1. (14 marks) Substances A and B were analysed chromatographically on two column

Question

1. (14 marks) Substances A and B were analysed chromatographically on two columns, X and Y, with the same stationary phase type and the same eluent. The outcome of the chromatography is shown in Table 1. Table 1. Chromatography of substances A and B on columns X and Y Column Column length (mm) 100 0.2 5.1 5.6 30 0.6 0.59 0.64 0.01050 0.01140 Flow rate (mL min) Retention time of A (min) Retention time of B (min) Half Height peak width of A (min) Half Height peak width of B (min) 0.1045 0.1145 (i) (4 marks) Calculate the resolution of both columns X and Y (ii) (4 marks) Calculate the number of theoretical plates in both columns X and Y (ii) (2 marks) Calculate the H for both columns X and Y (iv) (2 marks) State which column is most efficient and explain your choice. (v) (2 marks) With reference to the flow rate and column length explain how this efficiency has been achieved and why the resolution of the two columns is the same (2) (6 marks) Substances A and B give chromatographic peaks separated by 15 seconds and the retention time of A is 9.0 mins. Half height peak widths of A and B are approximately equal. Hovw many theoretical plates are required to give a minimum acceptable resolution of 1.50?

Explanation / Answer

(i) R= 2*(rtb-rta)/ (wb+ wa)

where, rta=retention time of A and rtb=ret. time of B

wband wa are peak widths of B and A resspectively.

for x

RX= 2*(5.6-5.1)(0.1145+0.1045)

= 4.56

similarly RY= 2*(0.64-0.59)/(0.01140+0.01050)

= 4.56

(ii) N= 5.55*(rt/w)2

for column X

N(for A)= 5.55*(5.1/0.1045)2= 13,218

N(for B)= 5.55*(5.6/0.1145)2= 13,275

for column Y

N(for A)= 5.55*(0.59/0.01050)2 = 17,523

N(for B)= 5.55*(0.64/0.01140)2= 17,492

(iii) lets just conisder substance A for simple comparision

H=L/N ; L= length of column, N= no. of theortical plates

HX = 100/13,218

=0.00756

HY= 30/17,523

=0.00171

(iv) resolution is the same for both columns. now, number of theoritical plates is more in column Y and less in X. Therotical plates means the hypothetical phases of the substance(liquid and gas) are in equlibirum and form plates(hypothetical). themore these plates the more distinguishable. so, a column with higher number of theortical plates is considered better. Also, in column Y peak width is less that is the peaks are narrower and sharp which is also good. Hence, Column Y is most efficient in this case

(v) for best efficiency the flow rates should be elevated. efficiency is directly proportional to column length. hence, a longer column length makes a more efficient column. In column X the flow rate is less but the column length is more while in Y if the flow rate is higher but the column length is short. although in Y the number of theoritical plates ismore and peak width is also less which gives it this better efficiency.

Resolution is same in both the columns becasue the ratios of retention time of A and B in column X and Y is 10. ratio of peak width for both A and B in both the columns is also 10. so the ratios are equal which gives the same resolution at the end of the day.

(vi) R= 2*(rtb-rta)/ (wb+ wa)

1.50= 2*15/2w

w= 10sec= 0.167 min

N= 5.55*(rt/w)2

= 5.55*(9/0.167)2

= 16,118 plates are required to give 1.50 resolution

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