Question

A peak with a retention time of 407 s has a width at half-height (w1/2) of 7.6 s. A neighboring peak is eluted 17 s later with a w1/2 of 9.4 s. A compound that is known not to be retained was eluted in 2.5 s. The peaks are not baseline resolved. How many theoretical plates would be needed to achieve a resolution of 1.5?

Answer 1

The number of theoretical plates needed to achieve a resolution of 1.5 can be calculated using the retention time and width at half-height of the peaks.

Step-by-step explanation:

The resolution (R) of two peaks in chromatography is defined as the distance between the centers of the peaks (Δt) divided by the average width of the peaks (w1/2). In this case, the resolution is given as 1.5. The resolution equation is: R = (Δt) / (2 * √(ln2) * w1/2). Rearranging the equation to solve for the number of theoretical plates (N): N = 16 * (R^2) * (w1/2^2).

Given the retention time of the first peak (407 s) and the width at half-height (7.6 s), and using the formula, we can calculate the number of theoretical plates needed to achieve a resolution of 1.5.

Answer 2

2.46 x 104

Step-by-step explanation:

Solution

Recall that:

The retention time of a peak = 407 s

with a width at half-height of = 7.6 s

A compound is retained in 2.5 s.

resolution to be achieved = 1.5

Thus,

The number of plates (theoretical)= 16(tr2 / w2)

The R Resolution R= 0.589 Δtr / w1/2av = 0.589(17s) / 1/2(7.6s + 9.4s) = 1.18

Supposed that applied column contains 10,000 theoretical plates and the resolution of two peaks is 1.18

So if the column is replaced to obtain 1.5 resolution, the number of theoretical plates is needed is stated below;

width at the base = 9.4 - 7.6 = 1.8; tr = 0.786

N = 5.55tr2 / w21/2 = 5.55 (0.7862/ 1.182) x 104

= 2.46 x 104

Therefore, required theoretical plates to achieve a resolution of 1.5 is 2.46 x 104