Читать книгу Chemical Analysis - Francis Rouessac - Страница 62

1 We have tR = tM(1+k). And tM = L/ū and k = KVS/VM, hence tR = (L/ū)(1+KVS/VM).

2 Knowing that VR = tR∙D, we get: α = (VR(2) – VM)/(VR(1) ‐ VM), i.e. α = 1.2.

3 To transform Eq. (P1.1) into Eq. (P1.2), we add k, from tM = tR/(1+k).

4 When peaks are neighbours, base widths are generally comparable. By posing ω1 = ω2 and by expressing ω2 with its value as a function of N2, we obtain R = ¼ √N2 (tR(2) – tR(1))/tR(2). Then, we add k, by using tR = tM(1 + k); finally, we introduce α = k2/k1;The relationship given in the problem is obtained by rearranging the relationship for R (as a function of k and α) and substituting the relationship for Neff from Eq. (P1.2) of problem 1.3.

5 From the base relationship R = 2(tR(2) – tR(1))/(ω1 + ω2), we can replace ω with its value depending on N, ω = 4tR /√N and since tR = tM(1 + k), we obtain Eq. (P1.5);We multiply the second member of (1), numerator and denominator by (k1 + k2) and we obtain α (α = k2/k1).

6 ω0.1 = a + b = 2A ∙ a; A = (a + b)/2a => b = a(2A − 1) => a/b = 1/(2A − 1) => where N = 53,000 plates, tR = 1.58 min and A = 1.07, and thus we get:a = 0.0142 min.b = 0.0162 min.ω0.1 = 0.0304 min.k = (tR − tM)/ tMtherefore, k4 = 4.43 and k5 = 6.62; α4‐5 = k5 /k4 = 1.47; (this assumes that the peak is Gaussian)β = 136.6;K = k β; K4 = 605 and K5 = 884; K5 > K4 means that solute 5 has more affinity for the stationary phase than solute 4; therefore, it is more retained and exits later.

7 %(ME) = 16.6; %(EE) = 16.6; %(PE) = 33.4; %(BE) = 33.4.

8 By adding N‐methylserotonin before extraction, we do not have to take into consideration any potential loss of product due to the various manipulations. We suppose that the extraction yield is the same for these two compounds, which are very similar;kS/NMS = 1.002;serotonin mass: 45 ng/ml.