Lab 4_ Gas Chromatography (GC) Analysis of Hydrocarbons (1)

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Temple University *

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2203

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Chemistry

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Jan 9, 2024

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Lab 4: Gas Chromatography (GC) Analysis of Hydrocarbons Goal: The focus of this experiment was to understand the gas chromatography analytical method while using the gas chromatography method to analyze and identify components present in a mixture. Background: Gas chromatography is an instrumental method of separation in which the components are separated and analyzed. This method is commonly used to test the purity of a substance, or even identify a compound. This instrument is very useful in providing information about unknown substances. The substance that is inserted into the chromatograph is separated between two phases, the moving gas phase and the stationary liquid phase. Reaction Equation: Chemicals and Equipment Used: Gas Chromatograph tool Known mixture Unknown mixture 84 GC Procedure:
A known mixture was analyzed in the GC1 instrument to obtain the chromatographs About 1mL of the mixture was injected into the instrument using the provided syringe Then, the same step was repeated using the GC2 instrument The unknown mixture was injected into the GC1 instrument using the same steps to obtain chromatographs Then, the unknown mixture was injected into the GC2 instrument The obtained data was used to calculate the % composition of each component Calculations: Results:
Retention Time Name of Comp. Area %Composition GC1 inst KNOWN MIX - 3.0 min - 6.17 min - 15.3 min - 6.7 min N-pentene N-hexane 2,2,4 - trimethylpentae 2,5- dimethylhexane - 7.4cm2 - 16.3cm2 - 15.3cm2 - 4.07 cm2 - 17.14 % - 37.7% - 35.4% - 9.4% GC1 inst. UNKNOWN MIX - 3.1 min - 8.2 min - 12 min N-pentene 2,2,4 - trimethylpentane 2,5 - dimethylhexane - 9.75cm2 - 19.8cm2 - 13.5c m2 - 22.6% - 45.9% - 31.4% GC2 inst. KNOWN MIX - 1.94 min - 2.15 min - 2.29 min N-pentene 2,2,4 - trimethylpentene 2,5-dimethylhexane - 3972.88pA - 7397.07pA - 3733.36pA - 26.3% - 48.9% - 24.7% GC2 inst. UNKNOWN MIX - 1.83 min - 1.94 min - 2.29 min N-pentene N-hexane 2,5- dimethylhexane - 8694.88pA - 5921.16pA - 4793.83pA - 44.8% - 30.5% - 24.7% Conclusion: Based on the procedure, the chromatography was performed twice for the known and unknown mixture, once in GC1 instrument and once in GC2. The unknown mixture used was labeled 84 GC. Based on the calculations and the number of peaks, the given compounds were identified above. As presented, the unknown mixture in GC1 had n-hexane absent, and the unknown in GC2 run had a 2,2,4-trimethylpentane peak missing. The percentage in the unknown and known mixture in GC2 for the 2,5-dimethylhexane came out same, indicating that the compound must be present in the unknown mixture. Changes: No changes were noted from the lab book Discussion: As seen in the results above, there could have been multiple miscalculations and measurements in the steps that gave an odd percentage composition. As the procedure was taking place, there were some technical difficulties taking place which may have altered the results of my experiment. The GC2 instrument had to be recalibrated and run a few times without any mixture as it was presenting too many peaks. This indicates that some left over mixture might
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have been remaining in the machine. Other than that, I would repeat this experiment again taking these precautions into consideration and injecting less than 1mL of the compound to reach more clear peaks. Post Lab Questions: 1. A.) A sample consisting of 1-bromopropane and 1-chloropropane is injected into a gas chromatograph equipped with a nonpolar column. Which compound has the shorter retention time? Explain your answer. - The compound with the shorter retention time is the 1-chloropropane. In comparison to the 1-bromopropane, which is more electronegative, the 1-chloropropane has less polar compounds thus making it have shorter retention time. B.) If the same sample were run several days later with the conditions as nearly the same as possible, would you expect the retention times to be identical to those obtained the first time? Explain - The retention times would not have been the same because over time there could be many environmental changes that can affect the study. Changes may be subject to temperatures and pressure as well as possible evaporation of the compound. 2. Using triangulation, calculate the percentage of each component in a mixture composed of two substances, A and B. - % of A = (0.037/0.057) x 100 = 64.9% - % of B = (0.020/0.057) x 100 = 35.1% - Ration = (64.9/35.1) = 1.84 3. Make a photocopy of the chromatogram in Figure 22.17. Cut out the peaks and weigh them on an analytical balance. Use the weights to calculate the percentage of each component in the mixture. Compare your answer to what you calculated in problem 2. - Compound 1 =0.059 g = 59 mg - Compound 2 = 0.038 g = 38 mg - 59+38= 97 total weight
- % weight of 1 = (59/97) x 100 = 60.8% - % weight of 2 = (38/97) x 100 = 39.1 % - (60.8/39.1) = 1.55 - Based on the calculations, the percentage and ratio are close to each other, however, it is not the same. This could have been due to errors in the process of weighing. 4. What would happen to the retention time of a compound if the following changes were made? a. If the flow is decreased - the retention time will increase b. If the temperature of the column is increased - the retention time will decrease c. If the length of the column is increased - retention time will increase