Question

Chromatography Calculations Short Answer: Answer the following short answer questions about Chromatography Calculations (SHOW WORKING):

A) Solute A is eluted through a 17 m chromatographic column with a retention time of 12.40 min and peak base width of 0.98 min, while methane elutes at 1.15 min on the same column.

Calculate the number of theoretical plates.

B) A solution containing an unknown compound was spiked with n-octane and n-decane and run on a GC using an isothermal oven temperature of 74 C. The retention times were found to be: n-octane 8.77 min, n-decane 13.50 min and unknown 9.96 min. The retention time of an unretained peak (methane) was 1.12 min.

Calculate the Kovat’s retention index for the unknown. (10)

C) A standard solution containing 3.21 x 10-8 M iodoacetone and 4.24 x 10-7 M p-diphenylbenzene (internal standard) gave peak areas of 272 and 582, respectively, in a gas chromatogram.

A 2.81 mL aliquot of sample, containing an unknown concentration of iodoacetone, was spiked with 0.259 mL of 2.67 x 10-5 M p-diphenylbenzene. The mixture was diluted to 100 mL. Gas chromatography gave peak areas of 725 and 384 for iodoacetone and p-diphenylbenzene, respectively.

Calculate the concentration of the original unknown, in nM.

Note: This experimental approach assumes the calibration graphs for both compounds are linear and pass through the origin. It is more usual nowadays to add the same concentration of internal standard to standard and sample solutions and then plot the area ratio of versus analyte concentration.

Answer 1

N, the number of theoretical plates, is one index used to determine the performance and effectiveness of columns, and is calculated using equation (

where tr = retention time, and W = peak width

no. of theoritical plates = 16 (12.40/0.98)^2 = 160.09 = 160

B) The Kovats index is given by the equation   

where I is the Kovats retention index, n is the number of carbon atoms in the smaller n-alkane, N is the number of carbon atoms in the larger n-alkane, and t_r^/ is the adjusted retention time.

I= 100 * [ 8 + (10 - 8)( log (9.96) - log( 8.77)) / (log(13.50) - log(8.77))] = 861.11