Question

This question applies to the separation of a solution containing pentanol (MM 88.15 g/mol) and 2,3-dimethyl-2-butanol (MM 102.17 g/mol). Pentanol is the internal standard. Separation of a standard solution containing 225 mg of pentanol and 233 mg of 2,3-dimethyl-2-butanol in 10.0 mL of solution led to a pentanol:2,3-dimethyl-2-butanol relative peak area ratio of 0.913:1.00. Calculate the response factor, F, for 2,3-dimethyl-2-butanol. F= Calculate the areas for pentanol and 2,3-dimethyl-2-butanol gas chromatogram peaks in an unknown solution. For pentanol, the peak height is 37.9 mm and the width at half-height is 1.44 mm. For 2,3-dimethyl-2-butanol, the peak height is 54.9 mm and the width at half-height is 3.07 mm. Assume each peak to be a Gaussian. pentanol: mm 2,3-dimethyl-2-butanol: mm? If the concentration of pentanol in the unknown solution of is 24.4 mM, what is the 2,3-dimethyl-2-butanol concentration? concentration: mM

Answer 1

O. Answer Can use the Pollowwing formula Por internal Staordaods: Avea 8kd Bignal Conc of signales Avea of analyte signal (Aa) Concentration of analyle@) wheve F= Reonse factove = 233gx103 102.17 9/mol XO:0looL [x] molot XV 0.233 1.0217 [a] = 0:2280 mol/L 225x103 88 153/mol* [S] = 1Mal yollotxv ,Xo.0100L = 0225 0.8815 [s] = 0. 2552 mol/L Scanned with CS Cảm Sicanner

.. F= Ag [s] [] Ag I [0•2552] [0-2280] [0913] O: 2552 O. 2081 1.2263 * Area of Gaussian peak = l·064 Kpeak headghtX Loidth at halb-height z1.064x37.9mmx1:4kmme = 58.0688 mm Pentanol 2,3-dimethyl-2-butanol = 1.064 x54.9mm x 3:0tun 179.3297mm? %3D cs Scanned with. CamScanner

(As F(t5) we know [x] F= 1:2263 As= 179.3297mm? Ag = 58 0688mm? %3D 24.4m12 .'. [a]= Ae [$] F xAg 179.3297 ma?244mi1) 1-2263 (58.0688 mxê) 4375.6446 71.2097 61.4473 m11 [*] Scanned with CS CamScanner
Thank you.