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 has a peak height of 0.072, elution time of 120 s, and a peak width at base 28 s Compound 2 has a peak height of 0.046, elution time of 255 s, and a peak width at base 70 s Compound 3 has a peak height of 0.061, elution time of 310 s, and a peak width at base 55 s Compound 4 has a peak height of 0.041, elution time of 608 s, and a peak width at base 98 s

The efficiency (number of plates) of the column for component 1 = blank3 - Numeric Answer

You are incorrect

The plate height for component 1 = blank5 - Numeric Answer

You are incorrect mm The efficiency (number of plates) of the column for component 2 = blank2 - Numeric Answer

You are incorrect

The plate height for component 2 = blank6 - Numeric Answer

You are incorrect mm The pate height for component 4 =

Solution

To calculate the efficiency (number of plates) and plate height for the given components, we use the following formulas:

1. Number of Plates (NN):

N=16×(tRWb)2N = 16 \times \left(\frac{t_R}{W_b}\right)^2 where:

  • tRt_R = retention time of the compound (elution time)
  • WbW_b = peak width at base

2. Plate Height (HH):

H=LNH = \frac{L}{N} where:

  • LL = length of the column (in mm, so convert 25 cm to 250 mm)
  • NN = number of plates

Data for the column:

  • Length of column, L=250mmL = 250 \, \text{mm}
  • Flow rate = 1.2 mL/min (not directly needed for these calculations)

Component 1:

  • tR=120st_R = 120 \, \text{s}
  • Wb=28sW_b = 28 \, \text{s}
  1. Calculate NN: N=16×(12028)2N = 16 \times \left(\frac{120}{28}\right)^2 N=16×(4.2857)2=16×18.3673=293.88N = 16 \times \left(4.2857\right)^2 = 16 \times 18.3673 = 293.88

  2. Calculate HH: H=250293.88=0.851mmH = \frac{250}{293.88} = 0.851 \, \text{mm}


Component 2:

  • tR=255st_R = 255 \, \text{s}
  • Wb=70sW_b = 70 \, \text{s}
  1. Calculate NN: N=16×(25570)2N = 16 \times \left(\frac{255}{70}\right)^2 N=16×(3.6429)2=16×13.2657=212.25N = 16 \times \left(3.6429\right)^2 = 16 \times 13.2657 = 212.25

  2. Calculate HH: H=250212.25=1.178mmH = \frac{250}{212.25} = 1.178 \, \text{mm}


Component 4:

  • tR=608st_R = 608 \, \text{s}
  • Wb=98sW_b = 98 \, \text{s}
  1. Calculate NN: N=16×(60898)2N = 16 \times \left(\frac{608}{98}\right)^2 N=16×(6.2041)2=16×38.4773=615.64N = 16 \times \left(6.2041\right)^2 = 16 \times 38.4773 = 615.64

  2. Calculate HH: H=250615.64=0.406mmH = \frac{250}{615.64} = 0.406 \, \text{mm}


Final Results:

  1. Component 1:

    • N=293.88N = 293.88
    • H=0.851mmH = 0.851 \, \text{mm}
  2. Component 2:

    • N=212.25N = 212.25
    • H=1.178mmH = 1.178 \, \text{mm}
  3. Component 4:

    • H=0.406mmH = 0.406 \, \text{mm}

Would you like me to calculate similar metrics for the other components or explain the steps further? 😊


Expansion Questions:

  1. How do variations in column length affect the plate height?
  2. Why is the plate height important for chromatography performance?
  3. How does flow rate impact retention times and peak widths?
  4. What role does the stationary phase (C-18) play in separation?
  5. Can this calculation method be applied to gas chromatography?

Tip:

Peak symmetry can also provide insights into column performance; consider calculating the asymmetry factor if peaks are not Gaussian.

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Math Problem Analysis

Mathematical Concepts

Chromatographic Efficiency
Retention Time
Peak Width
Number of Plates
Plate Height

Formulas

N = 16 × (t_R / W_b)^2
H = L / N

Theorems

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Suitable Grade Level

Undergraduate Chemistry or Analytical Science Students