Math Problem Statement

An unknown compounds was separated into its 4 components using liquid chromatography. The experiment was run using a C-18 column with a length of 25 cm and a diameter of 0.46 cm. The solvent eluted at 16.5 seconds and the experiment was run using a flow rate of 1.2 mL/min. The table above shows the resulting chromatogram parameters

Compound 1 had a peak height of 0.072, an elution time of 120 seconds, and a peak width at base 28 s. Compound 2 had a peak height of 0.046, an elution time of 225 seconds, and a peak width at base 70 s. Compound 3 had a peak height of 0.061, an elution time of 310 seconds, and a peak width at base 55 s. Compound 4 had a peak height of 0.041, an elution time of 608 seconds, and a peak width at base 98 s.

Solution

We can analyze the chromatography experiment based on the provided data. Here's the step-by-step breakdown:

1. Dead Time (tmt_m)

The dead time, tmt_m, is the time the solvent takes to elute, which is given as 16.5 seconds.


2. Capacity Factor (kk')

The capacity factor for each compound is calculated using: k=trtmtmk' = \frac{t_r - t_m}{t_m} where trt_r is the retention time (elution time).

  • For Compound 1: k1=12016.516.56.27k'_1 = \frac{120 - 16.5}{16.5} \approx 6.27

  • For Compound 2: k2=22516.516.512.64k'_2 = \frac{225 - 16.5}{16.5} \approx 12.64

  • For Compound 3: k3=31016.516.517.76k'_3 = \frac{310 - 16.5}{16.5} \approx 17.76

  • For Compound 4: k4=60816.516.536.85k'_4 = \frac{608 - 16.5}{16.5} \approx 36.85


3. Column Efficiency (Number of Theoretical Plates, NN)

The number of theoretical plates is calculated using: N=16(trwb)2N = 16 \left(\frac{t_r}{w_b}\right)^2 where wbw_b is the peak width at the base.

  • For Compound 1: N1=16(12028)22943N_1 = 16 \left(\frac{120}{28}\right)^2 \approx 2943

  • For Compound 2: N2=16(22570)21652N_2 = 16 \left(\frac{225}{70}\right)^2 \approx 1652

  • For Compound 3: N3=16(31055)25072N_3 = 16 \left(\frac{310}{55}\right)^2 \approx 5072

  • For Compound 4: N4=16(60898)26107N_4 = 16 \left(\frac{608}{98}\right)^2 \approx 6107


4. Selectivity (α\alpha)

The selectivity is calculated for consecutive peaks as: α=k2k1,k3k2,k4k3\alpha = \frac{k'_2}{k'_1}, \frac{k'_3}{k'_2}, \frac{k'_4}{k'_3}

  • Between Compound 1 and 2: α1,2=12.646.272.01\alpha_{1,2} = \frac{12.64}{6.27} \approx 2.01

  • Between Compound 2 and 3: α2,3=17.7612.641.41\alpha_{2,3} = \frac{17.76}{12.64} \approx 1.41

  • Between Compound 3 and 4: α3,4=36.8517.762.07\alpha_{3,4} = \frac{36.85}{17.76} \approx 2.07


5. Resolution (RsR_s)

The resolution between peaks is calculated as: Rs=ΔtrwbR_s = \frac{\Delta t_r}{w_b} where Δtr\Delta t_r is the difference in retention times between adjacent compounds, and wbw_b is the average of the two base widths.

  • Between Compound 1 and 2: Rs,1,2=22512028+702=105492.14R_{s,1,2} = \frac{225 - 120}{\frac{28 + 70}{2}} = \frac{105}{49} \approx 2.14

  • Between Compound 2 and 3: Rs,2,3=31022570+552=8562.51.36R_{s,2,3} = \frac{310 - 225}{\frac{70 + 55}{2}} = \frac{85}{62.5} \approx 1.36

  • Between Compound 3 and 4: Rs,3,4=60831055+982=29876.53.89R_{s,3,4} = \frac{608 - 310}{\frac{55 + 98}{2}} = \frac{298}{76.5} \approx 3.89


6. Peak Asymmetry (Optional)

If peak asymmetry data is needed, it would require additional information about the peak tailing factor.


Summary of Key Results:

  • kk' values: 6.27,12.64,17.76,36.856.27, 12.64, 17.76, 36.85
  • NN values: 2943,1652,5072,61072943, 1652, 5072, 6107
  • α\alpha values: 2.01,1.41,2.072.01, 1.41, 2.07
  • RsR_s values: 2.14,1.36,3.892.14, 1.36, 3.89

Would you like detailed steps for any calculation, or additional analysis such as peak areas or interpretation of chromatographic performance?


5 Questions for Deeper Exploration:

  1. How does the selectivity α\alpha influence the separation efficiency of the column?
  2. Why might Compound 2 have the lowest number of theoretical plates?
  3. Can we infer anything about the polarity of the compounds from their retention times?
  4. How would increasing the flow rate impact the resolution of the peaks?
  5. What other factors can affect the column efficiency in this experiment?

Tip:

Always ensure the column's performance metrics (like NN, RsR_s) meet the method's validation criteria to maintain accurate and reproducible separations.

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Math Problem Analysis

Mathematical Concepts

Chromatography
Retention Time
Capacity Factor
Column Efficiency
Selectivity
Resolution
Peak Asymmetry

Formulas

Capacity Factor (k') = (t_r - t_m) / t_m
Column Efficiency (N) = 16 * (t_r / w_b)^2
Selectivity (α) = k'_2 / k'_1
Resolution (R_s) = Δt_r / w_b

Theorems

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Suitable Grade Level

Undergraduate