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Ch. 20 Spectrophotometry

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Ch. 20 Spectrophotometry What is Spectroscopy? • Spectroscopy: a general term for the science that deals with the interactions of various types of radiation (electromagnetic radiation and other forms of energy) Ch. 18 – Ch. 20 with matter Spectrophotometry • Spectrometry and spectrometric methods: -a large group of analytical methods that are based on atomic and molecular spectroscopy -the measurement of the intensity of radiation with a photoelectric transducer or other type of electronic device. • Colorimetry: based on absorption of visible light. Properties of Electromagnetic Radiation Properties of Electromagnetic Radiation Classical sinusoidal wave model Particle model • Plane-polarized electromagnetic radiation of wavelength λ, propagating along the x axis. • When EM radiation is emitted or absorbed, a • The electric field of the plane-polarized light is confined to permanent transfer of energy from the emitting object a single plane. or to the absorbing medium occurs. • Ordinary, unpolarized light has electric field components in all planes. • It’s necessary to treat EM radiation not as a collection of waves but rather as a stream of discrete particles υ× λ =c c =2.998 × 108 m/s called photons or quanta. • The discovery of the photoelectric effect in the late nineteenth century. •υ: frequency (Hz) E= hυ h =6.626 × 10−34 J ⋅ s •λ: wavelength •c: the velocity of E = hcυ~ υ~ =1/λ (wave number) radiation in vacuum Electromagnetic Spectrum Example: How much is the energy of CO2 increased per mole when it absorbs IR at 2300 cm-1? c ΔE = hυ = h = hcυ~ λ =(6.626 × 10-34 J ⋅ s)(2.998 × 108 m/s)(2300 cm1- )(100 cm/m) =4.6 × 10-20 J (4.6× 10-20 J)(6.022×1023 ) = 28 kJ/mol E = hcυ~ υ~ =1/λ (wave number) 1 A Spectrum of White Light Line Spectrum A spectrum produced by a luminous gas or vapor and appearing as distinct lines characteristic of the various elements constituting the gas. Hydrogen Line Emission Spectrum Emission Spectrum The spectrum of bright lines, bands, or continuous radiation characteristic of and determined by a specific emitting substance subjected to a specific kind of excitation. Ground State Excited State The state of least possible energy in a Being at an energy level higher than the physical system, as of elementary particles. ground state. Also called ground level. 2 Methods of Interaction Absorption Spectrum • Light shinning on a sample causes electrons to be excited from the ground state to an excited state • The wavelengths of that energy are removed from transmitted spectra M + hν → *M (excited state) → + heatM*M (relaxation) •Absorption of light increases the energy of a substance. •Emission of light decreases its energy. Types of Molecular Transitions Interactions of Radiation and Matter Electronic transition – electron between two orbitals (UV-vis). hν = Energy Difference of Orbitals Vibrational transitions – vibrational states associated with bonds holding molecule. Rotational transitions – changes in rotational states, about its center of gravity. Overall change in Energy with an orbital of a molecule is: E = Eelectronic + Evibrational + Erotational Interactions of Radiation and Matter Single-beam Absorbance Experiment Vibrational Irradiance energy Irradiance out transferred in Lamp or laser grating, prism or filter Pathlength Irradiance (P): the energy per unit area in the light beam (W/m2). Relaxation by emitting heat Transmittance (T): the fraction of original light not absorbed by the sample. T = P/P0 Monochromatic light: consists of “one color” (one wavelength). Relaxation by emitting photons Chromophore: the part of a molecule responsible for light absorption 3 Derivation of Beer’s Law Derivation of Beer’s Law P dP b ∫= − ∫ βcdx P0 P 0 After integrate both sides: P P dP: the change in irradiance ln= 2.303 × log = −βcb P P dP = −βPcdx β: a constant of proportionality 0 0 c: concentration P β dP P dP b log 0 =cb = ε cb −logT = A= ε cb = −βcdx ∫= − ∫ βcdx P 2.303 P P0 P 0 Absorbance and Beer’s Law Relation between Transmittance and • Absorbance (A): the amount light absorbed by the sample is related to transmittance (T): Absorbance ⎛ P0 ⎞ A = log⎜ ⎟ = −logT ⎛ P0 ⎞ P A = log⎜ ⎟ = −logT ⎝ ⎠ ⎝ P ⎠ • Beer’s law relates the absorbance of a chemical to its concentration: P/Po %T A Aλ= ε λ bc 1 100 0 b : the pathlength, typically in cm, and c is the 0.1 10 1 concentration of the chemical species in M 0.01 1 2 ε : the molar absorptivity, the unit that tells how much light is absorbed for a given wavelength. ε has units of M-1 cm-1 Absorption Methods - Beer’s Law Absorption Methods - Beer’s Law A = abc = εbc • The light being shined on A = abc = εbc the sample must be monochromatic (one color where a => absorptivity or wavelength) b => path length c => concentration A • The analyte must not be ε => molar absorptivity participate in a concentration dependent c equilibrium 4 Limitations to Beer’s Law Spectroscopic Procedure • Real Limitations –high 1.0 -1 • Have a spectrometer and cuvette(s) 1 - 10 mg l NO 3-N concentrated solutions, 0.8 r = 0.8791 concentrated electrolyte – Single-beam instrument has one sample holder (swap blank solutions (proximity and sample) 0.6 alters molecular Abs – Double-beam instrument splits light output between two 0.4 absorption). holders (measure blank and sample) 0.2 –A baseline spectrum is a spectrum of a reference solution 0.0 • Chemical Limitations – 0246810 (solvent or reagent blank) -1 absorbing species [NO -N] (mg l ) 3 participate in association or dissociation reactions, e.g. weak acids in HNO + H O ←⎯→ H O+ NO − 2 2 3 2 concentrated solutions, complexation. Spectroscopic Procedure Spectroscopic Procedure • Try to do an analysis at the λmax • Keep the absorbance reading of the sample below 1. – Sensitivity is greatest at maximum absorbance – % transmittance is related logarithmically with concentration – Curve is relatively flat in case the monochromator drifts (from 1-99% transmittance you can detect ~ 2 orders of and is off by a little in wavelength magnitude in analyte concentration) λmax.= 620 nm – Any orders of magnitude greater than that will be detected in 0.5 550 nm the range of 0-1% T. 0.4 0.3 • Dilute the solution so that the transmittance reading is 510 nm not maxed out in that region (for accuracy) 0.2 0.1 0.0 Absorbance -0.1 -0.2 350 400 450 500 550 600 650 700 Wavelength (nm) Example: Find the absorbance and transmittance of a Example: Consider a solution of benzene in hexane 0.0220 M solution with a molar absorptivity of 15.5 (28.8 mg in 250 mL solution). If the absorbance of M-1 cm-1 in a 2.00 cm pathlength cell your 1 cm cell is 0.266 at λ=200nm, what is the ε value for benzene (FW: 78.11)? -1 -1 Aλ=ε λ bc = (15.5 M cm )(2.00 cm)(0.0220 M)= 0.68 logT = −A 28.8mg C6H6 1g 1000ml 1mol C6H6 −0.68 c = T =10 = 0.21 250 ml sol’n 1000mg 1L 78.11 g C6H6 -3 ε=A /cb =A/c 21% of the irradiated light emerges from the analyte c=[C6H6]=1.321x10 M sample ε=0.266/1.321x10-3 5 Wavelengths and Color Example: A 22.5 mg sample of anthracene (C14H10, MW=178.23) was dissolved in 100 mL ethanol and diluted by 50% four times. The absorbance reading of a 1.0 cm cell of this solution at 266 nm was still 0.93. What is ε? 4 (0.0225 g)/(178.22 g/mol) ⎛ 1 ⎞ −5 []CH14 10 = ×⎜ ⎟ =7.89 × 10 M L 1.0 L ⎝ 2 ⎠ A 0.93 ε = = =1.2 × 104 M -1 cm− 1 bc (1.0)(7.89× 10−5 ) Analysis of Mixtures Isosbestic points Mixture = ε b[x ]+ε b [ Y ]+ ...... Abs x Y • Good evidence of presence of only 2 species A= ε ()()λ c b+ ε λ c b which interchange between themselves (e.g. 1 1 1 1 2 1 2 indicator dye with 2 states) A2= ε 1()()λ 2c 1 b+ ε 2λ 2 c 2 b •What are the knowns and unknowns in the above Eq.? •Measure Abs at more wavelengths than there are components in the mixture. •More wavelengths increase the accuracy. Emission Processes Absorption vs. Emission Spectra • Luminescence – emission of light from any excited state of a molecule. more sensitive than absorption measurements – Fluorescence – emission of photon during transition between states with same spin quantum SS #’s (e.g. 1 → 0 ) – Phosphorescence - emission of photon during transition between states with different spin TS quantum #’s (e.g. 1 → 0 ) 6 Absorption vs. Emission Spectra Emission Spectroscopy Experiment Emission Intensity at Low Concentration Components of Optical Instruments ⎛ P0 ⎞ A = log⎜ ⎟ = ε exbc ⎝ P ⎠ Single Beam Instruments ' −ε exb1 c Irradiance striking central region : PP0 = 010 ' ' −ε exb2 c Additional distance b2 : P= P0 10 Emission intensity is proportional to the irradiance '' '' absorbed in the central region of the cell: I= k( P0 − P ) Double Beam Instruments Emission intensity emerging from the cell: '' −ε emb3 c ' '' −ε emb3 c ' −ε exb1 c −ε exb1 c −ε exb2 c −ε emb3 c I= I10 = k ( P0 − P )10= k ( P0 10− P0 10 10 )10 ' −ε exb1 c −ε exb2 c −ε emb3 c = k P010 (1− 10 )10 (ε b c ln10)2 (ε b c ln10)3 10−ε exb2 c = 1 − ε b c ln10 + ex 2 − ex 2 + ... Q ex 2 2! 3! ' I = kPc ∴At low concentration : k P0ε ex b 2 cln10 ⇒ 0 Intensity of a Tungsten Filament and a Instrumentation - Basic Components Deuterium Arc Lamp (1) Light source -Tungsten filament lamp (Visible: 320 - 2500 nm); Deuterium arc lamp (UV: 200-400 nm); Nernst glower, globar (IR region).
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