SPEC FE LAB REPORT

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University of Florida *

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3120L

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Chemistry

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Oct 30, 2023

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SPEC FE LAB REPORT CALCULATIONS 1. Concentration Fe ( 0 uL )= 0 ppm Concentration Fe ( 150 uL ) ( .13 M 100 )( 55.45 g mol )( 1000 mg g ) ( 150 E − 6 ) ( 1 0.01 L ) = 1.1 ppm Concentration Fe ( 300 uL ) ( .13 M 100 )( 55.45 g mol )( 1000 mg g ) ( 300 E − 6 ) ( 1 0.01 L ) = 2.2 ppm Concentration Fe ( 450 uL ) ( .13 M 100 )( 55.45 g mol )( 1000 mg g ) ( 450 E − 6 ) ( 1 0.01 L ) = 3.3 ppm Concentration Fe ( 600 uL ) ( .13 M 100 )( 55.45 g mol )( 1000 mg g ) ( 600 E − 6 ) ( 1 0.01 L ) = 4.4 ppm Absorbance ( 150 uL ) :0.0613,0.0605,0.0597,0.0595,0.0595 Avg = 0.0601 Absorbance ( 300 uL ) :0.1213,0.1225,0.1205,0.1208,0.1207 Avg = 0.1212 Absorbance ( 450 uL ) :0.1816,0.1815,0.1814,0.1811,0.1810 Avg = 0.1813 Absorbance ( 600 uL ) :0.2417,0.2417,0.2417,0.2416,0.2425 Avg = 0.2418 Absorbance ( unknown 1 ) :0.0143,0.0146,0.0144,0.0142,0.0142 Avg = 0.0143 Absorbance ( unknown 2 ) :0.0142,0.0142,0.0144,0.0146,0.0142 Avg = 0.01438 Absorbance ( cereal ) : 0.0035,0.0037,0.0025,0.0027,0.0049 Avg = 0.00346 Average Transmittance = 10 −( Averageabsorbance ) x 100 ¿ 87.1% , 75.6% , 65.9% , 57.3% , 96.77%, 96.73%, 99.21% 2. Unknown 1 Y = 0.055 x – 0.0001 where yisabsorbance ∧ x isconcentration X =( 0.0143 + 0.0001 )/( 0.055 )

¿ 0.262 ppm Unknown 2 Y = 0.055 x – 0.0001 where yisabsorbance ∧ x isconcentration X =( 0.0144 + 0.0001 )/( 0.055 ) ¿ 0.264 ppm 3. Absorbance of cereal = 0.00346, B = 1 cm,concentration ∈ ppm is X =( 0.00346 + 0.0001 )/ 0.055 = 0.0647 ppm E = A / BC = 0.05 liters / ppm x cm 0.0647 ppm = 6.5 E − 5 grams / liter Percent weight =( 6.5 E − 5 grams / liter / 6.28 g ) x 100 = 0.001 percent 4. Wewere given thatthe known valueis 18 mg ∨ .018 gof Iron ∈ a 30 g serving ¿ 100% RDA. Wereceived 0.001 percent Iron ∈ our 6.28 − gramsample . ( 30 grams )( 0.001% )= 0.0003 g Iron found ∈ 30 gsample of cereal 0.0003 g / 0.018 g Iron RDA = 1.7 percent recovery 5. Slope = E x Lwhere Lisonecm Forthat reason,Eof FeBipy = 0.055 x 1 cm = 0.055 liters

Questions 1. A = ebc Absorbance(none) = (molar absorptivity constant (L/mol x cm or L/ppm x cm))(path length (cm))(concentration (ppm or M)) When you have a single calibration standard you use a known concentration to measure absorbance and molar absorptivity. Once you have E you can use this to find absorbance and path length. However, with calibration point you can account for many variations through use of different graphs. It also provides a lot of information through the use of y = mx+b without the need for algebra and arithmetic. On top of that you can determine the concentration easily and more accurately than a single calibration standard. 2. A. − log ( 36.8 / 100 )= 0.434 B. − log ( 69.2 / 100 )= 0.160 C. 10 (− 0.876 ) x 100 = 13.3% D. 10 (− 0.255 ) x 100 = 55.6% 3. Source: Light source that emits broad spectrum of UV and Vis light to provide source of light for sample analysis Sample Holder: This was a cuvette in this experiment it is designed to allow light to pass through sample whether that is UV light or Vis light. This is due to the cuvettes transparent design. Monochromator: A monochromator isolates a range of wavelengths from and then direct that to the sample in the cuvette. This is essential for determining concentration and absorption.

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Slit: Opening in the optical path that controls width of beam that reaches sample. Chopper: Manipulates the intensity of light by blocking it or unblocking it. Amplifier: Increases the signal detected by the photodetector this makes the signal more useful for analysis. Detector: Converts optical signal into electrical signal to process the absorbance value. This is the most important part as no absorbance value would be read if it wasn’t for the detector. 4. The useful concentration range is 0.1 ppm to 10 ppm. The structure is (Fe(bipyridine)2)2+ 5. Fe(NH4)2(SO4)26H2O, Primary standards are highly pure but ferrous ammonium sulfate has water that affects the molar mass and therefore isn’t primary standard. The sulfuric acid buffers the solution to the desired pH so that the reaction can go to completion. 6. Hydroxylamine hydrochloride acts as a reducing agent producing after reacting with iron (III). Afterwards it reacts with sodium acetate which increases the pH allowing the reaction to finish. 7. Bromophenol blue is used with cereal often due to its initial color in the presence of iron. When in iron originally which is Fe(III) the condition is acidic turning the blue indicator yellow. Once the cereal sample is changed the iron reduces into Fe(II) increasing the pH changing the bromophenol blue to blue. This allows us to understand that the reaction has completed. 8. We should begin by finding the Molarity of FAS by dividing moles by the volume. Once we have molarity, we can find concentration which is the same in this case. Finally, we can convert to ppm by (Molarity x Molar mass x 10^3) Results/ Discussion

In this experiment, we used the (Fe(bipy)3)2+ complex to analyze the concentration of Iron present in unknown solutions as well as the iron present in common breakfast cereal to determine if the iron present in a 30- gram serving size really did contain your RDA of iron. To do this we measured the absorbances of the aforementioned unknown solutions as well as breakfast cereal solutions. We used spectroscopic techniques to measure the absorbances of unknown solutions, known concentrations of different Iron volumes, and the unknown breakfast cereal solutions. A graph was created showing the correlation between concentration and absorbance in standardized Iron complexes that went up in volume by 150uL starting from 0uL and ending in 600uL. Concentrations were found using stoichiometric techniques apparent in calculation 1. This not only allowed us to qualitatively understand their relationship but also allowed us to find the concentrations of both unknowns and the breakfast cereal solution by implementing our slope-intercept equation into beers law (A=ebc). Where within our slope intercept equation Y was allocated as our absorbance (A), x was allocated as our concentration (c) and the slope of our equation was allocated as our molar absorptivity constant(e). This is all apparent by the graph shown below as well as calculations 2-3 which shows us using the absorbance of our unknown/ breakfast cereal solutions and our slope intercept equation shown below to find the concentrations of all unknowns. Through our calculations of the unknown we received the concentrations of 0.262, 0.264, and 0.0647 ppm for our two unknowns as well as breakfast cereal solution respectively. When scaling the amount of iron, we recovered to a single serving size of breakfast cereal. We recovered 0.0003 grams of Iron when 0.018 grams was expected. This value is off most likely due to errors in the preparation process such as too much dilution or the chemical reaction not coming to completion before analysis of absorption. Our data indicated that one serving of breakfast cereal only contains 1.7 percent of your daily recommended iron content.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0 0.05 0.1 0.15 0.2 0.25 0.3 f(x) = 0.05 x − 0 R² = 1 Absorbance Vs Fe Concentration Fe Concentration (ppm) Absorbance Volume of Fe (uL) Concentration (ppm) Avg Transmittance (%) Average Absorbance 150uL 1.1 87.1 0.0601 300uL 2.2 75.6 0.1212 450uL 3.3 65.9 0.1813 600uL 4.4 57.3 0.2418 Unknown 1 (210 uL) 0.262 96.77 0.0143 Unknown 2 (210 uL) 0.264 96.73 0.0144 Iron Cereal 0.0647 99.21 0.00346

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