1. 11 Instrumental Analysis Tutorial 1

2. 22 Use mathematical formulae to calculate absorbance, transmittance of a sample and wave parameters. Determine factors affecting absorption of radiation. Identify exceptions to Beer’s law. Apply Beer’s law in chemical analysis Define complementary colors. Describe other wavelength selectors and detectors. Objectives

3. 33 Energy of a photon in joule Beer’s Law A= ε b c Absorbance A = – log P/P o = – log T Transmittance% Transmittance E = h = hc/λ =hc 10 6 photons 500 nm 0.7x10 6 photons Remember that

4. 44 For the UV radiation, the increase in energy is: ΔE = h = h c/ λ = (6.626 x 10 -34 J.s) (2.998 x 10 8 m/s) / (147 nm)(10 -9 m/nm) = 1.35 x 10 -18 J / molecule. (1.35 x 10 -18 J/molecule) (6.022 x 10 23 molecules/mol) = 814 kJ/mol This is enough energy to break the O=O bond in oxygen. For CO 2, the increase in energy is: ΔE = h = hc/ λ =hc (Recall that =1 / λ ) = (6.626 x 10 -34 J.s) (2.998 x 10 8 m/s)(2300 cm -1 )(100 cm/m) = 4.6 x 10 -20 J/molecule = 28 KJ/mol. Infrared absorption increases the amplitude of the vibration of the CO 2 bond. Photon energies By how many kilojoules per mole is the energy of O 2 increased when it absorbs ultraviolet radiation with a wavelength of 147 nm? How much is the energy of CO 2 increased when it absorbs infrared radiation with a wavenumber of 2300 cm -1 ?

5. 55 1- Fill in the blanks a- If you double the frequency of electromagnetic radiation, you __________ the energy. b- If you double the wavelength, you __________ the energy. c- If you double the wavenumber, you __________ the energy 2- How many kilojoules are carried by one mole of photon of violet light with = 400 nm? 3- Calculate the frequency (in hertz), wavenumber (in cm -1 ), and energy (in joules per photon and joules per mole of photons) of visible light with wavelength of 562 nm. 4- Which molecular process correspond to the energies of microwave, infrared, visible and ultraviolet photons? Exercise 1:

6. 66 Beer’s Law A = b c Path Length Dependence, b Readout Absorbance 0.00 Source Detector

7. 77 Beer’s Law A = b c Path Length Dependence, b Readout Absorbance 0.22 Source Detector b Sample

8. 88 Beer’s Law A = b c Path Length Dependence, b Readout Absorbance 0.44 Source DetectorSamples Of course, we are not introducing two cells in the light pathway, but let us assume that we doubled the path length of light through the absorbing medium

9. 99 Beer’s Law A = b c Path Length Dependence, b Readout Absorbance 0.66 Source DetectorSamples

10. 10 Beer’s Law A = b c Concentration Dependence, c Readout Absorbance 0.00 Source Detector

11. 11 Beer’s Law A = b c Concentration Dependence, c Readout Absorbance 0.42 Source Detector b Sample

12. 12 Beer’s Law A = b c Concentration Dependence, c Readout Absorbance 0.63 Source Detector b Sample

13. 13 Beer’s Law A = b c Wavelength Dependence, Readout Absorbance 0.30 Source Detector b The blue solution do absorb the red radiation

14. 14 Beer’s Law A = b c Wavelength Dependence, Readout Absorbance 0.00 Source Detector b The red solution can not absorb the red radiation but it can absorb radiation that is complimentary to red.

15. 15 It is clear that absorbance of a sample depends on: 1.Type and nature of sample. 2.Concentration of the absorbing species. 3.Path length of light through the absorbing medium. 4.Wavelength of radiation (the energy used for stimulation) If the effect of any of these factors is sought to be studied, the other factors should be kept constant. In order to study the effect of changing the concentration of the absorbing species on absorbance, wavelength of radiation as well as the path length of the light through the same absorbing medium should be constant. Thus Beer’s Law is obeyed only when monochromatic radiation is used (fixed ).     

16. 16 When Beer’s law fails? Beer’s law: Absorbance α concentration It applies to monochromatic radiation and works very well for dilute solutions (<0.01 M). Deviation from Beer’s law: 1- At high concentrations: the proximity of molecules to each others alters their absorptivity. 2- If the absorbing molecule participates in a concentration-dependent chemical equilibrium: HA A - + H + 3- The monochromator generally fails to choose only one λ, instead it projects a very narrow band on the sample (recall that Beer’s law is applied at one certain λ).

17. 17 (a)Find the molar absorptivity of benzene at this wavelength Solution: The concentration of benzene is: [ C 6 H 6 ]= {(0.0258 g) / (78.11 g/mol)} / 0.250 L = 1.32 x 10 -3 M We find the molar absorptivity form Beer’s law: Molar absorptivity = ε = A / b c = (0.266) / (1 cm)(1.32 x 10 -3 M) = 201 M -1 cm -1. (b) A sample of hexane contaminated with benzene had an absorbance of 0.070 at 256 nm in a cuvet with a 5 cm path length. Find the concentration of benzene in mg/L. Solution: Using Beer’s law with the molar absorptivity form part (a), we find [ C 6 H 6 ] = A / ε b = 0.07/ (201 M -1 cm -1 )(5 cm) = 6.9 x 10 -5 M. [ C 6 H 6 ] = (6.9 x 10 -5 mol/L)(78.11 x 10 3 mg/mol) = 5.4 mg/L Pure hexane has negligible UV absorbance above a λ of 200 nm. A solution, prepared by dissolving 25.8 mg of benzene in hexane and diluting to 250 mL, had an absorption peak at 256 nm and an absorbance of 0.266 in a 1 cm cell. Beer’s law in chemical analysis

18. 18 Solution: (a)A = – log P/P o = – log T = – log (0.45) = 0.347. (b)Absorbance is proportional to concentration, so the absorbance will double to 0.694 T = 10 -A = 10 -0.694 = 0.202 %T = 20.2% Example 1 a)What value of absorbance corresponds to 45% T? b) If a 0.01 M solution exhibits 45% T at a certain wavelength, what will be the percent transmittance for a 0.02 M solution of the same substance?

19. 19 Solution: ε = A/c b = (0.624 – 0.029) / (3.96 x 10 -4 M)(1 cm) = 1.5 x 10 3 M -1 cm -1. Example 2 A 3.96 x 10 -4 M solution of compound A exhibited an absorbance of 0.624 at 238 nm in a 1 cm cuvet; a blank solution containing only solvent had an absorbance of 0.029 at the same wavelength. Find the molar absorptivity of compound A. Value of absorbance corrected for blank

20. 20 Example 3 At 480 nm, a 4.0x10 -5 M solution of [FeSCN] 2+ has a transmittance of 50% when measured in 1.0 cm cell. What is the transmittance when measured in 4.0 cm cell? Solution: At first you have to convert Transmittance to absorbance A 1 = – log P/P o = – log T 1 = – log (0.5) = 0.30 Second, relate absorbance to pathlength T 2 = 10 –A 2 = 10 –1.2 = 0.063 = 6.3% Absorbance is the parameter that is linearly related to concentration, as well as pathlength, but not transmittance. Thus, it is not correct to assume that: Finally convert back from absorbance to the required transmittance

21. 21 Molecule absorbs a certain range of  of visible light  we see non-absorbed colors. The absorbed  are the  complementary wavelengths (complementary colors) of Max. Absorption 380-420 420-440 440-470 470-500 500-520 520-550 550-580 580-620 620-680 680-780 Color Absorbed violet violet-blue blue blue-green green yellow-green yellow orange red purple Color Observed green-yellow yellow orange red purple violet violet-blue blue blue-green green Complementary Colors

22. 22 Absorbance by Molecules: Carotene Carotene is the molecule responsible for the colour orange in carrots. It absorbs light in the blue (400-430 nm) and green (430-500 nm) ranges. Carotene transmits or reflects longer wavelengths, and appears orange. 400 430 480 560 630 590 750 Absorbance

23. 23 Exercise 2: 1- Explain the difference between transmittance, absorbance, and molar absorptivity. Which is proportional to concentration? 2- Describe, using diagram, what happens when a molecule absorbs UV-Vis radiation. 2- What is absorption spectrum? 3- Why does a compound whose visible absorption maximum is at 480 nm (blue-green) appear to be red? 4- why is it most accurate to measure the absorbance in the range of A = 0.2–1.5? 5- A compound (M. Wt = 180) absorbs 65% of the radiation at certain wavelength in a 1.0 cm cell at a concentration of 15  g/mL. Calculate its molar absorptivity.

24. 24 Wavelength selector – Filters Filters permit certain bands of wavelength (bandwidth of ~ 50 nm) to pass through. The simplest kind of filter is colored glass, in which the coloring species absorbs a broad portion of the spectrum (complementary color) and transmits other portions (its color) Disadvantage: They are not very good wavelength selectors and can’t be used in instruments utilized in research. This is because they allow the passage of a broad bandwidth which gives a chance for deviations from Beer’s law and allows interference with impurities.

25. 25 Wavelength selector – Grating Example: An echellette grating that contains 1450 grooves/mm was irradiated with a polychromatic beam at an incident angle 48 deg to the grating normal. Calculate the wavelengths of radiation that would appear at an angle of reflection of 20 and 10 deg. Solution:  To obtain d (constant spacing between grating’s grooves):  For reflection angle 20 deg: n = d (sin  i + sin  r ) Calculation of the first-order reflections are shown in the table Wavelength (nm) n =1  r (deg) 748 632 20 10 ii rr d

26. 26 Bandwidth Choice The size of the monochromator exit slit determines the width of radiation (bandwidth) emitted from the monochromator. A wider slit width gives higher sensitivity because higher radiation intensity passes to the sample but on the other hand, narrow slit width gives better resolution for the spectrum. In general, the choice of slit width to use in an experiment must be made by compromising these factors. Still, we can overcome the problem of low sensitivity of the small slit by increasing the sensitivity of the detector. What are the advantages and disadvantages of decreasing monochromator slit width?

27. 27 Detector Photodiode Array Spectrophotometers Ordered arrangement of photodiodes Multiple wavelengths detected simultaneously –no need to select single for analysis Sample Source Grating Polychromator Photodiode Array white light

28. 28 Advantages unique to array detectors 1- Fast spectral acquisition (in a fraction of a second). Dispersive instruments require several minutes. 2- No wavelength selection needed, all wavelength are recorded simultaneously. In a dispersive instrument only one narrow band reaches the detector at any time. 3- No source intensity lost because we do not use slits as in conventional spectrophotometers.

29. 29 1.Draw a schematic diagram of a double beam scanning spectrophotometer describing briefly the role of each component in the spectrophotometer. 2.Would you use a tungsten lamp or a deuterium lamp as a source of 300 nm radiation? 3.A solution with A = 0.76 (3.0 cm cell) is doubled in concentration and measured in a 2 cm cell; calculate the absorbance of the measured solution. 4.A solution of Cu(II) ion prepared by dissolving 0.4 g of CuSO 4.5 H 2 O in 100 ml distilled water has an absorbance of 0.576 at 790 nm in a 3 cm cell. Calculate the molar absorptivity of [Cu(H 2 O)] 2+ complex. (Atomic masses: Cu = 63.5; S = 32; O = 16; H = 1). Assignment 1