Researchers used gas chromatography to determine the % v/v methyl salicylate in rubbingalcohol [Van Atta, R. E.; Van Atta, R. L. J. Chem. Educ. 1980, 57, 230-231]. A set ofstandard additions was prepared by transferring 20.00 mL of rubbing alcohol to separate 25 -mL volumetric flasks and pipeting 0.00 mL, 0.20 mL, and 0.50 mL of methyl salicylate to theflasks. All three flasks were diluted to volume using isopropanol. Analysis of the three samplesgave peak heights for methyl salicylate of 57.00 mm, 88.5 mm, and 132.5 mm, respectively.Determine the % v/v methyl salicylate in the rubbing alcohol.

### Step 1: Understand the Principle of Standard AdditionsThe standard additions method involves…

Answered: Researchers used gas chromatography to…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter31: Introduction To Analytical Separations

Section: Chapter Questions

Problem 31.19QAP

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In a reverse-phase separation, a solute was found to have a retention time of 19.6 min, while an unretained peak had a retention time of 1.45 min when the mobile phase was 20/80 v/v% methanol/water. (b) Calculate the k’ for the solute. (c) Calculate the new k’ if the mobile phase is changed to 60/40 methanol/water assuming identical void volume. (d) Calculate a solvent composition that would bring k’ from initial value to a value of 1.20 using tetrahydrofuran/water mobile phase rather than 20/80 v/v % methanol/water.
(c) What is the minimum distribution ratio that permits removal of 99% of a solute from 50.0 mL of water with (i) two 20.0-mL extractions with an organic solvent; (ii) four 10.0-mL extractions with the organic solvent?
A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, tc, was found to be 209.6 s with a baseline peak width, w., of 14.4 s. The retention time for aspartame, ta, was 260.0 s with a baseline peak width, wa, of 21.2 s. The retention time for the unretained solvent methanol was 49.2 s. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 22.1 cm long column. H = um %3D Calculate the resolution, R, for this separation using the widths of the peaks. R = Calculate the resolution if the number of theoretical plates were to increase by a factor of 2. R2N =
A solute with a partition coeffcient (K-[S)s/S]h) of 5 is extracted from 20 mL of phase 1 into phase 2. What volume of phase 2 is needed to extract 90% of the solute in one extraction? ( (a) 0.44 mL (b) 12 ml (c) 36 ml. (d) 72 ml
4.Consider the peaks for pentafluorobenzene and benzene in the gas chromatogram shown here. The elution time for unretained solute is 1.06 min. The open tubular column is 30.0 m in length and 0.530 mm in diameter, with a layer of stationary phase 3.0 μm thick on the inner wall. a) Measruing the width, w, at the baseline on the chromatogram, find the number of plates for these two compounds b) Use your answer to (a) to find the resolution between the two peaks c) Using the number of plates N=sqrtN1*N2 with the values from (a) calcuate what the resolution should be and compare your answer with the measured resolution in b
Caffeine Benzoic acid y = 66.391x R2E0.9995 y= 23.633x R? =0:9999 0.5 Injection Volume Injection Volume Aspartame Saccrine y = 6.2204x.. R2 =0.9997 y = 21.43x R² = 0.9891 2.5 0.5 2.5 Injection Volume Injection Volume (Abs) Peak height (Abs) Peak height 2. (Abs) Peak height (Abs) Peak height
A student researcher performed a chromatographic separation of caffeine and aspartame. The retention time for caffeine, te, was found to be 200.8 s with a baseline peak width, we, of 16.1 s. The retention time for aspartame, ta, was 258.7 s with a baseline peak width, wa, of 20.6 s. The retention time for the unretained solvent methanol was 44.2 s. Calculate the average plate height, H, in micrometers for this separation, given that it was performed on a 22.1 cm long column. H = Calculate the resolution, R, for this separation using the widths of the peaks. R = 88.2 R₂5N = 3.16 Calculate the resolution if the number of theoretical plates were to increase by a factor of 2.5. 3.533 Incorrect μm
Chromatography involves the separation of two or more compounds/ions caused by their molecular interactions with two phases. True False
Tooth enamel consists mainly of the mineral calcium hydroxyapatite, Can(PO)(OH). Trace elements in teeth of archeological specimens provide anthropologists with clues about diet and disease of ancient people. Students at Hamline University measured the trace element strontium in cnamel from extract wisdom teeth by atomic absorption spectroscopy. Solutions were prepared with a constant total volume of 10.0mL. containing 0.750mg dissolved tooth enamel plus variable concentration of added Sr. \table[[Added Sr (ng/mL- ppb), Signal (arbitrary units)], [0, 28.0], [2.50, 34.3], [5.00, 42.8], [7.50, 51.5], [10.00, 58.6]] a) Prepare a graph to find the concentration of Sr in the 10mL sample solution in part per billion = n m L. (2%) b) Find the concentration of Sr in tooth enamel in part per million = μ (3%) 5. Tooth enamel consists mainly of the mineral calcium hydroxyapatite, Ca(PO)(OH). Trace elements in teeth of archeological specimens provide anthropologists with clues about diet and…
20 aspirin tablets labeled 80mg were dissolved in 100mL of 90% ethanol. A 10mL aliquot was taken and was used for assay. The analyte followed usual process and was treated with 50mL of 0.1000N NaH and was titrated with 35mL 0.1050N H2O4 until the solution achieved completion. Calculate the % content of the total aspirin capsules and the actual label claim
3. This is the standard procedure for determination of nicotine in urine. A 1.00mL sample was placed in a 12mL vial 5-aminoquinoline was added as the internal standard (5 ug). The sample was monitored using mass spectrometry at m/z values for nicotine (84) and the internal standard (144). Given the following data determine the concentration of the two unknown samples Nicotine (ug/L) Area Area 12 0.056 0.059 51 0.402 0.391 102 0.684 0.669 157 1.011 205 Unknown 1 Unknown 2 1.278 0.51 1.18 1.063 1.355 0.53 1.32
1

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