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Biochem Report1 - Spectrophotometry

Essay by   •  March 20, 2016  •  Lab Report  •  926 Words (4 Pages)  •  1,314 Views

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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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