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Gas Chromatography of 24

MODEL OF CHROMATOGRAPHS

700XA Natural Gas Chromatograph

The Danalyzer 700XA Gas Chromatographs provide the most accurate analysis of

natural gas available in a field mounted gas chromatograph (GC). The

700XA features a design that increases analytical capability, maximizes ease of use,

and widens the range of analysis options in a GC with an ambient temperature rating

of -4 to 140 °F (-20 to 60 °C). These enhanced features make the 700XA ideal for

natural gas custody transfer and applications requiring advanced analysis such as

C9+ (with hydrocarbon dew point and cricondentherm calculations) and C6+ with

hydrogen sulfide (H2S).

Engineered with a single-cast enclosure, the 700XA offers an efficient use of oven

space to accommodate both micropacked and capillary columns, as many as four 6-

port or 10-port valves, a rotary valve for liquid injections, up to two thermal

conductivity detectors, and an optional micro flame ionization detector (µFID). With a

significant reduction in internal cabling, the 700XA allows maximum access to valves

and internal components, making maintenance quick, easy, and cost effective.

The analytical power inherent within the 700XA ensures the precise, fast

measurement you need to measure up to C9+ and calculate the hydrocarbon dew

point of natural gas. The reliability and simplified design also makes this GC the best

choice for critical custody transfer C6+ energy measurement, relative density, and

Wobbe measurements.

500 Natural Gas Chromatograph

The Danalyzer 500 Gas Chromatographs offer the broadest range of analysis options

available today in a field-mounted GC. Whether it is heating value measurement,

trace contaminant monitoring, pipeline integrity, or product quality/process control, the

500 GC is flexible enough to meet your analysis needs. The proven technology and

software of the 500 GC series offers superior reliability and precision, lower

installation and operating costs, greater application flexibility, and unmatched

measurement performance.

The Danalyzer 500 Gas Chromatograph is available in two models:

570/571 BTU/CV Gas Chromatograph

The ideal choice for compositional analysis of pipeline gas, including C1 to C6+, N2,

and CO2.

590/591 Dual Oven Gas Chromatograph

Utilizes dual chromatography for advanced applications. Provides more complete

AGA 8 calculations. Typical applications include C9+ with optional hydrocarbon

dewpoint calculation and C6+ with trace H2S.

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

In the fluoride and

manganese

experiments, for

example, the

“detector” is

selective for the

component of

interest within the

matrix of o

particular s


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

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.

control panel

on/off switch

injection port

FID detector

detail of column in oven

Detail of column in oven

Figure 1. Front view of gas chromatograph

capillary column wound around holder

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

H2 inlet (detector)

Air inlet (detector)

N2 inlet (make-up gas)

heater fan

He inlet (carrier gas)

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 Si O

CH3

CH3

Si

R

R

O Si

CH3

CH3

CH3

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 re-

vaporization 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

n


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.)

Properties of liquids

and solutions are

covered in chapter 6

of Oxtoby. In

particular, section 6-

1 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.

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

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

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.

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

If your peaks are very far apart such that the analysis takes a long time, should you increase or decrease the column temperature?


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

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.

Figure 3. Sample chromatogram

The detector voltage (y-axis) is plotted as a function of time (x-

axis). 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

time (min)

resp

on

se (

µV

)

peak retention

time

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 well-

separated 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 non-

interfering 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.

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.

A similar

calibration

method is used

in the fluoride

and manganese

experiments.

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,

( )( )( )( )[ ]

%100ak

akp

ii

iii •=

∑ (2)

or

( )( )( )( ) ( )( ) ( )( ) ( )( )

%100akakakak

akp

44332211

111 •

+++= 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.


4

ii

k

kf = (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.

( )( )( )( )

( )( )( )( )

( )( )( )( )44

ii

ii

44

ii

ii

4

i

ak

ak

akak%100

akak%100

p

p=

•

•

=

∑

∑ (5)

==4

ii

k

kf (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.

p1 = etc. (7)

f a

f a f a f a f a + + +

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.)

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.

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 µL 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 µL of liquid into the syringe; then, holding the syringe

vertically with the needle pointing upward, push the

plunger in until it reads 0.1 µL (the volume to be injected).

With the syringe still held vertically, withdraw the plunger

to about 0.5 µliter; 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

Without careful cleaning between runs, contamination from one injection to the next can be a problem.


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 µL.

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

average value of each fi based on the three runs and determine

the relative 95% confidence interval of the mean for each.

4. 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.

NAME LAB SECTION

Chromatograph Number Sample Number

Date Report Submitted

GAS CHROMATOGRAPHY

Results for Standard Mixture

Final Column Temperature: ____________________

Run 1 Run 2 Run 3

Component ret. time area ret. time area ret. time area

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

Run 1

Run 2

Run 3

Avg.

95% CIm

f1

f2

f3

f4

increasing ret. time

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 Run 2 Run 3

Component ret. time area ret. time area ret. time area

1

2

3

4

Calculation of % Composition of Unknown

Component

Run 1 Run 2 Run 3

fi*ai % fi*ai % fi*ai %

1

2

3

4

Total

Component Avg. % Comp. Rel. 95% CIm

1

2

3

4

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.