tartrazine calculations including slope

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Chemistry

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Feb 20, 2024

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Experiment ii: Spectrophotometry: Preparing Standard Solutions and Plotting a Standard Curve (1) Abstract (2) Introduction The objective of the experiment is to determine the concentration of an “Unknown Tartrazine” solution. (3) Theory

When light is shown on a solution, a proportion of it may be absorbed and the remainder transmitted through the solution. A spectrophotometer measures absorbance of a solution by passing “light of a single wavelength” through the solution (1) . The absorbance of a solution is related to its concentration by the Beer-Lambert Law, where absorbance is equal to the product of absorption coefficient/absorptivity(ε), cell length (l), and concentration [C]. Cell length is typically 1 cm. (4) Experimental Materials 0.0983 g Tartrazine, FD&C Yellow #5 (Na 3 C 16 H 9 N 4 O 9 S 2 , MW=534.36 g/mol) (1). Distilled water Small beakers 1, 2, 3, 5, 10, 15-ml Volumetric pipettes 100-ml and 200 ml volumetric flasks Eyedropper Metal spatula KIMTECH wipes Cuvette UV-VIS spectrophotometer Spectronic 200 Analytical Balance Procedure 0.0983 g of tartrazine was transferred to a clean and dry 25 ml beaker with a metal spatula. The mass of the tartrazine was obtained using an analytical balance. A 100 ml solution was made using the tartrazine and this was called stock solution I. A 5ml aliquot of stock solution I was diluted to 100 ml with distilled water to form stock solution II. The concentrations of the stock solutions were calculated in PPMm. The stock solutions were used to make 5 standard solutions with mass concentrations in the range 4.915-19.66 ppm. Refer to Table 1 for the dilution scheme. Distilled water was used as the blanking matrix for the UV-VIS spectrophotometer Spectronic 200. The absorbances of the 5 standard solutions were found at the maximum wavelength of 430 nm. 4 readings of the absorbance of the unknown tartrazine solution were taken at the maximum wavelength.

(5) Results Data/Tables Table 1. Dilution Scheme for Standard Solutions from Stock Solutions Ci, ppm Vi, ml Cf, ppm Vf, ml 49.15 10 4.915 100 49.15 15 7.3725 100 983 1 9.83 100 983 3 14.745 200 983 2 19.66 100 The wavelength at the maximum absorbance was found to be λmax=430 nm. 4 6 8 10 12 14 16 18 20 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 f(x) = 0.04 x − 0.02 R² = 1 Absorbance vs. Concentration of Standard Solutions Concentration in PPMm Absorbance Figure 1. Absorbance vs. Concentration curve of Standard Tartrazine Solutions at λmax=430 nm. The slope of the graph represents absorptivity, ε=0.0418 cm -1 ppm -1 Table 2. Concentration and Absorbance of 5 Standard Tartrazine Solutions Stand Solutions Concentration, PPMm Absorbanc e 4.915 0.184 7.3725 0.294 9.83 0.398 14.745 0.610 19.66 0.799 Table 3. Absorbance of Solution with Unknown Tartrazine Concentration Trial Abs

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1 0.412 2 0.413 3 0.414 4 0.414 Table 4. Concentrations of all Solutions Solution Molarity, mol/l Mass %, % Mass Concentration , g/l PPMm Conversion Factor Stock I 1.84E-03 0.0983 0.983 983 Stock II 9.20E-05 0.004915 0.04915 49.15 0.05* Stock I Stand 1 9.20E-06 0.000492 0.004915 4.915 0.1* Stock II Stand 2 1.38E-05 0.000737 0.007373 7.3725 0.15* Stock II Stand 3 1.84E-05 0.000983 0.00983 9.83 0.01* Stock I Stand 4 2.76E-05 0.001475 0.014745 14.745 0.015* Stock I Stand 5 3.68E-05 0.001966 0.01966 19.66 0.02* Stock I Unknown 1.92E-05 0.001026 0.01026 10.26 Calculations a. Calculation of concentration of stock solution I in parts per million by mass, PPM(m). Assume a dilute solution with density of 1g/ml. Then 100 g(solution)=100ml (solution), can be substituted into the equation PPM ( m ) = m i m solution ∙ 10 6 PPM ( m ) = 0.0983 g 100 ml ∙ 10 6 = 0.0983 g 100 g ∙ 10 6 PPM(m)=983 ppm, concentration of stock solution I b. Dilution formula to calculate concentration of stock solution II, and every other dilution. V i C i = V f C f C f = V i C i V f C f = ( 5 ml )( 983 ppm ) 100 ml

C f = 49.15 ppm,stock solution II c. Use LINEST to find the slope of graph, which represents an absorptivity of 0.042 cm - 1 ppm -1 . Standard error of the slope is S m = 0.000656. The critical value for a 95% CI is found in excel to be 4.18. Then the confidence limit of the slope is CL = t critical ∙S m CL = 4.177 ∙ 0.000656 CL = 0.00274 cm -1 ppm -1 . The absorptivity as determined by the slope is ε = ( 0.042 ± 0.003 ) c m − 1 ppm − 1 , absorptivity. d. Solve for unknown concentration C using Beer’s Law. The equation of the standard curve, y = 0.0418x - 0.0155 . The slope of the curve gives absorptivity ε =0.0418 cm - 1 ppm -1 . The average absorbance of the unknown is A=0.4133. y = mx + b x = y − b m x = 0.4133 −− 0.0155 0.0418 X=10.26 ppm, unknown concentration A = εlC C = A εl C = 0.4133 0.042 cm 1 ppm − 1 ∗ 1 cm C = 9.89 ppm , unknown concentration a. The y-intercept and its confidence limit are calculated as in step c. With standard error of the y-intercept, S b =0.008, the confidence interval of the y-intercept is then b=-0.0155±0.03 cm -1 ppm -1 b. Calculation of Mass concentration of tartrazine in stock solution I in g/l. Mass conc = masssolute volume solution Mass conc = 0.0983 g 100 ml ( 1000 ml 1 l ) Mass conc=0.983 g/l c. Mass percent of tartrazine in stock solution I. As with PPMm calculation, assume a dilute solution with density of 1 g/ml so 100g=100ml.

Mass % = m i m solution ∗ 100% Mass % = 0.0983 g 100 g ∗ 100% Mass Percent=0.0984 % d. Molarity of stock solution I in mol/l. Molarity is defined as mass of solute in grams per liter of solution: M = massconc ( g / l ) molecular wt ( g mol ) M = 0.983 g / l 534.36 g / mol M = 0.00184 mol l ,molarity of Stock I Error Analysis The Q-test is performed on the extreme values of the 4 absorbance trials for the unknown concentration. According to the formula Q = | suspect value − nearest value | | largest value − smallest value | , Q is less than the critical value and no outliers exist. The mean and standard deviation of absorbance of the unknown solution was found to be 0.4133±0.0010. Confidence limit: the critical value for a 95% confidence interval and n=4 trials is t=3.18. The confidence limit of the absorbance is then CL = ±t ∙ s x √ n CL = ± 3.18 ∙ 0.0010 ml √ 4 CL = ± 0.0015 ml The confidence interval of the absorbance can then be reported as 0.4133±0.0015. Propagation of error on concentration calculation: The unknown concentration is found using a formula of the form y=mx+b, where x is the unknown concentration. X=(y-b)/m so propagation of error for concentration x must be a function of y,b, and m.

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m=0.0418, S m =0.000656 b=-0.0155, S b =0.0082 A=0.4133, S A =0.001 C=9.89 ppm Δ x = ± √ ( ∂ x ∂b ∆b ) 2 + ( ∂ x ∂ y ∆ y ) 2 + ( ∂ x ∂m ∆m ) 2 Δ x = ± √ ( − 1 m ∆b ) 2 + ( 1 m ∆ y ) 2 + ( −( y − b ) m 2 ∆m ) 2 Δ x = ± √ ( − 1 0.0418 0.008 ) 2 + ( 1 0.0418 0.001 ) 2 + ( −( 0.4133 + 0.0155 ) 0.0418 2 ∆ 0.000656 ) 2 ∆ x = ± 0.3 ppm δC = | C | ∙ √ ( δA A ) 2 + ( δε ε ) 2 + ( δl l ) 2 δC = | 9.887 ppm | ∙ √ ( 0.0010 0.4133 ) 2 + ( 0.0418 0.000656 ) 2 + ( 0.1 1.0 ) 2 δC = 1.0 ppm The concentration of the unknown solution can finally be reported as C=9.9±1.0 ppm . (6) Discussion (7) References https://www.geol.lsu.edu/jlorenzo/geophysics/uncertainties/Uncertaintiespart2.html