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Gas Chromatography Page 1 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 (FID). 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.
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
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 oparticular s
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depending on the type of sample involved. In this experiment, well 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.
detail of column in oven
Detail of column in oven
Figure 1. Front view of gas chromatograph
capillary column wound around holder
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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)
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 typespacked 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 Rs are methyl groups (-CH3) and 95% of the Rs are phenyl groups (-C6H5)
CH3 Si O
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
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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 its polarity matches that of the sample. When this is the case, the interaction and elution times can be rationalized according to Raoults 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
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?
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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
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
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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 youll 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 pis 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
( )( )( )( ) ( )( ) ( )( ) ( )( ) %100akakakak
+++= 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 referencewell choose component four, the component with the longest elution time.
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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 pis and ais.
( )( ) ( )( )( )( ) ( )( )
( )( )( )( )44
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 fis for each ratio of ks to obtain an expression for each pi as a function of appropriate fis and ais.
p1 = etc. (7)
f a f a f a f a + + +
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
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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 C injector = 200 C detector = 200 C
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 V. This sets the baseline for the chromatogram. (The range of the output is 5000 V 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 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
Without careful cleaning between runs, contamination from one injection to the next can be a problem.
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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.
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
Results for Standard Mixture
Final Column Temperature: ____________________
Run 1 Run 2 Run 3
Component ret. time area ret. time area ret. time area
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
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
Calculation of % Composition of Unknown
Run 1 Run 2 Run 3
fi*ai % fi*ai % fi*ai %
Component Avg. % Comp. Rel. 95% CIm
Identification of Peaks
Name of Component Structure
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.