Chemical Analysis. Francis Rouessac
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Kovats index of butanol is thus 645. Its McReynolds constant is: 645 – 590 = 55.
9 tM = L/u = 3,000/50 = 60 s, i.e. 1 minute. Peak No.1 probably corresponds to the solvent peak.D = VM/tM = π ∙ L ∙ (ID)2/4 tM = 40.46 mm3∙s−1, i.e. 2.4 ml/min.β = (ID)/4df = 321/(4 x 0.25) = 321.HETP for peak No.2: H = L/N with N = 5.54tR 2 /δ 2 = 27,555 plates => H = 1.09 mm.k = (tR ‐ tM)/tM k2 = 0.93 and k3 = 0.99.K = k ∙ β, thus K2 = 0.93 × 321 = 298.5; K3 = 0.99 × 321 = 317.8.α2‐3 = k3 /k2 = 1.06.R2‐3 = 1.18(tR3 – tR2)/(δ2 + δ3) = 1.2. The resolution between two peaks for a good separation must be at least 1.5. The two peaks are therefore not well separated. To improve this separation, retention times must be increased by reducing either the carrier gas flow rate or the oven temperature.The higher the Kovats index is, the longer the retention time. Therefore, tR values are in the following order: octane < toluene < methyl pentanoate.These retention times are in the order of growing polarity of the analytes; therefore, this is a polar column.
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