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Chemical Analysis. Francis Rouessac
Читать онлайн.Название Chemical Analysis
Год выпуска 0
isbn 9781119701347
Автор произведения Francis Rouessac
Жанр Химия
Издательство John Wiley & Sons Limited
5 In a GC analysis series, we seek to determine the influence of the column’s length on several parameters of the chromatogram. All experiments are conducted under the same temperature and carrier gas flow rate conditions.L (m)a = √LtR (min)tR/LRR/a153.72.05307.52.916015.34.15Complete the table.To the nearest measurement uncertainties, what simple relationship can we consider between retention times and column length?To the nearest measurement uncertainties, what simple relationship can we consider between the resolution and the square root of the column length?If we assume that the two peaks used to determine the resolution have practically the same full width at half‐maximum (FWHM), deduce from the previous questions a relationship between the FWHM and the column length as well as a relationship between the theoretical efficiency of the column and its length.The two peaks used to calculate R have retention times respectively of 8.3 min and 9.7 min with the column length equal to 60 m. Calculate the FWHM value of these peaks (same approximation as in question d), as well as the theoretical efficiency of the column for the solute whose retention time is 8.3 min.Calculate the theoretical efficiency of 15 and 30 m columns for the same solute.
6 The chromatographic analysis of an unleaded gas is conducted under the following conditions:Column: L = 150 m, ID = 0.25 mm, df = 0.25 μm, nonpolar stationary phase ‘PDH 150’. Injector, 100:1 split ratio, T = 200°C; FID detector, T = 200°C; oven T = 35°C; carrier gas, He; u = 20 cm/s; column head pressure, 2.5 bar; injected volume = 1 μL.In the following table, the variations in the free enthalpy of dissolution are shown, at the experimental temperature, for the solutes observed on the chromatogram.Solute2,3‐dimethyl pentane2,4‐dimethyl pentane2‐methyl hexane3‐methyl hexanebenzeneΔG0308 (kJ/mol)−17.79−16.80−17.73−18.05v17.27The retention times of the various solutes are given according to peak No.Peak No.1011121314tR (min)48.0055.1763.9064.5968.28Calculate the hold‐up time tM associated with this analysis. Since methane is not retained by the stationary phase, determine its free enthalpy variation at the temperature of the experiment.Assign each peak to its corresponding solute and justify your answer.From the 2,3‐dimethyl pentane peak, whose FWHM is 25 s, determine the theoretical plate number of the column.Calculate the phase ratio of this column.Calculate the distribution coefficient K of benzene in two different ways:By using its variation of free enthalpy of dissolution.From its retention factor.
7 The best‐known method for estimating the hold‐up time, tM, consists of measuring the retention time of a compound not retained on the column. Described here is another method for calculating the hold‐up time, which involves the relationship used in establishing retention factors. Knowing that for a homologous series of organic compounds if the column temperature is constant, we can write: (where tR represents the total retention time of a compound having n atoms of carbon, while a and b are constants that depend upon the type of solute and the stationary phase chosen).Recall the chromatographic parameters for which it is essential to know the hold‐up time tM. Which compound is used in general to determine tM?Calculate tM from the following experiment, employing the method above: a mixture of linear alkanes, possessing six, seven and eight atoms of carbon, is injected into the chromatograph. The retention times for these compounds are respectively, 271 s, 311 s, and 399 s, at a constant temperature of 80°C (column length 25 m, ID = 0.2 mm, df = 0.2 μm and a polysiloxane‐based stationary phase).If the Kovats index for pyridine on squalane is 695, what is the McReynolds constant of this compound on the column in question, if it is known that under the conditions of the experiment, its retention time is 346 s?
8 In a GC experiment, a mixture of n‐alkanes (up to n carbon atoms, where n represents a variable number) and 1‐butanol (CH3CH2CH2CH2OH) is injected onto a column maintained at a constant temperature and whose stationary phase is of the dimethylpolysiloxane type. The equation of the Kovats straight line derived from the chromatogram is: (where is expressed in seconds).The adjusted retention time of butanol is 168 s. If it is known that the retention index for butanol on a squalane column is 590 s, then deduce its corresponding McReynolds constant on this column.
9 We study a chromatogram obtained under the following conditions: Column: DB‐Wax, L= 30 m; ID = 0.321 mm; df = 0.25 μm; oven T = 210°C; mobile phase: H2, u = 50 cm/s; volume injected 0.2 μl; detector: FID, T = 220°C.The chromatogram has three peaks with the following characteristics:peak 1: tR1 = 1.01 min.; δ = 0.151 min.peak 2: tR2 = 1.95 min.; δ = 0.0277 min.peak 3: tR3 = 2.01 min.; δ = 0.0284 min.Calculate the dead time of the column. What can we conclude from this?From this, deduce the mobile phase flow rate in ml/min.Calculate the column phase ratio.Calculate the HETP (H) for peak No.2.Calculate retention factors k for peaks 2 and 3.From this, deduce the distribution coefficients K of the two corresponding compounds.Calculate the selectivity factor α between compounds 2 and 3.Calculate the resolution between these two peaks. Is it enough? What could be a solution to improve the resolution?Kovats retention indexes for octane, toluene, and methyl pentanoate on this same column are respectively 800, 850, and 985. Classify these three solutes by increasing retention times on this stationary phase.What is the polarity of this stationary phase? Justify your answer.
SOLUTIONS
1 We consider that volume V, leaving the column every second, is equal to the internal volume of the column over length ū. The section of the column is πID2/4 and thus V = ūπID2/4. Flow rate D expressed in ml/min (if ū and ID are in cm) is then D = 60V. If we choose to express ID in mm, we write D = 60ūπ(0.1)2 ID 2/4, or D = 0.47ūID 2.
2 The thermodynamic study of a chemical equilibrium leads to the formula: or or If we assume that enthalpy and entropy variations are constant in the temperature range in question and if we switch to log, we can write log k = ‐ A + B/T if we assume the entropy variation < 0 (loss of disorder when the solute is fixed on the stationary phase) and the enthalpy variation < 0 (exothermic reaction).Coelution => k1 = k2, hence log k1 = log k2 and thus T1 = 421 K or t = 148°CFor T < T1 , k2 > k1, therefore 2 elutes after 1.If α = 2, this means that k2 / k1 = 2, or log k2 ‐ log k1 = log 2, and thus T = 343 K.T = 423 K thus k1 = 0.163 and k2 = 0.161; K1 = 40.73 and K2 = 40.16.
3 The cancelation of the first derivative of H = f(u) enables us to find u corresponding to Hmin, thus u = 36.9 mm/s; Hmin = 0.295 mm.Nmax = L/Hmin = 40,674 plates.We are looking for the value of u that gives N= 0.95Nmax or N =38,641 theoretical plates, thus H= 0.310 mm; u = 50.2 mm/s or u = 27.3 mm/s.
4 The order of elution follows that of the increasing boiling points. The alkenes, which are more polar than the corresponding alkanes, are less retained, which is expected from a nonpolar column.α = k2/k: α = 1.12 (at –35°C), α = 1.11 (at 25°C) and α = 1.09 (at 40°C).Since the column is the same, the retention time decreases following the reduction in the partition factor K = CS/CM with temperature.N = 138,493 theoretical plates.Hmin theo = 0.113 mm.
5 L (m)a = √LtR (min)tR/LRR/a153.873.70.252.050.53305.487.50.252.910.53607.7515.30.254.150.53tR is proportional to length.R is proportional to .δ proportional to and N is proportional to L.δ = 0.0568 min, N60 = 148,166 plates.N30 = 74,083 plates and N15 = 37,041 plates.
6 tM = L/u = 750 s = 12.5 min. For methane, k = 0, therefore K = 0 and cannot be calculated.The higher K is, the more solute is retained and the larger the absolute value of ∆ is. Therefore, peak 10 → 2,4 dimethyl pentane, 11 → benzene, 12 → 2‐methylhexane, 13 → 2,3 dimethyl pentane and 14 → 3‐methyl hexane.N = 5.54 (tR/δ)2 = 133,126 plates.d.β = ID/4df = 250. , K = k ∙ β = 853. These are very close values.
7 Neff