Question

Gas Chromatography 1. A sample of a mixture of ethyl acetate (Rt = 1.67 min) and toluene (Rt = 2.65 min) was analyzed by gas chromatography and the following results obtained: Peak # Time [min] Area (uV sec) Area [%] 8.703 1.32 3708 1.67 16449 38.61 52.69 2.65 22449 a. Provide one explanation why three signals were obtained b. What is the normalized percentages of ethyl acetate and toluene.

Answer 1

1:32 Q: Gas Chromatography. 1) A Sample of a nuexture of ethyl acetate fR+= 1-67 min) and Toluene (R+=2.65 min) was analysed by... gas Chromatography and the following results obtained, peak. Time Acea (uv/sec) AreaC%) 3708 8-703 1.2 1.67 16449 38-61 1 3 2.65 22449 52-69 . a) provide one explanation why three Signals were obtained Ans Anexta There are threefee Three packs in two peaks are corresponding to analyte. ethyl acetate peak is around 1.67 min and Toluene peak is around 2-65 min .. the Question is arised to that third peak es for what which came around 2.65 mu 1-32 min An extra peak (1-32 min peak) is not related to the sample itself. Generally extent peak arises form one of these Sources D Jate elution from earlier speetea chromato gram 3) Garry ouer from prevoies evjection 3) Contamination of mobile phase (Ghost peal) rate elution of a peak from previous chromatog bem into clonot allow enough time to following samo all the peaks to be eluted injection.

that cremaining peaks which left in Coducerea Ut will come in next chromatogesim Extra peak Carryover ColztyAlls from a prelomerizetten you tou 322 peab exen you make a blank injection with no arrialute presentat Ghost peaks to be Wrigue to qadient elution Ghost peaks appear when a contaminant » present in the mobile phase (usually very les) is concentrated on the colum and then is released as the gradient runs Ghost peak will opear in blank injection and even in without injectura ob) What is normalized percentage of ethyl . acetate to and Toluene. Ans. Normalized percentage of ethyl acetate. is = 22-289% Normalized percentage of Toluene is 57.71 consider both ethyl acetate & Toluene peak Atea pecent as 100% then calculate indoiidutally for each . 38-61+52-69591-3 So for ethyl acedate so for Toluene 52-69 %100 -=57-777, 91-3 - 91-3 38.61 x 100 = 42.289%