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Report Template EXPERIMENT 2: ANALYSIS OF TOLUENE IN PETROL BY GAS CHROMATOGRAPHY-MASS SPECTROMETRY (GC-MS) Full Name: Student Number: Date of Submission: 02/10/2018 Date of Practical Session: 04/09/2018

Experimental GC column full description: DB5-MS (5% Phenyl + 95% PMS) Temperature program full description: The temperature starts at 40℃ and is held for 4 minutes, increasing to 180℃ at the rate of 30℃ per minute. The temperature is held at 180℃ for 1 minute Split ratio: 100:1 Injection volume: 1µL Injector temperature: 250℃ Helium flow rate: 1ml/min

Results Part 1: Standard blank

Copy and paste the GC mass chromatogram of the standard blank here. Make sure you caption the figure.

Figure 1: GC mass chromatogram of the standard blank

Part 2: Determination of SIM parameters and library search

Copy and paste the GC mass chromatogram of the TIC of the 80 ppm toluene standard here. Make sure you caption the figure.

Figure 2: TIC of the 80 ppm toluene standard

Copy and paste the EIC of the 80 ppm toluene standard here. Make sure you caption the figure.

Figure 3: EIC of the 80 ppm toluene standard

Copy and paste the mass spectrum of the toluene standard here. Make sure you caption the figure.

Figure 4: Mass spectrum of the toluene standard (80 ppm)

Copy and paste the GC mass chromatogram of the TIC of the unknown here. Make sure you caption the figure.

Figure 5: GC mass chromatogram of the TIC of the unknown (Sample A)

Copy and paste the EIC of the unknown here. Make sure you caption the figure.

Figure 6: EIC of the unknown (Sample A)

Which ions have you chosen for SIM mode? Identify the molecular ion and fragment ion. The ions chosen for SIM mode was ion 91 and 92. Where 92 is the molecular ion and 91 is the fragment ion.

Part 3. Construction of calibration curve and analysis of unknown Copy and paste the chromatogram of a representative SIM standard here. Make sure you caption the figure.

Figure 7: Chromatogram of the Representative SIM Standard

Copy and paste the chromatogram of SIM unknowns here. Make sure you caption the figure.

Figure 8: Chromatogram of SIM unknown (sample A)

Figure 9: Chromatogram of SIM unknown (sample B)

Part 4: Data processing and reporting Complete the following table, ensuring that you update the units, and caption the table. Table 1 Results from the GC- MS method

Sample Identity

Data File Name

toluene retention time (min)

toluene area

10 ppm toluene

Sample1

2.173

3088265

toluene height

(Abundance*s) (Abundance) 123868

20 ppm toluene

Sample2

2.171

3407839

136699

40 ppm toluene

Sample3

2.171

7675093

306794

80 ppm toluene

Sample4

2.172

15368313

615604

120 ppm toluene

Sample5

2.172

22136532

885700

Unknown Sample A

SampleA

2.177

3251494

204564

Unknown Sample B

SampleB

2.180

2644706

166288

Calibration curve Using the data in the table above, construct two calibration curves in Excel (peak area vs concentration, and peak height vs concentration), and fit trend lines. Copy and paste your calibration curves here. Ensure that you label all axes, include the calibration curve equation, correlation coefficient, graph title and caption.

Peak area vs Conc. 25000000 f(x) = 180455.02 x + 590637.35 R² = 1

Area (Ab*s)

20000000 15000000 10000000 5000000 0

0

20

40

60

80

100

Concentration (ppm)

Figure 10 Calibration curve of peak area vs concentration

120

140

Height (Ab)

Peak height vs Conc. 1000000 900000 800000 700000 600000 500000 400000 300000 200000 100000 0

f(x) = 7219.81 x + 23863.19 R² = 1

0

20

40

60

80

100

120

140

Concentration (ppm)

Figure 11 Calibration curve of peak height vs concentration

Determination of unknown Using the trend line equations determine the concentration of the unknowns in the diluted sample. Sample A dilution factor= 0.0005 Sample B dilution factor= 0.001 Peak Area vs Concentration curve Equation: y=180455x + 590637 Sample A When y=3251494, x=14.75 Conc.= 14.75 ppm Conc. after dilution factor = 14.75/0.0005 = 29500 ppm Sample B When y=2644706, x=11.38 Conc.=11.38 ppm Conc. After dilution factor = 11.38/0.001= 11380 ppm

Peak height vs concentration Equation: y=7219.8x + 23863 Sample A When y=204564, x= 25.03 Conc.=25.03 ppm Conc. after dilution factor = 50060

Sample B When y=166288, x=19.73 Conc.=19.73 ppm Conc. after dilution factor = 19730

Apply the dilution factors to calculate the concentration of toluene in the samples and express as % v/v. 1% v/v =1/100 1 ppm=1/10000000 Therefore, 1 ppm=0.0001% v/v Density of toluene=0.867 1% v/v of toluene= 1/0.867 = 1.153% v/v Hence, 1 ppm of toluene= 0.0001153% v/v Peak Area vs Concentration Curve Sample A 29500 x 0.0001153 = 3.40% v/v Sample B 11380 x 0.0001153 = 1.31% v/v Peak Height vs Concentration Curve Sample A 50060 x 0.0001153 = 5.77% v/v Sample B 19730 x 0.0001153 = 2.27% v/v Discussion Questions 1) Comment on any peaks present in the standard blank. Within the standard blank there is a peak at 1.700 min which indicates the solvent used in the standard. 2) Why is it important to perform the standard blank? It is important to perform the standard blank because it equilibrates the GC-MS instrument and identifies if there is any contamination within the solvent. It also identifies solvent peaks in the sample.

3) Comment on the quality of your calibration curve, e.g. is it linear, does it pass though the origin, were there any outliers, etc. If the calibration curve is inadequate, provide possible explanations. The quality of the curves displays a linear trend and when using the Grubbs test all values of the area and height were below the g crit value indicating that there is no outliers. The linear regression of the curves is almost 1.

4) Is there a difference in your unknown concentration using the two methods? Which method is superior, peak height or peak area? Comparing the values of the unknown concentration, we can see that there is small difference in the values. This may be from the fact that there is a small difference between the linear regression of the trendline equations of both peak height and peak. However, when asked to compare which is superior the peak area is superior for calculating the concentration as the linear regression is just slightly closer to 1 than peak height. 5) Why did you use SIM mode for quantification of the petrol? The SIM mode was used in the quantification of petrol as it has the highest sensitivity because the quadrupole only monitors the ions of interest and has more time to collect data across the peaks. It was also used to eliminate background ions 6) Do you think that it is acceptable to quantify in scan mode in combination with EIC? No, it is not acceptable to quantify in scan mode in combination with EIC as the scam mode is the least sensitive because the quadrupole is scanning across the whole spectrum only able to pick up the major ions, when combined with the EIC it will then be able to identify unique ions but cannot be used to quantify as it is less sensitive than SIM because it is a post-acquisition process. For better quantification it is best to use SIM mode as it is most sensitive compared to scan mode and EIC....