,

𝑑𝑚𝑎𝑥 = 1

2(4 𝑐𝑚−1)= 0.125 𝑐𝑚

(b) ∆𝜈 = (1000 𝑐𝑚−1 − 999.5 𝑐𝑚−1) =1

2𝑑𝑚𝑎𝑥

Solving for dmax we obtain,

𝑑𝑚𝑎𝑥 = 1

2(0.5 𝑐𝑚−1)= 1 𝑐𝑚

Problem 11.11: What would be the spectral resolution if the mirror in a Michelson interferometer moved a total distance of 7.5 mm?

Δ�̅� = 1

2𝑑=

1

(2)(0.75 𝑐𝑚) = 0.6667 cm-1 = 0.67 cm-1

Problem 11.12: Speculate on the reasons why it is more common to report UV-vis spectra using absorbance, whereas it is more common to report IR spectra using transmittance (recall that absorbance = –log(T)). Older scanning IR instruments recorded the data on a roll of paper called a strip chart recorder. The paper rolled past a pen at the same rate at which the spectrometer scanned the sample. The pen was attached to an arm which was in turn connected to a potentiostat. Amplified current from the detector moved the arm up or down as the paper rolled past thus generating a peak as a function of wavelength. What was being measured was proportional to the transmitted light through the sample. Early spectrometers lacked the electronic sophistication to convert transmittance to absorbance. So part of the reason we use transmittance is historical. Modern FTIR instruments can easily report the data in either %T or Abs. And absorbance has the advantage of being linearly proportional to concentration. The reason we have not shifted the convention to absorbance rather than %T is probably due to the fact that FTIR analysis is seldom used for quantitative analysis but rather is most often used for qualitative analysis.

Problem 11.13: Using the peak at 1,220 cm-1–1, in the transmittance spectrum of Figure 11.18, estimate the percent error in the percent T reading if your spectrometer has a ± 2 cm-1 variance in its precision. A variance of ± 2 cm-1 represents a Δ�̅� = 4 𝑐𝑚−1. Therefore

% 𝑒𝑟𝑟𝑜𝑟 = 4 𝑐𝑚−1

1220 𝑐𝑚−1 𝑋 100% = 0.33%

Problem 11.14: Define the term evanescent wave. The word evanescent comes from the Latin verb evanesce which translates as to vanish, to die or to fade away. The term is applied to wave fonts that exhibit an exponential decay in their intensity in at least one dimension (x, y or z). The term is more often seen in radiofrequency technology as it pertains to directional transmission antennae. Evanescent IR waves can be formed when IR light is incident on a reflective surface at a shallow grazing angle such that the angle is greater than the critical angle so that total internal reflection occurs between the reflective surface and the sample’s surface. The result is a wave who’s intensity diminishes exponentially in the z-direction and exists only in the xy-plane.

Problem 11.15: Under what conditions would an ATR accessory be ill suited? In other words, under what conditions might you get better data by using one of the trade tional sample introductory modes (KBr pellet or Nujol mull)? The ATR crystal attenuates the IR beam thus decreasing the S/N ratio. ATR accessories are not recommended for dilute samples that already have a low S/N ratio.

EXERCISE 11.1: Determine the number of vibrational modes for each of the following molecules.

The relevant equations are m = 3N – 6 Eq. 11.4a (non-linear) m = 3N – 5 Eq. 11.4b (linear) (a) Normal modes: (3N-6) = (3)(17)-6 = 45 normal modes (b) (3N-6) = (3)(15)-6 = 39 normal modes (c) (3N-5) = 3(4)-5 = 7 normal modes (d) (3N-6) = 3(10)-6 = 24 normal modes

EXERCISE 11.2: Convert each of the following frequencies into wave numbers. (a) 5.01 × 1013 Hz (d) 1.5 × 1014 Hz (b) 6.7 × 1013 Hz (e) 8.97 × 1013 Hz (c) 1.5 × 1013 Hz (f) 3.1 × 1013 Hz (a) c = λν

𝜆 =𝑐

𝜈=

2.99 𝑋 108 𝑚𝑠⁄

𝟓. 𝟎𝟏 𝑿 𝟏𝟎𝟏𝟑𝒔−𝟏= 5.97 𝑋 10−6𝑚 = 5.97 𝑋 10−4𝑐𝑚

�̅� = 1

𝜆(𝑐𝑚)=

1

5.97 𝑋 10−4𝑐𝑚= 1675.04 𝑐𝑚−1

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

EXERCISE 11.3: Convert each of the following wave numbers into (i) units of hertz (s–1) and then (ii) into wavelength in unit of micrometers. (a) 2,970 cm–1 (c) 1,699 cm–1 (e) 756 cm–1 (b) 43,810 cm–1 (d) 2,120 cm–1 (f) 1,597 cm–1 (a) 2,970 cm-1

(ii) μm = (2970 cm-1)-1 x (1 m/100 cm) x (106 μm / m) = 3.37 μm ≈ 3.4 μm

(i) 𝜈 =𝑐

𝜆=

2.998x108𝑚𝑠⁄

3.37 𝑋 10−6𝑚= 8.87 𝑋 1013𝐻𝑧

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

EXERCISE 11.4: Convert each of the following wavelengths into wave numbers. (a) 3.12 mm (c) 6.85 mm (e) 11.99 mm (b) 2.97 mm (d) 3.76 mm (f) 5.71 mm

𝟑. 𝟏𝟐 𝒎𝒎 𝑋 1𝑐𝑚

10𝑚𝑚= 0.312 𝑐𝑚

�̅� = 1

𝜆(𝑐𝑚)=

1

0.312= 3.2 𝑐𝑚−1

The remaining parts are solved in an analogous way by making appropriate substitutions at the highlighted section.

EXERCISE 11.5: Why are deviations from Beer’s law more common in IR spectroscopy than in UV-vis spectroscopy? Peaks are much sharper in IR spectroscopy than they are in molecular UV-vis spectroscopy. Therefore, deviations in the wavelength selection result in much larger deviations in absorbance in IR spectroscopy than what is seen in molecular UV-vis spectroscopy

EXERCISE 11.6: Determine the number of vibrational modes in each of the following molecules. (a) SO2 (d) SO3

–2 (g) H2O (b) NO2

+ (e) HCN (h) CO2 (c) NO2

– (f) CO (i) NO3–

The relevant equations are m = 3N – 6 Eq. 11.4a (non-linear) m = 3N – 5 Eq. 11.4b (linear)

Students will also need to draw correct Lewis structures, apply V.S.E.P.R.T. and determine the geometry of each molecule before they apply Equation 11.4.

EXERCISE 11.7: In your own words, discuss the considerations one must make when selecting a suitable source for FTIR spectroscopy? This is an open ended question so answers will vary. Students are encouraged to review section 11.4. Some of the key points that should be included in the discussion include:

• Black Body Emission

• Spectral Range of the source

• Transmission range of the optics

• Pros & cons of some of the more common sources used.

EXERCISE 11.8: Explain the relationship between the use of a Michelson interferometer and the time constant of the detector in FTIR spectroscopy. This is an open ended question and answers will vary. Students are encouraged to review the feature Compare and Contrast: Interferograms and Beat Patterns in the text. In summary, the response should include a discussion of how the interferometer shifts the time domain of the signal into a region that is compatible with modern FTIR detectors.

EXERCISE 11.9: Explain why a thermal detector is not suitable for use in an FTIR instrument. Thermal detectors were common in early scanning IR spectrometers however the response time of thermal detectors is too slow for use in a modern FTIR spectrometer.

EXERCISE 11.10: Pyroelectric detectors are much less sensitive than photomultiplier tubes (Chapter 6). Why do we use pyroelectric detectors rather than PMTs in FTIR instruments? PMT detectors utilize the photoelectric effect where the impinging photon on the PMT electrode dislodges an electron. The energy of the photons present in the infrared range is too low to dislodge electron from the PMTs electrode.

EXERCISE 11.11: In your own words, describe how a pyroelectric detector works. This is an open ended question and answers will vary. The answer should include a discussion of

• Crystalline polarization within pyroelectric materials

• How the light impinging upon a pyroelectric material can perturb the polarization vector within the crystal and how that can be detected as a voltage change by attaching leads to the material and passing the leads to a voltmeter.

EXERCISE 11.12: Discuss some of the challenges and considerations associated with sample preparation in FTIR spectroscopy. This is an open ended question and answers will vary. Students are encouraged to review section 11.8. In their discussion students should touch upon the following topics

• Why dissolving your sample in a solvent is problematic

• Why quartz and/or fused silica are bad choices for sample compartments

• Suitable optical materials used as a matrix during sample introduction (see Table 11.3)

• The pros and cons of using a KBr pellet vs. Nujol Mull.

• The challenges of gas phase IR spectroscopy

• The challenges of IR spectroscopy of a sample in solution.

EXERCISE 11.13: Construct a suitable circuit diagram for a pyroelectric detector. Be sure to include a means to drive an output reading device. A simple voltage follower would be an appropriate circuit for this device. See section 4.5 in Chapter 4

EXERCISE 11.14: The spectrum seen here was taken as a KBr pellet and belongs to a compound with the formula C2H6O. Determine the identity of the compound.

There are two possible isomers for the formula C2H6O, ethanol (CH3CH2-OH) and dimethyl ether (CH3-O-CH3). The strong peak seen at ≈ 1190 cm-1 is most likely a C-H bending. The strong peak seen at ≈ 2900 cm-1 is C-H stretching. If a hydroxide functional group were present, we would also expect to see a dominant broad peak centered at ≈ 3,300 cm-1. The absence of the peak at 3,300 cm-1 excludes the possibility of this molecule being ethanol. However the IR is consistent with dimethyl ether.

Unless the solvent used is rigorously dry, the ketone group in the compound [Pt(DPK)2Cl2][PF6]2 undergoes hydrolysis to a stable enol (see Figure 11.1). Suppose you wanted to study the kinetics of this hydrolysis reaction using IR spectroscopy. Propose a suitable solvent for this study and discuss how you would introduce the sample into the spectrometer. Defend your recommendations. This is an open ended question and answers will vary. In their discussion, students should touch on the following points. The ketone functional group has a strong and sharp peak at ≈ 1700 cm-1 so the choice of a solvent system would have to be a solvent that is transparent in the region of 1700 cm-1 and a solvent that will dissolve DPK, additionally, the solvent must be capable of solvating at least a trace amount of water because we wish to observe the hydrolysis reaction. Likewise the optical material we choose has to be transparent in the region of 1700 cm-1. and because we wish to observe the hydrolysis reaction, our optical material must also be somewhat inert towards water (i.e. not water soluble). A look at Table 11.4 produces three solvent candidates that are transparent in the region of 1700 cm-1. They include, carbon tetrachloride, ethyl methyl ketone and pyridine. Students who have had an advanced inorganic course should be able to comment on why pyridine is probably a bad choice because pyridine can act as a strong ligand to transition metal complexes and is likely to displace the chlorides on the molecule. Students should also be able to comment on the fact that carbon tetrachloride is non-polar and therefore probably a bad choice due to the low solubility of water in carbon tetrachloride. Leaving us with ethyl methyl ketone as the top choice for this experiment.

A look at Table 11.3 produces several good choices for an optical material. All of the optical materials listed in Table 11.3 are transparent in the region of 1700 cm-1. However there are two choices that are water soluble and will fog in the presence of trace water. Those two are KBr and NaCl. Neither of these materials would be a good choice.

Review your answer to Exercise 11.5 and comment on why it is more common to use UV-vis spectroscopy for quantitative work than IR spectroscopy. You might also want to review the definition of sensitivity. See Figure 11.22. This is an open ended question and answer will vary. A very detailed discussion of this was presented in a Chapter 1, Compare and Contrast: Ultraviolet-visible versus Fourier Transform Infrared in Quantitative Analysis. See page 8. EXERCISE 11.17: Determine the number of IR active vibrational modes in each of the following molecules. Note: You will have to use symmetry arguments to determine which vibrations result in a net change in the molecular dipole moment. (a) SO2 (d) SO3

-2 (g) H2O (b) NO2

+ (e) HCN (h) CO2

(c) NO2- (f) CO (i) NO3

- Students will need to have had some training in the use of group theory and character tables to answer this question.

(a) The point group for SO2 is C2v. The character table for the C2v point group is given below in the gray shaded region of the table. The basis set (ISO)for this analysis is the S=O bond and is given in row 5. Only those irreducible representations that transpose like x,y or z are active in the IR so we are only concerned with the A1, B1 and B2 normal modes. Convolution of the basis set with the IR active normal modes are given in rows 6, 7 & 8 and are labeled aA1, aB1 and aB2 respectively.

C2V E C2 (xz) (yz)

A1 1 1 1 1 z

A2 1 1 -1 -1 Rz

B1 1 -1 1 -1 x Ry

B2 1 -1 -1 1 y Rx

ICO 2 0 2 0

aA1 2 0 2 0 = 4

aB1 2 0 2 0 = 4

aB2 2 0 -2 0 = 0

The order of the C2v point group is 4. Therefore, for the A1 normal mode we should expect to see 4/4 = 1 peak in the IR spectrum and for the B1 normal mode we should expect to see 4/4=1 peak in the IR spectrum and for the B2 normal mode we should expect to 0/4 = 0 peaks in the IR spectrum for a total of 2 peaks in the SO region of the IR spectrum. Parts (b) (i) are performed in an analogous way.

Program a spreadsheet and plot the beat pattern created from the sum of each of the frequencies in Problem 11.2. See activity Creating a Beat Pattern on p. 361.

Program a spreadsheet and perform a Fourier transform on the data set created by Exercise 11.18.

See activity Performing a Fourier Transform on p. 361.

The ketone peak in Figure 11.1 is centered at 1,714.8 cm–1. What is the force constant for the C=O bond? Predict the wave number of the carbonyl peak if the carbon atom is a 14C isotope and the oxygen atom is an 17O isotope.

In Problem 11.2, we found the reduced mass for CO as 1.139 x 10-26 kg

Recall from Chapter 4 that ; we need to calculate the frequency (ν) then use that to get the

angular frequency (ω).

λ = (1714.8 cm-1)-1 x 1 m/100cm = 5.8316 x 10-6 m

c = λν (2.998 x 108 m/s) = (5.8316 x 10-6 m)(ν)

so ν = 5.14097 x 1013 s-1

ω = 2πν = (2)(π)(5.14097 x 1013 s-1) = 3.2302 x 1014 rad/s

𝜔 = √𝑘

𝜇

(3.2302 𝑥 1014 𝑟𝑎𝑑

𝑠) = √

𝑘

1.139 𝑥 10−26𝑘𝑔

k = 1188.4 kg/s2 = 1188 kg/s2

Second part:

We can estimate the reduced mass (assuming simple molar masses of 14 and 17) as:

μ = 1.275 x 10-26 kg

so ω = 3.053 c 1014 rad/s

and ν = 4.8592 s-1

and λ = 6.1697 x 10-4 cm

so finally �̅� = 1621 cm-1.

This makes sense – the heavier masses should vibrate at a lower frequency.

2oscillator oscillator

EXERCISE 11.21: Calculate a theoretical absorption frequency in wave numbers for a C–H bond, assuming a force constant of 4.89 × 102 N/m. How would the frequency change if you substituted deuterium for hydrogen? The relationship between a force constant and vibrational frequency was discussed in Chapter 2. The reduced mass can be found using 12.011 for the mass of carbon and 1.0094 as the mass of hydrogen. C-H

𝑢 = 𝑚1𝑚2

(𝑚1+𝑚2)=

(12.011)(1.0079)

(12.011+1.0079) = 0.9248 g/mol

in kg/molecule: 0.9248 𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 1.536 x 10-27 kg

And,

𝜔 = √𝑘

𝜇= √

4.89 × 102 𝐾𝑔𝑠2⁄

1.536 𝑋 10−27𝐾𝑔= 5.64 𝑋 1014 𝑟𝑎𝑑

𝑠⁄ = 2πν

∴ 𝜈 = 5.64 𝑋 1014 𝑟𝑎𝑑

𝑠⁄

2𝜋= 8.98 𝑋 1013𝐻𝑧

𝜆 =𝑐

𝜈=

2.99 𝑋 108 𝑚𝑠⁄

8.98 𝑋 1013𝑠−1= 3.32 𝑋 10−6𝑚 = 3.32 𝑋 10−4𝑐𝑚

�̅� = 1

𝜆(𝑐𝑚)=

1

3.32 𝑋 10−4𝑐𝑚= 3004.89 𝑐𝑚−1

C-D

𝑢 = 𝑚1𝑚2

(𝑚1+𝑚2)=

(12.011)(2)

(12.011+2) = 1.715 g/mol

in kg/molecule: 1.715 𝑔

𝑚𝑜𝑙 𝑥

1 𝑘𝑔

1000 𝑔 𝑥

1 𝑚𝑜𝑙

6.022 𝑥 1023 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 = 2.847 x 10-27 kg

And,

𝜔 = √𝑘

𝜇= √

4.89 × 102 𝐾𝑔𝑠2⁄

2.847 𝑋 10−27𝐾𝑔= 4.144 𝑋 1014 𝑟𝑎𝑑

𝑠⁄ = 2πν

∴ 𝜈 = 4.144 𝑋 1014 𝑟𝑎𝑑

𝑠⁄

2𝜋= 6.60 𝑋 1013𝐻𝑧

𝜆 =𝑐

𝜈=

2.99 𝑋 108 𝑚𝑠⁄

8.98 𝑋 1013𝑠−1= 4.531 𝑋 10−6𝑚 = 4.531 𝑋 10−4𝑐𝑚

�̅� = 1

𝜆(𝑐𝑚)=

1

3.32 𝑋 10−4𝑐𝑚= 2207.11 𝑐𝑚−1

EXERCISE 11.22: Compare and contrast TXRF to ATR-IR (see Chapter 10.8). What optical principle do both tech-niques exploit? The contrasts between TXRF and ATR-IR are significant. First off, TXRF uses x-rays and ATR-IR uses EM radiation in the infrared region of the spectrum. The energy differences between these two types of radiation are significant. In TXRF, the source radiation ejects a core electron and in ATR-IR the source radiation promotes a molecular bond into a higher vibrational state. In TXRF the quantum event giving rise to the analytical signal is fluorescence as an electron falls into a vacant position in a core electron shell. In ATR-IR, the quantum event giving rise to the analytical signal is absorption of a photon matching the energy difference of the vibrational states. The only similarity in the two techniques is the use of grazing angles. In TXRF, a steep grazing angle is used to ensure that the x-ray reflects off of a surface support. The analyte is placed on top of the reflective surface and as x-rays pass through the surface and are reflected, they pass through the sample a second time, thus increasing the signal. In ATR-IR the steep grazing angle causes the source radiation to reflect off of the diamond support’s surface multiple times. Part of the source radiation penetrates the sample on the surface and thus increases the absorbance path length resulting in an increased signal. EXERCISE 11.23: Using Table 11.1 and Example 11.1 as a guide, analyze the following spectra.