Inductively Coupled Plasma – Atomic Emission Spectroscopy

Name: Bulose Sihle Student Number: 209504496
Date of Practical: 11 April 2013

Inductively Coupled Plasma – Atomic Emission Spectroscopy
Aim
The aim of this practical was to use an ICP-OES to perform a multi-elemental analysis of different types of tea. The elements that were analysed in the tea were copper, iron, manganese and magnesium.

The sample is introduced into the plasma as an aerosol. Argon gas flows through three concentric quartz tubes in the plasma torch. This argon gas transports the sample from the nebulizer, acts as a cooling gas and is also a source of electrons and ions for the plasma [1]. A magnetic field is created around the plasma torch. The ionisation process is started by the ignition of the argon gas from a spark produced by a tesla coil [1]. During this process the plasma reaches a very high temperature and the sample is atomised. The atoms are then excited to a higher state. Since this is an emission analysis, the analysis is performed as the atoms emit energy and return to a lower energy state or their ground state. A spectrometer or monochromator is used to select the wavelength that is being analysed. The multi-element detector then gives us a readout that can be understand for each element that is analysed
Explain the benefits and limitations of plasma over a conventional flame used for FES with photometry and AAS.
The benefits of using plasma are that the chemical interferences are decreased due to the high temperatures that the plasma reaches and refractory elements can also undergo excitation. There are also many wavelength for different that can be chosen for analysis so you don’t have two elements that have wavelengths that are close to each other as this would interfere with the analysis. A limitation is that spectral overlap can occur and the preparation of the samples is a very long process compared to that of the flame spectrometry

Data
Show the calculations for you multi-element standard.
Instrument: Perkin Elmer OES (optima 5300 DV)
The preparation of multi-element standard
1000ppm solutions of (Cu), (Fe),(Mg) and (Mn ) were given. A working standard solution containing the entire four elements was prepared in the following way;
1 ml, 10 ml and 5 ml of Fe, Mg, and Mn were respectively added in one 100 ml volumetric flask. A solution of Cu was prepared separately (intermediate solution) by placing 1 ml of the 1000ppm Cu solution in a 100 ml volumetric flask and made up to the mark with distilled water, the resulting concentration was 10ppm. The formula that was used to calculate the volume of the 1000ppm Cu solution needed to prepare 10ppm solution in a 100 ml volumetric flask is; C1V1 = C2V2 …… equation 1

Where C1 is the initial concentration C2 is the final concentration V1 is the volume of the initial concentration V2 is the total volume of the initial concentration and the solvent
The volume of the barium required was calculated as follows
C1V1 = C2V2 1000ppm x V1 = 1ppm x 100ml V1 = 10 ppm x 100 ml 1000 ppm = 1 ml
Then 1 ml of the 10 ppm Cu solution was placed in the 100 ml volumetric flask that contained the other three elements, and made up to the mark with distilled water, this was the working standard solution. Using equation 1 it is found that the concentrations of the elements in the working standard solution are as follows;

Table 1: Concentrations of the elements in the working standard solution.

Element | Volume used (mL) | Cu | 1 | Fe | 5 | Mg | 25 | Mn | 10 |

Then 0.5 ml, 2 ml, 4 ml, 8 ml and 10 ml of the working standard were placed in five separate 100 ml volumetric flasks and made up to the mark with distilled water. Standards | Element concentration/ ppm | | Cu | Fe | Mg | Mn | 1 | 0.05 | 0.25 | 1.25 | 0.5 | 2 | 0.2 | 1 | 5 | 2 | 3 | 0.4 | 2 | 10 | 4 | 4 | 0.8 | 4 | 20 | 8 | 5 | 1.0 | 5 | 25 | 10 |

Present a sample calculation for one of the teas and one element e.g. Cu for Five Roses. Present the rest of the calculated values in a table. (Present your results in μg element / g sample).
Table 2: showing masses of green tea used and concentrations of the elements. sample | Mass/g | Cu/ppm | Fe/ppm | Mg/ppm | Mn/ppm | 1 | 0.499 | 0.125 | 4.076 | 10.85 | 3.934 | 2 | 0.507 | 0.146 | 2.517 | 10.82 | 3.979 | 3 | 0.505 | 0.126 | 3.021 | 10.68 | 3.782 |

Table 3: Showing masses of five roses used and concentrations of the elements. Sample | Mass/g | Cu/ppb | Fe/ppm | Mg/ppm | Mn/ppm | 1 | 0.5135 | 0.144 | 1.838 | 14.77 | 2.296 | 2 | 0.5101 | 0.120 | 1.741 | 14.48 | 2.163 | 3 | 0.5012 | 0.183 | 2.156 | 15.58 | 2.289 |

Table 4: Showing masses of rooibos used and concentrations of the elements. Sample | Mass/g | Cu/ppb | Fe/ppm | Mg/ppm | Mn/ppm | 1 | 0.5164 | 0.039 | 1.235 | 8.919 | 0.192 | 2 | 0.5242 | 0.024 | 1.007 | 9.121 | 0.195 | 3 | 0.5086 | 0.039 | 1.041 | 8.428 | 0.187 |

Calculation for Cu in the sample 1 of green tea
Mass = 0.5135 g
[Cu] = 0.125 mg/L
Volume used = 50 mL
Concentration of Cu in sample 1 of green tea = 0.125 mg/L x 0.05 L = 0.00625 mg x 1000 = 6.25 µg ÷ 0.499 g = 12.53 µg/ g

All calculations were carried out in a similar manner and the results are tabulated below:

Table 5: Showing concentrations of elements in green tea. Sample | Cu/µg element/ g sample | Fe/µg element/ g sample | Mg/µg element/ g sample | Mn/µg element/ g sample | 1 | 12.53 | 408.4 | 1087.2 | 394.2 | 2 | 14.40 | 248.2 | 1067.1 | 392.4 | 3 | 12.48 | 299.1 | 1057.4 | 374.5 | Mean ± std.dev | 13.14 ± 1.09 | 318.57 ± 81.85 | 1070.57 ± 15.20 | 387.03 ± 10.89 | % RSD | 8.33 | 25.69 | 1.42 | 2.81 |

Table 6: Showing concentrations of elements in five roses. Sample | Cu/µg element/ g sample | Fe/µg element/ g sample | Mg/µg element/ g sample | Mn/µg element/ g sample | 1 | 14.02 | 178.97 | 1438.2 | 223.6 | 2 | 11.76 | 170.6 | 1419.3 | 212 | 3 | 18.26 | 215.1 | 1554.3 | 228.4 | Mean ± std.dev | 14.68 ± 3.30 | 188.22 ± 23.65 | 1470.60 ± 73.10 | 221.33 ± 8.43 | % RSD | 22.48 | 12.56 | 4.97 | 3.81 |

Table 7: Showing concentrations of elements in rooibos. Sample | Cu/µg element/ g sample | Fe/µg element/ g sample | Mg/µg element/ g sample | Mn /µg element/ g sample | 1 | 3.776 | 119.58 | 863.6 | 18.59 | 2 | 2.289 | 96.05 | 870 | 18.6 | 3 | 3.834 | 102.34 | 828.6 | 18.38 | Mean ± std.dev | 3.30 ± 0.88 | 105.99 ± 12.18 | 854.07 ± 22.29 | 18.52 ± 0.12 | % RSD | 26.54 | 11.49 | 2.61 | 0.67 |

Carry our statistical analysis of your data and determine if the experimental values you have obtained are different from the expected/reported values for your sample.

Comparison of the five roses sample and the green tea sample using statistical means:
F calculated = S FR2 /S GT2 = (1.09)2 / (3.30)2 = 1.09
F table = 5.390
F calculated < F table, therefore there is a significant difference between Cu in five roses and green tea.
Interpretation
Explain why the type of sample preparation carried out was necessary.
The Acid digestion was suitable because a complete transfer of analyte into the solution in order for the determination step to be introduced in liquid form is highly desirable and this method completely transfers the analyte into solution. This thus means that the digested sample is a complete solution of the analyte and has a complete decomposition of the matrix however with minimal loss or contaminated of the analyte
Explain what matrix matching is and what problems may arise if this is not carried out.
Matrix matching involves preparing solutions in which the major chemical compositions of the standards, blanks and samples are made identical thereby cancelling out the effect of the sample matrix on the analysis results. While matrix matching involves matching the solvents, it also involves matching the concentrations of acids and other major solutes. In case where the standard and sample matrices are quite different or cannot be matched and interference occurs as a result, internal standards can be used.
Comment on the correlation coefficient of your calibration graphs.
The correlation coefficient measures the strength in the linear relationship between two variables. A correlation coefficient of 1 would mean a very strong linear relationship between two variables, which means the points form a perfect straight line. The correlation coefficients for the calibration graphs were very good. The calibration graphs for copper, iron, magnesium, and manganese gave correlation coefficients of 0.991539, 0.991005, 0.999874, and 0.999952 respectively. This shows that the samples were prepared very well and that there is a strong linear relationship between the emission and the concentrations of the elements.
Compare the different elements in the different types of tea. Which tea would you recommend and why?
The tea that seems to be most essential for human consumption is the green tea owing to the high content of each and every element present in it. It has high concentration of manganese which is a vital substance in the body as it is an enzyme activator; it keeps bones strong and healthy, and also maintains the health of our nerves.
References
1. Dr L Pillay, Chem 340, Instrumental Analysis, ICP-OES notes 2. http:// www.whfoods.com/genpage.php?tname=nutrint& dbod=77(accessed 16/04/2013) 3. http://www.chemplex.com/petrochemical/multielementMetal.aspx(accessed 16/04/2013)

References: 1. Dr L Pillay, Chem 340, Instrumental Analysis, ICP-OES notes 2. http:// www.whfoods.com/genpage.php?tname=nutrint&amp; dbod=77(accessed 16/04/2013) 3. http://www.chemplex.com/petrochemical/multielementMetal.aspx(accessed 16/04/2013)