Unit 33 Detection of substances Suggested answers to in-text activities Discussion (page 248)
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- Bu səhifədəki naviqasiya:
- Problem solving (page 253)
- Problem solving (page 257)
- Internet Search and Presentation (page 258) Uses of modern chemical techniques Infrared spectroscopy
- Nuclear magnetic resonance spectrometry
- Chemical shifts for some types of protons Type of proton Chemical shift, §
- Mass spectra of propane, propanone and propanal
- Colour of solid Deduction
- Soluble in water Insoluble in water
- Cations Flame test Dilute sodium hydroxide solution Ammonia solution
- Gas / water Test Observation
- Anion Test Observation
- Option Compound Flame colour
- Cation Adding NaOH(aq) to sample solution Adding NH 3 (aq)
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Unit 33 Detection of substances Suggested answers to in-text activities Discussion (page 248) 1 a) White precipitate (calcium sulphate) b) Green precipitate (copper(II) carbonate) 2 Potassium ion It gives a purple colour in flame test and gives no precipitate with sodium hydroxide solution. Problem solving (page 253) Solution D contains calcium ion because it gives a white precipitate (calcium sulphate) with dilute sulphuric acid. Ca2+(aq) + SO42- (aq)  CaSO4(s)
Solution A contains zinc ion because it gives a white precipitate (zinc hydroxide) with dilute sodium hydroxide solution. Zn2+(aq) + 2OH-(aq) Zn(OH)2(s)
Solution B contains copper(II) ion because it gives a blue precipitate (copper(II) hydroxide) with dilute sodium hydroxide solution. Cu2+(aq) + 2OH- (aq) Cu(OH)2(s)
Solution C contains iron(III) ion because it gives a brown precipitate (iron(III) hydroxide) with dilute sodium hydroxide solution. Fe3+(aq) + 3OH-(aq) Fe(OH)3(s)
Problem solving (page 257) a) Carbonate ion is present in solution A. CO32-(aq) + 2H+(aq) CO2(g) + H2O(l)
b) Adding aqueous chlorine to solutions B and C separately. The solution containing bromide ion becomes orange / light brown due to the formation of bromine.
Internet Search and Presentation (page 258) Uses of modern chemical techniques Infrared spectroscopy Infrared (IR) spectroscopy is concerned with exciting the vibrations that are occurring in molecules. Atoms in molecules are continuously vibrating. A simple diatomic molecule is like two heavy spheres connected by a spring: the spheres vibrate backwards and forwards about their equilibrium position alternately stretching and compressing the spring. 
In polyatomic molecules such as water, ammonia and methane there are several different vibrations occurring simultaneously: bonds are being stretched and compressed, called stretching vibrations, and bond angles are being increased and decreased in what are called bending vibrations. Some stretching and bending vibrations for a bent triatomic molecule such as water are shown below. 
The vibrational energy of molecules is quantized. In the ground vibrational state, the atoms are vibrating with a certain amplitude. When the molecule absorbs a quantum of vibrational energy, it vibrates with greater amplitude. The ‘spring’ is stretched to longer lengths and compressed to shorter ones. When bending vibrations are excited, the change in angle is greater. Quanta of vibrational energy correspond to infrared radiation. When molecules absorb IR radiation, they become hot. Molecules absorb IR radiation at particular wavelengths, depending upon which chemical bonds are present in the molecules. The instrument used is called infrared spectrometer. An IR spectrum is obtained by scanning through the whole wavelength range and plotting the % absorption (or % transmission) as a function of wavelength. The IR spectra of three different compounds with the same molecular formula (C4H8O) are shown below. Each compound has a characteristic spectrum that is determined by the types and arrangements of the bonds in its molecules. 
Because IR spectra of compounds are quite complex, they can be used as ‘fingerprints’ for identifying compounds. It is often possible to identify a sample just from its IR spectrum by comparing it with standard spectra. Today catalogues (and computer files) of IR spectra of thousands of compounds are readily available. This makes the job of identifying such compounds relatively easy. IR spectroscopy is particularly useful for identifying organic substances. Nuclear magnetic resonance spectrometry When certain atoms are placed in a strong magnetic field, their nuclei behave like tiny bar magnets and align themselves with the field. Electrons behave like this too. Therefore both electrons and nuclei are said to possess ‘spin’ since any spinning electric charge has an associated magnetic field. Just as electrons with opposite spin pair up together, a similar thing happens with the p 12 6
13 6
1 1
If the substance is now placed in an external magnetic field, the nuclear ‘magnets’ line up with the field, in the same way as a compass needle lines up with a magnetic field. The nuclear ‘magnet’ can have two alignments, of low and high energy. To make the nucleus change to the high energy alignment, energy must be supplied. 
It happens that the energy absorbed corresponds to radio frequencies. The precise frequency of energy depends on the environment of the nucleus, in other words, on the other nuclei and electrons in its neighbourhood. So by placing the sample being examined in a strong magnetic field and measuring the frequencies of radiation it absorbs, we can obtain information about the environments of nuclei in the molecule. The technique is called nuclear magnetic resonance (NMR). The figure below shows a simplified diagram of an NMR spectrometer. 
The technique of NMR is particularly useful for identifying the number and type of hydrogen atoms (1H) in a molecule. It is also used to find the positions of carbon atoms. Consider the NMR spectrum of ethylbenzene, C6H5CH2CH3. This is a proton NMR spectrum: the frequencies correspond to the absorption of energy by 1H nuclei, which are protons. Notice that there are three major peaks, of different heights. Each peak corresponds to an H atom in a different molecular environment. The area under each peak is proportional to the number of that type of H atom in the molecule. The largest peak (A) corresponds to the 5 H atoms in C6H5, the benzene ring. The second largest (C) corresponds to the 3 H atoms in the CH3 group, and the third peak (B) corresponds to the 2 H atoms in the CH2 group. The H atoms in a particular type of environment have similar positions in the NMR spectrum. Normally, this position is measured as a chemical shift, d, from a fixed reference point. The reference point normally used is the absorption of a substance known as TMS. The chemical shift of TMS is set at zero. TMS stands for tetramethylsilane, Si(CH3)4. This non-toxic and unreactive substance is chosen as the NMR reference because its protons give a single peak that is well separated from peaks found in the NMR spectra of most organic compounds. 
Consider the simplified proton NMR spectrum for ethanol (CH3CH2OH) shown below. It has been simplified by removing some of the detail, so the peaks appear single. The spectrum shows three single peaks. The smallest peak corresponds to the single OH proton; the middle peak corresponds to the two CH2 protons and the largest peak corresponds to the three CH3 protons. 
Nuclear magnetic resonance is also being used in medical diagnosis. It is the principle behind magnetic resonance imaging, a diagnostic scanning tool, which allows visualization of damage to various organs. Mass spectrometry Mass spectrometry is a form of analysis that separates and identifies substances on the basis of the masses of the positive ions formed by substances when they are bombarded by high energy particles (usually electrons) in a high vacuum. The instrument used is called mass spectrometer. In a mass spectrometer, the vaporized sample is bombarded with high energy electrons. These high energy electrons knock electrons off atoms or molecules in the sample, producing positive ions. The positive ions are accelerated through an electric field and then passed through a magnetic field. The magnetic field causes the moving charges to follow circular paths. With fixed accelerating voltage, V, and magnetic field strength, B, the radius of the path is dependent upon the mass-to-charge ratio of the ion. If the values of V and B are such that the radius, r, for argon (say) corresponds to that of the apparatus, then argon ions will pass through the collector slit and reach the collector. Krypton atoms with greater mass will follow a path of greater radius and will not pass through the collector slit; neon, of smaller mass, will follow a path of smaller radius and again will not be collected. 
The ion detector is usually linked through an amplifier to a recorder. As the strength of the magnetic field is slowly increased, ions of increasing mass will be detected and a mass spectrum is traced out by the recorder. The relative heights of the peaks in the mass spectrum give a measure of the relative amounts of the different ions present. (Strictly speaking, it is the areas under the peaks and not the peak heights which give the relative amounts or abundances.) In practice, a reference peak using a known substance is first obtained on the mass spectrum. The relative masses of other particles can then be obtained by comparison with this. 
Determination of atomic masses was the major use of mass spectrometry when the technique was first developed in the 1910s. Francis Aston, in 1919, found that the mass spectrum of neon contained not just the one expected line at mass 20, but three lines (as shown below from a modern instrument). This directly demonstrated the existence of isotopes. In the next few years many elements were shown to be mixtures of two or more isotopes. 
Cracking patterns and relative molecular masses The mass spectra of molecular substances are generally quite complex: they contain peaks corresponding to many different masses as shown below for propane, propanone and propanal. Multiple peaks appear from one compound because the parent ion formed by direct electron bombardment of the molecule is generally unstable and breaks up in a variety of ways to form ions of smaller masses. 
Mass spectra of propane, propanone and propanal When propane is admitted to the mass spectrometer, ion bombardment produces the parent ion, CH3–CH2–CH3+ (mass 44). This easily loses one or two H atoms to form CH3–CH2–CH2+ and CH3–CH2–CH+ (masses 43 and 42). In addition the parent ion can split off a CH3 group, forming CH3–CH2+ (mass 29) and by loss of further H atoms it forms masses 28 and 27. These three compounds are relatively simple ones and lead to fairly simple mass spectra. As molecules become more complex, their mass spectra also become more complex because more and more fragmentation paths are possible. As a result, compounds containing more than about 8 to 10 atoms have unique mass spectra. Consequently mass spectra can be used as ‘fingerprints’ for compounds. The relative molecular mass of a compound can often be deduced from the mass spectrum of the compound. Although ions fragment after electron bombardment in mass spectrometers, generally some of the parent ion survives and so there is a peak in the spectrum corresponding to it. Our first thought is that the peak with the highest mass in the spectrum would be from the parent ion and so its mass would be the molecular mass of the compound. However, some parent ions can ‘take’ an H atom from a neutral molecule and this can produce a peak one unit above the parent mass. Consequently in selecting the peak corresponding to the parent ion, we need to take this into account. Consider the mass spectra of propane, propanone and propanal shown above. We identify the peaks at masses 44, 58 and 58 as the parent ion peaks, recognizing that smaller peaks at 45, 59 and 59 correspond to a 13C atom or an extra H atom being present. Gas chromatography In all forms of chromatography, a mobile phase passes over a stationary phase. The stationary phase is either a solid with a high surface area, or it may be a liquid which is coated onto a finely divided solid. The mobile phase is moved over the surface of the stationary phase and carries the mixture to be separated with it. Of all the forms of chromatography, GC is the most sensitive, being able to detect quantities as small as 10-12 g from a microlitre sample. A gas chromatograph has three main sections: a preheated injector into which the sample is injected; a chromatographic column inside an oven, the temperature of which is carefully controlled; a detector. 
In operation, a stream of inert gas such as nitrogen carries the vaporized sample through the column. The column contains a high-boiling-point liquid, coated onto fine granules of an inert solid. It sorts out a mixture into its components. After passing through the column, the components are detected by a suitable device and recorded on a chart as a series of peaks. The position of the peak identifies each component, while the height (or more correctly, the area under each peak) identifies the amount. This instrument is capable of both qualitative and quantitative analysis. Making use of gas chromatography The wine industry makes frequent use of gas chromatography to analyze the ethanol content of wines. As ethanol is easily vaporized and can be separated quite easily from other wine components using an appropriate column and operating conditions, gas chromatography is a quick and accurate technique for such analysis. In other areas of industry, it is used to check the composition of petrols, oils, drugs and pesticides. In analyzing a wine sample, a number of reference samples of known ethanol concentration are first run through the machine to calibrate it. The position of the ethanol peak can thus be established for the particular operating conditions. The height of the peak can also be related to the concentration of ethanol. When a wine sample is subsequently injected, the ethanol peak is easily identified and the amount of ethanol can be determined by its height. A typical result is shown below. 
Suggested answers to exercise 1
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9 D Zinc hydroxide dissolves in excess dilute sodium hydroxide solution and ammonia solution due to the formation of soluble complex salts. 10 C 11 C 12 B The blue precipitate is copper(II) hydroxide. Cu2+(aq) + 2OH- (aq) 13 D 14 D 15 A
16 a) Add dilute hydrochloric acid to the samples. Sodium hypochlorite gives a gas that turns moist blue litmus paper red and then white (the gas is chlorine). Sodium chloride gives no observable change. b) Add dilute sodium hydroxide solution to the sample solutions. Copper(II) chloride solution gives a blue precipitate while iron(II) sulphate solution gives a green precipitate. c) Add ammonia solution to the sample solutions. Iron(III) chloride solution gives a reddish brown precipitate while there is no observable change with calcium chloride solution. d) Use flame test. Sodium nitrate gives a golden yellow flame while potassium nitrate gives a purple flame.
Potassium sulphite gives a gas that turns filter paper moistened with acidified potassium dichromate solution from orange to green (the gas is sulphur dioxide). Potassium sulphate gives no observable change. 17 a) Blue b) i) Sodium ion and ammonium ion ii) Use flame test to detect the presence of sodium ion in filtrate A. It gives a golden yellow flame. Heat the filtrate with dilute sodium hydroxide solution. A gas that turns moist red litmus paper blue would be given off (the gas is ammonia). This shows the presence of ammonium ion in the filtrate.
deep blue solution forms while the white precipitate remains undissolved. 18 a) Calcium ion is present in X. Sodium ion is present in Y. b) i) Calcium carbonate ii) CaCO3(s) + 2H+(aq) Ca2+ (aq) + H2O(l) + CO2(g)
c) i) Sodium chloride ii) Ag+ (aq) + Cl-(aq) AgCl(s)
d) Add water to Y. Obtain the insoluble reddish brown solid by filtration. Add a little dilute hydrochloric acid to dissolve the solid. Add dilute sodium hydroxide solution to the mixture until a reddish brown precipitate (iron(III) hydroxide) appears. 19 a) i) Chloride ion ii) Ag+ (aq) + Cl-(aq) AgCl(s)
b) i) Iron(II) ion ii) The green precipitate is iron(II) hydroxide. c) i) Acidified potassium permanganate solution ii) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)
20 a) i) Potassium sulphite and sodium hypochlorite ii) Test the gas given off with dilute hydrochloric acid. Potassium sulphite gives a gas that turns filter paper moistened with acidified potassium dichromate solution from orange to green (the gas is sulphur dioxide). Sodium hypochlorite gives a gas that turns moist blue litmus paper red and then white (the gas is chlorine). b) i) Calcium chloride and zinc sulphate ii) Add dilute sodium hydroxide solution. Test whether the precipitates formed can dissolve in excess dilute sodium hydroxide solution. Zinc hydroxide dissolves to give a colourless solution. Calcium hydroxide does not dissolve in excess dilute sodium hydroxide solution. Yüklə 1,62 Mb. Dostları ilə paylaş:
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