consider a separation performed on a 35 mm long n tubular column with a 0.50 mm
ID: 695752 • Letter: C
Question
consider a separation performed on a 35 mm long n tubular column with a 0.50 mm diameter, and Consider a separation performed on a 35.0 mm a 2.0 pm thick stationary phase. Compound A el min. What are the adjusted retention times, t and retention factors, k for compounds A and B? Number Number min Tools x 102 Number min What is the relative retention, a, for this separation? Number Scroll down to answer the rest of this question The width at half-height, wy2, of the peak for compounds A and B are 0.142 min and 0.285 min, respectively. Find the number of theoretical plates, N, and plate height, H, for each compound. HintExplanation / Answer
Column separation of compound A and compound B
Compound A elutes at 12.70 min and Compound B elutes at 13.46 min, unretained solvent elutes at 1.28 min
adjusted retention times for,
t'r(A) = 12.70 - 1.28 = 11.42 min
t'r(B) = 13.46 - 1.28 = 12.18 min
Retention factor for,
k(A) = 11.42 min/1.28 min = 8.922
k(B) = 12.18 min/1.28 min = 9.516
Relative retention (alpha),
alpha = k(B)/k(A) = 9.516/8.922 = 1.0666
Width at half-height for,
Compound A (w1/2) = 0.142 min
Compound B (w1/2) = 0.285 min
Number of theoretical plates N(A) = 5.54(11.43/0.142)^2 = 35900.585 (or 35900)
Number of theoretical plates N(A) = 5.54(12.18/0.285)^2 = 10119.96 (or 10120)
Plate height (H) = L/N
Length of column (N) = 35 mm
H(A) = 35 mm/35900.585 = 0.000975 mm
H(B) = 35 mm/10119.96 = 0.00346 mm
Number of theoretical plates for the column = sq.rt.(N(A).N(B))
= sq.rt.(35900.585 x 10119.96) = 19060.758
Resolution (Rs) = [sq.rt.(N)/4][(alpha-1)/alpha][k/(1 + k)]
k = k(B)
feeding values from above calculations,
Rs = [sq.rt.(19060.758)/4][(1.0666 - 1)/1.0666][9.516/(1 + 9.516)]
= 1.95
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