Biochem Report1 - Spectrophotometry
Essay by qt pie • March 20, 2016 • Lab Report • 926 Words (4 Pages) • 1,314 Views
Introduction
Spectrophotometry 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 Result
TABLE 1.1 Absorbance of duplicate standards solution for standard curve
Test tubes | Mass of phosphate (µg) | Volume of standard phosphate solution (mL) | Volume of added water (mL) | Absorbance at 660nm | ||
Set A | Set B | Average | ||||
Blank | 0 | 0.0 | 5.0 | 0.000 | 0.000 | 0.000 |
1 | 10 | 0.1 | 4.9 | 0.032 | 0.030 | 0.031 |
2 | 20 | 0.2 | 4.8 | 0.057 | 0.051 | 0.054 |
3 | 30 | 0.3 | 4.7 | 0.090 | 0.080 | 0.085 |
4 | 40 | 0.4 | 4.6 | 0.116 | 0.102 | 0.109 |
5 | 50 | 0.5 | 4.5 | 0.141 | 0.126 | 0.134 |
Calculation; standard phosphate solution 1mL = 100µg
e.g For test tube 1 10µg = 0.1mL of standard phosphate solution
Table 1.2 Absorbance of unknown samples
Unknown | Volume of solution (mL) | Volume of water added (mL) | Absorbance at 660nm |
X1 | 5 | 0 | 0.033 |
X2 | 5 | 0 | 0.034 |
Y1 | 5 | 0 | 0.106 |
Y2 | 5 | 0 | 0.105 |
FIGURE 1.1 Standard curve for estimation of phosphorus mass
[pic 1]
Calculation; Absorbance= 0.0027(mass of phosphorus) + 0.002
e.g. X1 absorbance = 0.033
0.033 = 0.0027(mass of phosphorus) + 0.002
Mass of phosphorus = (0.033-0.002) ÷ 0.0027
= 11.481µg
Concentration of X1 = 11.481 ÷ 5
= 2.296 µg/mL
Table 1.3 Mass and concentration of unknown samples
Unknown | Mass of phosphorus (µg) | Concentration of phosphorus (µg/mL) |
X1 | 11.481 | 2.296 |
X2 | 11.852 | 2.370 |
Y1 | 38.519 | 7.704 |
Y2 | 38.148 | 7.630 |
Mean of X = (2.296+2.370) ÷ 2 ± 0.052
Mean of Y = (7.704+7.603) ÷ 2 ± 0.071
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