Experiment #4: Vibrational-Rotational
Spectra of HCl
Devin Baynes
Lab Partners: Luisa Isla, April
Rains, William Alvarez
CHM 3411L Section 1
3/20/2019
Abstract:
In this lab the FTIR spectrum was taken of HCl gas and through an analysis of the spectrum the
atomic molecular properties were able to be determined. The sample was analyzed and had a ω0
of 2885.4 cm-1, a αe of 0.2828 cm-1, a Be 10.553cm-1, a De of 6.5 x 10-4. From this information
the moment of inertia was determined to be 2.652 x 10-47 kg˙m2, and the bond length of the
molecule was found to be 1.28 Å.
Introduction:
The function of the IR gas is to encase the sample during the moment of phase transition
while the liquid vaporizes at room temperature within the cell. The spectrophotometer can
measure the gas peaks, which are sharper than liquid peaks due to decreased intermolecular
interaction. FTIR was used because HCl35 and HCl37 can reach excited vibrational and
rotational energy states when exposed to infrared radiation due to the permanent dipole moment
on both molecules. Also, the reduced masses of 35Cl and 37Cl are so similar that a highresolution instrument is required to identify where splitting occurs. Therefore, FTIR is used
because it is a high-resolution instrument and works quickly to find the absorption of each
molecule while having an appropriate spectral range for mid-range infrared measurements as
required for HCl35 and HCl37. Also, HCl37 is used in order to induce a difference in the
chlorine isotopic, rotational, and vibrational aspects. Peaks are identified by branch, though the
forbidden Q branch is not shown as a peak. Rather, it manifests in the absence of a peak between
the P and R branches. The absorbance band of a molecule is identified by the zero peak of the R
branch.
From the vibrational quantum numbers, vibrational frequency, and the planks constant,
the energy levels allowed in harmonic oscillator can be determined by using the following
equation.
1
𝐸(𝓋) = ℎ𝒱(𝓋 + 2)
Here, h is the plank constant, 𝓋 is the vibrational quantum number having integral values
0,1,2…, and 𝒱 is the vibrational frequency.
In this experiment the rotational energy of a molecule was also taken to account. The
simplest way of explaining the rotational energy of a diatomic molecule is a rigid rotor model
where two molecules with different mass rotates around a common center of mass. The atom
with heavier mass rotates closer to the center of mass and the atom with lighter mass rotates
further from the center of mass. There is a fixed bond length when the rotation occurs. the
allowed energy levels for a rigid rotor can be determined by this equation:
ℎ2
E(𝐽)=8𝜋2 𝐼 𝐽(𝐽 + 1)
Where I stands for moment of inertia and j stands for rotational quantum number. The
moment of inertia can be obtained from another equation 𝐼 = 𝑢𝑟 2 where 𝑢 is the reduced mass
of the molecules and r is the intermolecular distance.
To obtain the total energy of the molecule the one has to take in account both of the
vibrational and the rotational energies for this molecule. In order to combine both of the
energies, an equation is obtained, T (𝒱, 𝐽) =
𝐸(𝒱,𝐽)
ℎ𝑐
1
1 2
= ν̃e(𝒱 + 2) − ν̃exe(𝒱 + 2) + 𝐵𝑒 𝐽(𝐽 + 1) −
1
𝐷𝑒 𝐽2 (𝐽 + 1)2 −αe(𝒱 + 2)𝐽(𝐽 + 1)
Where c is the speed of light in cm s-1, ν̃e is the frequency in cm-1, ν̃exe is the constant of
anharmonicity, 𝐵𝑒 is the rotational constant, 𝐷𝑒 is the centrifugal distortion constant and αe is
rotational-vibrational coupling constant.
Procedure:
The experiment was performed as written in the lab manual on p.38-41 and no deviations were
made.
Experimental Data:
HCl35
M value
Wave Number (cm^-1)
12
3097.711
11
3072.869
10
3059.328
9
3045.073
8
3030.093
7
3014.425
6
2998.096
5
2980.945
4
2963.329
3
2945.034
2
2925.944
1
2906.287
-1
2865.112
-2
2843.691
-3
-4
-5
-6
-7
-8
-9
-10
2821.61
2798.948
2775.824
2752.046
2727.781
2703.022
2677.746
2651.978
Wave Number vs M for HCl35
Wave Number
3150
3100
3050
3000
2950
2900
2850
2800
2750
2700
2650
2600
-15
-10
-5
y = -0,2828x2 + 20,496x + 2885,4
R² = 0,9998
y = 0,0026x3 - 0,291x2 + 20,297x + 2885,6
R² = 0,9998
Ряд1
Полиномиальная
(Ряд1)
0
M Value
HCl37
M value
9
8
7
6
5
4
3
2
1
-1
-2
-3
-4
-5
Wave Number ( cm^-1)
3042.745
3027.79
3012.144
2995.8
2978.766
2961.079
2942.756
2923.75
2904.142
2863.044
2841.604
2819.548
2796.965
2773.836
5
10
15
-6
-7
-8
-9
2750.151
2725.947
2701.202
2675.976
Wave Number Vs M For HCL37
y = -0,3028x2 + 20,433x + 2883,9
R² = 1
3100
3050
Wave Number
3000
y = -0,0021x3 - 0,3028x2 + 20,548x + 2883,9
R² = 1 Ряд1
2950
2900
2850
Полиномиальная
(Ряд1)
2800
2750
Полиномиальная
(Ряд1)
2700
2650
-10
-5
0
5
10
Re
ω0
A
cm-1
M value
H35Cl
Be
αe
De
cm-1
cm-1
cm-1
Calculated
10.553
0.2828
6.5E-4
1.28
2885.4
Literature
10.5934
0.30718
5.319E-4
1.27
2885.3
% Error
.37
7.91
22.4
.78
0.0034
Discussion:
The purpose of this laboratory was to produce a FTIR spectrum for a hydrochloric gas
sample and analyze the spectrum to determine the initial wave number, ω0 and the constants Be
and αe. To accomplish this the first thing that had to be done was assigning the m values to the
spectrum. Once the m values were all determined a table was made showing the m values and
their corresponding wave numbers. From the table a graph was made of the wave numbers vs
values. The data on the graph was analyzed and two different equations for two different trend
lines were determined. When the trend line was first analyzed as a polynomial to the second
power, the equation of the line was determined to be, y= -0.2828x2 + 20.496x + 2885.4. Then
when the trend line was produced for a polynomial to the third power the equation that was
produced was, y = 0.0026x3 - 0.291x2 + 20.297x + 2885.6.
When using the first equation obtained from the graph the ω0 and the constants Be and αe
were able to be determined. This can be done because the general equation for the wave number
as it depends on m is as follows from the manual:
ṽ(m)= ω0+(2Be-2αe) m- αem2
This shows that the equation obtained from the graph fits directly into this form and
shows that the fundamental wavenumber is equal to 2885.4 cm-1, examining the produced
spectra it can be seen that this is in fact true between m=-1 and m=1 and the rest of the results
can be taken in good faith. Next we can see that the αe value is equal to 0.2828 and can therefore
determine the value of Be. which by using algebra results in being 10.530. These two numbers
are constants in the large energy equation and really don’t yield any further important
information.
From the second equation obtained from the graph which raises the polynomial to the 3rd power,
the De constant can be determined. In the equation the 3rd power value is equal to the De constant
while all the other values in the equations represent the same as they did in the last equation.
Therefore, the De value was determined to be 6.5 x 10-4 which suggests why the value was
omitted from the earlier derivation. The number is so small that it really has no impact on the
equation as a whole and that is why it is left out in previous calculations. As shown from the rest
of the equation the only number it’s presence changes is the Be constant and it would only chage
this value form 10.530 to 10.576. since the wave number and the alpha constant stay the same it
can be said that the De constant is needed to obtain perfect accuracy in calculations, however for
basic calculations it can be omitted from derivations and has a negligible effect on the equation.
Now from using the information gathered, the moment of Inertia (I) can be obtained by plugging
in the information into the equation for Be.
Be= h/8𝜋2Iec
I=2.652 x 10-47 kg˙m2= 2.652 x 10-40g˙cm2(using the avg. value for Be)
The moment of inertia can further be used to determine the internuclear distance r, by plugging
the needed information into the equation:
I=ur2 where m1= 1 and m2= 35.45
10−3 𝑘𝑔
mh= 6.02223
mcl=
u= 1.61 x 10-27 kg
0.03545𝑘𝑔
6.02223
r= 1.28 Å
where the bond distance is equal to the two atoms in the system, hydrogen and chloride. Finally a
comparison of HCl37 was made using the equations:
𝐵𝑒 ∗
𝐵𝑒
ω0∗
ω0
𝑢
u=1.626 x 10-27 kg
= 𝑢∗
𝑢 1/2
= (𝑢∗)
u*=1.629 x 10^-27kg
ω0∗ =2883.21 cm^-1
𝐵𝑒 ∗ =10.537 cm^-1
As expected, the results are extremely similar as the difference in mass between the two
isotopes are very small. The small percent error throughout this experiment shows that the
experiment was highly accurate, and the results can be taken in good confidence.
All the information obtained from the spectrum is useful in identifying different qualities
and characteristics of the molecule. For this experiment HCl gas was used but many other
molecules can be used if they are in the gaseous form. The information can also be used to
compare different samples of the same gas to determine if its properties have changed or to see if
the gas has been contaminated in any way. The FTIR technique is useful tool in studying
different gaseous molecules and their properties with relative ease. Although the calculations
associated with the spectrum can be tedious at points, they yield a wide array of information and
is a useful tool in lab sample analysis.
References:
Jones, J. (n.d.). Physical Chemistry 2 Laboratory Manual.
Atkins, P. W., & Paula, J. D. (2010). Physical chemistry. New York: W.H. Freeman and.