Map Sapling Learning macmilan learning t researcher performed a chromatographic

Question

Map Sapling Learning macmilan learning t researcher performed a chromatographic separation of caffeine and aspartame. The retention A studen time for caffeine, le. was found to be 200.9 s with a baseline peak width, We, of 14.1 s and the retention t for aspartame, a, was 259.6 s with a baseline peak width, wa. of 19.1 s. The retention time for the unretained solvent methanol was 41.7s Calculate the average plate height in micrometers for this separation, given that it was performed on a 20.7- long column. Number Calculate the resolution, R, for this separation using the widths of the peaks. Number R= Calculate the resolution if the number of theoretical plates were to increase by a factor of 2 Number O Previous GHiisat&3/ten sei ttt! O Check Answer 0 Next Exit Hint

Explanation / Answer

(a) Average plate height is the average of the plate height of aspartane and caffeine

plate height (H) can be calculated usong formula H = L/ N where L is length of column (given 20.7 cm) and N is the number of theoretical plates

N can be calculated as 16 X (tR/W)2

N of aspartame=16(259.6/19.1)2=16X184.6881=2955

So, H of aspartame = 20.7 cm/2955=0.007005 cm = 70 micro meter

now, N of caffeine can be calculated as 16 X (200.9/14.1)2 =16 X 203.06=3249

so, H of caffeine = 20.7 cm/3249=0.006371 cm = 63.71 micro meter

now, average plate height =(plate height of aspartame + plate height of caffeine)/2

=(70 + 63,71)=133.71/2=66.855 mcro meter

(b) Resolution R can be calculated as R=(tR2 - tR1)/1/2 (w1+w2)

tR2 is the retention time of aspartame = 259.6 s

tR1 is the retention time of caffeine = 200.9 s

w1 is the baseline peak width of aspartame =19.1 s

w1 is the baseline peak width of caffeine =14.1 s

R=(259.6-200.9)/1/2 (19.1+14.1)

R=58.7/16.6

R=3.536

(c) As resolution R is inversly proportional to baseline peak width (w) and number of plates N is also inversly proportional to w. If N increases by a factor of 2, w will reduce to half and if w is reduced to half, resolution R will become double.. So increase in number of theoretic plates N will also increase the resolution by a factor of 2

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