Lab 5 Report

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Kennesaw State University *

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2800L

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Chemistry

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Feb 20, 2024

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1 Lab 5: Calibration Methods and Introduction to UV-Visible Spectroscopy Prelab Questions: 1. a. (5) You will create 1.000L of 100.0 ppm stock solution of Acetylsalicylic acid (ASA, from the pure powder) in a 2 (v/v) % HCl in this lab. 2 (v/v) % HCl means the volume of concentrated HCl and the volume of the diluted HCl is 2:100. Determine the mass of the solute and volume of conc. HCl you are going to weigh out or measure with a graduated cylinder. Show your calculation. 1.00 𝐿 × 0.02 = 0.0200 𝐿 𝐻𝐶? → 0.0200 𝐿 𝐻𝐶? × 1000 ?𝐿 = 20 ?𝐿 𝐻𝐶? → 𝑆????𝑒 1 ??? = 1𝑚𝑔 𝐿 → 100 ??? = 100.0 𝑚𝑔 𝐿 × 1.00 𝐿 = 100.0 ?𝑔 ?𝑟 0.1000 𝑔 𝐴𝑆𝐴 b. (5) Describe exactly how you will make the 1.000 L of 100.0 ppm ASA. When this is weighed out, you must get the mass you have found exactly! Rinse the 1.00 L volumetric flask with DI water. Then, add .1000 g, as nearly as possible, of Acetylsalicylic Acid to a weigh boat. Fold the weigh boat into a funnel-shape and use the methanol to wash all the ASA powder into the volumetric flask. Add DI water to the flask until it is half full. Then, add 20 mL of the HCl (aq), and mix it by swirling the flask.to ensure the solutions dissolve completely. Once dissolved and mixed well, add DI water to the flask’s calibration line. One cup, cap the flask and fully invert it 3 times to ensure proper mixing. 2. (5) Create dilution tables to create your standards for the external calibration method given a 100.0 ppm solution and a 10.00 mL volumetric flask and a micropipette with a range from 100.0 to 1000. µL. Use C 1 V 1 =C 2 V 2 . Concentration (ppm) Calculation Volume 0 ppm (100.0 ???)(𝑉) = (0.000 ???)(10.00 ?𝐿) 0.000 mL 4 ppm (100.0 ???)(𝑉) = (4.000 ???)(10.00 ?𝐿) 0.400 mL 6 ppm (100.0 ???)(𝑉) = (6.000 ???)(10.00 ?𝐿) 0.600 mL 8 ppm (100.0 ???)(𝑉) = (8.000 ???)(10.00 ?𝐿) 0.800 mL 10 ppm (100.0 ???)(𝑉) = (10.00 ???)(10.00 ?𝐿) 1.000 mL 15 ppm (100.0 ???)(𝑉) = (15.00 ???)(10.00 ?𝐿) 1.500 mL 20 ppm (100.0 ???)(𝑉) = (20.00 ???)(10.00 ?𝐿) 2.000 mL 100 ppm (100.0 ???)(𝑉) = (100.0 ???)(10.00 ?𝐿) 10.00 mL 3. (5) Create a procedure and dilution for a 325 mg/tablet ASA to be placed in a 1.000 L volumetric flask and then diluted to 10.00 ppm in a 10.00 mL volumetric using a micropipette. 325 ?𝑔 1𝐿 = 325 ??? → (325.0 ???)(𝑉) = (10.00 ???)(10.00?𝐿) → .308 ?𝐿 ?𝑟 308 𝜇𝐿

2 Write the concentration of ASA in the tablet of Aspirin given. Then, put one tablet into a mortar and pestle and begin to crush it into a fine powder. Prepare this in a 1.000 L flask by washing it from the mortar and pestle into the flask using methanol. Fill the flask halfway with DI water, then slowly add 20 mL of HCl and swirl to dissolve. Finish adding DI water until the calibration mark is reached. Put on the cap and invert the flask 3 times to ensure proper mixing. To prepare the dilute 10.00 ppm test solution, pipette .308 mL with a micropipette into a 10 mL volumetric flask. Then, fill the rest of the flask with 2% (v/v) HCl to the calibration line. Cap and invert 3 times to mix the solution, and then pour it into a labeled test tube. Our Data Table: Concentration (ppm) Absorbance 0 ppm -0.0030 4 ppm 0.1470 6 ppm No Data 10 ppm 0.2145 15 ppm 0.3402 20 ppm 0.4443 100 ppm 0.6794 Unknown ASA 2.4217 Unknown ASA + Spike 0.5092 Results and Discussion 1. Grubbs Test: I tested one questionable value each from 6 ppm and 100 ppm. G critical for 19 samples at a 95% confidence level is 2.409. |. 2298 − .1996| . 0502 = .6016 < 2.409; Outlier can be kept with 95% confidence |. 02799 − 2.2347| 0.5168 = 3.783 > 2.409;Outlier can be rejkected with 95% confidence

3 The Raw Data has a different R 2 value, but it is still close to that of the averaged data, with the averaged R 2 at 0.9921 and the raw R 2 at 0.9737. The raw data had a different linear equation (slope and intercept), but it is only evident mathematically and is visually unnoticeable. The average equation was y = 0.0228x + 0.091, and the raw equation was y = 0.0229x + 0.0903. The most notable change is the intercept; the slope was greater in the raw data by .0001. Therefore, the results of averaging the raw data for calibration are not significantly affected by using the raw data. y = 0.0228x + 0.091 R² = 0.9921 -0.5000 0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 80.00 90.00 100.00 Absorbance Concentration of ASA Solution (ppm) External Standard Calibration (0-100 ppm) y = 0.0229x + 0.0903 R² = 0.9737 -0.5000 0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 0.00 10.00 20.00 30.00 40.00 50.00 60.00 70.00 80.00 90.00 100.00 Absorbance Concentration of ASA Solution (ppm) Raw Data for ASA Standard

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4 Concentration LoQ LoD 0 ppm 0.5131 0.1693 4 ppm 17.58 5.801 6 ppm 21.93 7.236 8 ppm 24.35 8.036 10 ppm 26.6 8.778 15 ppm 42.86 14.14 20 ppm 44.95 14.83 100 ppm 93.21 30.76 A more appropriate linear range for this graph would be a 0-20 ppm range. The distance between 20 and 100 is far greater than the other points and the LOD and LOQ for 100 ppm is much higher than those for 0-20. This means that, for 100 ppm, the smallest amounts of analyte to be detected and quantified are far greater than those for 0-20 ppm. The standard deviations for all concentrations were between .001~.1, with the exception being about .2 for 100 ppm, meaning 100 ppm had a greater standard deviation than the other concentrations. The precision decreased as ppm increased, as the standard deviation rose with each ppm. Group # Conc. By External Standard Conc. By Standard Addition 1 13.32 13.14 2 10.79 10.19 3* 11.63 28.22 4 N/A N/A 5 11.1 8.066 6 11.64 9.816 7 12.6 11.13 8 10.82 9.724 9 11.37 9.409 10 14.34 13.21 11 17.46 14.32 12 12.53 11.9 13 N/A N/A 14* 23.7 17.21 15 12.89 10.11 16 N/A N/A 17 N/A N/A 18 18.34 18.81 19 12.87 9.886 20 13.08 11.64 21 12.44 10.19

5 *Red indicated eliminated value through the Grubbs Test. Values listed as N/A were excluded from the sample data size (i.e., the sample size was 17 instead of 21). The average for the external standard is 13.58, and the average for the standard addition is 12.76. The values in groups 3 and 14 were outliers and rejected. The math for this is shown below: |23.7−13.58| 3.481 = 2.959 > 2.475 ; Reject outlier with 95% confidence. |28.22−12.76| 4.889 = 3.162 > 2.475 ; Reject outlier with 95% confidence. Sample calculation for Group 2 of the data set – External Standard: ? = 0.0228? + 0.091 → 0.3370 = 0.0228? + 0.091 → 0.246 0.0228 = 0.0228? 0.0228 → 10.79 ± 3.35 Sample Calculation for my contribution to the data set – Standard Addition: [x] f = [x] i ( v i v i +v s ) = [x] i = 9.692 mL 9.692 mL+ .308 mL = 0.9692[x i ] → y i y f = [x i ] 0.9692[x i ]+10 ppm 0.5092 0.7642 = [x] i 0.9692[x] i +10 ppm = (0.5092) × (0.9692[x] i + 10 ppm) = 0.7642[x] i 0.4935[x] i + 5.092 ppm = 0.7642[x] i → 5.092 ppm = 0.2707[x] i → [x i ] = 18.81 ppm ± 4.89 Analysis of Paired t-Test: The t-Test P-Value for paired data gave a p-value of 6.33 x 10 -5 . This is a very small p-value, and this represents that for most confidence percentages, the method of external standards and the method of standard addition will result in different concentrations. Standard addition directly accounts for the use of a known spike alongside the unknown. The purpose of standard addition is to minimize matrix effects that interfere with analyte measurement signals. This reduces potential errors, which is something that external standards do not account for. The calibration methods, external standards, internal standards, and standard addition, each have times where their usage is more appropriate than the others. The method of internal standards, where a substance that is chemically similar to the analyte is added in a fixed quantity to the blank, is most useful to compensate for losses during sample prep. An example would be serum being stuck in a tube during centrifugation, and thus a portion of analyte being lost. The addition of a primary standard could account for the small quantity lost in the tube. External standards are standard materials, separate from the analyte, that are used as a basis of quantification. This generates a calibration curve to assess how much analyte was lost in the preparation process. This method is ideal for analyzing a series of samples without generating multiple curves, as we did in this lab. 21 groups tested 8 concentrations of a known standard and all that data was pooled to generate a curve that would allow for comparison. Standard addition is most ideal for when a sample being analyzed has multiple components that could be subject to matrix effects, or interference with the analyte absorbance signal. For example, testing the level of sodium in filtered bottled water versus sodium in tap water. The bottled water has been filtered for most extraneous particles, while the tap water has not and may have matrix effects.