Your colleague has collected data from injecting a mixture of compounds into a c

Question

Your colleague has collected data from injecting a mixture of compounds into a chromatography column and has asked for your help with the analysis. The column length is 25.7 cm. A chromatogram from a mixture of A, B, C, and D produced the following data: Please calculate the number of theoretical plates for each peak. Use the average number of theoretical plates to calculate the plate height, H, for the column. Please calculate the retention factor for each peak. Calculate the resolution between species B and C. Calculate the selectivity factor for species B and C. Calculate the length of column necessary to give a resolution of 1.5.

Explanation / Answer

N = 16(tr/W)2

Where tr is the retention time and W the width of the peak base

Retention time(min) =tr

Width of the peak base=W

N = 16(tr/W)2

Non- retained

3.1

-

-

A

5.4

0.41

N = 16(5.4/0.41)2     =2775.60

B

13.3

1.07

N = 16(13.3/1.07)2 =2472.08

C

14.1

1.16

N = 16(14.1/1.16)2 = 2364.29

The performance and the effectiveness of columns are indicated by N

      b) The Average number Nav and the plate height H are related as,

Nav= L/H, where L is the length of the column packing (given=25.7cm)

Find out the Nav

Nav =[(2775.60+2472.08+2364.29)/3]=2537.32

Now we thatNav x H=L substituting the values we have

=2537.32 x25.7= 65209.124

N and H are the quantitative measure ofchromatographic efficiency

C) Retention factor for each peak

The retention factor k’n is the extent of retention of a compound under a given set of conditions. The retention factor for each peak is given by the formula

K’n=Vn-V0/V0 or k’n= tr – t0/t0

Where Vo is the elusion volume and to is the corresponding time of the component proceeding through the column without any interaction( Here it is given as= 3.1min). we have to use the time factor as the volume is not given. Substituting the values we have

k’n= tr – t0/t0

Retention time(min) =tr

k’n= tr – t0/t0

Non- retained

3.1(t0

-------

A

5.4

5.4-3.1/3.1=0.742

B

13.3

13.3-3.1/3.1 = 3.290

C

14.1

14.1-3.1/3.1 = 3.548

d)Resolution (Rs): (between B and C- asked)

Retention time(min) =tr

Width of the peak base=W

Non- retained

3.1

-

A

5.4

0.41

B

13.3

1.07

C

14.1

1.16

Retention time(min) =tr

Width of the peak base=W

N = 16(tr/W)2

Non- retained

3.1

-

-

A

5.4

0.41

N = 16(5.4/0.41)2     =2775.60

B

13.3

1.07

N = 16(13.3/1.07)2 =2472.08

C

14.1

1.16

N = 16(14.1/1.16)2 = 2364.29

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