Matthew Marthaler
February 19, 2012
Instrumental Lab
Laboratory 3: IR/ IR Microscopy/ Raman
Introduction:
In this lab, I will be working with the FT-IR and Raman spectroscopy. The first instrument that I
will talk about is the Fourier Transform Infrared Spectrometer (FT-IR). There are two beams whose path
lengths can be varied to give interference patterns which establish coding. The FT is then used for data
processing. IR measures absorption, emission, and reflection spectra from molecular species by
changing the energy of the molecules from one vibrational or rotational energy state to another. The
states are changed by infrared radiation passing through the sample. This works well for spectra in the
range of 10-500 um. It can be used to determine components of a mixture or different solvents.
Raman spectroscopy quantifies scattering from the same type of vibrational changes as IR
absorption. Radiation passes through a transparent medium, the species scatters a fraction of the beam
in multiple directions, and the difference in wavelength between the incident and scattered visible
radiation corresponds to wavelengths in the mid-IR region. Fluorescence is avoided when measuring the
radiation by removing the excited wavelengths from an absorption band of the analyte. Raman is great
for the study of biological species. There are also less peak interlaps than in FT-IR which results in
simpler quantitative measurements. The Raman can look at smaller samples than the FT-IR as well.
Purpose:
The purpose of this lab is to become better acquainted with the FT-IR and Raman Spectroscopy
and learn the techniques on how to use these instruments. For the FT-IR, we will look at different
solvents and then identify their components. We will also run various mixtures and try to identify them.
For Raman, we will create a calibration curve from three solvents and determine the concentration of
components in different mixtures.
Method (day one):
1)
2)
3)
4)
Set up the FT-IR.
Run three organic solvents on the FT-IR (2-propanol, toluene, and 2-butanone).
Identify unique components on each spectrum.
Run a mixture of the solvents and determine each ones unique components at the different
concentration levels. This is done by using ratios of 1:1:1, 1:2:1, and 2:1:2 of toluene, 2
butanone, and 2-propanol, respectively. With these ratios, you find the percent transmittance at
100%, 40%, 33.33%, and 25% composition to make a calibration curve for each compound.
5) Run an unknown mixture and use the calibration curve to find the % composition of 2-propanol,
toluene, and 2-butanone.
Method (day two):
1) Run a sample of acetaminophen and methyl salicylate on the Raman (One solid and one liquid
sample).
2) Collect the graphs of these two samples.
3) Try to run the IR microscopy.
4) Obtain good spectra for the known fibers.
5) Collect unknown fibers to see if any may match the known fibers.
Results:
The spectra for this experiment are in our lab notebooks. Our calibration curves show all the
data points in the first graph and we took out the outlier in the second graph. The Raman spectra are in
the lab notebook and no calculations were asked for these spectra.
2-Propanol
% Transmittance at 3320 𝑐𝑚−1 (OH Peak)
18
39
58
48
Percent Composition
100
40
33.3
25
2-Propanol
% Transmittance cm-1
70
y = -0.4537x + 63.241
R² = 0.8281
60
50
40
30
20
10
0
0
20
40
60
Percent Composition (%)
80
100
2-Propanol
% Transmittance (cm-1)
70
y = -0.501x + 67.277
R² = 0.8459
60
50
40
30
20
10
0
0
20
40
60
80
100
120
Percent Composition (%)
2-Butanone
% Transmittance at 1720 𝑐𝑚−1 (C=O Peak)
2
1
8
1
Percent Composition
100
50
33.3
20
% Transmittance (cm-1)
2-Butanone
9
8
7
6
5
4
3
2
1
0
y = -0.02x + 4.0162
R² = 0.0432
0
20
40
60
80
Percent Composition (%)
100
120
% Transmittance (cm-1)
2-Butanone
9
8
7
6
5
4
3
2
1
0
y = -0.0206x + 4.7192
R² = 0.0544
0
20
40
60
80
100
120
Percent Composition (%)
Toluene
% Transmittance at 1500 𝑐𝑚−1 (C=C Peak)
8
18
33
25
Percent Composition
100
40
33.3
25
Toluene
% Transmittance (cm-1)
35
30
y = -0.2627x + 34.025
R² = 0.7154
25
20
15
10
5
0
0
20
40
60
80
Percent Composition (%)
100
120
Toluene
% Transmittance (cm-1)
35
y = -0.2926x + 36.567
R² = 0.7293
30
25
20
15
10
5
0
0
20
40
60
80
100
120
Percent Composition (%)
Unknown
Peak
O-H
C=O
C=C
% Transmittance
48
1
17
Calculations:
We used our second linear regression equation that took out the outlier for our calculations.
2-Propanol:
y = -0.501x + 67.277
48 = -0.501x + 67.277
X = 38.48%
2-Butanone:
y = -0.0206x + 4.7192
1 = -0.0206x + 4.7192
X = 180.54%
Toluene:
y = -0.2926x + 36.567
17 = -0.2926x + 36.567
X = 66.87%
Conclusion:
Overall, I would say this experiment did not go too well. Our measurements of the known
standards were not entirely accurate. The 100% standards for each solvent seemed to work out well and
yielded good transmittance readings for the peaks as well as the standards of 1:1:1 and the 2:1:2. The
1:2:1 standard came out horribly. The percent transmittance for each peak was well below where they
should be and the peaks did not look correct. This could be due to errors in making the standard or
there was contamination that came into contact with this standard at some point. We could have
measured incorrectly or not used a perfectly clean pipet when putting the standard in the IR. We were
able to find all the different peaks for each solvent on the spectra, which was an accomplishment.
However, some of the peaks were not as sharp and one could tell that the peaks had some sort of
contamination or error. For example: the O-H alcohol peak was pointy on the 1:2:1 spectrum instead of
rounded and the C=O ketone peak was broadened on most of the spectra. The C=C peak was good
though except for the 1:2:1 spectrum.
Our percent transmittance came out adequately for 2-Propanol and Toluene, but not so good
for 2-Butanone. We decided to use a linear regression line that discarded the 1:2:1 standard, as it
yielded a much better r2 coefficient. Our percent transmittance for 2-butanone was 180.54%, which is
completely off. This could be due to bad 2-Butanone standards (maybe the 2-Butanone was
contaminated); the IR might not be very good for this type of solvent, or a miscalculation on our part in
the measurements of the standards. The last one does not seem as likely, as the other two solvents
came out well (besides the 1:2:1) using the same measurements.
The Raman took a very long time to get a centerburst. I do not know why, as you, Dr. Foy
worked on it with Dr. Taylor and it randomly started working after about an hour of messing with it.
Once the centerburst was working, the Raman worked well and was easy to work with.