Chem ch8

Worked solutions to student book questions Chapter 8 Combining and choosing analytical techniques

Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

Q1. Use the data to calculate the relative atomic mass of lead. Relative isotopic mass Abundance (%)

203.97 1.48 205.97 23.6 206.98 22.6 207.98 52.3

A1. Average mass of 100 atoms of lead = (203.97 1.48) + (205.97 23.6) + (206.98 22.6) + (207.98 52.3) = 301.88 + 4860.89 + 4677.75 + 10 877.35 = 20 717.87 Average mass of 1 atom = 207.1787 = 207.18

Q2. What is the relative molecular mass of the parent molecular ion of dichloromethane due to: a CH235Cl37Cl? b CH237Cl2?

A2. a Parent molecule ion = CH235Cl37Cl+, m/e = 86 b Parent molecule ion = CH237Cl37Cl+, m/e = 88

Q3. What ions cause the peaks at m/e 49 and 35 in the mass spectrum of dichloromethane?

A3. Peak at 49 = CH235Cl+; Peak at 35 = 35Cl+



Q4. Complete the table below. Ion fragment m/e CH3+ 15 OH+ CO+ CH3CH2+ Cl+ C2H4+ COCH3+ COOH+ C6H5+

A4.

Ion fragment m/e CH3+ 15 OH+ 17 CO+ 28 CH3CH2+ 29 Cl+ 35 or 37 C2H4+ 28 COCH3+ 43 COOH+ 45 C6H5+ 77

Q5. A compound of oxygen and hydrogen is analysed and the following fragments measured. Note: There are two isotopes of hydrogen, 1H and 2H. There are also two isotopes of oxygen, 16O and 18O.

m/e Relative intensity (%) Possible fragment 1 < 0.1 H+ 16 1.0 17 21 18 100 19 0.08 20 0.22

a Complete the column Possible fragment in the table above. b What is the m/e of the base peak? c What is the m/e of the parent molecular ion? d Draw the structure of the parent molecular ion. e Draw the structure of the ion at m/e 20.



A5.

a m/e Relative abundance Possible fragment 1



Q9. Name one or more analytical technique that could be used to: a determine the concentration of aspirin in a headache remedy b find the concentration of the large organic molecule vitamin A c find the blood alcohol content of a motorist suspected of being above the legal

limit d determine the molecular structure of an organic compound

A9. a high performance liquid chromatography, UVvisible b high performance liquid chromatography c infrared spectroscopy, fuel cell d mass, infrared and NMR spectroscopy

Q10. Modern laboratories use a range of instrumental techniques to analyse substances. Briefly outline the advantages of using instrumental techniques.

A10. Instrumental techniques are quick, have a high degree of accuracy and can be used with very small samples. It is possible to combine techniques. Analysis can be automated and linked to computer data bases for a more accurate and rapid result.



Chapter review

Q11. Compare and contrast the techniques of mass spectrometry and infrared spectroscopy under the following headings: a The energy source with which the sample interacts b The chemical or physical basis of the analysis c The type of sample that can be analysed

A11.

Mass Spectrometry Infrared Spectroscopy Source with which sample interacts

The sample is bombarded with a beam of electrons

The sample is irradiated with infrared light

Chemical or physical basis for analysis

Different molecules for ions with a different mass/charge ratio

Different molecules have bonds of differing strengths requiring differing amounts of energy to cause vibration

Samples analysed A huge range of elements and organic molecules. Samplese must be able to be vapourised, e.g. identification of an explosive in a bomb

A very wide range of organic and molecules including solids, liquids and gases, e.g. determination the composition of soils or food materials

Q12. The mass spectrum and molecular structure of 1,2-ethanediol are shown in Figure 8.14 and Table 8.2. a Identify the molecular ion peak in the spectrum. b Identify the base peak in the spectrum. c What is the structure of the base peak ion? d Find out the common name for 1,2-ethanediol. e What is the best way to analyse quantitatively for this substance in a sample of

waste water?

Figure 8.14 Mass spectrum for 1,2-ethanediol.



Table 8.2 Data for 1,2-ethanediol mass spectrum.

m/e Relative intensity 15.0 9.6 29.0 13.2 31.0 100.0 33.0 34.9 48.0 8.3 62.0 3.5

A12. a m/e = 62 b m/e = 31 c CH2OH+ d anti-freeze e gas chromatography or high-performance liquid chromatography

Q13. The mass spectrum for a hydrocarbon is shown in Figure 8.15. Table 8.3 gives the relative intensities of the peaks.

Figure 8.15 Mass spectrum for a hydrocarbon.

Table 8.3 Data for mass spectrum of unknown hydrocarbon.

m/e Relative intensity 27.0 22.2 29.0 20.3 41.0 50.5 42.0 78.3 43.0 100.0 57.0 20.6 72.0 17.8 a What is the formula of the hydrocarbon?



b Copy and complete the following table for the peaks shown.

m/e Possible ion

c What fragment was lost to form m/e 57? d Write an equation illustrating this fragmentation. e Write an equation showing the formation of the base peak.

A13. a C5H12 b

m/e Possible ion 27 C2H3+ 29 C2H5+ 41 C3H5+ 42 C3H6+ 43 C3H7+ 57 C4H9+ 72 C5H12+

c CH3, methyl group, was lost to form ion m/e = 57 d CH3CH2CH2CH2CH3 CH3CH2CH2CH2+ + CH3 e Base peak = m/e 43 CH3CH2CH2+

CH3CH2CH2CH2CH3 CH3CH2CH2+ + CH2CH3

Q14. Name one or more analytical techniques that could be used to: a analyse for low concentrations of lead in blood b determine the concentration of a silver nitrate solution c verify the identity of a sample of an organic compound d identify the components in a mixture of alkanes

A14. a atomic absorption spectroscopy b gravimetric analysis, volumetric analysis, atomic absorption spectroscopy c infrared, mass or NMR spectroscopy d gasliquid chromatography

Q15. a Distinguish between thin-layer chromatography and atomic absorption

spectroscopy in terms of the following features: i the type of sample analysed, for example metals, inorganic compounds,

organic compounds, anions, cations, etc. ii whether atoms or compounds are determined iii whether separation of the components of the sample occurs iv quantitative or qualitative analysis v cost of the test



b Thin-layer chromatography and atomic absorption spectroscopy involve very different chemical principles. Below are four key terms that can be used in an explanation of each of the techniques. For each of the four terms, identify whether it applies to thin-layer chromatography or atomic emission spectroscopy and use the key term in a sentence to explain its meaning. i desorption ii energy levels iii excited electron iv solvent front

A15.

a Thin-layer chromatography Atomic absorption spectroscopy i organic compounds, especially coloured

components Atoms

ii compounds are analysed atoms of particular elements in compounds are analysed

iii components are separated components not separated iv primarily qualitative analysis qualitative and quantitative analysis v cheap more expensive

b i Desorbs/desorption thin-layer chromatography

The components of a mixture adsorb to the stationary phase and desorb into the mobile phase many times during a chromatography separation. (Or any other sensible answer.)

ii Energy levels atomic absorption spectroscopy. Atoms absorb energy from the flame and electrons are promoted to higher

energy levels. iii Excited electron atomic absorption spectroscopy. An excited electron is unstable and will soon return to a lower energy level. iv Solvent front thin-layer chromatography. The solvent front is the leading edge of the mobile phase, which sweeps the

components of the sample along the paper.

Q16.

A sample of the energy drink High Caff was analysed for the concentration of caffeine it contained. Details on the pack indicated that it should contain 12 mg caffeine per 100 mL. A chemist decided to analyse a sample of the drink, without dilution, by HPLC. 100 mL of 10.0 mg mL1 stock standard solution of caffeine was prepared from pure caffeine tablets. The chemist decided to prepare standards of the following concentration: 5, 10 and 20 mg/100 mL caffeine. Available were 10 mL and 20 mL pipettes and 100 mL and 200 mL volumetric flasks. a Describe the dilutions the chemist would need to carry out in order to prepare the

standard solutions using the equipment available. b The chemist could have used a 1 mL pipette and 100 mL volumetric flask to

obtain a 10 mg / 100 mL standard from the stock solution with only one dilution. Why is this a less accurate method?



A16.

a Stock solution = 10.0 mg/mL 1000 mg/100 mL Dilute 10 mL of 1000 mg/100 mL solution to 100 mL = 100 mg/100 mL Dilute 10 mL of 100 mg/100 mL solution to 200 mL = 5 mg/100 mL Dilute 10 mL of 100 mg/100 mL solution to 100 mL =10 mg/100 mL or 20 mL of 100 mg/100 mL solution to 200 mL = 10 mg/100 mL Dilute 20 mL of 100 mg/100 mL solution to 100 mL = 20 mg/100 mL b There is a greater error in a dilution taking 1 mL and diluting to 100 mL than in

two serial dilutions of 10 mL to 100 mL.

Q17. Condys crystals, a traditional remedy composed of potassium permanganate, can act as a fungicide. The amount of potassium permanganate in a sample of Condys crystals can be determined by a number of analytical techniques. In the table below, for each technique indicate: a the chemical species being analysed b any chemical reagents other than water and the sample needed to carry out the

analysis c whether or not a calibration curve is needed d the sensitivity: low sensitivity is used when samples are concentrated; high

sensitivity is used when samples are very dilute.

Technique a Species analysed

b Other reagents

c Calibration curve

d Sensitivity

UVvisible Volumetric analysis

Atomic emission spectroscopy

A17.

Technique Species analysed

Other reagents Calibration curve

Sensitivity

UVvisible MnO4 ion None Yes Lowhigh concentrations needed

Volumetric analysis

MnO4 ion A suitable reductant, e.g. Fe2+ ion; no indicator needed

No Lowhigh concentrations needed

Atomic emission spectroscopy

K atom Standard solutions of K+ ions

Yes High



Q18. Low concentrations of gold can be extracted economically by dissolving the tiny gold particles in the rock with sodium cyanide. An equation for the reaction is:

4Au(s) + 8CN(aq) + 2H2O(l) + O2(g) 4Au(CN)2(aq) + 4OH(aq) a Is this an acidbase or redox reaction? Give an example of an acid/conjugate base

or an oxidant/conjugate reductant redox pair in this equation to support your answer.

b What mass of gold could be theoretically extracted using 1.00 L of 0.540 M sodium cyanide solution?

c What technique can be used to analyse gold solutions in the concentration range mg Au per litre?

A18. a Redox reaction: Au/Au+ or O2/OH b n(NaCN) = 0.5 4 M 1.0 L = 0.54 mol

n(Au) = 2mol 54.0 = 0.27 mol

m(Au) = 0.27 mol 197 g/mol = 53 g c AA spectroscopy

Q19. Nuclear magnetic resonance spectrometers and mass spectrometers both employ a magnet. Describe the function of the magnet in each instrument.

A19. NMR the magnet provides an intense magnetic field with which the protons in the nucleus can orient themselves Mass spectrometer the magnet attracts the positively charged ions and changes their path; the ions with the lowest m/e ratio are attracted most.

Q20. High performance liquid chromatography and mass spectrometry both involve the movement and separation of the sample components in the instrument. Describe and explain how, for each: a the components of the sample move through the instrument b the components of the sample are separated

A20.

a HPLC the sample is swept through the column by a stream of solvent, the mobile phase. Mass spectrometer the positively charged particles of the sample are accelerated along the length of the spectrometer by their attraction to charged metal plates.

b HPLC the components of the sample have different degrees of attraction for the solvent and the stationary phase; the components attracted most strongly to the stationary phase move more slowly down the column.



Mass spectrometer the path of the fast moving ions are bent as the positively charged ions are attracted to the field generated buy the magnet. The paths are bent to different extents depending on the m/e of the ion.



Unit 3 Area of Study 1 review

Multiple-choice questions

Q1. A sample of nitrogen (Mr = 28.0) has a mass of 10 g and a volume of 8 L. Which of the following values can be calculated for the nitrogen sample from these data? I moles II % w/v III density A I only B I and II only C I and III only D I, II and III

A1. B. 1%mv = 1 g per 100mL = 10g per L 1000 ppm = 1g per L 0.017M = 0.017 mole per L = 0.017 58 1 g per L

Q2. Which concentration of sodium chloride is much higher than the others? A 1 g L1 NaCl B 1% m/v NaCl C 1000 ppm NaCl D 0.017 M NaCl

A2. B. All concentrations must be expressed in the same units, say mole L1 in order to make comparison.

In A, 1 g L1 NaCl = 5.58

1 mole L1 = 0.017 M

In B, 1% m/v NaCl = 1 g in 100 mL = 10 g L1 = 5.58

10 mole L1 = 0.17 M

In C, 1000ppm = 1000g per 1 000 000 g = 1 g per 1000g = 1 g L1 = 0.017 M The concentration in alternatives A, C and D are all equivalent to 0.017 M. The concentration in B is ten times this.

Q3. What volume of 2.0 M stock sodium iodide solution is needed to prepare 500 mL of 0.3 M sodium iodide solution? A 25 mL B 75 mL C 150 mL D 300 mL



A3. B. Number of moles of sodium iodide in 500 mL 0.3 M solution: 0.5 L 0.3 M = 0.15 mol Volume of 2 M sodium iodide that contains 0.15 mol:

M 2mol 15.0

= 0.075 L = 75 mL

Q4.

A 20 cm3 sample of which solution contains the greatest number of chloride ions? A 0.2 M NaCl solution B 0.1 M KCl solution C 0.2 M CaCl2 solution D 0.1 M AlCl3 solution

A4. C. nCl1 = 2 0.2 0.020 = 0.008 mol In A, nCl1 = 0.2 0.020 = 0.004 mol In B, nCl1 = 0.1 0.020 = 0.002 mol In D, nCl1 = 3 0.1 0.020 = 0.006 mol

Q5. A conjugate acidbase pair for this reaction is: HSO3(aq) + OH(aq) SO32(aq) + H2O(l) A HSO3 and SO32 B HSO3 and OH C OH and SO32 D SO32 and H2O

A5.

A. An acid has one more proton than its conjugate base.

Q6.

Potassium hydroxide reacts with carbon dioxide as follows: 2KOH(aq) + CO2(g) K2CO3(aq) + H2O(l)

Which expression correctly represents the volume of 0.1 M KOH solution that will react with 56 mL of CO2 at STP?

A 056.02

1.04.22 L

B 1.04.22

056.02

L

C 4.222

056.01.0 L

D 2 0.056 22.4 0.1 L



A6.

B. nCO2 = 4.22056.0 mol

nKOH = 2 nCO2

= 4.22056.02 mol B.

VKOH = 0.1 22.4

0.056 2

L

Q7.

Consider the following reaction. K2Cr2O7(aq) + 4H2SO4(aq) + 6HCl(aq)

Cr2(SO4)3(aq) + K2SO4(aq) + 3Cl2(g) + 7H2O(l) Which statement about this reaction is not true? A Dichromate ions have been reduced. B Chloride ions have been oxidised. C Hydrogen ions have been reduced. D Sulfate ions have not been oxidised or reduced.

A7. C. The oxidation state of H has not changed.

Q8.

In gasliquid chromatography the liquid stationary phase is coated on very small particles of solid silica or alumina. Small particles are used in preference to large particles to: A make the flow of gas through the column easier B reduce the overall mass of the column C increase the surface area of the liquid phase available D make it easier to fill the column

A8. C. Small particles would have the effect of reducing gas flow, not making it easier. Overall the mass of the column might be slightly higher as there would be a smaller dead volume caused by spaces between particles. It is slightly harder to pack a column evenly with very fine particles.

Q9. The retention time can be used in gas chromatography to determine: A the identity of a chemical B the concentration of a chemical C the amount of a chemical in the sample D all of the above



A9. A. The retention time in GLC and HPLC is used to determine the identity of a compound. The concentration and hence amount can be determined by comparison of peak areas or peak heights.

Q10.

The concentration of a substance is most accurately determined in HPLC by measuring: A peak area B peak height C retention time D Rf value

A10. A. Peak area is a more accurate measure of concentration than peak height. Peak height is acceptable for narrow peaks.

Q11. Chromatographic analysis of an unknown sample of an angina preparation containing nitroglycerine resulted in a narrow peak of height 25.3 cm. A standard solution containing 9.5 g/mL nitroglycerine gives a narrow peak of 10.5 cm. What is the concentration of nitroglycerine in the sample? A 23 g/mL B 27 g/mL C Cannot be determined as peak area is the only acceptable measure of

concentration. D Cannot be determined as a range of standards must be tested to determine

concentration.

A11. D. At least one standard must have a concentration greater than the unknown sample and at least three standards should be used in order to ascertain that the standard curve plot is linear.

Q12. Sodium salts give a distinctive yellow colour in a flame test. The colour is due to: A electrons near the nucleus of the sodium atom absorbing energy B excited electrons returning to lower energy levels and releasing energy C sodium atoms losing an electron to form sodium ions D sodium ions being converted to sodium atoms in the reducing flame

A12.

B. In a flame test, sodium ions are converted to sodium atoms in a reducing flame. The sodium atom absorbs energy from the flame and electrons move to a higher energy level. The electrons are unstable in this state and as they return to their ground state, the energy is released as light.



Answer A is true, but does not cause the coloured light. Sodium ions are converted to sodium atoms in the reducing flame but it does not cause the light.

Q13.

UV and atomic absorption are forms of spectroscopy that involve: A a prism or grating to separate the light B a light that is passed through the sample C emission of light by the sample D a flame to excite the atoms

A13. A and B. Light is absorbed by samples in both techniques. Only AAS uses a flame to excite the atoms in the sample being analysed.

Q14. The function of the flame in atomic absorption spectroscopy is to: I evaporate the solvent II convert metal ions to atoms III excite the metal atoms A I B I and II C I, II and III D II and III

A14. B. The monochromatic light from the lamp, not the flame, excites the atoms in AAS.

Q15. The concentration of low levels of the heavy metal cadmium in a sediment sample is best determined by: A gas chromatography B high pressure liquid chromatography C colorimetry D atomic absorption spectroscopy

A15.

D. Chromatography is primarily used to determine organic compounds. Colorimetry requires the preparation of a soluble coloured compound and is suitable for relatively high concentrations of the metal.

Q16. More energy is required to vibrate: A CC compared to C=C B CCl compared to CBr C CO compared to CH D C=C compared to CC



A16. C. More energy is required to vibrate a bond attached to a heavier atom.

Q17.

The number of peaks expected in the 1H NMR and 13C NMR spectra of methyl ethanoate, CH3COOCH3, are:

1H NMR spectrum (low resolution)

13C NMR spectrum

A 2 3 B 3 2 C 6 3 D 1 3

A17.

A. There are two distinct proton environments: CH3COO and OCH3. There are three carbon atom environments, each generating a different peak.

Q18. The following events occur in a mass spectrometer but not in the order indicated. I acceleration of particles in an electric field II vaporisation of the sample III deflection of particles in a magnetic field IV bombardment of the sample with high energy electrons V production of an electric current when charged particles enter the detector

The correct sequence of events is: A IV, II, I, III, V B II, I, IV, III, V C IV, I, III, II, V D II, IV, I, III, V

A18.

D. The sample is vaporised then ionised by a beam of high energy electrons. The ions are accelerated through an electric field. They are then deflected in a magnetic field. The extent of deflection depends upon the mass to charge ratio. The ions generate an electric current in the detector.

Q19. Which of the following techniques does not use the mass spectrometer as a detector? A gas chromatography B high performance liquid chromatography C nuclear magnetic resonance spectroscopy D atomic absorption spectroscopy

A19. D. The mass spectrometer is used in conjunction with other instruments such as GLC, HPLC and NMR.



Short-answer questions

Q20. Methyl salicylate, a compound of hydrogen, carbon and oxygen, is commonly known as oil of wintergreen. It is often used in medical rubs. If a 20.00 g sample of methyl salicylate was burned in air and it produced 9.46 g of water and 46.30 g of carbon dioxide, show that the empirical formula of this compound must be C8H8O3.

A20.

n(H) = 2n(H2O) = 1846.9 = 0.1.05 mol

m(H) = 1.05 1 = 1.05 g

n(C) = n(CO2) = 4430.46 = 1.05 mol

m(C) = 1.05 12 = 12.63g m(O) = 20.00 m(C) m(H) = 20.00 12.63 1.05 = 6.32 g C : H : O Mass 12.63 : 1.05 : 6.32

No. mole 12

63.12 : 105.1 :

1632.176.1

= 1.05 : 1.05 : 0.395 = 2.66 : 2.66 : 1

= 38 :

38 : 1

= 8 : 8 : 3 Empirical formula is C8H8O3

Q21.

A student wishes to prepare a standard solution of a base for an acidbase titration. She chooses to weigh the required amount of anhydrous sodium carbonate, dissolve the solid in water and make up the volume in a volumetric flask. The 0.0500 M standard sodium carbonate solution is then used to determine the amount of hydrochloric acid in a sample of concrete cleaner. 20.00 mL of the diluted concrete cleaner is pipetted into the conical flask. After several titrations a 23.45 mL titre is obtained and the concentration of the hydrochloric acid is calculated. a Distinguish between the terms primary standard and standard solution. b List three features that make anhydrous sodium carbonate a suitable choice of

solid to use to prepare the standard solution. c What is the concentration of hydrochloric acid in the diluted concrete cleaner? d Determine the effect of the following errors on the final calculated value of the

hydrochloric acid concentration. i The burette was rinsed with water. ii The conical flask was rinsed hydrochloric acid.

Note

2.66 = 232 =

38



A21. a A primary standard is a substance of such purity that the amount in moles can be

determined accurately from its mass. A standard solution is a solution of accurately known concentration.

b A primary standard should be: obtained pure of known formula able to be stored without deterioration in the atmosphere.

c n(HCl) = 2 n(Na2CO3)

1000

20 c = 2 1000

45.23 0.05

c = 0.117 M

d i Higher. The concentration of base in the burette will be lower so more will be needed to neutralise the acid. The calculated number of moles of base and so moles of acid will be higher than the true value.

ii Higher. More acid will be in the burette and so the calculated titre and number of moles of base and acid will be higher.

Q22. The active ingredient in a certain brand of antacid is magnesium carbonate. A 1.30 g tablet was crushed and placed in a conical flask. A small volume of water was added, the mixture stirred, and methyl orange indicator added. The indicator changed colour permanently after 27.75 mL of 1.00 M hydrochloric acid had been added. a Describe the changes that would be observed in the colour of the indicator. b Write an equation for the reaction that occurs during the titration. c Calculate the amount, in mol, of hydrochloric acid used to reach the end point. d Calculate the percentage by mass of magnesium carbonate in the tablet. e What is the function of the substances that make up the remaining mass of the

tablet? f Why do people often burp after taking an antacid tablet?

A22. a yellow to pink b 2HCl(aq) + MgCO3(aq) MgCl2(aq) + H2O(l) + CO2(g) c n(HCl) = 1.00 M 0.02775 L = 0.02775 mol = 0.0278 mol (three significant figures)



d Step 1 From the equation, 1 mol MgCO3 reacts with 2 mol HCl.

(HCl)

)(MgCO3n

n =

21

n(MgCO3) = 2mol 0.02775

= 0.01388 mol Step 2 Convert this amount to the mass of MgCO3. m(MgCO3) = 0.01388 mol 84.32 g mol1 = 1.1704 g Step 3 Calculate the % MgCO3 to the correct number of significant figures.

% MgCO3 = 1.3

1.1704 100%

= 90.03% = 90.0% e binder and flavouring f Burping after the consumption of antacid is caused by the build-up of carbon

dioxide gas in the stomach; the gas is produced by the reaction between the antacid and stomach acid.

Q23. Health experts are encouraging us to consume unsaturated oils, rather than saturated oils (or fats) in our diet. Unsaturated oils contain one or more double carbon-to-carbon bonds. Iodine reacts readily with these double bonds. Chemists measure the degree of unsaturation by determining the iodine number of the oil (see table). The iodine number is the number of grams of iodine that reacts with 100 g of oil. The oil is put in a stoppered flask, a known excess of iodine is added, and the flask shaken to allow the iodine to react. The remaining iodine is determined by titration with sodium thiosulfate solution. The reaction of iodine with an unsaturated oil is illustrated.

Reaction of iodine with an unsaturated oil.



Oil or fat Iodine number Peanut oil 93 Butterfat 36 Sunflower oil 126 Canola oil 99 Olive oil 81 Coconut oil 10

10 g of an unknown oil sample is shaken with 100 mL of solution containing 15 g of iodine. The mixture is shaken vigorously and 10.00 mL of the unreacted iodine solution is removed and titrated with 0.250 M sodium thiosulfate solution. A titre of 21.70 mL of the sodium thiosulfate is required. a Consider the table, which shows iodine numbers of a range of oils. Using the

iodine number, list the oils in the table from most saturated to least saturated. b Write half equations and an overall equation for the reaction between iodine and

sodium thiosulfate, given that the thiosulfate ion (S2O32) is oxidised to S4O62 and I2 is reduced to I.

c Calculate the number of moles of iodine in the 10.00 mL aliquot. d Calculate the number of grams of iodine that reacted with the 10 g of oil. e What is the identity of the oil in the sample? f Why is a back titration sometimes used instead of a direct titration?

A23. a coconut oil, butterfat, olive oil, peanut oil, canola oil, sunflower oil b I2(aq) + 2e 2I(aq)

2S2O32(aq) S4O62(aq) + 2e I2(aq) + 2S2O32(aq) 2I(aq) + S4O62(aq)

c Step 1 Calculate the amount of S2O32. n(S2O32) = 0.250 M 0.02170 L = 0.005425 mol Step 2 From the equation, 1 mol of I2 reacts with 2 mol of S2O32. Calculate the

amount of I2 that was in excess after reaction.

)O(S

)(I2

32

2n

n = 21

n(I2) in 10.00 mL aliquot = 20.005425 mol

= 0.0027125 mol = 0.00271 mol (three significant figures) d Step 1 Calculate the amount of I2 in 100 mL of solution that remained after

reaction.

n(I2) in 100 mL = 0.00271 00.10100 mol

= 0.0271 mol Step 2 Calculate the mass of I2 in 100 mL of solution that was in excess. m(I2) in excess in 100 mL = 0.0271 mol 253.8 g mol1 = 6.87798 g



Step 3 Calculate the mass of I2 that reacted with 10 g of oil. m(I2) reacted with oil = 15 6.87798 g = 8.12202 g = 8.12 g e olive oil f A back titration may be used for reactions that do not produce a sharp end point.

Q24. HA is a monoprotic acid and MOH is a base. The titration curve shown is obtained when 20.0 mL of 0.200 M HA is titrated with MOH.

Titration curve for HA and MOH.

a Is HA a strong or a weak acid? Explain your answer. b What is meant by the term monoprotic acid? c Methyl orange changes colour in the pH range 3.1 to 4.5 while phenolphthalein

changes colour between pH 8.2 and 10. Which indicator would be the most suitable to use for this titration? Give reasons for your answer.

d Determine the concentration of the base.

A24.

a HA is a weak acid as the initial pH 4.0 indicating that HA is partly ionised. If 0.2 M HA was fully ionised, pH 0.7.

b Monoprotic acids have only one proton that can be donated to a base. c Phenolphthalein has a sharp end point in this titration; it changes colour when a

very small volume of MOH is added at the end point. The end point as indicated by phenolphthalein is close to the equivalence point. Methyl orange has a broader end point that is not close to the equivalence point for this reaction.

d HA(aq) + MOH(aq) MA (aq) + H2O(l) n(MOH) = n(HA)

= 0.2 0.020 mol = 0.004 mol

c(MOH) = 025.0004.0 M

= 0.16 M



Q25. a The amount of alcohol, CH3CH2OH, in your breath can be determined by

blowing into a tube containing acidified potassium dichromate, K2Cr2O7. Which substance has been oxidised and which has been reduced?

b Explain how an IR spectrometer may be used to determine the amount of alcohol in a drivers breath.

A25. a Dichromate ions have been reduced and ethanol has been oxidised. b The bonds in ethanol absorb IR radiation at particular frequencies. The amount of

radiation absorbed is used to determine the blood alcohol.

Q26. The structures of propan-2-ol and propanone are shown below.

The infrared spectra of these compounds are shown in below and are labelled IR Spectrum A and IR Spectrum B.

IR spectrum A.



IR spectrum B.

a Identify the bonds that produced the IR absorption, marked with a (*), in each spectrum.

b Determine which spectrum represents which molecule. c The 1H NMR spectrum of propanone is shown below.

1H NMR spectrum of propanone.

Explain why there is only a single peak although there are six hydrogen atoms in the molecule.

d Explain why there are two peaks in 13C NMR spectrum of propanone, shown below, when the molecule contains three carbon atoms.

13C NMR spectrum of propanone.

e How many sets of peaks would you expect in 1H NMR and 13C NMR spectra of propan-2-ol? Provide an explanation for your answers.



f The mass spectrum of propanone is shown below.

Mass spectrum of propanone.

i What is the mass of the molecular ion? ii What information does this provide about the molecule? iii Suggest a formula for the molecular fragment that produces the peak at

m/e 43.

A26.

a Spectrum B Propanol Peak at ~3350 cm1 due to OH Peak at ~3000 cm1 due to CH Spectrum A Propanone

Peak at ~3000 cm1 due to CH Peak at ~1700 cm1 due to C=O b Spectrum B is propan-2-ol; Spectrum A is propan-2-one. c Different proton environments will generate peaks at different chemical shifts. In

propanone, all the environment of the protons in each CH3 group are identical. Hence only a single peak is produced at a chemical shift of 1.26 ppm.

d The chemical shift in 13C NMR depends on the carbon environment. The two CH3 carbons are identical and will only produce a single peak. The two peaks are produced by CH3 groups (30.8 ppm) and the C=O group (206.6 ppm).

e The 1H NMR spectrum of propan-2-ol will show three sets of peaks produced by protons in the CH3, CH and OH environments. The two sets of CH3 protons are in the same environment. The 13C NMR shows two peaks due to carbon atoms in a CH3 environment and COH environment. The CH3 carbons are in identical environments.

f i 58 ii molar mass of propanone is 58 iii The peak at m/e 43 indicates that a fragment of molar mass 15, CH3, has been

removed leaving a CH3CO+ ion having mass of 58 15 = 43.



Q27. The diagram shows a thin-layer chromatogram of amino acids in a medicine.

TLC of some amino acids.

a What amino acids are in the medicine? b Amino acids are colourless. What technique could be used to visualise the amino

acid spots? c Calculate the Rf value of taurine. d Which amino acid is bonded least strongly to the mobile phase?

A27. a Taurine, glycine and an unknown. Two of the three spots produced by the

medicine match those produced by taurine and glycine. The third spot does not match any of the amino acid standards and represents an unknown substance.

b There are several ways of visualising the spots: Viewing the chromatogram under UV light Spraying the finished chromatogram with a compound that causes the amino

acids to fluoresce Spraying the finished chromatogram with sulfuric acid that causes a brown

spot for organic compounds.

c Rf = solventby origin from moved distance

componentby origin from distance = Rf = 56.0 = 0.12

d Leucine. The component in a mixture bonded least strongly to the stationary phase will move the greatest distance from the origin.



Q28. Indicate how each of the parts of the electromagnetic spectrum listed below interacts with matter. For each part of the electromagnetic spectrum listed, name a spectroscopic technique that utilises this interaction. a radio frequency b infrared c visibleUV

A28.

When electromagnetic radiation interacts with matter energy is transferred to the atoms, ions or molecules of the chemicals involved. As energy is absorbed changes occur within the atom or molecules. The nature of the change depends upon the chemical and amount of energy involved. Radio frequencies. Under the influence of an external magnetic field, the absorption of electromagnetic radiation in the radio frequency range causes the nuclei of some atoms to move to a higher energy level. The strength of the external magnetic field experienced by these nuclei is modified by the molecular arrangements around it. Use of this is made in NMR where the absorption of EMR in the radio frequency ranges is measured and used to determine an atoms environment and the number of nuclei of that type within the molecule. This information is used to determine molecular structure. Infrared radiation. The absorption of IR radiation causes changes in the vibrational energy levels of bonds. The frequency of IR radiation absorbed provides information about the bond, the identity of the atoms forming the bond and if the bond is a single double or triple bond. Infra red spectroscopy is used to identify the functional groups in organic molecules. The fingerprint region of an IR spectrum can be used to identify a compound by comparison of the spectrum with those contained in database. Visibleultraviolet radiation. The absorption of visible or ultraviolet radiation causes electrons to jump to higher energy. Energy is emitted as electromagnetic radiation and is emitted when the atom returns to its ground energy state. The frequencies of radiation absorbed or emitted is characteristic of the atoms of an element. Atomic absorption spectroscopy is based on the principle that an atom will absorb characteristic frequencies of EMR from the visible/UV region of the spectrum.

Chem ch8

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