Chapter 17 spectroscopy

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Transcript of Chapter 17 spectroscopy

Chapter 17-18: Spectrophotometry

Spectroscopy – the interaction of radiation and matter

Spectroscopic methods measure the amt of radiation produced or absorbedElucidation of molecular structureQual./Quant. Detn. of inorganic and organic compds

Classify

Region of the electromagnetic spectrumX-rayUVVisibleIR

Electromagnetic radiation – as a wave

Waves – properties of wavelength, frequency, velocity, amplitude

Particles – discrete packets of energy called photons

Important Equations

E = h

= c

V = 1/

E = hc/

E = hcV

c = 2.998 x 1010 cm/s h = 6.626 x 10-34 Js

UV: 180- 380 nmVis: 380 – 780 nmNear IR: 0.78 – 2.5 mFar IR: 2.5 – 50 m

Memorize

Memorize

wavenumber

What happens when a molecule absorbs a photon of light?

Energy increases

GS

ES

E = hc/

M + h M*

Energy absorbed = exactly the energy difference betweenthose states

Three Basic Transitions1- rotational (lower energy)2- vibrational3- electronic (higher energy)

Pure Vibrational IR regionPure Rotational Microwave

UV-Vis: move bonding (outervalence electrons)

* E large (<150 nm) n* (halogens, N, O, S) E smaller (=150-250 nm)* n* E small (=200-700 nm)

UV-Vis: move bonding (outer valence electrons)

GS

ES

E = hc/

M + h M*

Know this.

OrganicChromophores

From Skoog, West, Holler

Clicker questions

If E = 600 KJ/mol , what is the wavelength in nm?

If E = 160 KJ/mol , what is the wavelength in nm?

Hint: 6.023 x 1023 photons/mol

Instrumentation

BasisRadiation goes through the sample, certain frequenciesare removed via absorption

Plot A vs determine what frequencies are absorbed

A

Broad: why?

From Skoog, West, Holler

Why are lines narrower in vapor phase?

When radiation interacts with matter

1. Some transmitted through sample2. Some absorbed by the sample3. Some reflected at each surface4. Some scattered by dust….

Absorption - Transmission

P0 P

T = P/P0 %T = P/Po x 100%

A = -log T

A = 2 – Log %TMemorize

Quantitative Chemical Analysis

Beer’s Law: A = abc

a = absorptivityb = pathlengthc = concentration

If C has units of M and b has units of cm,A = bc

= molar absorptivity (M-1cm-1)

Memorize

Beer’s Law is additive:

1. Chemical

A. Dilute Solutions

B. Analyte dissociates/associates….HIn = H+ + In-

color 1 color 2

Limitations of Beer’s Law1. Instrumental2. Chemical

2. Instrumental

A. Polychromatic radiation

Beer’s law valid for monochromatic radiation

From Skoog, West, Holler

B. Stray lightScattered radiation… “stray”Usually a different and may not have passedthrough the sample

ExampleA solution contains 1.00 mg of K3Fe(CN)6 (FW 328.26)

in 100.0 mL. It transmits 70.0% of incident light comparedto a blank in a 1.00 cm cell. Calculate molar absorptivity?

Clicker Question:

Applications and Fluorescence

Qualitative AnalysisSupplemental to other techniques (lacks structure)Detect certain chromophoric groupCompare to other spectra

Quantitative AnalysisMolecule must absorb UV-Vis radiationBeers Law must be obeyed

Moderately sensitive, 10-4 – 10-6 MModerately selectiveGood accuracyRelatively easy, convenient, rapid

Details of Analysis

Properly select wavelengthProperly clean and handle sample cells

Chose standard solutions carefullyProper concentration and composition

Standard Addition Method

Match the overall composition of the sampleskeep the matrix constant

Impt for solns with complex composition

C

ASingle Point MethodMultiple additions

Extension of the calibration curve methodInvolves addition of known quantity of std to unk

Photometric Titrations

S + t = Ps t p

s = 0 t > 0 p = 0

A

Vol

Used to locate equiv. PointBeers Law must be obeyedCorrect A for volume changes

Know how sketch thesefor different species

S= substrate, analyteT = titrantP = product

Fluorescence

Emission processMolecules excited by absorption of electromagnetic radiation,lose excess energy via photon emission

Very sensitive (ppb)Limited number of compds fluoresce (aromatic)

M + h M*

M* M + heatM* M + h

Emits at a longer wavelength than it absorbs

Energy level Diagram Vibrational relaxationInternal conversionIntersystem crossingFluorescencePhosphorescence

Fluorescence

Know these terms…

Longer wavelength- Why?

Instrument componentsSource (mercury arc lamp)MonochromatorSample (right angles)Photomultiplier tube

900: why?

Quantitative Chemical AnalysisF = kC (at constant P0)Linear at low concentrations

Why drop in I ?

A compound with a molecular weight of 125.0 has a molar absorptivity of 2.5 x 105 M-1cm-1. How many grams of thiscompound should be dissolved in 1.00 L such that after a 200-fold dilution the resulting solution will give an absorbanceof 0.60 in a 1.0 cm cell

Clicker question

Cytochrome c has a molar absorptivity of 106,000 M-1cm-1. 100 uL of a solution cyt. C is diluted to 1.00 mL. The Absorbance of the diluted solution is 0.30 in a 1.0 mm cell. Calculate the concentration of cyt. C in the original soln.

A 2.00 mL specimen was treated with reagents togenerate color with phosphate following which thesample was diluted to 100.0 mL. Photometric measurement for the phosphate in a 25.0 mL aliquot yielded an absorbance of 0.428. Addition of 1.00mL of a solution containing 0.0500 mg of phosphateto a second 25.0 mL aliquot resulted in an absorbanceof 0.517. Calculate the mgs of phosphate in each milliliter of the specimen.

Clicker question

Quinine in a 1.664 g antimalarial tablet was dissolved insufficient 0.10 M HCL to give 500.0 mL of solution. A 15.00 mL aliquot was then diluted to 100.0 mL with the acid. Thefluorescent intensity for the diluted sample at 347 nm provideda reading of 288 on an arbitrary scale. A standard 100.0-ppmQuinine solution registered 180 when measured under identical conditions. Calculate the mgs of quinine in the tablet

Clicker question (fluorescence)