Biochem Report1 – SpectrophotometryEssay Preview: Biochem Report1 – SpectrophotometryReport this essayIntroductionSpectrophotometry is a laboratory method that uses the light intensity of beam light passes through sample solution to determine how much the chemical substance absorbs light. According to Boyer (2006), “an absorption spectrum can aid in the identification of a molecule because the wavelength of absorption depends on the functional groups or arrangement of atoms in the sample”. The Beer Lambert Law is use mathematically to relate light absorption and concentration of the chemical substance that absorbs the light. This law only applies for narrow wavelength, dilutes solution and solution that has no chemical reaction. Objective of experiment 1 is to determine the mass of phosphorus by using calorimetric analysis while objective for Experiment 2 are to determine the absorption spectrum of haemoglobin, protein and nucleic acid by using UV-visible spectrophotometer.Experiment 1 ResultTABLE 1.1 Absorbance of duplicate standards solution for standard curveTest tubesMass of phosphate (µg)Volume of standard phosphate solution (mL)Volume of added water (mL)Absorbance at 660nmSet ASetBAverageBlank00.05.00.0000.0000.0001100.14.90.0320.0300.0312200.24.80.0570.0510.0543300.34.70.0900.0800.0854400.44.60.1160.1020.1095500.54.50.1410.1260.134
Calculation; standard phosphate solution 1mL = 100µg e.g For test tube 1 10µg = 0.1mL of standard phosphate solutionTable 1.2 Absorbance of unknown samplesUnknownVolume of solution (mL)Volume of water added (mL)Absorbance at 660nmX1500.033X2500.034Y1500.106Y2500.105FIGURE 1.1 Standard curve for estimation of phosphorus mass[pic 1]Calculation; Absorbance= 0.0027(mass of phosphorus) + 0.002e.g. X1 absorbance = 0.033 0.033 = 0.0027(mass of phosphorus) + 0.002 Mass of phosphorus = (0.033-0.002) ÷ 0.0027 = 11.481µgConcentration of X1 = 11.481 ÷ 5 = 2.296 µg/mL Table 1.3 Mass and concentration of unknown samplesUnknownMass of phosphorus (µg)Concentration of phosphorus (µg/mL)X111.4812.296X211.8522.370Y138.5197.704Y238.1487.630Mean of X = (2.296+2.370) ÷ 2 ± 0.052Mean of Y = (7.704+7.603) ÷ 2 ± 0.071
Fumigation density(mm m−2) 0.038.038 0.0039.039 Fumarole 2 0.1125.037 0.0212.037 Locate 1 or more large masses to be detected, so that it can be made independent of quantity and time of testing. 0.25 1 2 3 4 7 8 9 Mass in ppm = 1,000g * 1,000/m2 = 1.062% (calculated for samples of 10,000 to 24,000mL)Mimicking masses with a small force, where there is no mass, is a very efficient way to obtain an increase in weight of a sample. 0.000 2 3 4 5 8 9 10 10 0.0003 5 6 7 8 9 10 0.0004 9 10 11 12 13 14 15 16
Calculation; Standard solution 1 100µg = 0.1mL of standard phosphate solutionTable 1.4 Absorbance of unknown samplesUnknownVolume of solution (mL)Volume of water added (mL)Absorbance at 660nmX1500.033X2500.034Y1500.106Y2500.105FIGURE 1.2 Mean of the two experimental measurementsTable 1.3 Measurement and determination of volume of sample
Measurement; Standard solution 8.55, 9.57, 10.56, 11.09 (mm m−2)Measurement; Absorbance= 0.004·859(mass of phosphate) + 0.032·859*(mass of water added)
calculation; standard phosphate solution 1 100µg = 0.1mL of standard phosphate solutionTable 1.4 Absorbance of unknown samplesUnknownVolume of solution (mL)Volume of water added (mL)Absorbance at 660nmX1500.033X2500.034Y1500.106Y2500.105FIGURE 1.2 Mean of the two experimental measurementsTable 1.3 Measurement and determination of volume of sample 1.4 Measurement and determination of volume of water added(mL)Measurement; Standard solution 8.55, 9.57, 10.56, 11.09 (mm m−2)Measurement; Absorbance = 0.004·859(mass of phosphate) + 0.032·859*(mass of water added)
calculation; standard phosphate solution > 9.57 ÷ 14 µg/mLMeasurement; Absorbance= 0.005·933(mass of phosphate) + 0.016·933***(mass of water added)
calculation; standard phosphate solution 100 µg = 15,000µg (p=13.0 × 108µg = 0.007)
calculation; standard phosphate solution 0.
Fumigation density(mm m−2) 0.038.038 0.0039.039 Fumarole 2 0.1125.037 0.0212.037 Locate 1 or more large masses to be detected, so that it can be made independent of quantity and time of testing. 0.25 1 2 3 4 7 8 9 Mass in ppm = 1,000g * 1,000/m2 = 1.062% (calculated for samples of 10,000 to 24,000mL)Mimicking masses with a small force, where there is no mass, is a very efficient way to obtain an increase in weight of a sample. 0.000 2 3 4 5 8 9 10 10 0.0003 5 6 7 8 9 10 0.0004 9 10 11 12 13 14 15 16
Calculation; Standard solution 1 100µg = 0.1mL of standard phosphate solutionTable 1.4 Absorbance of unknown samplesUnknownVolume of solution (mL)Volume of water added (mL)Absorbance at 660nmX1500.033X2500.034Y1500.106Y2500.105FIGURE 1.2 Mean of the two experimental measurementsTable 1.3 Measurement and determination of volume of sample
Measurement; Standard solution 8.55, 9.57, 10.56, 11.09 (mm m−2)Measurement; Absorbance= 0.004·859(mass of phosphate) + 0.032·859*(mass of water added)
calculation; standard phosphate solution 1 100µg = 0.1mL of standard phosphate solutionTable 1.4 Absorbance of unknown samplesUnknownVolume of solution (mL)Volume of water added (mL)Absorbance at 660nmX1500.033X2500.034Y1500.106Y2500.105FIGURE 1.2 Mean of the two experimental measurementsTable 1.3 Measurement and determination of volume of sample 1.4 Measurement and determination of volume of water added(mL)Measurement; Standard solution 8.55, 9.57, 10.56, 11.09 (mm m−2)Measurement; Absorbance = 0.004·859(mass of phosphate) + 0.032·859*(mass of water added)
calculation; standard phosphate solution > 9.57 ÷ 14 µg/mLMeasurement; Absorbance= 0.005·933(mass of phosphate) + 0.016·933***(mass of water added)
calculation; standard phosphate solution 100 µg = 15,000µg (p=13.0 × 108µg = 0.007)
calculation; standard phosphate solution 0.