MT1-Sp10-v4 - Name________________________...

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Unformatted text preview: Name________________________ GSI________________ Chemistry 4B, Spring 2010 Name: February 12, 2010 (50 min., closed book) SID: GSI: Section: Write your name and that of your GSI on all 7 pages of the exam. Equations, constants, and data are provided below. In order to maximize your score on the exam: Do the questions you know how to do first. Circle your final answer. Go back and spend more time on the questions you find more challenging. Budget your time carefully ‐ don't spend too much time on one problem. Show all work for which you want credit and don't forget to include units. For the explanation sections, try keep your answers brief as well as complete. Topic Question Page Points Score ICP‐AES of Cigarettes Q1‐4 2 20 ICP‐AES of Cigarettes Q5‐7 3 20 HPLC of Vanillin Q8 4 5 HPLC of Vanillin Q9‐12 5 25 EDTA Q13‐18 7 30 Useful Equations and Constants: % % pH = ‐ log[H3O+] % ∆ % pX = ‐ log X Constants: E = h N0 = 6.02214 x 1023 mol‐1 = c kb = 1.380 × 10‐23 J K‐1 ∗ ∗ ∆ h = 6.62607 × 10‐34 J s Analytemobile ⇌ Anlaytestationary K = Cs/Cm 16 me = 9.101939 × 10‐31 kg c = 2.99792 × 108 m s‐1 1 nm = 10‐9 m = 10‐6 Page 1 of 7 Name________________________ GSI________________ ICP‐AES of Cigarette Ash Calibration Curves for Metals in Cigarette Ash Intensity of Emission 6.00E+04 y = 91270x + 4608.7 R² = 0.9995 5.00E+04 4.00E+04 y = 68035x + 4304.3 R² = 0.9982 3.00E+04 Cu 2.00E+04 Zn 1.00E+04 0.00E+00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 ppm (mg/L) ICP‐AES Instrument Setup Copper (Cu) Zinc (Zn) Detected wavelength (nm) 324.754 213.856 Cigarette ash was tested for the presence of heavy metals using ICP‐AES. A cigarette was lit and allowed to smolder for several minutes. A mass of 0.1054 g of ash was collected and dissolved in 2 mL of concentrated nitric acid, HNO3. The solution was filtered and then transferred to a 25.00 mL volumetric flask and diluted to the mark with distilled water. A series of metal standards were prepared and the response to ICP‐AES was collected. Data for the standards is shown above. 1) The cigarette unknown had an emission intensity of 1.631 × 105 for copper. What is the concentration of copper in the final solution of the cigarette ash? 2) What is the amount of copper (g) per gram of ash? 3) The calibration curve was made from a purchased standard solution with a Cu2+(aq) concentration of 100.0 ppm ± 1%. What is the concentration of a solution 0.500 mL (±0.006mL) of standard diluted to a final volume of 100.00mL (±0.08mL)? Include associated uncertainty. 4) Do you trust the value you calculated for the concentration of copper in the ash solution? Why or why not? Page 2 of 7 Name________________________ GSI________________ 5) ICP‐AES detects emission of light from atoms and ions heated to extremely high temperatures in a plasma. Shown below are two electronic energy levels for copper for the detected wavelength. Sketch the energy levels for zinc for the detected wavelength. (No rigorous calculations are needed.) E* E Energy levels for Cu Energy levels for Zn 6) This experiment does not have equal sensitivity for the metals. a) What is the relative sensitivity of the Cu and Zn? b) Explain why the sensitivity for Zn is different than for Cu over the same concentration range. (Assume they have the same degeneracy ratio.) 7) At high concentrations the detected emission from Cu actually decreases, data shown below. Explain why this happens. Intensity of Emission Observed AES Signal for Cu at 324.754 nm 2.50E+05 2.00E+05 1.50E+05 1.00E+05 5.00E+04 0.00E+00 0 20 40 60 80 ppm (mg/L) Page 3 of 7 Name__________ ______________ GSI________________ HPLC Analysis of Vanilla The main flavoring in vanilla is a compound called vanillin. Su rprise! Ortho‐vanillin or o‐vanillin is a structural isomer of vanillin. Structural isomers have the sam e chemical fo ormula but the atoms are connected differently. The structures of both compounds are shown below. HO O HO O O O vanillin o-vanillin Vanilla tastes and smells delicious so it is often used as a flavo ring. In contrast o‐vanillin does not have a pleasant aroma and is harmful if ingested. A student in Chem 4B wanted to determine if the inexpensive vanilla purchased in Mexico contained both com pounds so she performed an HPLC analysis of the Mexican vanilla. Vanilla and o‐vanillin are both sensitive to UV light. The UV s pectra of the compounds are shown below. 8) Which wavelength would you chose for detection as the compounds elute from the HPLC column? Why would you choose that wavelength? Page 4 of 7 Name________________________ GSI________________ In order to optimize the HPLC separation of the compounds, known standards were used. The stationary phase was C18 modified silica. This means that the silica particles have long hydrocarbon chains attached to them. The mobile phase is a mixture of methanol and water. H3C OH Mobile Phase: methanol SiO2 Stationary Phase: C18‐silica Two chromatograms with different run times are shown on the next page for different mixtures of the mobile phase. The peak at 4 minutes is due to an unretained solute. Vanillin exits the column first, then o‐vanillin. 9) Calculate the resolution for both chromatograms. a) Resolution for 90% methanol: 10% water b) Resolution for 30% methanol: 70% water 10) Which separation is more successful? 11) Based on the data for the vanillin, which experiment represents a larger number of theoretical plates for the column? 12) The first sample to exit the HPLC column is vanillin. Why does changing the ratio of methanol to water change the separation? Page 5 of 7 Name________________________ GSI________________ 90% Methanol and 10% water UV Absorbance at fixed wavelength 12.00 10.00 8.00 6.00 4.00 2.00 0.00 3.00 4.00 5.00 6.00 7.00 8.00 9.00 10.00 11.00 12.00 time (minutes) 30% Methanol and 70% water UV Absorbance at fixed wavelength 12.00 10.00 8.00 6.00 4.00 2.00 0.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00 time (minutes) Page 6 of 7 Name________________________ GSI________________ H O EDTA O H H H H C H C O H H pKa1 = 0.0 O C C N C H pKa2 = 1.5 H C N C C O pKa3 = 2.0 H H H HH H C pKa4 = 2.69 O pKa5 = 6.13 H C O pKa6 = 10.37 O H In lab you used EDTA to complex to Co(II) ions in solution. The EDTA solution was in an ammonium acetate (NH4+CH3COO‐) solution at a pH of 7.5. 13) In the structure above, draw an X over all the hydrogens that would be lost at a pH of 7.5. In lab you used the method of back titration to determine the concentration of Co(II) in your unknown after the cobalt was separated from copper in an anion‐exchange column. An excess of EDTA was added to the samples, then the titration was completed with the addition of 9.180 mg/mL Co(II) standard. Data for the reference and unknown titrations are shown below. Sample Volume (mL) of standard Co2+(aq) to reach endpoint Color at endpoint Reference Trial #1 Trial #2 Trial #3 Trial #4 11.74 15.36 15.37 15.31 12.80 blue blue blue blue blue 14) Because it took more titrant to reach the endpoint for the unknown, the unknown has more Co than the reference sample. 15) Trial #4 is so different from the other samples, the data should be discarded. No other reason is required. 16) There is good precision between Trials 1‐3. TRUE FALSE TRUE FALSE TRUE FALSE 17) All the samples contained the same initial moles of EDTA. TRUE FALSE Determine if the statements below are true or false. Circle your answer. Hannah, a Chem 4B student, thinks that for Trial #4 the sample preparation was the problem. She thinks that the copper and the cobalt did not separate well using chromatography. She thinks that the presence of copper and cobalt could have led to the lower volume of titrant required. 18) Do you agree with Hannah’s assessment? Why or Why not? Page 7 of 7 ...
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This note was uploaded on 03/16/2012 for the course MATH 1B taught by Professor Reshetiken during the Spring '08 term at University of California, Berkeley.

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