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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