Q1. Potassium chloride can be used instead of salt by people suffering from high blood pressure. Suppose, while cooking, someone spilt some potassium chloride in the flame of a gas stove. a What colour would the flame be? b Suggest why the presence of chloride ions has no effect on the flame colour. A1. a lilac b The energy required to promote an electron in a chlorine atom is much higher than for the potassium atom. As a consequence, the flame of a Bunsen burner excites relatively few chlorine atoms. Furthermore, energies emitted as electrons in the excited atoms return to their lowest energy levels are mainly outside the energies of visible light. Q2. Barium and calcium are both Group II elements. Account for the observation that samples of barium and calcium compounds produce different colours when they are held in a flame. A2. As a consequence of the presence of different numbers of protons in the nuclei of barium and calcium atoms, the energies of electrons in the shells of the two atoms are not the same. Different amounts of energy are released when an electron moves from a higher energy level to a lower one in a barium atom compared with when the same process occurs in a calcium atom. These different energies are seen as light of different colours. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1
Q3. Why does an emission spectrum contain a number of lines of different colours? A3. When atoms absorb energy, it is often possible for electrons to be promoted to various higher energy levels. Electrons can return to the ground state from these excited states by undergoing a number of transitions of different energy. (This is shown in the diagram.) Each transition results in a line of specific energy in the emission spectrum. Q4. Explain why atomic emission spectroscopy is regarded as a superior method of analysis to flame tests. A4. Atomic emission spectroscopy (AES) is regarded as superior to flame tests because, in AES, more elements produce emission spectra in the hotter flame, and a much smaller quantity of an element needs to be present in order for it to be detected. Using AES, the amount of the element present in a sample can be accurately determined, even in the presence of larger amounts of other elements that would mask the flame colour of the element to the naked eye. E1. What are the advantages of using inductively coupled plasma rather than a flame to excite atoms in an atomic absorption spectrometer? AE1. At the high temperature in the plasma all atoms in a sample are excited and emit electromagnetic radiation when they return to the ground state. ICP can be used to identify most elements. The technique is suitable for concentrations as low as ppb. It can also identify many elements at once and is very fast. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 2
E2. Plasma is often referred to as the fourth state of matter. Conduct a web search to find out how the plasma state differs from the other states of matter. AE2. At high temperatures in plasma atoms lose one or more electrons. Plasma is a gas containing many positively charged ions and free moving electrons. Plasma conducts electricity and responds to magnetic fields. The properties of plasmas are determined by the interaction between neutral and charged particles. Plasma TV screens and fluorescent light tubes are examples of everyday applications of plasma technology. It has been estimated that 99% of the visible universe is in a plasma state. Q5. Iron is essential to our health. To determine the iron content in a Milo milk drink, a 5.0 ml sample was diluted to 50.0 ml. The absorption of the diluted solution and of several standard solutions was measured using AAS. The results are shown in Table 7.3. Table 7.3 Atomic absorption spectroscopy measurements Solution concentration (ppm) Absorbance 0.00 0.010 1.00 0.080 2.00 0.150 3.00 0.220 4.00 0.290 Sample 0.190 a b c d e f g Plot a graph of absorption against concentration of iron. What is the concentration of iron, in ppm, in the diluted Milo? Calculate the concentration of iron, in ppm, in the undiluted Milo. What mass of iron would you consume by drinking a 250 ml glass of Milo? The recommended daily allowance (RDA) of iron for people over the age of 11 years is 18 mg. What percentage of your daily needs does a 250 ml glass of Milo provide? The 0.00 ppm standard, which contained no added iron, gave a small absorption reading. Suggest an explanation for this. Suggest why the sample of Milo was diluted in order to measure its absorption. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 3
A5. a b 2.57 ppm c The 5 ml sample was diluted to 50 ml. This is a factor of 10. c(fe) in the undiluted Milo = 25.7 ppm d Step 1 Rewrite the concentration in ppm. 25.7 ppm means 25.7 g Fe per 10 6 g Milo Step 2 Assume the density of Milo is 1 g ml 1 and calculate the mass of Fe in 250 g Milo. 25.7 g Fe per 10 6 ml Milo =? g Fe per 250 ml Milo. m(fe) 250 = 25.7 6 g 10 = 6.4 10 3 g (two significant figures) e % Fe 6.4 mg = 100% 18 mg = 36% f The water used to prepare the standards may have contained traces of iron. g If it had not been diluted, the absorbance reading for the Milo would have been outside the range of the calibration curve. The liquid might also have been too thick to spray into the flame. Q6. Two samples of copper sulfate solution, with concentrations of 0.080 M and 0.30 M, were analysed using a spectrophotometer. a Which sample concentration would allow most light to pass through to the detector? b Which sample concentration would show the strongest absorption of the light? A6. a 0.080 M solution b 0.30 M solution Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 4
Q7. Why would red light be used in a colorimeter to measure the concentration of a blue copper sulfate solution, rather than blue light? A7. Copper sulfate is blue because it transmits blue light and absorbs light of other frequencies. Since a colorimeter measures the amount of light absorbed by a sample, light of a colour other than blue must be used when measuring the concentration of a copper sulfate solution. Q8. Compare the analytical techniques of AAS and UV visible spectroscopy. In what ways are they: a similar? b different? A8. a Both involve the absorption of electromagnetic radiation of particular wavelengths characteristic of the substance under investigation. The amount of radiation absorbed is measured by a detector. The amount of light absorbed is proportional to the amount of the light-absorbing substance in the sample. b Atomic absorption spectroscopy The sample is sprayed into a flame and light of a particular wavelength is passed through the flame. Atoms of the element being analysed absorb some of the radiation and the amount of light absorbed indicates the amount of the element present in the sample. AAS can be used to detect most metals UV Spectroscopy Two lamps are used to produce radiation covering the visible and UV spectrum. The light is passed through a prism to produce the desired wavelength; for UV analysis a silica sample holder is used; UV spectroscopy can be used to analyse many colourless compounds as well as coloured ones; it can give information about the structure of a substance. UV visible spectroscopy can be used to determine the concentration of atom ion and molecules include complex organic substances. Q9. The absorption spectrum of chlorophyll is shown in Figure 7.16. a b At what wavelengths is there maximum absorbance of light? What wavelength would you select if you were required to determine the concentration of chlorophyll in a leaf extract using UV visible spectroscopy? Provide an explanation for your answer. A9. a There are two strong absorption peaks at 420 nm and 660 nm. b Either 660 nm or 420 nm wavelengths could be used as chlorophyll absorbs strongly at both. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 5
Q10. The absorption spectrum of a commercial dye is shown in Figure 7.19. What is the colour of the dye? Figure 7.19 Absorption spectrum of a dye. A10. The dye absorbs in the yellow and red regions of the visible spectrum and transmits light in the violet-blue region. The dye would be a violet-blue colour. Q11. The absorption spectrum of benzene is shown in Figure 7.20. Figure 7.20 Absorption spectrum of benzene. a In what part of the electromagnetic spectrum does benzene absorb? b Explain why humans see benzene as colourless. A11. a Benzene absorbs in the ultraviolet region of the electromagnetic spectrum. b The human eye only detects electromagnetic radiation in the visible region of the electromagnetic spectrum. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 6
Q12. The phosphate content of a detergent may be analysed by UV visible spectroscopy. In one analysis, a 0.250 g sample of detergent powder was dissolved in water and the solution made up to 250 ml. The solution was treated with a small volume of sodium molybdate solution to form a blue-coloured phosphorus compound. The absorbance of the solution at a wavelength of 600 nm was measured as 0.17. The absorbances of five standard phosphate solutions were measured in a similar fashion and the calibration graph shown was obtained. a b c What is the concentration of phosphorus in the 250 ml detergent solution? Determine the percentage by mass of phosphorus in the detergent powder. Why was a wavelength of 600 nm selected for this analysis? A12. a 32 mg L 1 b Step 1 Using the answer from part a, calculate the mass of P in the 0.250 g sample of detergent powder, assuming that the density of the solution is 1 g ml 1. 32 mg L 1 is 32 mg per litre (1000 ml) =? mg per 250 ml solution m(p) 250 = 32 1000 = 8.0 mg Step 2 Calculate the percentage of P in the 0.250 g sample of detergent powder. % P 8 mg = 250 mg 100% = 3.2% (two significant figures) c Orange light of wavelength 600 nm is strongly absorbed by a blue solution. Q13. Consult Table 7.8 on page 93 of the student book. What bond is undergoing a vibrational change in ethanol at 2950 cm 1? A13. C H stretching Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 7
Q14. Look at the IR spectrum and structural formula of propanone (Figure 7.36). Use Table 7.8 on page 93 of the student book to identify Peak A and Peak B. Figure 7.36 Infrared spectrum of propanone. A14. Peak A: C H stretching Peak B: C=O stretching Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 8
Q15. Ethanol belongs to a group of organic compounds called alkanols (alcohols). Its IR spectrum is shown in Figure 7.29. The broad absorption between 3200 cm 1 and 3600 cm 1 is due to the O H group of atoms common to all alcohols. The absorption just below 3000 cm 1 is caused by the C H stretch while the C O stretch is at 1050 cm 1. IR spectra of two liquids, A and B, are shown in Figure 7.37. Figure 7.37 IR spectra of sample A and sample B. a Explain how these IR spectra confirm that both samples belong to the alkanol group of compounds. b Are A and B samples of the same compound? Explain your answer. A15. a The IR spectra of both samples A and B shows broad absorption between 3300 cm 1 and 3400 cm 1 indicative of the presence of the O H bond in alkanols. The absorption at 2950 cm 1 is indicative of C H bonds which occur in all organic compounds. That at about 1000 cm 1 is indicative of C O bond. However specific troughs in this region are often difficult to pick out. b Although the absorptions at 3300 cm 1 and 3000 cm 1 are similar indicating similar types of bonds the pattern of absorptions below 1500 cm 1, the finger print region, are different indicating that samples A and B belong to different members of the alkanol family. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 9
Q16. Explain how radio waves interact with the nucleus of an atom. How is this phenomena applied to NMR? A16. Atomic nuclei produce a magnetic field as a result of proton and neutron spin. The orientation of these magnetic fields depends on the direction of spin. Where the number of protons plus neutron is an odd number the nucleus will have an over all spin. The magnetic fields generated by nuclei will align with or against an external field. Nuclei aligned against the external filed are at a higher energy level than those aligned with the external field. Electromagnetic radiation in the radio wave frequencies range causes nucleus to move to a higher energy level. The external magnetic field experienced by an atom within a molecule is modified by the magnetic field of neighbouring atoms. The energy required to excite a nucleus of an atom to a higher energy level depends on the atoms environment or the arrangement of atoms within a molecule. The peaks in a NMR spectrum represent different chemical environments and the area under the peaks indicates the ratio of atoms in each environment. Chemical shift is the energy required to change energy states and is measured against the absorption produced by a standard which is assigned an arbitrary zero. The chemical shift is characteristic of an atom s environment. The peaks in high resolution proton NMR are split into a cluster of peaks. The amount of splitting provides further formation about the number of hydrogens attached to adjacent carbon atoms. The interactions between protons on adjacent atoms that cause the peaks to split is called spin coupling. The interaction of an external magnetic field with the magnetic field of the nucleus of an atom is used in nuclear magnetic resonance spectroscopy to determine the identity and structure of a compound. Analysis of an NMR spectrum involves consideration on the number of peak sets and their chemical shift, the relative area under each peak set and peak splitting. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 10
Q17. a Samples analysed by nuclear magnetic resonance are not dissolved in water but instead dissolved in CD2Cl2 or D2O or similar solvents. Explain why water is not used. b Explain why tetramethylsilane is used a standard to set the zero point in a NMR spectrum c What is meant by the term chemical shift? How are chemical shifts used in NMR? d What does the area under each peak indicate? A17. a Water, with two hydrogen atoms, gives a strong signal in proton NMR and so would mask the response of the sample. b TMS produces a strong,single absorption signal well away from the signals produced by other compounds. c Chemical shift is the position of signals relative to those produced by TMS. It is a measure of the gap between low and high energy states. d The area under the peaks indicate the relative number of atoms generating each peak. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 11
Q18. Ethyl ethanoate is a solvent commonly used in nail varnish remover. The proton NMR and semistructural formula of ethyl ethanoate is shown in Figure 7.51. To answer this question you may need to refer to the chemical shift data in Table 7.9 on page 98 of the student book. 3 Chemical shift (ppm) Figure 7.51 The proton NMR spectrum of ethyl ethanoate. a How many hydrogen or proton environments are there in ethyl ethanoate? b Refer to Table 7.9 and identify the expected chemical shift for the groups present in the ethyl ethanoate molecule. c What is the relative number of protons in each peak set: A, B, and C? d Explain why: i peak set A is split into a quartet (four peaks) ii peak set B is a single peak iii peak set C is a triplet. e Identify the protons responsible for each peak set and the carbon atoms to which they are bonded. f How many peaks would you expect in the 13 C NMR spectrum of ethyl ethanoate? Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 12
A18. a There are three peak sets, each representing a different proton environment. b From Table 7.9, the expected chemical shifts are: COCH 3 2.0 2.9, RCH 3 0.7 1.7, OCH 2 R 3.3 4.3. c Relative number of protons for the peak sets A : B : C = 2:3:3 d Proton NMR peaks are split due to interaction of the magnetic fields on adjacent atom. The number of peaks associated with each proton environment is shown in the figure in part b. Using the n+1 rule: The quadruple peaks at A indicate that there are three hydrogens attached to an adjacent atom. The single peak at B indicates that there are no hydrogens attached to the adjacent atom. The triplet at B indicates that there are two hydrogens attached to the adjacent atom. e A = CH 2 group, B = methyl group of ester, C = methyl group of CH 3 CH 2. f There will be four peaks in the 13 C NMR spectrum as there are four different carbon environments in the ethyl ethanoate molecule. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 13
Chapter review Q19. Figure 7.5 shows a flame test being performed. a What colour would the flame be if copper were present in the sample? b Before performing the test it would be necessary to heat the wire strongly for several minutes. Why? c Why would copper wire be unsuitable for use in flame tests? d Why are flame tests rarely used by modern chemists for qualitative analysis? A19. a green b It is necessary to heat the wire strongly before performing the flame test to vaporise traces of other substances that could lead to a false result. c Copper wire would give the flame a green colour, and so interfere with the test. The wire must be made of an element that does not change the colour of the Bunsen burner flame. d Flame tests are rarely used for analysis by today s chemists because relatively few elements may be analysed in this way; the presence of one element (sodium, for example) may mask the presence of another, and colours emitted by some elements are similar and difficult to distinguish, and because a relatively large amount of sample is needed to give a clear colour. Q20. When electrons are excited they lose energy by jumping from the shell they are in to a lower energy shell. Examine the spectrum of calcium in Figure 7.9a. Which line represents the electron jump of the largest energy? Explain. A20. The violet line on the far right-hand side represents the biggest jump. The further apart the electron shells, the greater the difference in energy. Violet light has the highest energy in the visible spectrum. Q21. Atoms will emit as well as absorb light of a characteristic wavelength when they are sprayed into a flame. Use Figure 7.13 to explain how the emitted light is prevented from interfering with the absorption measurement. A21. Use of a pulsed light beam permits comparison of the intensity of the light coming from the flame when the light beam is not present with the intensity of the beam passing through the vaporised sample. The light absorbed by the sample can then be determined. Q22. Explain the difference between inductively coupled plasma atomic emission spectroscopy and atomic absorption spectroscopy in terms of the origin of the light being measured and the changes in energy levels. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 14
A22. Inductively Coupled Plasma Atomic Emission Spectroscopy Light originates from the atoms in the sample which are excited by the very high temperatures generated in the plasma. Electrons move to a higher electronic energy level, and then relax back to their former level emitting light as they do so. The amount of light emitted by the excited atoms of the sample is measured Atomic Absorption Spectroscopy Light originates from a hollow cathode lamp. The metal filament of the lamp is heated. Electrons move to a higher electronic energy level, and then relax back to their former level emitting light as they do so. The light emitted is passed through the sample. The light is absorbed by the atoms in the sample. The amount of light absorbed by the sample is measured. Q23. Lead exposure can cause permanent brain damage in infants and young children, even at very low levels. The amount of lead in infant milk formula can be measured by atomic absorption spectroscopy. A 2.5 g sample of milk powder was dissolved in 50 ml of distilled water. A very small volume of this solution was analysed and gave an absorbance of 0.130. Four standard solutions were analysed in the same way and the calibration graph in Figure 7.52 was obtained. Figure 7.52 a Determine the concentration of lead (ng ml 1 ) in the diluted milk powder solution. b Calculate the concentration of lead (ng g 1 ) in the dried milk powder. c Express your answer to part b in ppm. d What is the percentage by mass of lead in the milk powder? e The zero standard, distilled water, gave an absorbance reading that was not zero. Suggest a reason for this. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 15
A23. a 2.45 ng ml 1 b Step 1 Calculate the mass of lead in the total volume of milk, using the answer from part a. This is the same mass of Pb as is present in 2.5 g of milk powder. 2.45 ng Pb per 1 ml milk = (2.45 50) ng per 50 ml milk = 122.5 ng per 50 ml milk Step 2 Convert into concentration of lead in ng g 1. c(pb) 122.5 ng = 2.5 g = 49 ng g 1 c Convert concentration to ppm. 49 ng Pb per 1 g of milk powder = 49 10 9 g Pb per 1 g of milk powder = 49 10 9 10 6 g Pb per 10 6 g milk powder = 49 10 3 g Pb per 10 6 g milk powder = 0.049 ppm d % Pb mass of Pb = 100 mass of milk powder 0.049 = 6 100% 10 = 4.9 10 6 % e The distilled water still contains minute traces of contaminants from the air, glassware or any other material the water has come into contact with. This arises because of the extreme sensitivity of the technique. Q24. The mineral cobaltite is mined for the production of cobalt. Ore containing cobaltite may also contain trace quantities of nickel. A sample of ore was analysed by AAS to determine the concentration of nickel present. 5.0 g of the ore was dissolved in 25 ml concentrated nitric acid, then diluted to 100 ml. A concentrated stock solution containing 1000 ppm nickel was also prepared. 10 ml of the ore sample solution was pipetted separately into four 100 ml flasks and 1, 2, 4 and 6 ml of the concentrated stock nickel solution was added to the flasks. All five flasks were then made up to the mark. A standard addition calibration curve was prepared using 10, 20, 40 and 60 ppm added nickel, giving the standards listed in the table: Concentration of nickel (ppm) Absorbance Standard 1 x + 10 0.25 Standard 2 x + 20 0.34 Standard 3 x + 30 0.44 Standard 4 x + 40 0.53 Sample x 0.15 The absorbance readings for the standards and sample were determined by using light of wavelength 325.4 nm. The value of x, the concentration of nickel in ppm, was found from the intercept of the calibration curve with the x-axis at absorbance 0. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 16
a Copy Figure 7.53 and plot the absorbance values for the standards and sample. Insert a line of best fit through the points and determine the concentration of nickel in the sample solution. Figure 7.53 b c d e What mass of nickel was present in 5.0 g of ore? What is the percent by mass of nickel in the ore sample? Why is it preferable to use a standard addition calibration technique instead of preparing the standards in distilled water? Arsenic is present in equal concentrations to that of cobalt in cobaltite. Why does it not interfere in the calculation of the cobalt or nickel? A24. a Concentration of nickel in solution = 16 ppm. b 16 ppm Ni 16 µg/ml Ni. 100 ml solution contains 100 16 µg Ni = 1600 µg 100 Ni from 10 ml of the stock solution = 1.6 mg Ni 10 = 16 mg Ni from 5.0 g of ore. c 3 16 10 g Percentage by mass = 5 g 100% = 0.32% w/w d In sample solutions with high concentrations of other atoms present, the concentration of metal atoms formed in the flame is reduced compared to standards prepared in distilled water. If standards are prepared with the same matrix or background as the sample, these changes are cancelled out. e The wavelength of light required to excite arsenic atoms is different from that which excites cobalt or nickel atoms; the presence of arsenic in the sample does not interfere with the analysis of cobalt or nickel. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 17
Q25. The wavelength selector on a UV visible spectrophotometer is adjusted to give light of wavelength 450 nm. a What is the colour of the light that is passed through the sample? b What colour would you expect the sample solution to be? Explain your answer. A25. a blue b orange; blue light of wavelength 440 470 nm is most strongly absorbed by samples of its complementary colour, orange. Q26. a Name a substance, other than copper sulfate solution, whose concentration can be determined directly by UV visible spectrometry. b Describe the purpose of the monochromator in the spectrometer (Figure 7.17 on page 87 of the student book). c Refer to Table 7.4 on page 85 of the student book. What wavelength of light would be best used to analyse a sample of copper sulfate solution? d Describe the changes on the atomic level that occur during UV spectroscopy. A26. a Any coloured solution can be chosen, for example chlorophyll solution. b The monochromator is used to select the optimum wavelength for absorption by the sample. c Electrons in the atom, ion or molecule are excited by the absorption of UV light, 580 620 nm. d Electrons move from a low energy level to a higher one and subsequently return to the ground state. Q27. a Which of the following molecules will not absorb infrared radiation: water, methane, oxygen or carbon dioxide? b Explain why the molecule you have chosen doesn t absorb infrared radiation. A27. a Oxygen b Bending or stretching will not change the symmetry of the oxygen molecule so there is no change in the overall dipole moment. Q28. Show, using labelled drawings of the water molecule: a symmetrical stretching b asymmetrical stretching c scissoring. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 18
A28. Q29. Examine the structure and infrared spectrum of butan-2-ol (Figure 7.54). Which bonds correspond to to the absorption peaks labelled A and B? Figure 7.54 Structure and IR spectrum of butan-2-ol. A29. a C H peak = 2970 cm 1 b O H peak = 3600 cm 1 Q30. The infrared spectrum of the nitrile (C N) group shows an absorbance at 2300 cm 1. Calculate: a the wavelength of this absorbance (in m) b the frequency of this absorbance (Hz). A30. a Wave number = 1/wavelength Wavelength = 1/2300 cm 1 = 0.000 4347 cm = 0.000 004 347 m = 4.4 10 6 m b Frequency (Hz) = speed light (m s 1 ) wavelength (m) 2.998 108 ms 1 4.347 10 6 m = 6.897 1014 Hz = 6.9 1014 Hz Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 19
Q31. Order the following molecules from highest to lowest energy required to cause stretching of the bond: HCl, HI, HBr. A31. HCl > HBr > HI Q32. Figure 7.55 shows the NMR spectrum of ethanol (CH 3 CH 2 OH). a How many hydrogen environments are there in this compound? b What is the relative number of protons at each peak set? c Identify the groups that produce the peak sets labelled A, B and C. d Explain why: i peak A is split into a quadruplet. ii peak C is split into a triplet. Chemical shift (ppm) Figure 7.55 High resolution NMR spectrum of ethanol. A32. a There are 3 peak sets; each set corresponds to a different hydrogen environment b Relative number of protons peaks A:B:C = 2:1:3 c The chemical shift of peak set A is consistent with that generated by the protons in O CH2 CH3 environment (shown in red). Peak set C is consistent with proton in a CH2 CH 3 group. The O H group generates the single peak at a chemical shift of 2.6. d i Peak A has 4 fine peaks so the adjacent C atom must have 3 hydrogens attached. ii Peak B is a triplet due to the influence of two hydrogen atoms bond to the adjacent C atom. For reasons beyond the scope of this text, peak C generated by the proton on the O H group is not split and this proton does not interact with neighbouring protons to cause peak splitting. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 20
Q33. The semi-structural formulas of propane, propan-1-ol and tartaric acid are shown in Figure 7.56. Figure 7.56 Semi-structural formulas of a propane, b propan-1-ol and c tartaric acid. Each of the samples is analysed by nuclear magnetic resonance spectroscopy. For each molecule state: a the number of different types or environments of protons in the molecule b the number of peaks into which the signal due to the proton marked with (*) would be split. Molecule Number of types of 1H Number of peaks H(*) spilt into Propane Propan-1-ol Tartaric acid A33. a and b Molecule Number of types of 1H Number of peaks H(*) spilt into Propane 2 7 Propan-1-ol 4 3 Tartaric acid 3 1 Protons that are chemically equivalent do not split their own signal. For example, ClCH 2 CH 2 Cl does not show splitting. Q34. Place in order of the energy required to cause a change in quantum energy level (lowest to highest); vibrational change in a molecule; electronic change in an atom, rotational change of 1H (proton), rotational change of molecule. A34. Spin change of proton (NMR) < Vibrational change of molecule (IR) < Electronic Change in an atom (UV/AA/AES) Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 21
Q35. Write a definition of each of the following terms that you met in this chapter: emission spectrum, quantum of energy, complementary colour, homonuclear molecule, opaque, transmitted light, chemical shift, wave number. A35. emission spectrum A plot of the energy emitted by a substance against wavelength (or frequency), appearing as coloured lines on a black background. quantum of energy A discrete quantity of electromagnetic radiation complementary colour Colours which mix together to give white light homonuclear molecule A molecule with only one type of element, e.g. H2, S8 opaque Light cannot pass through the material transmitted light Light that has passed through the sample chemical shift Variation of nuclear magnetic resonance frequency due to the electronic environment of the molecule wavenumber The inverse of wavelength, i.e. 1/λ Used in IR spectroscopy as a measure of wavelength. Q36. Atomic absorption spectroscopy (AAS) and UV visible spectroscopy both involve absorption of light. Both can be used to determine the amount of copper in a solution. a What species absorbs the light when copper nitrate is analysed by: i UV visible spectroscopy? ii AAS? b Which technique would be simplest for the analysis of 0.5 M copper nitrate solution? Explain your answer. c How is the light of the required wavelength selected in: i UV visible spectroscopy? ii AAS? A36. a i 2+ The copper hexaaqua ion, Cu(H 2 O) 6 ii Cu atoms b UV visible spectroscopy would be the easiest technique to determine the concentration of a 0.5 M copper nitrate solution; AAS would require several dilutions in order to bring the concentration of copper into the linear range of the calibration curve. c i White light is passed through a monochromator or wavelength selector. ii A hollow cathode lamp containing a cathode composed of the element under investigation produces light of the required wavelength. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 22
Q37. A copper mine can produce copper at a profit if the average concentration of ore is 0.4% w/w copper or greater. A 2.0 g sample of ore was ground, the copper dissolved in 10.0 ml aqua regia (a mixture of concentrated hydrochloric and nitric acids) and diluted to 200 ml with distilled water. Copper standards containing 2.0, 4.0. 6.0 and 12.0 µg/ml copper were also prepared and the UV visible absorbance measured (Figure 7.57). Figure 7.57 Calibration curve for the analysis of copper. a What was the concentration of copper in the sample solution? b What was the mass of copper in the 2.00 g sample of ore? c What mass of copper would be found in a tonne of ore based on the concentration in this sample? Is it commercially viable to extract the copper from this ore? A37. a 7.2 µg/ml Cu b 7.2 µg/ml 200 ml = 1400 µg Cu = 1.4 10 3 g Cu c 1.4 10 3 g Cu in 2.0 g ore 7.0 10 4 g/g Cu = 7.0 10 4 10 6 g Cu/tonne ore = 700 g Cu/tonne ore 700 g % Cu in ore = 100% = 0.07% Cu 6 10 g Yes, it is commercially viable to extract copper from this sample. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 23
Q38. Carrots contain the pigment carotene. The absorption spectrum of carotene is shown in Figure 7.58. Figure 7.58 Absorption spectrum of carotene. a Explain why carrots have an orange colour. b What wavelength should be used to determine the concentration of carotene in carrot juice using a UV visible spectrometer? c Draw a flow chart summarising the steps involved in analysing the concentration of carotene in carrot juice using UV visible spectrsocopy. A38. a Carotene absorbs light in the violet-blue region of the visible spectrum and transmits light in the yellow-red region resulting in an orange colour. b Carotene absorbs strongly at about 425 nm. This wavelength should be selected. c Flow chart should include the following steps: 1 Measure the absorbance over a range of wavelengths and plot the absorption spectrum. 2 Identify the wavelength at which carotene absorbs strongly about: 425 nm. 3 Prepare a set of standard solution of known concentration of carotene. 4 Measure the absorbance of each standard at 425 nm. 5 Draw a calibration graph by plotting absorbance against concentration. 6 Measure the absorbance of the caroten sample at 425 nm. 7 Read off the concentration of carotene form the calibration graph. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 24
Q39. Compare and contrast the techniques of infrared spectroscopy and ultraviolet spectroscopy under the following headings: Energy changes that occur in the sample Main features of the spectrometer Examples of samples analysed with the technique. A39. Energy changes Features of the instrument Samples analysed Q40. Infrared spectroscopy Changes in vibrational energy levels due to bending and stretching of bonds in molecules Refer to Figure 7.26 on page 91 of the student book. A very wide range of organic and molecules including solids, liquids and gases Ultraviolet spectroscopy Changes in electronic energy levels due to excitation of electrons in molecules Refer to Figure 7.17 on page 86 of the student book. Relatively low molecular weight organic molecules which are coloured or have conjugated double or triple bonds eg caffeine in soft drink, sunscreen in a skin cream Ethyl methanoate, propanoic acid and methyl ethanoate have the same molecular formula, C 3 H 6 O 2. The structural formulas of these compounds are shown in Figure 7.59. The proton NMR spectra of these compounds, labelled A, B and C, are shown in Figure 7.60. The peak sets are labelled X, Y and Z. Figure 7.59 Semi-structural formulas of I ethyl methanoate, II propanoic acid and III methyl ethanoate. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 25
Chemical shift (ppm) Chemical shift (ppm) Chemical shift (ppm) Figure 7.60 Proton NMR spectra of compounds A, B and C. a For each NMR spectrum, indicate: i the the number of proton environments in the molecule ii the relative number of hydrogens in each environment iii the approximate value of the chemical shifts for each peak set iv the protons responsible for each peak set and the carbon to which they are bonded. b Use the information from part a to match each compound with its NMR spectrum. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 26
A40. Compound A: a i 2 proton environments (2 peak sets) ii equal number of protons in each environment. X:Y = 1:1 iii chemical shifts Y =2.0, X = 3.7 iv b The NMR spectrum of compound A corresponds to the structure of methyl ethanoate. Compound B: a i 3 proton environments (3 peak sets) ii the ratio of protons generating peaks X:Y:Z = 1:2:3 iii chemical shifts X = 8.0, Y = 4.2, Z = 1.3 iv b The NMR spectrum of compound B corresponds to the structure of ethyl methanoate. Compound C: a i 3 proton environments (3 peak sets) ii equal number of protons in each environment. X : Y : Z =1 : 2 : 3 iii chemical shifts X = 11.8; Y = 2.4; Z =1.1 iv b The NMR spectrum of compound C corresponds to the structure of propanoic acid. Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 27