Procedure A. Sample Preparation 1. Clean two 2-mL vials with ethanol and allow t

Question

Procedure A. Sample Preparation

1. Clean two 2-mL vials with ethanol and allow to dry.

2. Pipet exactly 1.0 mL of the alkane standard solution into a clean vial, add exactly 0.5 mL of the decane internal standard solution, cap and mix well.

3. Pipet exactly 1.0 mL of the your unknown solution into a clean vial, add exactly 0.5 mL of the decane internal standard solution, cap and mix well.

B. Quantitative Analysis of an Unknown Mixture

1. Your instructor will demonstrate how to inject samples into the gas chromatograph and how to collect and print data using the computer interface.

2. Follow the instructions posted by the gas chromatograph to select the sequence 321 and to analyze your samples . The GC method in the 321 sequence uses temperature programming to achieve an efficient separation of the analytes in a reasonable time. Your instructor will describe the temperature profile used. The total run time is 4.5 minutes, and the carrier gas flow is 1.0 mL/min. Prepare the GC for injection and inject approximately 1-L of the spiked standard solution. Record the name of the file where the data are saved. Print out a copy of the chromatogram and a report listing retention times and peak areas for all of the components. Identify each peak in your chromatogram.

3. Make at least two additional injections of the spiked standard solution and print out the chromatogram and report for each run. Assess the reproducibility of your measurements by calculating the ratio of the octane peak area to the decane peak area for each injection. This ratio should vary by only 2-3%.

4. Make several injections of your spiked unknown solution using the same approach used for the standard in B.2 above. Obtain printouts of the chromatogram and report for each run.

IV Calculations

A. Column Efficiency (N) Using data from the standards run in III. B, calculate the number of theoretical plates (N) for octane using: N=16(tR/w)^2. where W is the magnitude of the base of the triangle defining the octane peak. Make sure that the units of tR and W are the same.Chemistry 321L Manual Page 37

B. Unknown Mixture Multiple injections of a sample will usually result in different peak areas for a given component because it is difficult to reproduce the injection volume. These variations are compensated for by adjusting (normalizing) the component peak area by dividing it by the peak area of the internal standard (decane). In this way the concentration of each component in your unknown can be determined by a simple comparison of normalized peak areas: where (Component AREA/Decane AREA)unk is the normalized peak area for a particular component in the unknown solution, (Component AREA/Decane AREA)std is the average normalized peak area for a particular component in the standard solution, and Cunk and Cstd are the unknown and standard concentrations of that component, respectively. Report the mean concentration and its RMD for each component of the unknown. The precision of your results should be 2-3%.

Explanation / Answer

A)

Here, the number of theoretical plates (N) for octane using: N=16(tR/w)^2

where tR = retention time and W is the magnitude of the base of the triangle defining the octane

So, respectively:

For octane piking the data from serial no 2.

2.) tR = 2.445 min = 2.445 * 60 s = 146.7 s
     , Area = (1/2) * H *W
   W = (2 * area) / H
   = 2 * (Area/ Height ) s
   = 2 * 1.6770 s
   = 3.354 s
  

Therefore,  number of theoretical plates (N) for octane using:
     N=16(tR/w)^2
= 16*( 146.7 / 3.354) ^2
   = 30609 ( approx)

B).Since,

[(Component AREA/Decane AREA)unk / (Component AREA/Decane AREA)std] = Cunk/Cstd

where (Component AREA/Decane AREA)unk is the normalized peak area for a particular component in the unknown solution, (Component AREA/Decane AREA)std is the average normalized peak area for a particular component in the standard solution, and Cunk and Cstd are the unknown and standard concentrations of that component, respectively.

Here, Standard solution is decane

Then, then.
   4210975.59 / 676082.01 = Cunk/Cstd
As you have not mentioned the dtrenght of std solution. Let take the std ,soluion as 1 M.

Then, Cunk =   (4210975.59 / 676082.01 ) * 1
= 6.228 M

You can further refer to the below refrence for solving the the Cunk if you have any doubt, it will really help you, as you have not provided the full information about the question.

Refrence : http://www.chemistry.adelaide.edu.au/external/soc-rel/content/int-std.htm

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