Lab Report 9- Qualitative Identification of Hydrocarbons in Gasoline by Gas Chromatography PDF

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Lab Report 9- Qualitative Identification of Hydrocarbons in Gasoline by Gas Chromatography...

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Qualitative Identification of Hydrocarbons in Gasoline by Gas Chromatography

Hannah Strickland

March 30, 2018

Chemistry 355 Section JW

Experiment 9

Introduction: In this experiment, gas chromatography was used to analyze a gasoline sample composed of the hydrocarbons: hexane, heptane, octane, and nonane. Gas chromatography is an accurate analytical technique for analyzing and identifying volatile mixtures, such as gasoline composed of different hydrocarbons.1 Gas chromatography works by the interactions of the volatile samples with the machine. A sample is injected into the machine and vaporized, and it is carried throughout the capillary tube by a stream of gas, this is known as the mobile phase. This mobile phase is usually composed of helium or nitrogen. The capillary tube that the sample moves through contains particles, this is known as the stationary phase. Different molecules move through the column at different times, this is called retention time. At the end of the run, a chromatogram is obtained that contains the retention times and peaks for each chemical in the sample.1 The retention time depends on several things such as the chemical structure of the sample, the column packing for the column, the temperature, and the rate flow of the mobile phase. The retention time in the chromatogram allows for the identification of the chemical. It is important that each individual peak is separated from each other. This can be calculated using the formula for retention time, which is shown as equation 1. In order for the peaks to be accurate and fully separated, R should equal 1.5 or greater.1 Equation 1: R = tR2 — tR1 / 0.5 (W1 + W2)

(1)

Experimental: Data was given for this lab. Gas chromatograms were given for the following hydrocarbons: hexane, heptane, octane, and nonane. A gas chromatogram for a mixture of these hydrocarbons was also given. With the data of each hydrocarbon, a calibration curve was constructed using Excel. This was done by plotting the number of hydrocarbons against the retention time. Results: In this experiment, gas chromatography was used to analyze different hydrocarbons. The hydrocarbons analyzed were hexane, heptane, octane, and nonane. Also, an unknown mixture was given. Figure 1 shows the gas chromatogram for hexane. Figure 2 shows the gas chromatogram for heptane. Figure 3 shows the gas chromatogram for octane. Figure 4 shows the gas chromatogram for nonane. Figure 5 shows the gas chromatogram for the unknown mixture. Finally, table 1 shows the different hydrocarbons with the number of carbons they contain and their retention times.

Figure 1: Figure 1 shows the gas chromatogram for hexane, with a retention time of about 3.83 minutes.

Figure 2: Figure 2 shows the gas chromatogram for heptane, with a retention time of about 4.74 minutes.

Figure 3: Figure 3 shows the gas chromatogram for octane, with a retention time of about 6.23 minutes.

Figure 4: Figure 3 shows the gas chromatogram for nonane, with a retention time of about 7.68 minutes.

Figure 5: Figure 5 shows the gas chromatogram for the unknown mixture of hydrocarbons. Table 1: Data for Given GC Spectra Molecule

Retention Time (min)

Number of Carbons

Hexane

3.83

6

Heptane

4.74

7

Octane

6.23

8

Nonane

7.68

9

Discussion: Throughout the course of this experiment, gas chromatography was used to analyze hydrocarbon samples and a mixture of hydrocarbons in a gasoline sample. The molecules are identified by their retention times, which can be related to the number of carbons they contain. This was done by creating a calibration curve by plotting the retention time versus the carbon number. This is shown as graph 1. The linear equation obtained from this graph is: y = 0.104 x + 0.475. The R2 value is 0.9996. Because this value is close to 1.0, it indicates that there is little variance in the data obtained.2 The unknown gasoline sample was analyzed by looking at the major peaks and the corresponding retention times. Figure 6 shows the unknown gas mixture labeled. Table 2 contains the labels and their analysis that correspond to figure 6. Because nonane has the highest retention time, it was estimated that peak number 4 was nonane. Because octane had the second highest retention time, it was determined that peak number 3 was octane. Because heptane had the second smallest retention time, it

was determined that peak number 2 was heptane. Because hexane had the smallest retention time, it was determined that peak number 1 was hexane.

Number of Carbons v. Retention Time 0.9 0.8

y = 0.104x + 0.475 R² = 0.9996

Log (time)

0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 6

7

8

9

Number of Carbons

Graph 1: Graph 1 shows the calibration curve of the various hydrocarbons by comparing the number of carbons to the log of the retention time.

Figure 6: Figure 6 shows the gas chromatogram for the unknown gas sample with labels.

Table 2: Analyzation of Unknown Gasoline Sample Label

Retention Time (min)

Estimated Molecule

1

1.8 Hexane

2

2.7 Heptane

3

3.7 Octane

4

4.0 Nonane

Conclusion: In this lab, the components of an unknown gasoline sample were identified by first running a gas chromatogram for each of the four hydrocarbons: hexane, heptane, octane, and nonane. Then, creating a calibration curve and comparing the retention time of these hydrocarbons to the retention time and peaks in the unknown gas chromatogram. Overall, gas chromatography is a useful analytical technique for identifying volatile compounds. Gas chromatography has several advantages such as it can separate a complex mixture, it gives an accurate determination, and it has a multiple detector with high selectivity. Although it has these advantages it also has disadvantages such as it is limited to thermally stable volatile components, and the sample is usually destroyed.1

Citations: 1. Vyazovkin, Sergey. Department of Chemistry. CHEM 355: Quantitative Analysis Student Laboratory Manual. 2. Frost, J. Regression Analysis: How Do I Interpret R-squared and Assess the Goodness-of-Fit? http://blog.minitab.com/blog/adventures-in-statistics-2/regressionanalysis-how-do-i-interpret-r-squared-and-assess-the-goodness-of-fit (accessed Mar 30, 2018). !

! Questions: 1. A homologous series of primary alcohols is to be determined by GC. Given that the retention time for ethanol is 1 min, predict the retention times for 1-propanol, 1butanol, and 1-pentanol. The retention times for 1-propanol, 1-butanol, and 1-pentanol can be calculated by subtracting the void time from each of the peaks that correspond to the alcohol. The elution order depends on the boiling points of the alcohols which are in the order of: ethanol < 1-propanol < 1-butanol < 1-pentanol. So, the retention times will also follow this order. So, 1-propanol will have a retention time of ~2 minutes, 1-butanol ~4.5 minutes, and 1-pentanol ~8.3 minutes.

2. Which type of column would you recommend for the separation of the alcohols in Question 1? The column should be selected by looking at the samples and they should somewhat match the column stationary phase. For alcohols, the usual recommended column should have a mid-polar phase. ! 3. List several factors that affect resolution and describe how they can be controlled. Resolution of a spectrum is how well the peaks are separated from each other. ` The farther apart peaks are, the easier they are to distinguish from each other and thus they have better resolution. A longer and narrower column allows for better resolution. Also, as the flow rate of the carrier gas increases, the temperature will send the vapors through the column faster, which lowers the retention time of the spectrum, causing the resolution to be poor. So, when the temperature is lowered, the flow rate increases, the retention times increase, and the peaks are broadened. ! 4. What length of column would be required to just resolve two peaks with retention times of 92 and 107 s, respectively, using a column with an H of 1.60 cm/plate? (This can be answered from Dr. Vyazovkin’s notes.) Relative retention (𝛾) = 107 / 92 = 1.16 Height of plate: 1.60 cm Solving for number of plates with a resolution of 1.5: N = ((4)(1.5) / 1.16 - 1)2 N = 1406.25 Column length = (number of plates) x (plate height) Column length = (1406.25) x (1.6 cm) = 2250 cm...