# Experiment on analysis of copper in brass by UV-Visible spectroscopy

Published: 2021-06-27 21:15:05  Category: Chemistry

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GET MY ESSAY By the end of this experiment, the student should be able to demonstrate the following proficiencies: 1. To determine the percentage of copper in brass by UV-visible spectroscopy. 2. Properly calibrate and use a spectrophotometer. 3. Convert percent transmittance to absorbance, and vice versa. 4. Construct a calibration curve relating absorbance and concentration for solutions of known concentrations. 5. Use a calibration curve to determine the concentration of an unknown solution. 6. Convert a molar concentration to a mass percent value.
INTRODUCTION Electromagnetic radiation, of which ultraviolet and visible light are but two examples, has properties of both waves and particles. When light acts as a particle, called a photon, each light particle possesses a discrete amount of energy called a quantum. When a molecule is exposed to electromagnetic energy it can absorb a photon, increasing its energy by an amount equal to the energy of the photon. The energy of the absorbed photon can be calculated if the frequency, ? , of the light is known according to Equation 1. E = h? Equation 1
Where h is a constant known as Planck’s constant after Max Planck, the German scientist who first proposed it. Planck’s constant has a value of 6. 63 X 10-34 J. s. Frequency is measured in units of 1/s or Hertz (Hz). The frequency of a light wave is inversely proportional to its wavelength, ? , which is typically measured in meters. The product of ? and ? is the speed of light, c as shown in Equation 2, c has a value of 2. 998 X 108 m/s. c = ?? Equation 2 Molecules are highly selective in the wavelengths of light they can absorb. The photons absorbed depend on the on molecular structure and can be measured by instruments called spectrometers.
The data obtained from a spectrometer are very sensitive indicators of molecular structure and concentration. UV-visible spectrophotometry (UV-VIS) uses only the ultraviolet and visible regions of the electromagnetic spectrum. UV light ranges from approximately 10 nm to about 400 nm. The visible region of the spectrum ranges from 400 nm to 700 nm. Red light lies at the low Page 2 of 9 energy end of the visible spectrum and violet lies at the high energy end. UV-VIS spectroscopy depends on transitions of electrons in a molecule from one electronic energy level to another.
It is used mostly in studying transition metal complexes and conjugated ? systems in organic molecules. One of the principal uses of UV-VIS spectrometry is in determining the concentration of an absorbing (coloured) molecule. The amount of light absorbed, and thus the intensity of the colour of the solution, depends on the concentration of the absorbing species in the solution. UV-VIS absorption peaks are typically quite broad and are often spoken of as bands rather than as peaks. The wavelength at maximum absorption is referred to as ?
max and is the optimal wavelength for a pure solution. For an impure solution, a “blank” solution can be prepared which contains all components of the solution except the one being analyzed. This solution is used to subtract out absorbance due to interfering species. When a beam of light with intensity, Io, passes through a solution, a coloured species, or analyte, will absorb some of the light energy. The beam of light that passes through (or is transmitted through) the solution will have a lower intensity, I, than the incident light, Io because some of the light will be absorbed.
Spectrometers typically measure either transmittance, T, which is the amount of transmitted light, or absorbance, A, which is a measure of the light absorbed. Both transmittance and absorbance are measures of the amount of light that is absorbed by the analyte. Transmittance is calculated by dividing the transmitted light by the incident light (Equation 3). Experimentally, however, transmittance is usually measured as the percent of the incident light or %T and is defined as shown in Equation 4. T = I/Io Equation 3 %T = (I/Io) X100% Equation 4 Absorbance is defined by Equation 5.
A = -log(T) Equation 5 But since %T rather than T is actually measured, Equation 5 is changed to Equation 6. A = -log(%T/100) = log(100/%T) = log 100 – log(%T) = 2-log(%T) Equation 6 Each chemical species absorbs a different quantity of light at any given wavelength. The amount of light absorbed by a compound depends on the structure of the compound and the solvent. However, for any chemical species in a given solvent, the amount of light absorbed at a given wavelength will be a constant called the molar absorptivity, ? (also referred to as Page 3 of 9 the molar extinction coefficient).
The absorbance, A, is a function of the concentration, c, of the absorbing species and the distance the light travels through the solution, that is the path length, l. Absorbance can be calculated for any solution for which the molar absorptivity is known (Equation 7). A = ?. c . l Equation 7 Equation 7 is often referred to the Beers-Lambert Law or Beers’ Law for short. Since molar absorptivity and path length are both constants for a given instrument, one may safely assume that the absorbance is directly proportional only to the concentration of the analyte.
The concentration of a solution can be determined from a Beers’ Law calibration plot prepared by measuring the %T (or A) for several solutions of known concentration (standard solutions). The Beers’ Law plot uses concentration as the independent variable (x-axis) and absorbance as the dependent variable (y-axis). When the absorbance of an unknown sample is determined, the concentration can then be determined by graphical interpolation from the prepared calibration graph. Brass is an alloy composed of zinc and copper. In order to determine the copper content in a sample of brass we must first dissolve the sample.
To achieve this we will use nitric acid, a strong oxidizing acid. The reaction can be described by the following equation. 3Cu(s) + 8H+(aq) + 2NO3-(aq) J 2NO(g) + 4H2O(l) + 3Cu+2(aq) Equation 8 The nitric oxide, NO, gas evolved in this reaction reacts immediately with oxygen to form brown NO2 and N2O4 gases. NO, NO2, and N2O4 are all toxic. The dissolution process therefore must be done in a working fume hood. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. Brass foil Copper foil Distilled water 8M nitric acid 150ml beaker 5 conical flasks Electronic balance; model: adventurer Pro AV 812; S/N: 1126492837 Cuvette
Fumehood 25ml burette 100ml measuring cylinder Funnel 100ml volumetric flasks Shimadzu UV-160 spectrophotometer CHEMICALS AND EQUIPMENT Page 4 of 9 PROCEDURE ACTIVITY An amount of brass and copper were measured into two separate clean beakers 20ml of 8M nitric acid was added to the brass in fumehood and warmed. 20ml of 8M nitric acid was added to the copper also in fumehood and warmed. The contents of each beaker was transferred into a 100ml volumetric flask and topped to the mark. 10ml, 20ml and 30ml of the prepared copper solutions were measured into three conical flasks and diluted with 30ml, 20ml and 10ml of
distilled water respectively. OBSERVATION INFERENCE The masses obtained were 1. 5000g and 2. 0000g for brass and copper respectively. There was an evolution of a brown smoke and a formation of a green solution after warming. There was an evolution of The copper present was copper a brown smoke and a (II) formation of blue solution. The absorbances of the samples were determined with the help of a UV-160 spectrophotometer at a wavelength of 760nm. The results are as follows: Distilled water was used as the blank solution. SAMPLE ABSORBANCE 10 ml of copper standard solution and 30 ml of distilled water
0. 175 20 ml of copper standard solution and 20 ml of distilled water 0. 870 30 ml of copper standard solution and 10 ml of distilled water 1. 778 40 ml of copper standard solution only 2. 524 Brass solution 1. 505 Page 5 of 9 Further details of the results can respectivefully be found on the endorsed sheet at the extreme end of this document. CALCULATIONS AND EVALUATION OF DATA 6. Mass of copper = 2. 000g, volume of copper solution = 100ml = 0. 1dm3, Molar mass (Cu(NO3)2) = 63. 55 + (14 + 16 x 3) x2) = 187. 55g/mol Concentration = mole but mole = mass Volume molar mass Concentration = 2. 000
= 0. 1066M 187. 55 x 0. 1 i). C1 = 0. 1066M, V1 = 10ml, C2 = ? , V2 = 40ml C1V1 = C2V2 C2 = 0. 1066 x 10 40 = 0. 0267M ii). C1 = 0. 1066M, V1 = 20ml, C2 = ? , V2 = 40ml C2 = 0. 1066 x 20 40 = 0. 0533M iii). C1 = 0. 1066M, V1 = 30ml, C2 = ? , V2 = 40ml C2 = 0. 1066 x 30 40 = 0. 07995M = 0. 0800M 7. Concentration/M 0. 0267 0. 0533 0. 0800 0. 1066 Absorbance 0. 175 0. 870 1. 778 2. 524 Page 6 of 9 GRAPH 3 y = 29. 862x – 0. 6536 2. 5 Absorbance 2 1. 5 absorbance Linear (absorbance) 1 0. 5 0 0 0. 05 0. 1 Concentration/M 0. 15 From the graph above, Concentration of copper in brass is 0. 0723M 9.
Concentration of copper = 0. 0723, molar mass of copper = 63. 55g/mol, volume = 100ml=0. 1dm3 Mass of copper = conc. x molar mass x volume = 0. 0723 x 63. 55 x 0. 1 = 0. 4595g Percentage of copper in brass = 0. 4595 x 100% 1. 5000 = 30. 63%. DISCUSSION From the result above, the experiment tries to shows the quantity of copper in brass, since brass is an alloy made of cooper and zinc. By so doing nitric acid which can dissolve most metals is being used. Foils of brass and copper are dissolved in nitric acid, brass react with the acid by liberating a brown poisonous gas as the foil dissolves.
After dissolution pale blue solution was obtained the pale blue colour was as a result of copper component in the brass, Page 7 of 9 the solution was diluted by adding 20ml of distilled water to it before transferred to 100ml volumetric flask. Zinc nitrate is colourless. Brass is an alloy of copper in brass and zinc which contains a small quantity of lead. Alloys are actually uniform mixtures of two or more solid substances. For example, brass is a solid solution of primarily copper and zinc. Alloys often possess properties which are unique and distinct from those of the individual components.
It is also true, however, that the individual solids still retain more or less the same individual properties. Thus, copper atoms in a brass sample have more or less the same atomic properties, such as ionization energy, as in a copper metal sample. Alloys can usually possess any ratio of the component solids, not just stoichiometric ratios as would be typically found in chemical compounds. Some brasses also have small amounts of other types of metals, such as tin. Brass’ properties make it easy to form and bend. Copper foils reacted in the nitric acid and liberated a brownish gas (NO2) gas as the foil completely dissolves.
After the dissolution a sea blue solution was obtained, the solution was diluted by adding 20ml of distilled water in the fumehood before sent to the laboratory bench. It was transferred into 100ml volumetric flask and was labelled as the standard copper solution. 10ml, 20ml and 30ml of the standard copper solution was measured into three different beakers, 30ml, 20ml and 10ml volume of distilled water was added respectively to each beaker. The various beakers and its content were sent to the spectrophotometric room for measurement.
The instrument had being switched on for about 60 minutes before measurement, the wavelength was set at 760nm. Distilled water was used as the blank solution to zero the machine. The absorbance of the various solutions were then recorded. A graph of absorbance versus concentration of copper is plotted. This graph is used to find the concentration of copper in the brass solution by tracing the absorbance to the concentration axis, since zinc nitrate is colourless. Mass of copper in the sample was calculated and was used to calculate the percentage of it in brass using spectroscopy.

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