Lab 7
Liquid Chromatography-Mass Spec
INTRODUCTION
The LC-MS id a chromatography instrument that takes a liquid mixture and separates the compound
based on mass and different physical properties. This instrument is using a column in which the mobile
phase pulls the sample through and it uses a mass spec analyzer. Once the sample is separate in the
column, as the sample comes off the column, it is ionized so that the mass spec can further analyze the
sample. Thus, it will yield a mass spec spectra that provides additional information about the structure
and what structures are in the sample. This instrument also provides a chromatograph that can be used
to note retention times and resolution.
PURPOSE
The purpose of this experiment is to learn the instruments of chromatography. These
instruments are the HPLC and LC-MS. We will run standard samples of caffeine made at different
concentrations of ppm. With these concentrations, they will serve as our standard samples. Running
these samples on the LC-MS and HPLC and analyze them by evaluating the spectra. Noting the
retentions times and by creating a standard curve we can analyze the concentration and separations of
an unknown sample.
LC-MS is Liquid Chromatography–Mass Spectrometry. This instrument utilizes a similar column
separation technique employed in HPLC, but with mass analysis in the form of mass
spectrometry. After separation, at the end of the column, solvent is removed and the sample is
ionized. A mass detector then scans the molecules (by mass) and produces a spectrum
identifying the compounds within the sample.
PROCEDURE
(i)
(ii)
(iii)
(iv)
(v)
(vi)
Power on the instrument and make sure all reading are well
Turn on the nitrogen gas tank in the gas room and on bench top.
Before the instruments is setup and ran, make sure are samples are prepared and ready to
go.
Turn on the software for the LC-MS
Make sure all parameters and setting are set and LC-MS is working properly.
Set all Standards and unknown to a set vial position and run samples.
Sample Preparation
(i)
Weighted 1.0006g of caffeine and dissolve in 1L of deionized water. This will made 1000ppm
solution. Most of the caffeine was dissolve there may have been a couple little pieces of
caffeine that weren’t dissolved and that tiny fraction was left behind when the solution was
decanted.
(ii)
Make dilutions using the M1V1= M2V2 and the desired volume was 100mL. The desired
concentrations were 50ppm, 100ppm, 150ppm, 200ppm, 250ppm, 300ppm. Once the
standards were made, they were assigned a vial and the vial was filled with sample and
placed in the LC-MS tray.
The unknown samples were collected by heating a tea bag of decaffiened and caffeinated.
Once the tea bags were fully brewed 1mL of solution was diluted into a 100mL volumetric
flask with deionized water. Assigned it to a vial, filled the vial with sample and placed in the
LC-MS tray.
(iii)
Vial Number #
A
B
C
D
E
F
G
H
L
Sample Preparations of Standards and Unknown
(mL) Of Stock
Dilution Flask (mL)
Desired Concen.
Solution
(ppm)
5 (mL)
100 (mL)
50 ppm
10 (mL)
100 (mL)
100 ppm
15(mL)
100 (mL)
150 ppm
20(mL)
100 (mL)
200 ppm
25(mL)
100 (mL)
250 ppm
30 (mL)
100 (mL)
300 ppm
1 (mL)
100 (mL)
(x)
1 (mL)
100 (mL)
(x)
1 (mL)
100 (mL)
(x)
Retention Times
(min)
1.8480 (min)
1.8457 (min)
1.8575 (min)
1.8602 (min)
1.8606 (min)
1.8540 (min)
1.7056 (min)
1.7579 (min)
1.6862 (min)
RESULTS
LC-MS Results For Caffeine Standards (A-F) and Unknown Tea Samples (G-H)
Vial Number #
Concentration (ppm)
Area (Counts)
A
50 (ppm)
5495900
B
100 (ppm)
2853700
C
150 (ppm)
20375000
D
200 (ppm)
23934000
E
250 (ppm)
31307000
F
300 (ppm)
33924000
G
unknown
528110
H
unknown
480140
I
unknown
515490
When running the samples on the LC-MS the previous results were obtained. In order to analyze our
results we had to make a standard curve of the data of Area Vs. Concentration. The standard curve
could be further used to find a concentration for the unknown samples given the equation y = mx + b.
thus, with this equation the following information can be gather
(i)
(ii)
The concentration of our unknown species
Where the unknown sample concentration fall within our standard curve graph.
Graph
Standard Curve for Caffiene Standards (LC-MS)
Concentration Vs. Area Under Peak
40000000
35000000
Area (counts)
30000000
25000000
20000000
15000000
10000000
y = 117476x - 505400
R² = 0.9804
5000000
0
0
50
100
150
200
250
300
350
Concentration (ppm)
CALCULATIONS
All o f the calculation done to develop (ppm) concentration were done using the M1V1= M2V2. The
desired concentrations were chosen by making an educated guess about what the concentration of the
unknown (which was the caffeine composition in the tea). A sample calculation is shown below.
For Sample A:
(M2) Desired concentration: 50 ppm
(V2) Desired Volume: 100 mL Volumetric Flask
(M2) Known (Stock) Concentration: 1000ppm
(1000 𝑝𝑝𝑚)( 𝑥 ) = (50 𝑝𝑝𝑚)(100 𝑚𝐿)
X = 5 mL
This means that 5 mL of the stock solution needs to be diluted into a 100 mL volumetric flask to achieve
a concentration of 50 ppm. These calculations were done for all standard concentration.
The other calculation that had to be done was using the equation of the line obtain from the standard
curve graph above to determine the concentrations of the three unknown samples. Simply, you plug in
the obtained (y) values (Area) that were obtained from the LC-MS spectra and solve for x. The following
calculations were made.
For (1) Unknown Sample G:
(528110) = 117476 (𝑥) − 505400
𝑥 = 8.797626749 ppm
The following concentration was determine using the equation of the line obtained for the sample A and
the procedural calculation was repeated for the two additional samples.
Sample H:
(480140) = 117476 (𝑥) − 505400
x = 8.39892880247880 ppm
Sample I:
(515490) = 117476 (𝑥) − 505400
x = 8.69020055160 ppm
CONCLUSION
We determined that the concentrations of all three caffeine samples were 8.79 ppm, 8.39 ppm, and
8.69 ppm respectively. One of the samples was suppose to be decaffeinated. Thus, for sample H there
wasn’t supposed to be a caffeine peak. However, there could have been a miss up because the first two
dilutions that were made the protocol were to have unknown G be the caffeinated tea sample and
unknown H be the decaffeinated sample. Thus, When I was making the dilutions I wasn’t sure if I did
just that because I got side tracked while I was making the dilutions. Thus, the peak that appeared on
the data output for the LC-MS for all unknown were 1.71 (min), 1.757 (min), and 1.68 (min) for unknown
G, H, and I respectively. This was very different then what we obtained from the standards which were
about 1.85 on average for all standards. Thus, we were consistent with picking our peaks for the
unknown for around 1.70 because that was the characteristic peak that appeared for Unknown G, which
was suppose to be the caffeine sample and it was somewhat close to 1.85. Also, I think that since the
concentrations for the unknowns were so much lower and possibly a little more diluted that the
retention time was so much lower. Also, It could be a possibly that the tea bag that was supposed to be
decaffeinated had small amounts of caffeine because when doing the HPLC experiment that was done
consecutively with the experiment the same data came about and all unknowns were shown to have
caffeine. Thus, we could have also made the mistake of grabbing 3 samples that were caffeinated. To
improve this experiment it was be best to repeat the experiment being a more cautious and patient.