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(Solved): Lab Question 4. GLC Quantitative Analysis, Peak Table. Build a peak table to organize the mixed stan ...
Lab Question
4. GLC Quantitative Analysis, Peak Table. Build a peak table to organize the mixed standard chromatogram data. Such a table records the masses of the mixed standard components and resulting peak areas from the mixed standard chromatogram (see the example on page 15). Appropriately title the table and label all columns. Include the molecular compound names. Be certain correct units are included (the peak area can be left unitless). Sample Information \begin{tabular}{|r|r|r|r|} \hline \multicolumn{1}{|c|}{ Peake } & Ret Time & \multicolumn{1}{|c|}{ Areal } & \multicolumn{1}{|c|}{ Arcallo } \\ \hline 1 & \( 0.416 \) & 3991 & \( 0.3751 \) \\ \hline 2 & \( 0.505 \) & 1237 & \( 0.1162 \) \\ \hline 3 & \( 0.593 \) & 1056233 & \( 99.2858 \) \\ \hline 4 & \( 1.125 \) & 2370 & \( 0.2228 \) \\ \hline Total & & 1063831 & 1000000 \\ \hline \end{tabular} Signal Minutes \begin{tabular}{|r|r|r|r|} \hline Peakf & Ret Time & \multicolumn{1}{|c|}{ Area } & \multicolumn{1}{|c|}{ Area\% } \\ \hline 1 & \( 0.423 \) & 3042 & \( 0.3414 \) \\ \hline 2 & \( 0.621 \) & 2186 & \( 0.2453 \) \\ \hline 3 & \( 0.690 \) & 884611 & \( 99.2840 \) \\ \hline 4 & \( 1.135 \) & 1152 & \( 0.1293 \) \\ \hline Total & & 890991 & \( 100.0000 \) \\ \hline \end{tabular} Sampie Information \( \begin{array}{ll}\text { Analysis Date \& Time } & : 3 / 18 / 2020 \text { 10:44:46 AM } \\ \text { Sample Name } & \mathrm{SG} \\ \text { Sample ID } & : \mathrm{p} \text {-Tolualdehyde } \\ \text { Data Name } & \mathrm{C}: \text { GCsolutioniDatalData Files Dry Lab DatalGClion } \\ \text { Method Name } & \text { C:GCsolutionWatalMethod FilesiGC-1 GC-1 CHEM 143.gem } \\ \text { Report Name } & \text { C:lGCsolutionWatalReport FilesiCHEM 143143 Report 070218.gcr }\end{array} \) Sample Information \begin{tabular}{|r|r|r|r|} \hline Peak# & Ret Time & \multicolumn{1}{|c|}{ Area } & \multicolumn{1}{|c|}{ Area } \\ \hline 1 & \( 0.413 \) & 2337 & \( 0.2568 \) \\ \hline 2 & \( 0.500 \) & 1199 & \( 0.1317 \) \\ \hline 3 & \( 0.697 \) & 288743 & \( 31.7268 \) \\ \hline 4 & \( 0.965 \) & 3081 & \( 0.3385 \) \\ \hline 5 & \( 1.151 \) & 189756 & \( 20.8502 \) \\ \hline 6 & \( 1.379 \) & 417201 & \( 45.8416 \) \\ \hline 7 & \( 2.374 \) & 7775 & \( 0.8543 \) \\ \hline Total & & 910092 & \( 100.0000 \) \\ \hline \end{tabular} Signal Sample Information Signal Minutes
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solution :- On Area normalisation basis means .it will be calculated in % Area of Impurity/Unreacted/reactant divided by Area of main peak/Analyte pe
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