Chromatography Lab Report

GC-1 (Gas Chromatography)

Experiment 1, 2, 3 & 4

Experiment 1:- Determination of ethanol content of a mouthwash using an internal standard

Experiment 2:- Determination of oxygenates in gasoline

Experiment 3:- Qualitative and Quantitative analysis of BTEX (Benzene, Toluene, Ethylbenzene and Xylene)

Experiment 4:– Determination of volatile compounds by headspace analysis

• Anil Kumar

Introduction & Theory

The experiments performed in this lab were based on Gas chromatography – specifically gas-liquid chromatography. This technique involves a sample being vapourised and injected onto the head of the chromatographic column. The sample is transported through the column by the flow of inert, gaseous mobile phase. The column itself contains a liquid stationary phase which is adsorbed onto the surface of an inert solid.

GC: Schematic diagram (http://teaching.shu.ac.uk/hwb/chemistry/tutorials/chrom/gcdiag.gif )

In Experiment 1: the ethanol content in a mouthwash was determined. This was done by using an internal standard of butanol. Then, solutions of internal standard with unknown solution and ethanol were made and injected into the GC. The areas obtained for each compound were then used to calculate the % alcohol in the sample.

In Experiment 2: ethanol content in gasoline mixture was calculated using a standard curve of ethanol made with pure ethanol. The gasoline mixture was extracted with water to extract the ethanol in it for further analysis by GC. The GC method is used industrially to find ethanol content in gasolines, plus oxygenates like butyl ether that can contaminate drinking water..

In Experiment 3: qualitative analysis of BTEX (Benzene, Toluene, Ethylbenzene and Xylene) was performed by injecting pure solutions of these compounds in GC. The retention times obtained were recorded for each. In the second part, standard solutions of toluene and xylene were made and calibration curves were made for each. Next, the unknown sample of BTEX was injected to GC. The retention areas obtained were recorded and substituted in equations of standard curves to find the actual content of toluene and xylene in sample.

In Experiment 4: the technique of headspace analysis was used to determine the volatile compounds in a contaminated soil sample. The soil sample was also deliberately contaminated with toluene and xylene and placed in a vial for 10 minutes. Then using a syringe, the headspace was sucked and injected to GC for performing a qualitative analysis (the retention times of BTEX from Experiment 3 were used).

GC is continuing to be used in a number of fields as an analytical tool due to certain advantages like:

• Shorter run times
• Greater sample throughput
• Cheaper columns
• Higher signal to noise ratio
• Lower bleed (thinner films)
• High resolution power compared to others. Complex mixture can be resolved into its components by this GC method. The separation, determination and identification of many compounds withnegligibledifferences in boilingpoints is possible by this technique.
• Sensitivity in detection is very high with thermal conductivity detectors. One can detect upto 100 ppm, while flame detectors, electron capture and phosphorus detectors can detect ppm, parts per billion or picograms respectively.

(http://www.sge.com/support/training/fast-gc-analysis/advantages-/-disadvantages-of-fast-gc)

Some industrial applications of GC include:

• Identification and quantification of ubiquitous pollutants in the environment: analysis of various classes of persistent organic contaminants in air, water, soils, sediments and biota
• GC Analysis of Antioxidants
• Determination of ethanol in gasoline
• Analysis and quality assessment of alcoholic beverages –
• Quantitative and qualitative assessment of

• Alcohols in blood
• Aromatics (benzene, toluene, ethylbenzene, xylene)
• Flavors and Fragrances
• Permanent gases (H₂, N₂, O₂, Ar, CO₂, CO, CH₄)
• Hydrocarbons
• Pesticides, Herbicides, PCBs, and Dioxins
• Solvents

(http://www.med.cmu.ac.th/dept/vascular/alcho/research/res_out/Application%20of%20gas%20chromatography.pdf)

As we can see, the varied applications of GC in industry and its advantages over other methods, performing of these GC experiments is industrially justified.

Experiment 1:- Determination of ethanol content of a mouthwash using an internal standard

Procedure

1. The instrument was set to the following parameters:

Injector temperature: 200 degree C

Detector temperature: 250 degree C

Oven temperature: 80 degree C

Attenuation: between 4 and 16

Carrier gas pressure: 8-10 psi

FID range: 1

Valve: Split 1 on

2. Dilute the unknown ethanol sample and mouthwash provided with water in a 1:10 ratio. 10 ml of each was diluted to 100 ml in volumetric flask.
3. Next, ethanol standard solution and butanol (internal standard) (each2 ml to 100 ml water) were made.
4. Equal volumes of ethanol and internal standard were mixed. And equal volumes of sample solution and internal solution were made (each 5 ml).
5. Now, three injections (0.3 uL each) were made of each of these solutions into the GC.
6. From the chromatographs, the % of alcohol in sample were calculated.

Observations, Calculations and Results

1. Instrument: GC SST
2. Instrument settings:

Injector temperature: 200 degree C

Detector temperature: 250 degree C

Oven temperature: 80 degree C

Attenuation: between 4 and 16

Carrier gas pressure: 8-10 psi

FID range: 1

Valve: Split 1 on

Using the formula,

Rspl

RI.S. = Cspl

Rstd Cstd

RI.S.

where, R=response (peak area); spl=sample, I.S.=internal standard

C= Concentration

For Unknown ethanol sample,

we have, Rspl = 874.3, R I.S= 2878.8, Rstd = 1341.9, R I.S. = 2226.7

Cstd = 2 % (From observation table # 1 & 2)

Putting these values in equation,

874.3 / 2878.8 / 1341.9 / 2226.7 X 2 = Cspl

On solving we get, Cspl = 1.00 %

For mouthwash,

we have, Rspl = 2079.5 , R I.S= 3373.4, Rstd = 1341.9, R I.S. = 2226.7

Cstd = 2 % (From observation table # 1 & 2)

Putting these values in equation,

2079.5 /3373.4 / 1341.9 / 2226.7 X 2 = Cspl

On solving we get, Cspl = 0.98 %

So, % alcohol in Mouthwash is = 0.98 %

in Unknown ethanol sample = 1.00 %

Experiment 2:- Determination of oxygenates in gasoline

Procedure

1. Develop a set of operating conditions that will satisfactorily separate ethanol from hexane. To do this equal volume of hexane and ethanol in a small vial were combined and injected. Inject this mixture into GC and ensure two resolved peaks. Inject pure hexane to establish its identity.

The oven temperature was decreased from 80 degree C to 70 and then to 60 degrees C to separate the two peaks.

2. Prepare ethanol standards: 0.2 ml, 0.5 ml, 1.0 ml and 2.0 ml in 25 ml DI water.

Concentration, 0.2 ml = 0.2/25 = 0.008 %,

0.5 ml = 0.5/25 = 0.02 %

1.0 ml = 1.0/25 = 0.04 %

2.0 ml = 2.0/25 = 0.08 %

3. These standards were injected into GC and a calibration curve was prepared using the peak area data obtained.
4. The unknown gasoline sample (Unknown Sample D: Ethanol in hexane) was taken and 5.0 of it was transferred to a vial. 5.0 ml of water was added and mixed thoroughly in the vial. It was allowed to stand for 5 minutes.
5. The water layer was taken using a Pasteur pipette and injected into GC.
6. The % ethanol was determined using calibration curve data and peak area data from step 5.

Observations, Calculations and Results

1. Instrument: GC SST
2. Instrument settings:

Injector temperature: 200 degree C

Detector temperature: 250 degree C

Oven temperature: 60 degree C (Earlier 80 degree C and 70 degree C)

Attenuation: between 4 and 16

Carrier gas pressure: 8-10 psi

FID range: 1

Valve: Split 1 on

6. Calculating % of ethanol in sample

Equation of line from standard curve : y = 89994x + 1605.1

where, y = peak area, x = concentration of ethanol in %

From Observation table # 3 we have,

Area of sample = 1982.2 and 2138.1

Putting these values in equation in place of y we get,

1982.1 = 89994x + 1605.1 & 2138.1 = 89994x + 1605.1

On solving for x we get, x = 0.00418 % and x = 0.00592 %

Averaging the two values, we get x = 0.00505 %

So, the % of ethanol in the given Unknown Ethanol in Hexane Sample D is 0.00505 %.

Experiment 3:- Qualitative and Quantitative analysis of BTEX (Benzene, Toluene, Ethylbenzene and Xylene)

Procedure

Part A: Qualitative Analysis

1. Take 1 ml each of Benzene, Toluene, Ethylbenzene and Xylene in separate vials.
2. Run the individual standards and record the retention times.

Part B: Quantitative analysis

1. Prepare a series of standard of toluene and para-xylene using hexane as a solvent.

Make 2,4,6,8 and 10 % solutions of each of toluene and para-xylene in 50 ml volumetric flasks.

For 2 %= 1 ml each of toluene or para-xylene (separate), for 4 % = 2ml

For 6 % = 3 ml, for 8 % = 4 ml, and for 10 % = 5 ml to 50 ml with hexane.

2. Prepare a calibration curve based on the peaks area data obtained.
3. Analyse an unknown sample of BTEX provided and find the % of toluene and para-xylene in it using the standard curve data.

Observations, Calculations and Results

1. Instrument: GC SST
2. Instrument settings:

Injector temperature: 200 degree C

Detector temperature: 250 degree C

Oven temperature: 80 degree C

Attenuation: between 4 and 16

Carrier gas pressure: 8-10 psi

FID range: 1

4. From individual injections of hexane, toluene and xylene, it was seen that that the retention times for each of these were 0.500, 0.933 and 1.550 respectively.

From the sample analysis, we find that the two biggest peaks are seen at Rt = 0.733. This value neither corresponds to value of Rt for toluene (around 0.900) nor xylene (around 1.350). So we cannot identify or quantitate them.

The other peak (although small) which is seen occurs at Rt = 1.600 is close to Rt for xylene (Rt for 10 % xylene is 1.600). So this peak would be for xylene.

We can find its concentration by making a standard curve for xylene from the data in Observation table # 7.

6. Calculating amount of xylene in sample.

From the standard curve (Graph # 2), we have the equation of line

y = 217.04x + 22.45

where, y = peak area, x = concentration of ethanol in %

From Observation table # 7 we have,

Area of sample = 863 and 181

Putting these values in equation in place of y we get,

863 = 217.04x + 22.45 & 181 = 217.04x + 22.45

On solving for x we get, x = 3.87 % and x = 0.73 %

Averaging the two values, we get x = 2.30 %

Therefore, the sample (Unknown B) contains 2.30 % of xylene and no toluene.

Experiment 4:– Determination of volatile compounds by headspace analysis

Procedure

1. Soil was taken in a sealed vial and 1-2 drops each of toluene and p-xylene were added to it.
2. Let the sample rest in the vial for about 10 minutes so that the volatile components gather in the headspace of the vial.
3. After 10 minutes, using a syringe, suck out 0.3uL of the headspace keeping in mind that the syringe doesn’t touch the soil sample itself and headspace sample is taken from just the midway of the vial.
4. Inject this to the GC. Perform duplicate injections if both components can not be seen at one go.

Observations, Calculations and Results

1. Instrument: GC SST
2. Instrument settings:

Injector temperature: 200 degree C

Detector temperature: 250 degree C

Oven temperature: 80 degree C

Attenuation: between 4 and 16

Carrier gas pressure: 8-10 psi

FID range: 1

From Observation Table # 7, last experiment, we know retention times of both toluene and p-xylene. They were 0.950 and 1.550. So on comparing these Rts with the Rts obtained in these chromatograms, we can identify the peaks as toluene or xylene as done in Observation Table # 8.

So, using headspace analysis, Toluene and p-xylene could be identified in the soil sample.

Discussion and Conclusion

In this lab, using the technique of Gas Chromatography, four different experiments were performed.

In Experiment 1: the ethanol content in a mouthwash was determined. This was done by using an internal standard of butanol. Then, solutions of internal standard with unknown solution and ethanol were made and injected into the GC. The areas obtained for each compound were then used to calculate the % alcohol in the sample. The amount of ethanol present in the unknown solution of ethanol was found to be 1.00 % and the ethanol content in mouthwash was found to be 0.98%.

In Experiment 2: ethanol content in gasoline mixture was calculated using a standard curve of ethanol made with pure ethanol (An unknown ethanol in hexane sample was used). The gasoline mixture was extracted with water to extract the ethanol in it for further analysis by GC. The amount of ethanol that was found to be present in this sample was 0.00505%. Firstly, we injected pure hexane and ethanol to find out their retention times and then an equal mixture was injected to see if the two peaks can be resolved or not. After doing this, the experiment was performed. The calculations for determining the % of ethanol were based on equation that was derived from the standard curve from ethanol standards. The value of peak area obtained in the chromatogram for the unknown sample was substituted in the equation from standard curve to find the ethanol concentration.

This GC method is even used industrially to find ethanol content in gasolines, plus oxygenates like butyl ether that can contaminate drinking water.

In Experiment 3: qualitative analysis of BTEX (Benzene, Toluene, Ethylbenzene and Xylene) was performed by injecting pure solutions of these compounds in GC. The retention times obtained were recorded for each.

In the second part, quantitative analysis of toluene and p-xylene was done. Standard solutions of toluene and xylene were made and calibration curves were made for each. Next, the unknown sample of BTEX was injected to GC. The retention areas obtained were recorded and substituted in equations of standard curves to find the actual content of toluene and xylene in sample. It was found that the unknown sample had not retention that matched with the retention times of toluene which was close to 0.9550. No peaks were seen at this retention times even on duplicate runs of the sample. So it is concluded that the sample had no toluene in it. Nevertheless, a peak for xylene was seen (determined by comparing the Rt with the Rt of toluene from Experiment 3). A standard curve was made from the data obtained by running xylene standards. Now with the equation of calibration curve and the peak area of sample for xylene, the concentration of xylene present in the sample was calculated. It was found to contain 2.30 % xylene.

In Experiment 4: the technique of headspace analysis was used to determine the volatile compounds in a contaminated soil sample. The soil sample was deliberately contaminated with toluene and xylene and placed in a vial for 10 minutes. Then using a syringe, the headspace was sucked and injected to GC for performing a qualitative analysis (the retention times of BTEX from Experiment 3 were used). The peaks for both toluene and xylene could be detected by GC. This conclusion was based on the fact that the peaks had comparable retention times as toluene and xylene.

To conclude, we can say that we used GC for quantitative analysis like analysis of BTEX in chemicals, food, etc. or BTEX in soil. And qualitative analysis like determination of ethanol content in gasoline, mouthwash, etc.

References

1. Page # 67-73 Chromatography Laboratory Manual, Durham College 2012
2. Advantages of GC http://www.sge.com/support/training/fast-gc-analysis/advantages-/-disadvantages-of-fast-gc)
3. Industrial applications of GC

(http://www.med.cmu.ac.th/dept/vascular/alcho/research/res_out/Application%20of%20gas%20chromatography.pdf)