GAS CHROMATOGRAPHY
The objective of this experiment is to separate, by gas
chromatographic techniques, a mixture of the four isomeric butyl
alcohols and to determine the percentage of each an unknown
mixture.
Upon completing this experiment, you should:

know the basic components of a chromatography instrument

understand the importance of component separation to
chemical analysis

understand the mechanism by which components are separated
on a GC column and the variables that affect separation

understand the basic methods of calibration common in
chromatographic analysis
PRINCIPLES
Chromatography is a very important analytical tool because it
allows the chemist to separate components in a mixture for subsequent
use or quantification. Most samples that chemists want to analyze are
mixtures. If the method of quantification is selective for a given
component in the mixture, separation is not required. However, it is
often the case that the detector is not specific enough, and a separation
must first be performed. There are several types of chromatography
depending on the type of sample involved. In this experiment, we’ll
use gas chromatography.
The gas chromatograph makes it possible to separate the
volatile components of a very small sample and to determine the
amount of each component present. The essentials required for the
method are an injection port through which samples are loaded, a
"column" on which the components are separated, a regulated flow of
In the fluoride and
manganese
experiments, for
example, the
“detector” is
selective for the
component of
interest within the
matrix of our
particular samples.
Gas Chromatography
Page 1 of 23
a carrier gas (often helium) which carries the sample through the
instrument, a detector, and a data processor. In gas chromatography,
the temperature of the injection port, column, and detector are
controlled by thermostatted heaters. Figures 1 and 2 are pictures of the
instrument from the front and rear respectively, with important
components labeled. The following sections describe in detail the
function of each component.
FID detector
control panel
injection port
on/off switch
capillary column
wound around holder
fan
Detail of column in oven
Figure 1. Front view of gas chromatograph
Air inlet
(detector)
H2 inlet
(detector)
N2 inlet
(make-up gas)
He inlet
(carrier gas)
heater fan
Figure 2. Rear view of gas chromatograph
INJECTION PORT
The sample to be analyzed is loaded at the injection port via a
hypodermic syringe. The injection port is heated in order to volatilize
the sample. Once in the gas phase, the sample is carried onto the
column by the carrier gas, typically helium. The carrier gas is also
called the mobile phase. Gas chromatographs are very sensitive
instruments. Typically samples of one microliter or less are injected
on the column. These volumes can be further reduced by using what is
Gas Chromatography
Page 3 of 23
called a split injection system in which a controlled fraction of the
injected sample is carried away by a gas stream before entering the
column.
COLUMN
The column is where the components of the sample are
separated. The column contains the stationary phase. Gas
chromatography columns are of two types—packed and capillary.
Capillary columns are those in which the stationary phase is coated on
the interior walls of a tubular column with a small inner diameter. We
will use a capillary column in this experiment.
The stationary phase in our column is a polysiloxane material.
The basic structure of the polymeric molecules is shown below, where
n indicates a variable number of repeating units and R indicates an
organic functional group. In our columns, 5% of the “R’s” are methyl
groups (-CH3) and 95% of the “R’s” are phenyl groups (-C6H5)
CH3
H3C
Si
CH3
CH3
R
O
Si
O
Si
R
CH3
CH3
n
This polymeric liquid has a high boiling point that prevents it from
evaporating off the column during the experiment.
The components in the sample get separated on the column
because they take different amounts of time to travel through the
column depending on how strongly they interact with the stationary
phase. As the components move into the column from the injection
port they dissolve in the stationary phase and are retained. Upon revaporization into the mobile phase they are carried further down the
column. This process is repeated many times as the components
migrate through the column. Components that interact more strongly
with the stationary phase spend proportionally less time in the mobile
phase and therefore move through the column more slowly. Normally
the column is chosen such that it’s polarity matches that of the sample.
When this is the case, the interaction and elution times can be
rationalized according to Raoult’s law and the relationship between
vapor pressure and enthalpy of vaporization. The rule of thumb is that
retention times correlate with boiling points. (Do not expect an exact
quantitative correlation, i.e. one with an R-value close to one, for this
simple model. You will be using a non-polar column and the
interaction between an alcohol molecule and the stationary phase will
be dominated by weak van der Waals forces.)
In this experiment, you will use a gas chromatograph to separate and
quantify mixtures containing various isomers of butyl alcohol (n-butyl
alcohol, sec-butyl alcohol, iso-butyl alcohol, and t-butyl alcohol). Look
up the structures of these compounds, their normal boiling points, and
their enthalpies of vaporization. (There are reference books in the
library that contain thermodynamic data for organic compounds.
Another source of thermodynamic data is NIST's webpage, which can
be accessed via the lab homepage. Quick information on structure
and boiling points can be obtained from a variety of sources including
chemical catalogs, the Merck Index, and Chemfinder.com) Based on
the relationship between vapor pressure and enthalpy of vaporization,
which component do you expect to travel fastest through the column?
Boiling points are another indicator of intermolecular forces. Do your
Properties of liquids
and solutions are
covered in chapter 6
of Oxtoby. In
particular, section 61 covers the
relationship between
molecular structure
and vapor pressure.
Section 6-6 covers
Raoult's law, which
relates the vapor
pressure of a
solution to the vapor
pressure of the pure
components.
predictions based on the trends in Hvap agree with the trends in
boiling points?
Structure
Hvap
bp(K)
n-butyl alcohol
isobutyl alchol
sec-butyl alchol
t-butyl alchol
Gas Chromatography
Page 5 of 23
As described above, the rate at which compounds move
through the column depends on the nature of the interaction between
the compound and the stationary phase. Other variables that affect this
rate are column temperature and carrier gas flow rate. In this
experiment, you will be provided a set of initial column conditions to
analyze your samples. Based on the results of your first run, you will
then vary the column temperature in order to achieve good separation
of the peaks in the shortest possible time. One should avoid
experimental conditions that lead to excessively long elution times.
Not only do you waste valuable resources (your time and chart paper)
but broadening of the peaks and loss of resolution will become evident
when the elution times are too long. This broadening is an inevitable
consequence of diffusion. The theory of diffusion shows that the
width of a peak is roughly proportional to the square root of elution
time. Thus the optimum conditions are those that result in complete
separation of the peaks in the shortest possible time.
If your peaks are very far apart such that the analysis
takes a long time, should you increase or decrease the
column temperature?
DETECTOR
If the column conditions are chosen correctly, the components
in the sample will exit the column and flow past the detector one at a
time. There are several different types of detectors common to gas
chromatography instruments. The choice of detector is determined by
the general class of compounds being analyzed and the sensitivity
required. Our gas chromatographs are equipped with flame ionization
detectors (FIDs)—the most widely used detectors for organic samples.
FIDs use an air/hydrogen flame to pyrolyze the effluent sample. The
pyrolysis of the compounds in the flame creates ions. A voltage is
applied across the flame and the resulting flow of ions is detected as a
current. The number of ions produced, and therefore the resulting
current, depends on the flame conditions and the identity of the
molecule in question. (As a rough approximation, the current is
proportional to the number of reduced carbons in the molecule.) In
other words, the detector shows a different response to each
compound. For this reason, separate calibrations must be performed
for each compound analyzed.
INTEGRATING RECORDER
response (V)
The output of the detector (converted from current to voltage)
is sent to an integrating recorder that plots, stores, and analyzes the
data. A typical chromatogram is shown in Figure 3.
peak retention
time
time (min)
Figure 3. Sample chromatogram
The detector voltage (y-axis) is plotted as a function of time (xaxis). Each peak corresponds to a separate component. The time it
takes for a given peak to appear after injection is called the retention
time. If the column conditions are kept constant, the retention time for
Gas Chromatography
Page 7 of 23
each component is quite reproducible from one sample and injection to
the next. The identity of each peak can be determined by injecting
pure samples of the individual components of the mixture and noting
their retention times.
The voltage from the detector is proportional to the number of
molecules passing through the detector at any given time. For wellseparated peaks, the total number of molecules of each component
reaching the detector is then proportional to the area under the peak.
The recorder determines the area of each peak by integration and
reports this in the results table. The proportionality factor between
area and amount must be determined by a calibration experiment.
Note that the integrator will also determine areas for peaks that are not
well-separated by dropping a vertical line where the slope changes
sign (Figure 4). The results will be in error, however, because the
voltage read at the beginning of peak 3 is actually the sum of the
response due to the third component plus the response due to the
second component still exiting the column.
Figure 4. Chromatogram of non-separated peaks
There are several methods by which gas chromatographs are
typically calibrated. One method is to inject standard samples
containing varying concentrations of the compound to be analyzed and
creating a calibration curve (area vs. concentration). As you’ll
discover, however, it is very difficult to reproducibly inject the same
volume onto the column each time.
A more advanced calibration method is to use something called
an internal standard. In this method, a constant concentration of a noninterfering compound is added to each sample before it is analyzed.
The ratio of the areas of the added compound and analyte are then
used to construct the calibration curve. Using an area ratio instead of
an absolute area compensates for varying injection volumes.
The calibration method we will employ is similar to the use of
an internal standard in that it corrects for variability in the injection
volume. It differs in that it determines relative response factors based
on the results of one standard sample of known composition. (This is
possible because our standard sample contains only the four isomers of
butyl alcohol, i.e. the sum of all the components adds up to 100%).
The next section guides you through the calculation of relative
response factors from a single standard mixture.
A similar
calibration
method is used
in the fluoride
and manganese
experiments.
RELATIVE RESPONSE FACTORS
As described above, if the detector were equally sensitive to
each component in a mixture, the peak areas could be used directly to
give the percentage composition of the mixture by dividing the area of
each peak by the total area under all of the peaks. Since the detector is
not equally sensitive to the different components, each peak area must
be multiplied by a suitable factor (called the response factor, k) to
correct for this difference. The corrected areas are then used for the
calculation of the percentage composition of the mixture. We will use
something called a relative response factor, f, which ratios each
response factor to that of a chosen component. The relative response
factors are determined by measuring the peak areas for a mixture of
known composition. These relative response factors can then be used
to determine the percent composition of an unknown mixture of the
same components.
Gas Chromatography
Page 9 of 23
The following paragraphs walk you through the derivation of
relative response factors in terms of chromatogram peak areas and
percent compositions. You will need to fill in the boxes to complete
the derivation.
The following symbols will be used in this derivation:
ai = area of peak for ith component (from chromatogram)
pi = percent composition of ith component (known for standard
sample, unknown for unknown sample; but in both cases the
sum of the pi’s is 100%)
qi = quantity of ith component reaching the detector
ki = response factor of ith component
fi = relative response factor of ith component
The quantity of each component passing through the detector is
proportional to the area of the chromatogram peak for that component.
qi = ki *
(1)
Since the percent composition for each component is the quantity of
that component divided by the sum of all the components,
pi 
k i a i   100%
 k i a i 
(2)
or
p1 
k1 a1 
 100%
k1 a1   k 2 a 2   k 3 a 3   k 4 a 4 
etc.
(3)
Using the four known values of p and the four measured areas yields
four equations and four unknowns that can be solved simultaneously.
However, to simplify the analysis we can define a relative response
factor fi, which compares each response factor to a common
reference—we’ll choose component four, the component with the
longest elution time.
fi 
ki
k4
(4)
The beauty of this approach can be seen when we express the
percentage of each component relative to the percentage of component
four and then re-arrange to solve for fi in terms of pi’s and ai’s.
100%  k i a i 
pi
 ki a i   ki a i 

k 4 a 4 
p4 100%  k 4 a 4 
 ki a i 
fi 
(5)
ki

k4
(6)
Now each relative response factor can be calculated directly from the
known percent composition of the standard mixture and the
experimentally measured peak areas.
As mentioned above, the relative response factors can then be
used to calculate the percent composition of an unknown mixture of
these components. To determine the percent composition of a given
component in the unknown mixture, pi, divide both the numerator and
denominator of the appropriate equation 3 by k4 and substitute in the
appropriate fi’s for each ratio of k’s to obtain an expression for each pi
as a function of appropriate fi’s and ai’s.
f a
p1 =
f a
+ f a
+ f a
+ f a
etc.
(7)
Gas Chromatography
Page 11 of 23
EXPERIMENTAL PROCEDURE
The following paragraphs outline the procedure to follow in
analyzing liquid samples on the gas chromatograph. Your first task is
to experimentally determine a set of column conditions that yield a
good separation of the four isomers in your standard sample in a
reasonable time frame. Once you have determined a good set of
conditions, repeat this analysis three times. You will use these
chromatograms to determine an average relative response factor for
each component. Next, obtain three chromatograms of your unknown
mixture (which will contain 3 of the 4 isomers in varying percent
compositions) from which you will determine the percent composition.
Finally, you will identify each peak by injecting pure samples of each
component and noting the retention time.
1.
Column. In order to separate the four isomers of butyl alcohol
you will use a 5% diphenyl, 95% dimethyl polysiloxane capillary
column that is 15 feet long and has a 0.25 mm inner diameter. The
column is already installed in the gas chromatograph.
2. Chromatograph Conditions. You will be given a set of initial
conditions for your standard sample. Your task is to vary the
conditions to achieve good separation of the peaks. Although
many parameters affect the separation (e.g. column type, column
length, carrier gas flow rate, column temperature) to simplify the
experiment you will only make changes to the column
temperature.
The following conditions will be set prior to your arrival.

Gas Pressures and Flows:
H2 (to FID detector) = 60 kPa
Air (to FID detector) = 50 kPa
N2 (make-up gas) = 60-70 kPa
He (carrier gas) = 0.5 mL/min
split ratio = 100

Temperatures:
column = 100 ºC
injector = 200 ºC
detector = 200 ºC
After making your first run, you will adjust the column
temperature to improve the separation of your peaks. To do
this, press col on the instrument, enter the desired temperature
using the numerical keypad, and press enter . The LCD panel
displays both the setpoint and the actual temperatures. When
the column reaches the setpoint and has stabilized, the green
ready light will appear indicating that the instrument is ready
for an injection.
3. C-R8A Recorder.
a. Turn the recorder switch ON at the back left.
b. Use the monit button to monitor the voltage from the
detector when no sample is injected. Use the zero
button on the gas chromatograph to set the output voltage
to approximately + 100-200 µV. This sets the baseline for
the chromatogram. (The range of the output is –5000 µV
to +1V)
c. Set the recorder attenuation using the Atten button on
the recorder. Try a value of 7 to start. You may have to
adjust this if your peaks are too small or too large.
(Choosing a larger number makes the peaks appear
smaller.)
Gas Chromatography
Page 13 of 23
4. Sample Injection. The liquid sample is injected into the helium
gas stream by inserting the needle of a special expensive syringe
through a heavy-wall rubber septum.
Without careful
cleaning
between runs,
contamination
from one
injection to the
next can be a
problem.
a. DO NOT PULL THE METAL PLUNGER OUT OF
THE GLASS BODY OF THE SYRINGE. IN FACT,
DON'T EVEN COME CLOSE TO DOING IT.
b. With the plunger pushed all the way into the syringe, insert
the needle into the liquid and then withdraw 0.3 - 0.5 L of
liquid; eject this into a Kimwipe. Repeat this rinsing
operation several more times. Without careful cleaning
between runs, contamination from one injection to the next
can be a problem. When changing to a new sample (e.g.
from standard to unknown), check carefully for
contamination peaks. Your standard contains four isomers
whereas your unknown contains only three isomers.
Therefore the presence of a “fourth” peak in your unknown
is clear indication of contamination and the run should be
repeated.
c. When you are ready to make an injection, withdraw 0.3 0.5 L of liquid into the syringe; then, holding the syringe
vertically with the needle pointing upward, push the
plunger in until it reads 0.1 L (the volume to be injected).
With the syringe still held vertically, withdraw the plunger
to about 0.5 liter; this will leave a protective air space at
the tip of the needle.
d. Insert the needle through the rubber septum until stopped
by the protective tube around the needle and inject the
liquid by pushing the plunger all the way in. Immediately
push the start button on the instrument to start the run.
The recorder should start automatically. Then remove the
needle from the septum. The above sequence of operations
must be carried out as quickly as possible.
e. When all components of the liquid have emerged from the
column (including the tail from the last peak), press the
start1/ stop 1 button on the recorder and the stop button
on the instrument to stop the run. Label the chromatogram
and record the retention times and peak areas in your
notebook. When you are ready to run the next sample,
repeat steps (a) - (e) above.
f. When analyses have been completed, carefully tear the
paper from the recorder. Cut out each chromatogram with
their associated peak analysis charts and tape to a sheet of
paper for inclusion in your report.
5. Sequence of Measurements— Summary.
a. Measure a series of chromatograms for the standard
mixture as a function of the column temperature.
Determine a column temperature that yields good
separation within a reasonable length of time. Obtain three
chromatograms of the standard under these optimal
conditions.
The relative response factors will be
determined from these runs. The injection volume for all
runs should be 0.1 L.
b. Obtain three chromatograms of your unknown under the
same conditions.
The percent composition of your
unknown will be determined from these runs.
c. Run each of the pure alcohols under the same conditions (3
out of 4 will suffice). The retention time for each alcohol
will be determined from these runs.
CALCULATIONS
1. Label each peak on the chromatograms of your standard and
unknown mixtures with the name of the alcohol associated with
that peak.
2. Using the runs for your standard mixture, calculate the relative
95% confidence interval of the mean for the area of the first
peak.
3. For each run of your standard mixture, calculate the relative
response factor (fi) for each of the butyl alcohols. Calculate the
Gas Chromatography
Page 15 of 23
average value of each fi based on the three runs and determine
the relative 95% confidence interval of the mean for each.
Use the average values for each fi obtained in 3 and the areas of the
peaks in the chromatogram of your unknown to determine the percent
composition of your unknown sample for each run. Use the values
determined for each run to calculate average percent compositions and
the associated relative 95% confidence intervals of the mean.
4. NAME
LAB SECTION
Chromatograph Number
Sample Number
Date Report Submitted
GAS CHROMATOGRAPHY
Results for Standard Mixture
Final Column Temperature:
____________________
Run 1
Component
ret. time
Run 2
area
ret. time
Run 3
area
ret. time
area
increasing ret. time
1
2
3
4
Avg. retention time of 1st peak _______________ Rel. 95% CIm __________________
Avg. area of 1st peak __________________
Rel. 95% CIm __________________
Calculation of Relative Response Factors for Standard Mixture
Rel.
Response
Factor
f1
f2
f3
f4
Run 1
Run 2
Run 3
Avg.
95% CIm
Results for Unknown Mixture
(Use same component designation as above; note that one line will be blank as there are
only 3 components in your unknown.)
Run 1
Component
ret. time
Run 2
area
ret. time
Run 3
area
ret. time
area
1
2
3
4
Calculation of % Composition of Unknown
Run 1
Component
fi*ai
Run 2
%
fi*ai
Run 3
%
fi*ai
1
2
3
4
Total
Component
1
2
3
4
Avg. % Comp.
Rel. 95% CIm
%
Identification of Peaks
Name of Component
Structure
1.
2.
3.
4.
Question
Discuss the difference in magnitude between the rel. 95% CIm for areas vs. relative
response factors. Why are they so different?
Show sample calculations on back side of this sheet.
Attach chromatograms.