Model and starting values
Import the three detector-trace CSVs. The earlier component is 1 and the later component is 2. Each Gaussian has a standard deviation σ of 0.35 min. The detector signal includes a constant baseline of 0.040 AU. The response factor k is integrated signal per nanomole of a component. The modeled sample has the same two component amounts in all three runs.
| Case | tR,1 (min) | tR,2 (min) | k1 (AU·min/nmol) | k2 (AU·min/nmol) |
|---|---|---|---|---|
| A · Separated | 3.00 | 5.10 | 1.00 | 1.00 |
| B · Overlap | 3.00 | 3.80 | 1.00 | 1.00 |
| C · Unequal response | 3.00 | 5.10 | 1.00 | 2.00 |
1. Plot the traces and calculate resolution
Make an XY plot for each case with time in min and detector signal in AU. Use common axis limits. Record how many maxima you can distinguish in each combined signal. Calculate the model baseline widths and Rs for each case. Use the supplied component retention times for the calculation.
wb = 4σ for the ideal Gaussian model.
Rs = 2(tR,2 − tR,1) / (wb,1 + wb,2).
Explain how Case B can contain two modeled components while the combined trace has fewer than two distinct maxima. [IUPAC definition of peak resolution]
2. Test the integration boundary
Subtract 0.040 AU from every detector reading. Integrate the corrected signal by summing trapezoids in the two windows, 0.00 min to the split and the split to 8.00 min. The split is a shared endpoint. Use each time interval once.
Area of one interval = (ti+1 − ti) × (yi + yi+1) / 2, where y is the corrected signal.
In a spreadsheet with headers in row 1 and raw data in columns A and B, enter =B2-0.040 in C2 and fill down through C802. Enter =(A3-A2)*(C2+C3)/2 in D3 and fill down through D802. Each D-row area ends at the time in that row. For the left area, sum through the D row at the split. For the right area, start with the next D row. Adjust decimal separators if your spreadsheet locale requires it.
For Case A use a split at 4.05 min. For Case B run the calculation twice, at 3.30 and 3.50 min. Record the left area AL, right area AR, total area and left-area percentage.
| Case | Split (min) | Left (AU·min) | Right (AU·min) | Total (AU·min) | Left area (%) |
|---|---|---|---|---|---|
| A | 4.05 | _____ | _____ | _____ | _____ |
| B | 3.30 | _____ | _____ | _____ | _____ |
| B | 3.50 | _____ | _____ | _____ | _____ |
| C | 4.05 | _____ | _____ | _____ | _____ |
How many percentage points does the left share move between the two Case B splits? Explain why the total area stays the same. What would you call the two reported areas in this overlapping case?
3. Correct for detector response
For Case C, repeat the integration with a split at 4.05 min. Compare its left-area percentage with Case A. Using Case C’s known response factors, estimate the amount in each window and calculate the estimated mole fraction of component 1. Treat the left and right windows as approximate areas of components 1 and 2 in the separated traces.
Estimated amount ni = assigned area / ki.
Estimated mole fraction of component 1 = n1 / (n1 + n2).
Report the uncorrected left-area percentage and the response-corrected mole percentage. What else would a real sample require before you could report an amount or composition? [NIST, Principles of Quantitation]
4. Write a result note
Submit your plots, the integration table and a two- to three-sentence note. State the baseline and split, identify what the Case B result measures, and describe the calibration and separation assumptions behind Case C’s estimated mole percentage.