Gas Chromatography lab report PDF

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written Gas Chromatography lab report for organic chem 2...

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Introduction: The term chromatography applies to the separation of chemical constituents in a sample so they can be either detected or utilized individually. Gas chromatography (GC) is a method of separating “volatile” compounds (those with a high‐vapor pressure or a relatively low boiling point) so that they may be detected individually in complex mixtures. Compounds are separated based on differences in their vapor pressures and their attraction to solid materials inside the instrument (a gas chromatograph or GC). Because the vapor pressure of a given compound is a function of the intermolecular forces between molecules, GC takes advantage of differences in at least one of the properties of matter discussed in lectures and in the text.

Important Definitions:  TCD is known as thermal conductivity detector. It is chemical specific detector commonly used in gas chromatography. It senses changes in the thermal conductivity of the column effluent and compares it to a reference flow of carrier gas. 

MS is known as mass spectrometer and it produces charged particles from the chemical substances that are to be analyzed. It uses electric and magnetic fields to measure the mass of the charged particles.



FID is known as flame ionization detector and it is a scientific instrument that measures analytes in a gas stream.

Procedure for running samples: 1. Inject the standard hydrocarbon mixture in three of the machines listed. 2. Inject your unknown in the instrument of your choice. 3. Make sure you include an analysis table for each of the sample runs. 4. Prepare a table for each machine (3 for STD and one for Unknown) 5. Calculate the percent difference between each percent are (difference*100) 6. Write out sample calculation for each column in the table.

n-pentane

n-hexane

2,2,4- trimethylpentane

2,5- dimethylhexane

36

69

98

107

Results: 580FID (peak #) 1 2 3 4

Retention Time (min) 0.22 0.32 0.56 0.73

Reported Area 90333 283093 583745 546643

Unknown #2 (peak #) 1 2 3 4 5 6

Retention Time (min) 0.16 0.21 0.32 0.47 0.55 0.71

Reported Area 39448 316549 326040 44259 553875 521707

GOW-MAC

Retention Time (min) 0.201 0.296 0.352 0.449 0.650

Reported Area 4096 51159 15921 192429 30787

1 2 3 4 5

% Area

Area by Triangulation

% Area

Area by Triangulation

% Area

Area by Triangulation

% Area

6.007 18.825 38.818 36.350

% Area 2.189 17.568 18.089 2.456 30.739 28.954

% Area .372 4.65 1.45 17.50 2.80

6 7

0.785 1.010

454966 350032

41.38 31.84

Agilent 7820 (peak #) 1 2 3 4 5 6 7

Retention Time (min) 1.771 1.880 1.947 2.082 2.168 2.219 2.304

Reported Area 302.42 1190.11 3.82 2883.74 1.89 2305.35 1.99

% Area

Area by Triangulation

% Area

4.52 17.79 0.06 43.11 0.03 34.46 0.03

Sample Calculations:

Differences in the lab vs the book: 1. We did not do GC‐MS in the lab last week. 2. We did not get to do the one technique because the machine was not cooperating so it was given to us instead. 3. I don’t believe we calculated relative area in the lab last week.

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 a. Two compounds 1‐bromopropane and 1‐chloropropane are halides of propane. The compound bromine has a higher atomic weight than chlorine. This means 1‐ bromopropane will have a higher molecular weight and therefore the boiling point will be higher. Since the greater the molecular weight the more time the compound spends in the column. 1‐chloropropane will have a 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. b. If the sample was run several days later with the conditions nearly the same as possible, we would expect the retention time to be identical to those obtained the first time. 2. Using triangulation, calculate the percentage of each component in a mixture composed of two substances, A and B. The chromatogram is shown in figure 22.17. a. Area Peak A =h X w(1/2) = 55mm X 7mm = 385mm2 Area Peak B = h X w(1/2) = 36 mm X 6mm = 216mm2 Total Area is A + B which is 601 mm2 %A = 385/601 (100) = 64.1% %B = 216/601 (100) = 35.9% 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.

4. What would happen to the retention time of a compound if the following changes were made?

a. Decrease the flow rate of the carrier gas i. The rate of flow of the gas has a huge impact on the rate of movement of a compound through a gas chromatograph as the components of a mixture gets distributed between a mobile gas phase and a static liquid phase. If the rate of flow of the carrier gas is decreased then the bands become too broad and it results the poor resolution. If means that elution is not done fast hence the retention time was increased. b. Increase the temperature of the column i. If the temperature was raised the retention time decreases. This is because the absorbent material on the column is attracting particles easier and the elution through mobile phase is faster. c. Increase the length of the column i. If two compounds have some similarities then we need to use a longer column to separate them. An increase in the length of the column promotes better resolution of the components as they travel through a longer path in a larger column. Therefore, the retention time is increased....