Presentation on theme: "Spectrophotometry Spectroscopy is the study of"— Presentation transcript:
1 Spectrophotometry Spectroscopy is the study of interaction of spectrum of light witha substance to be analysed, for itsidentification (i.e qualitative analysis)as well as determination of itsamount (i.e quantitative analysis).
2 LightIt is one of the different forms of radiant energy, such as, sunlight, radio waves and X-rays .Thy are called electromagnetic radiations (EMR) due presence of both electric and magnetic components
3 Dual nature of lightLight exhibits wave property during its propagation and energy particle during its interaction with matter. The double nature of light (waves and particles) is known as dualism.1. Wave propertyEMR display the property of continuous waves and can be described by the characteristics of wave motion.Such wave motion in conveniently classified according to the wavelength.
4 Wavelength ()It is the linear distance measured along the line of propagation between crest of one wave to the next.Unit of LengthMicron () = 1 x10-6m = 1 x 10-4cm = 1x 10-3 mm.Millimicron (m) = or nanometer (nm)= 1 x 10-9m = 1 x 10-7 cm = 1 x 10-6mm.Angstrom (Ao) = 1 x 10-10m = 1 x 10-8cm = 1 x 10-7 mm.
5 Frequency  Relation between & = C / Wave number ( `) is the number of waves/ second or number of cycles occurring/ second (CPS) or Hertz (Hz) or waves vibration/ second.Relation between & C = = C / C= 3.0 x 1010 cm/swhere c is the velocity of light cm/s.Wave number ( `)is the number of waves/ cm`= 1/ Its unit is cm-1
6 2-Particle property (light as energy) Light consist of energy packets, known as photons.The energy (E) of photons is proportional to the frequency i.e related to c and . It can be expressed by max plank relation:E = h where h = max plank constant = 3.63 x erg., sec.)( = C / )i.e E or E 1/ Therefore energy of a beam of EMR increases as wave length decreases.
7 The shorter the wave length, the greater the energy of the photons and the more powerful the radiation.U.V range (200nm-400nm) which contain shorter , carrying more energy photon than a beam of visible range .Visible range (400nm-800nm) having high energy more than I.R range (>800nm).
9 Example:The λ of the sodium D line is 589nm . What are the frequency( ) and wave umber ( `)Solution: = C/λC=3.0 x 108 m/s λnm----λm (nm x 10-9 ) = 3x108m/s / 589nm x 10-9 = 5.09x1014 s-1`= 1 / λ (`= cm-1 )λ nm----- λ cm ( nm x 10-7 cm )`= 1 / 589 x10-7 = 1.7 x 104 cm-1
11 For analytical purposes we use the region of I.R, visible and U.V radiations .-UV radiation region is classified into :far UV from (10nm-200nm) and near UV from ( 200nm-380nm--Visible radiation region ( nm) consist of colored radiations,which are, violet, indigo, blue, green, yellow, orange and red .-IR radiation region ( 0.75um-1000um ) is classified into :near IR , mid IR and far IR .
12 Interaction of a substance with EMR When a molecule interact with radiant energyThe molecule is said to be excited , because the outer valence electrons undergo transition from original energy level ground state (E g) to an excited state (Es).Excited state EsGround state E gThe transition energy is given by the following equation :E = Es E g = h
13 When a molecule in the ground state absorbs EMR, 3 energy state transition will take place. These types of transition are :1) Electronic ) Vibrational ) Rotational .1-When the molecule absorb Visible and U.V region .Raising electrons to a higher energy level(Electronic transition energy )2-When the molecule absorb I.R region.Raising the Vibration of molecule(Vibrational transition energy)3-When the molecule absorb F.I.R and Microwave regions.Increasing rotation of the molecule(Rotational transition energy)
14 The total energy E total = E electronic + E vibrational + E rotational The relative value of E electronic : E vibrational: E rotational is :: : 10(1) Photons of Far IR and Microwave spectrum (E↓) cause only rotation of the molecules.(2) Photons of IR spectrum (E >E far IR and microwave ) cause vibration and rotation of the molecule(3) Photons of UV-Visible spectrum (E↑>EIR >E far IR) cause electronic transition which is accompanied by vibrational and rotational transion .
15 The relation between structure and energy absorbed The energy absorbed by a molecule in the ground state is dependent on the nature of the bonds within a molecule-The outermost electrons in organic molecules may be:strong (σ) bonds, weaker () bonds or nonbonding (n)1-Sigma () electrons: they are bonding electrons posses the lowest energy level ( the most stable single bonds).2-Pi () electrons :the bonding electrons constituting the pi bonds (the weaker double or treble bonds)3-Non-bonding (n) : electrons: don’t participate in bonding, they usually occupy the highest level of ground state (hetero atoms like N,O,S )
16 In the excited state ;- electrons occupy an anti bonding energy level denoted as * and the transition is termed -* transition.--electrons occupy the anti bonding * level, and the transition is termed - * .-while n electrons occupy either * or *,and the transition is termed n- * or n- *
17 absorb at <180nm in the far UV. -Compounds containing only -electrons ( →σ* transition) are the saturated hydrocarbons whichabsorb at <180nm in the far UV.-They are transparent in the near UV ( nm) making them ideal solvents for other compounds to be studied in this region.-n-electrons absorption in saturated compounds (n→σ* transition ) containing hetero atoms or halogens.-The majority of these compounds show no bsorption in the near UV,eg. Methanol at 177nm, tri ethylamine at 199nm or chloroform at 173nm.-Alcohols and ethers absorb at wavelength shorter than185nm and so they are useful as common solventsat > 200nm.
18 -- * transition in unsaturated compounds containing double or triple bonds are easily exited than in single covalent bonds.-Also , the energy required for - * in conjugated compounds is smaller than in non-conjugated ones.-So conjugated compounds absorb at longer wave length than un-conjugated one .-n→ * transition occur in compounds containingnon-bonding electrons adjacent to unsaturated centers(double or triple bonds)
19 Suggest which of these compounds absorb in UV-Visible spectrum ? Compound: Type of transition: Answer1) CH3-CH3 :2) CH2=CH2 :3) CH3-O-H :4) CH3-C=O :|CH3
20 Lecture III 1) Absorption Spectrum . 2) Chromophores and Auxochromes . 3) Bathochromic shift and Hypsochromic shift .
21 1) Absorption spectrumaccording to the electronic transition that occur in each organic molecule, absorption spectrum is obtained by plotting Absorbance (A) as a function of wavelength ().It has characteristic shape with the of maximum absorbance (max).It is characteristic for each molecule according to its structure and the type of transitional energyTherefore it is used for identification of a chemical substance(qualitative analysis). Also max is used for quantitative measurement, in order to increase sensitivity and to minimize error of the analytical method.
22 2) Chromophores and Auxochromes Chromophres :Are unsaturated groups responsible for - * and n→ * electronictransitions. e.g. C=C , C=O , N=N and N=O ( 200nm-800nm)3)Bathochromic & Hypsochromic shift-Bathochromic shift (or red hift)It is the shift of max to a longer wavelength due to substitution with certain functional groups (e.g. –OH and –NH2), when two or more chromophores are present in conjugation, change in pH and effect of the medium (solvent).- Hypsochromic shift (or blue shift)It is the shift of max to a shorter wavelength due to removalof conjugation by changing pH or polarity of the solvent.
23 - Hypochromic effect -Auxochromes - Hyperchromic effectan increase in the intensity of absorption usually due tointroduction of an auxochrome- Hypochromic effectIt involves a decrease in the intensity of absorption-AuxochromesAre saturated groups posses unshared electrons, and doesnot absorb in near UV or visible radiations e.g. OH,NH2.But when attached to chomophoric molecule, increase both itswave length and intensity of absorption maximum .Because auxochrome inters into resonance interaction with the chromophore , thus increase the extent of conjugation, shift the absorption maximum to longer wave length
24 Example 1): effect of conjugation on absorption spectrum Changes inAbsorption spectrumExample 1): effect of conjugation on absorption spectrumIncrease in conjugation, increas absorbance of light to higher , bathochromic shift with hyperchromic effect.
25 Lecture IV Effect of pH on absorption spectrum. 2)Polychromatic and Monochromatic light.3)Theory of light absorption
26 Example 2) Effect of pHThe spectra of compounds containing acidic (phenolic-OH) orbasic (-NH2) groups are dependent on the pH of the medium.The U.V spectrum of phenol in acid medium, benzenoid form while in alkaline medium is the phenate anion ,quinonoid formThe free pair of of electrons of O2 increasing the elocalization of the -electrons, leading to the formation of conjugated system.So , electrons become more energetic and need less energy to be excited, therefore absorb longer bathochromic shift ; red shift) with hyperchromic effectPhenol
27 Its spectrum exhibits bathochromic shift and hyperchromic effect in alkaline medium due to its conversion to the quinonoid speciesAnilineWhile in acid medium (anilinium ) lost the free pair electrons of N decrease the conjugation .Its spectrum in acid medium exhibit hypsochromic shiftand hypochromic effect due to its conversion tothe benzenoid species.
28 Polychromatic light Monochromatic light A beam of light containing several wavelengths , e.g. white lightMonochromatic lightA beam of light containing radiation of only one discrete wavelength
29 Therefore, under experimental conditions. Theory of light absorption :When a monochromatic light having intensity (Io) is allowed to pass through absorbing medium ;Some is absorbed (I a), reflected (I r), transmitted (It), refracted (If) and scattered (Is).Is = zero for clear solution, while If and Ir may be canceled by means of control cuvette containing the solvent in which the substance to be anaylsed is dissolvedTherefore, under experimental conditions.Io = Ia It or Ia = Io - It
31 % Transmittance %T = 100 x T Transmittance T = It / Io The diagram below shows a beam of monochromatic radiationof radiant power Io, directed at a sample solution. Absorptiontakes place and the beam of radiation leaving the samplehas radiant power I.% Transmittance %T = 100 x TTransmittanceT = It / IoAbsorbanceA = log10 I0 / ItA = log10 1 / TA = log10 100 / %TA = 2 - log10 %T
32 The relationship between absorbance and transmittance is illustrated in the following diagram: So, if all the light passes through a solution without any absorption, then absorbance is zero, and percent transmittance is 100%. If all the light is absorbed, then percent transmittance is zero, and absorption is infinite.Path length / cm0.20.40.60.81.0%T1005025220.127.116.11Absorbance0.30.91.21.5
33 log Io/It b (thickness ) lamberts’ law:When a monochromatic light enter absorbing medium, its intensity is decreased exponentially with the increase of thickness of the absorbing medium (i.e solution) (b) at constant concentration (C)log Io/It b (thickness )log Io / It = K bCbItCbLog Io/ItA
34 Beers’ law log Io/It C (concentration) log Io / It = K C When a monochromatic light enter absorbing medium, its intensity is decreased exponentially with the increase of concentration of the absorbing medium (C) at constant pathlength (b)log Io/It C (concentration)log Io / It = K CbCLog Io/ItbCItA
35 a: is a constant, known as absorptivity Beer-lambert’s law:Absorbance (A ) = log Io/Itlog Io/It b Clog Io/It b Clog Io/It = a b CA = a b Ca: is a constant, known as absorptivitywhich is the absorbance, when thickness of solution is unity (1cm) and concentration is unity (1gm/L)
36 = a x mol. wt a = / mol.wt Molar absorptivity or epsilon () If the unit of concentration is 1Mi.e. conc.= mol/L(a) is known as molar absorptivity or epsilon ()As, a=A/b c =A/1cm . gm/L = L.gm-1.cm-1molar absorptivity , =A/b c= A/1cm.mol/LIts unit is L mol-1 cm-1(mol=gm/ mol. weight)i.e. Its unit = L.MWt.cm-1gm-1. = a x mol. wta = / mol.wt
37 = A (1% - 1cm) x mol. Wt / 10 A (1% - 1cm): If unit of concentration is 1% ( i.e. 1gm/100 ml )A (1% - 1cm) = A/ 1cm.gm% = 100. cm-1.gm-1.a= L.gm-1. cm-1A (1% - 1cm) = a x 10a= A (1% - 1cm) / 10(a = / mol. Wt= A (1% - 1cm) / 10 ) = A (1% - 1cm) x mol. Wt / 10A (1% - 1cm) = x 10 / mol. Wt
38 ε = A 1% 1cm x M.wt / 10 = 1030 x 147 / 10 = 15141 litre.cm-1.mol-1 Both and A 1%, 1cm are characteristic for each substance at the same max , pH and type of solvent and are used for quantitative purposeProblem 1: A sample solution give absorbance equals 0.6 ,A1% 1cm = 1030 and its molecular weight = 147.Calculate the sample concentration in Mol/LSolution 1:ε = A 1% 1cm x M.wt / 10 = x 147 / 10= litre.cm-1.mol-1A =ε b c = x 1 x C =15141 x CC = 0.5 / = x 10-5 mol / L
40 Isosbestic point-At different pH, the spectrum will be shifted to different maxbut all spectra intersect at certain which is known as isosbestic point- At isosbestic point, the same absorbance is given for the same concentration at different pH,i.e. absorbance is not pH dependent but concentration dependent-Thus solution ; its max affected by pH , must be buffered at specific pH or measurements are carried out at the isosbestic point.
41 ColorimetryWhen white light passes through a colored substance, a characteristic portion of the mixed wavelengths is absorbed.The remaining light will then assume the complementary color to the wavelength(s) absorbed.e.g orthophenanthrolene which reacts with ferrous (Fe2+) in buffered medium (acidic pH) to produce intense red colour.Complementary colors are diametrically opposite each other. Thus, absorption of nm light renders a substance yellow, and absorption of nm light makes it red.
42 wavelength region, nmcolorcomplementary colorVioletYellow-greenBlueYellowBlue-greenOrangeGreen-blueRedGreenPurpleColored substances appear colored because they selectively absorbed some of wavelengths of visible light and transmitted other wavelengths or colors (apparent color),Red substances absorb the blue- green wavelengths from the visible region, so the transmitted light appears redBlue substances absorb the yellow wavelengths, so the transmitted light appears blue.
43 Requirements for substances to be measured colorimetricaly: Substance must be coloured e.g CuSO4, organic dyes,….2-If the substance to be analysed is colourless, it must react with certain reagent (known as chromogen) to produce equivalent coloured product.3-If there is no suitable chromogen, the substance must be converted to a certain derivative which has a suitable chromogen.
44 Quiz I)Discuss shortly -Interaction of a substance with EMR -Factors affecting absorption spectrumII) Solve the following problem1-A 5.00x10-4M sample solution is measured in a cell with 1 cm bath length ; its absorbance at 592nm equalsa-What is the molar absorptivity at 582nm.If a solution of unknown concentration of the same sample has an absorbance at the same wave length.b-What is its concentration
45 III) Complete the answer in Exercise 1 2-Calculate the wave legnth in um , and in Angestron3-Calculate the frequancy and energy of this wave lengthIII) Complete the answer in Exercise 1
46 2-If the substance to be analysed is colourless, it must react with certain reagent (known as chromogen) to produce equivalent coloured product.Orthophenanthrolene reacts with ferrous (Fe2+)in buffered medium (acidic pH) to produce intense red color.2+3
47 3-If the sample is colorless and there is no suitable chromogen, the substance must be converted to a certain derivative which can be react with suitable reagent producing color .-Esters are first converted to hydroxamic acid derivative through the reaction with hydroxylamine. Hydroxamic acid derivative gives purple color on addition of ferric (Fe3+) due to the formation of iron chelate .OOR –C– O Et+ H2N – OHR –C– NH – OH+ Et OHHydroxamic acid derivative + Fe3+then measuring the absorbance at 520nm.
48 Requirements for ideal chromogen Chromogen isa compound containing chromophoric groupRequirements for ideal chromogen1-Should be colorless or easily separated from the colored product2-It Should be selective.3-Its reaction to produce colored product, should be of known mechanism and proceed stoichiometrically.4-The full development of color must be rapid.5-It must produce only one color of specified max.
49 Requirements for coloured product 1-Should be of intense color, to increase the sensitivity2-Should be unaffected by pH or the pH must be specified and maintained by suitable buffer or the measurement is carried out at of isosbestic3-Should be stable with time4.The reaction of its formation, must be rapid and quantitative.5-The colored product, should obey Beer-lambert’s law, i.e on plotting A versus C at fixed b, we obtain straight line passing through the origin.
50 InstrumentationThe instrument used, usually consists of 5 basic components1-Radiant energy source.2-Dispersing system (or monochromator).3-Sample compartment (cuvette).4-Detector5-Recorder (meter).
51 1. Radiant energy source (source of light) In visible rangeIn U.V rangeTungsten lampDeuterium lamp (D2)(or hydrogen lamp.)2. Dispersing system (monochromator)It convert polychromatic light to monochromatic light (definite range of .)a) Filtersb) PrismsC) Gratingact byselective absorption of unwanted and transmit the complementary color, which is needed to be absorbed by the sample to be analysed.Filters may be :gelatin, liquid and tinted glassDisadvantages:Filter transmit a wide band of nm. which is not exactly monochromatic and used only in visible range .
52 The dispersion power of the prism is ∝ 1/ b) PrismsAct by refraction of lightIn visible rangeIn U.V rangeglass prismquartz or fused silica prism.The dispersion power of the prism is ∝ 1/ .As they are transparent to UV lightCannot be used for UVas it absorb all UV lightThe dispersion power of the prism is ∝ 1/ i.e. angle of refraction ↑ as ↓
53 The grating disperses the light beam into almost single . C) GratingGrating consists of a large number of parallel lines ruled very close to each other on a highly polished surface e.g aluminum or aluminized glass (600 line/mm).Each ruled groove functions as a scattering center for light rays falling on its edge.The grating disperses the light beam into almost single .
54 It works in both UV and visible spectral region by dispersion. The scattering power of a grating is ↑ as the number of groves ↑. The resolution power of grating is > prism > filter.d) Associated opticsare used to control light intensityCollimating lensesslit of variable widthmirrors and diafragmshelps to narrow band widthalignment of the beam.act as condensersThey should be suitable for the spectral range, i.e glass for visible range and quartz or fused silica for U.V range.
56 3.Sample compartment (Cuvette) Transparent surfaceOpaque surfaceglass for visible rangeIt is made of →quartz or fused silica for U.V range.4. Detectora) photocell (Photovoltaic cell) e.g Barrier layer cellb) Phototube (photomultiplier or photoemissive tube)
57 a) photocell (Photovoltaic cell) e.g Barrier layer cell Light falling on cellTransparent metal layer of Ago(Collecting electrode)Photosenitive semiconductor of seleniumMetal base Plate of iron-+
58 b) Phototube (photomultiplier or photoemissive tube)
59 5. Recorder (meter) Electric signal produced in detector is fed to a sensitive galvanometer,its scale is graduated in absorbance or/and transmittance units.Commercial instruments1. Filter photo-electric colorimeter
60 2. Compensating two-photocell colorimeter In this type, fluctuation in intensity of EMR source is automatically cancelled.