A standard solution of containing 6.3 x 10-8 M iodoacetone and 2.0 x 10-7 M p-dichlorobenzene(internal standard) gave peak areas of 395 and 787, respectively in a gas chromatogram. A 3.00-mL unknown solution of iodoacetone was treated with 0.100 mL of 1.6 x 10-5 Mp-dichlorobenzeneand the mixture was diluted to 10.00 mL Gas chromatography gave peak areas of 633 and 520 foriodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00mL of original unknown.

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Answered: A standard solution of containing 6.3 x…

Appl Of Ms Excel In Analytical Chemistry

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Chapter4: Least-squares And Calibration Methods

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A standard solution of containing 6.3 x 10-8 M iodoacetone and 2.0 x 10-7 M p-dichlorobenzene (internal standard) gave peak areas of 395 and 787, respectively in a gas chromatogram. A 3.00-mL unknown solution of iodoacetone was treated with 0.100 mL of 1.6 x 10-5 M p-dichlorobenzene and the mixture was diluted to 10.00 mL Gas chromatography gave peak areas of 633 and 520 for iodoacetone and p-dichlorobenzene, respectively. Find the concentration of iodoacetone in the 3.00 mL of original unknown.
Internal standard. A mixture containing 12.8 mM analyte (X) and 44.4 mM standard (S) gave chromatographic peak areas of 306 for X and 511 for S. A second solution containing an unknown quantity of X plus 55.5 mM S had peak areas of 251 for X and 563 for S. Find [X] in the second solution.
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Next, 2.00 mL of 9.63 mM S was added to 10.00 mL of a solution of unknown concentration of X, and the mixture was diluted to 25.0mL. Chromatographic analysis of this solution produced peak areas of 4822 for X and 4099 for S. What was the concentration of 5 (in mM) of the 25.0mL solution?
2. Thin layer chromatography was performed to find the most effective mobile phase needed to separate a mixture containing three compounds (using silica gel as the stationary phase). TLC analyses were performed using five different mobile phases: (A) pure petroleum ether; (B) 25:75 ethyl acetate:petroleum ether; (c) 50:50 ethyl acetate:petroleum ether (D) 75:25 ethyl acetate:petroleum ether; and (E) pure ethyl acetate. Using the TLC results provided, indicate whether an effective separation was achieved for each of the five mobile phases. Be sure to provide a brief explanation for your answer and identify which mobile phase(s) could be used to offer an effective separation of this mixture using column chromatography. A D E
In a chromatographic analysis, a solution containing 0.0837M toluene and 0.0666M xylene resulted in peak areas of Atoluene= 423 and Axylene= 347. To analyze an unknown sample, 10.0mL of 0.146M xylene was added to 10.0mL of an unknown toluene-containing solution, and the mixture was diluted to 25.0mL. This mixture gave a chromatogram with peak areas of Atoluene= 553 and Axylene= 582. What was the concentration of toluene in the unknown solution?
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1 24) The method of internal standard was used to analyze and measure the concentration of chemical X by GC A known mixture of compounds X and Y (internal standard) gave the following chromatographic results: Compound X Y Concentration (M) 6.3 x 10-8 2.0 x 10-7 a) 2.0 x 10-6 M b) 8.4 x 10-7 M c) 6.6 x 10-7 M d) 4.1 x 10-7 M e) 2.9 x 10-7 M Peak Area 395 787 Then, an unknown was analyzed as follows. 3.0 mL of unknown X was treated with 0.100 mL of 1.6 x 10-5 M Y and the mixture diluted to 10.0 mL final volume. Peak areas were 633 (for X) and 520 (for Y). What is the concentration of X in the unknown. (The original 3.00 sample?)
2. Different marker inks were analyzed using a simple paper chromatographyexperiment. Two marker inks were dissolved separately in a solvent and spottedin the chromatogram. The mobile phase used was 1% NaCl and three inkstandards (A-copper-phthalocyanine blue, B-diarylide yellow,C- lithol rubine red)were spotted beside the two sample markers. After the chromatography, thechromatogram showed a separation of colors in the two inks (D and E).a. Calculate the retention factor for each spot from lanes A to E .b. Which ink standard was the most polar? least polar?c. What is the composition of ink D ?
Internal standard. A solution containing 3.47 mM X (analyte) and 1.72 mM S (standard) gave peak areas of 3 473 and 10 222, respectively, in a chromatographic analysis. Then 1.00 mL of 8.47 mM S was added to 5.00 mL of unknown X, and the mixed solution was diluted to 10.0 mL. This solution gave peak areas of 5 428 and 4 431 for X and S, respectively. (a) Find the response factor for X relative to S in Equation 5-9. (b) Find [S] (mM) in the 10.0 mL of mixed solution. (c) Find [X] (mM) in the 10.0 mL of mixed solution.(d) Find [X] in the original unknown.
A mixture of three components is separated using paper chromatography, and each component is analyzed using absorption. With standards containing 100 ppm of each component run simultaneously, the following data were obtained: cmpd A, 0.72 absorbance; B, 0.55; C, 0.80. For the unknown mixture, the data obtined when run under identical conditions were as follows: cmpd A, 0.15 absorbance, B, 0.84; C, 0.75. a) Calculate the relative percentage composition of the unknown mixture. b) The distances traveled by the three components were 8.5, 6.3, and 5.5 cm, respectively. Sketch the appearance of the paper chromatogram. Label each spot. c) Calculate the Rf values if the solvent front is 18 cm.
In a preliminary experiment, a solution containing 0.0837 mg benzene and 0.0666 mg toluene was analyzed by gas chromatography and gave peak areas of 495 for benzene and 367 for toluene. After this, the amount of unknown was determined by adding 0.0561 mg of benzene to a solution containing an unknown amount of toluene. The peak areas determined for benzene and toluene in this analysis were 643 and 515 respectably. How many mg of toluene are present in the sample?

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