Quantitative Determination of Food Colouring in Jelly Crystals using UV/Vis Spectroscopy Essay Sample

Quantitative Determination of Food Colouring in Jelly Crystals using UV/Vis Spectroscopy Pages
Pages: Word count: Rewriting Possibility: % ()

Introduction

Food colouring are mainly used in the food processing industry today as colour gives the food product certain flavours as people associate colours with certain flavours. Some is to stimulate a colour that is perceived by the consumer as natural food products. Food colouring also provides an identity to foods, to mask natural variations in colour, decorative or artistic purposes or to protect flavours and vitamins from being damaged by light.

Electron transition occurs when valence electrons in a molecule are excited from one energy level to a higher energy level (Silberberg,2008). The energy change associated with this transition provides information about molecular properties such as colour. Ultraviolet-visible spectrophotometry or UV-Vis refers to absorption spectroscopy in the UV visible spectral region. The absorption in the visible range affects the perceived colour of chemicals used in food products involved. In this region of the electromagnetic spectrum, molecules undergo electronic transitions from the ground state to the excited state

The Beer-Lambert law states that the absorbance of a solution is directly proportional to the concentration of the solution and the path length. Thus, for a fixed path length, UV/VIS spectroscopy can be used to determine the concentration of a solution. It is necessary to know how quickly the absorbance changes with concentration. Therefore the equation used is A=Ecl where A is the absorbance, E is the molar extinction coefficient, c is the sample concentrations in moles/litre and l is the length of light path through the sample. The purpose of this experiment is to determine the concentration of the unknown sample.

Method :

1g of jelly crystals was weighed using a laboratory balance and then the crystals was transferred into a 100ml conical flask and dissolved with 50ml of deionised water. The crystals was stirred and then left to be heated awhile to ensure all the crystals in the conical flask was dissolved completely. After heated, the solution of jelly crystals was transferred into a 100ml volumetric flask. The conical flask was rinsed with deionised water several times to ensure all the jelly crystals solution is in the volumetric flask. The filter funnel used was also rinsed a few times to ensure that no jelly crystals remained in the filter funnel.

The volumetric flask was then make up to the mark with deionised water. Then a colouring agent with known concentrations was measured for 1ml,2ml,3ml,and 5ml and was placed in 50ml volumetric flask respectively. Each of the 50ml volumetric flask was then made up to the mark. The absorbance of each solution was measured using a spectrophotometer. The spectrophotometer was blanked with a cuvette filled with deionised water to ‘zero’ the spectrophotometer. The absorbance was first tested with the sample as the cuvette was filled to 2/3 of the capacity, then triplicate measurements was made with the same cuvette. This procedure was repeated for several times using other 4 conical flask with 1ml,2ml,3ml and 5ml of the colouring agent.

Results

Mass of jelly crystals weighed : 1.007g

Table 1: Results obtained for absorbance of unknown sample of different concentrations and sample.

Scan 1

Scan 2

Scan 3

Average

Standard Deviation

Absorbance

Absorbance

Absorbance

Absorbance

Absorbance

Standard 1 (0.01nm)

1.85 x 10-1

1.93×10-1

1.92×10-1

1.90×10-1

4.36E-03

Standard 2 (0.02nm)

3.88 x 10-1

4.00×10-1

4.01×10-1

3.96 x 10-1

6.93E-03

Standard 3 (0.03nm)

5.83×10-1

5.98×10-1

6.02×10-1

5.94×10-1

1.00E-02

Standard 4 (0.05nm)

9.88×10-1

9.93×10-1

9.94×10-1

9.92×10-1

3.21E-03

Sample

1.12×10-1

1.16×10-1

1.18×10-1

1.73×10-1

3.06E-03

Figure 1 : Graph of absorbance against concentration.

Equation of best fit line : y=19.80x

R2 = 0.999

Calculations :

Average of absorbance of sample = 0.112 + 0.116 + 0.118

3

= 1.73 x 10-1

Substitute y = 1.73 x 10-1 into y = 19.80x

x=1.73 x 10-1

19.80

Concentration of jelly solution= 8.74 x10-3M

Discussion :

The regression line in the graph fits the graph as the line of best fit passes through all the points shown. The relationship that can be seen from this graph is linear. The linear graph is related to Beer Lambert law as his law states that the absorbance is directly proportional to the concentration. Besides that, the value of R2 was 0.999 which is close to the value of 1. Therefore a positive correlation can be seen from the graph.

The equation used to measure the absorbance was A=Ecl when the wavelength was set at a constant of 510nm in a with a light path of l (Anderton, et al, 2004) . As seen in Figure 1, it is possible to conclude that the results obtained was concordant to the actual value as absorbance is directly proportional to the concentration of the solution .Therefore, by using the equation from the graph , the concentration of the jelly crystal was 8.74×10-3M.

From the value of the R2 which was 0.999, there was some errors that caused the graph from coming close to obtain a perfect linear regression. This could be due to the defects of some cuvette that disrupted the value ofabsorbance reading which could not be seen with the naked eye. Besides that , there could be an accumulation of air bubbles in the cuvette which would also affect the result obtained. Moreover , we could assume that not all the jelly crystals dissolved completely when diluted leaving small remnants that affected the value of concentration from the actual value. The spectrophotometer used could also have fluctuated reading over time (Silberberg,2008).

Conclusion :

The value of the absorbance increases directly proportional to the concentration of solution, thus obeying Beer Lambert law. The unknown concentration of the food colouring jelly solution is 8.74 x10-3M based from the linear regression obtained from the graph.

References :

1. Martin S.Silberberg, 2008, Chemistry: The Molecular Nature of Matter and Change ( 4th edition), Mc Graw Hill, New York.

2. J.D. Anderton, P.J. Garnett, W.R.Liddelow, R.L. Lowe, I.J. Manno, 2004, Foundations in Chemistry (2nd Edition), Pearson, Australia.

Questions :

1. The value of E124 is obtained based on the ?max determined from the absorbance spectrum is red.

2. Based on Beer Lambert law which is A=Ecl

When A = 0.173 and cl = 8.74×10-3

E = 19.803

3. The value for the molar extinction coefficient is are made assuming that absorption measured is solely based on the colour of food dye and that other substances will not affect the absorbance reading.

4. Concentration of food dye in jelly solution counted above = 8.74×10-3M

y= 19.80x

0.04 = 19.80x

x=0.002020

Concentration in ppm = 0.002020 x 10-3 X 1000 X 604.47

=1.221ppm

5. Concentration of jelly solution is 8.74×10-3M for 1.00g of jelly crystals.

M = cv

= 8.74×10-3M x 0.1 L

= 8.74 x 10-4 M

In 85g of jelly crystals ,

= 8.74 x 10-4 M x 85

= 7.43 x 10-2M

Theory Test.

1. The definition of ?max for a molecule absorbing in the UV/Vis region is the wavelength where the maximum absorption is obtained from a fraction of light.

2. When colour intensity increases in colour of jelly, ?max also increases showing ?max is directly proportional to food colour.

3 A) ?max=430nm is yellow or lemon yellow.

B) ?max values of 650nm and 430nm and will show blue and yellow colour.

4. The gradient corresponds to the of molar extinction coefficient and is measured in mol-1dm3cm-1.

5.To show exact wavelength that the absorbance value for the solutions is the highest ?max to precisely determine the concentration of the solution.

6. Standard deviation is the deviation of a distribution from the mean.

7. Error bar is used to determine the range of standard deviations that deviates from the mean. The magnitude of the error bar can be determined by the standard deviations of the distribution.

8. An outlier is usually eliminated as outliers affects the overall results of the data.

9. a ) 4 decimal place = 62.4567

b) 3 significant figures = 6.25 x 101

10. ppm is defined as one milligrams in one kilogram of the solution.

Search For The related topics

  • solutions
  • Olivia from Bla Bla Writing

    Hi there, would you like to get such a paper? How about receiving a customized one? Check it out https://goo.gl/3EfTOL

    sample
    Haven't found the Essay You Want?
    GET YOUR CUSTOM ESSAY SAMPLE
    For Only $13.90/page