Worked solutions to textbook questions1
Chapter 8 Combining and choosing analytical techniques
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:
aCH235Cl37Cl?
bCH237Cl2?
A2.
aParent molecule ion = CH235Cl37Cl+, m/e = 86
bParent 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/eCH3+ / 15
OH+
CO+
CH3CH2+
Cl+
C2H4+
COCH3+
COOH+
C6H5+
A4.
Ion fragment / m/eCH3+ / 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.
m/e / Relative intensity (%) / Possible fragment1 / < 0.1 / H+
16 / 1.0
17 / 21
18 / 100
19 / 0.08
20 / 0.22
aComplete the column ‘Possible fragment’ in the table above.
bWhat is the m/e of the base peak?
cWhat is the m/e of the parent molecular ion?
dDraw the structure of the parent molecular ion.
eDraw the structure of the ion at m/e 20.
A5.
a
m/e / Relative abundance / Possible fragment1 / <0.1 / H+
16 / 1.0 / O+
17 / 21 / OH+
18 / 100 / OH2+
19 / 0.08 / O18H+
20 / 0.22 / O18H2+
b18
c18
dH2O
eH218O
Q6.
Infrared spectrophotometers are sometimes used to determine a motorist’s blood alcohol content. What is the functional group that is used as the reference peak for quantitative analysis of ethanol?
A6.
The broad peak due to the stretching of the OH (hydroxyl) group is used as the reference peak for quantitative analysis of ethanol.
Q7.
Blood alcohol content can also be determined by using a fuel cell. Write half equations describing the oxidation of ethanol and reduction of oxygen in an acidic electrolyte.
A7.
CH3CH2OH + H2O CH3CHOOH + 4H+ + 4e–
O2 + 4H+ + 4e– 2 H2O
Q8.
The mass spectrum of ethanol is shown in Figure 8.11. Identify the fragment ions causing the peaks marked on the ethanol spectrum.
A8.
m/e / Possible fragment15 / CH3+
27 / C2H3+
29 / C2H5+
31 / CH3O+
45 / C2H5O+
46 / C2H6O+
Q9.
Name one or more analytical technique that could be used to:
adetermine the concentration of aspirin in a headache remedy
bfind the concentration of the large organic molecule vitamin A
cfind the blood alcohol content of a motorist suspected of being above the legal limit
ddetermine the molecular structure of an organic compound
A9.
ahigh performance liquid chromatography, UV–visible
bhigh performance liquid chromatography
cinfrared spectroscopy, fuel cell
dmass, 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:
athe energy source with which the sample interacts
bthe chemical or physical basis of the analysis
cthe 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 differeing streths 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 rangeof 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 Figure8.14.
aIdentify the molecular ion peak in the spectrum.
bIdentify the base peak in the spectrum.
cWhat is the structure of the base peak ion?
dFind out the common name for 1,2-ethanediol.
eWhat 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 intensity15.0 / 9.6
29.0 / 13.2
31.0 / 100.0
33.0 / 34.9
48.0 / 8.3
62.0 / 3.5
A12.
am/e= 62
bm/e = 31
cCH2OH+
danti-freeze
egas 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 intensity27.0 / 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
aWhat is the formula of the hydrocarbon?
bCopy and complete the following table for the peaks shown.
m/e / Possible ioncWhat fragment was lost to form m/e 57?
dWrite an equation illustrating this fragmentation.
eWrite an equation showing the formation of the base peak.
A13.
aC5H12
b
m/e / Possible ion27 / C2H3+
29 / C2H5+
41 / C3H5+
42 / C3H6+
43 / C3H7+
57 / C4H9+
72 / C5H12+
cCH3, methyl group, was lost to form ion m/e=57
dCH3CH2CH2CH2CH3 CH3CH2CH2CH2++CH3
eBase peak = m/e 43 CH3CH2CH2+
CH3CH2CH2CH2CH3 CH3CH2CH2++CH2CH3
Q14.
Name one or more analytical techniques that could be used to:
aanalyse for low concentrations of lead in blood
bdetermine the concentration of a silver nitrate solution
cverify the identity of a sample of an organic compound
didentify the components in a mixture of alkanes
A14.
aatomic absorption spectroscopy
bgravimetric analysis, volumetric analysis, atomic absorption spectroscopy
cinfrared, mass or NMR spectroscopy
dgas–liquid chromatography
Q15.
aDistinguish between thin-layer chromatography and atomic absorption spectroscopy in terms of the following features:
ithe type of sample analysed, for example metals, inorganic compounds, organic compounds, anions, cations, etc.
iiwhether atoms or compounds are determined
iiiwhether separation of the components of the sample occurs
ivquantitative or qualitative analysis
vcost of the test
bThin-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.
idesorption
iienergy levels
iiiexcited electron
ivsolvent front
A15.
a
Thin-layer chromatography / Atomic absorption spectroscopyiorganic compounds, especially coloured components / Atoms
iicompounds are analysed / atoms of particular elements in compounds are analysed
iiicomponents are separated / components not separated
ivprimarily qualitative analysis / qualitative and quantitative analysis
vcheap / more expensive
biDesorbs/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.)
iiEnergy levels – atomic absorption spectroscopy.
Atoms absorb energy from the flame and electrons are promoted to higher energy levels.
iiiExcited electron – atomic absorption spectroscopy.
An excited electron is unstable and will soon return to a lower energy level.
ivSolvent 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 mL–1 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.
aDescribe the dilutions the chemist would need to carry out in order to prepare the standard solutions using the equipment available.
bThe 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.
aStock 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
bThere 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.
Condy’s crystals, a traditional remedy composed of potassium permanganate, can act as a fungicide. The amount of potassium permanganate in a sample of Condy’s crystals can be determined by a number of analytical techniques. In the table below, for each technique indicate:
athe chemical species being analysed
bany chemical reagents other than water and the sample needed to carry out the analysis
cwhether or not a calibration curve is needed
dthe 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 SensitivityColorimetry
Volumetric analysis
Atomic emission spectroscopy
A17.
Technique / Species analysed / Other reagents / Calibration curve / SensitivityColorimetry / MnO4– ion / None / Yes / Low–high concentrations needed
Volumetric analysis / MnO4– ion / A suitable reductant, e.g. Fe2+ ion; no indicator needed / No / Low–high 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)
aIs this an acid–base 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.
bWhat mass of gold could be theoretically extracted using 1.00 L of 0.540 M sodium cyanide solution?
cWhat technique can be used to analyse gold solutions in the concentration range mg Au per litre?
A18.
aRedox reaction: Au/Au+ or O2/OH–
bn(NaCN) = 0.5 4 M 1.0 L = 0.54 mol
n(Au) = = 0.27 mol
m(Au) = 0.27 mol 197 g/mol = 53 g
cAA 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:
athe components of the sample move through the instrument
bthe components of the sample are separated
A20.
aHPLC – 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.
bHPLC – 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 thefast 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?
Imoles
II% w/v
IIIdensity
AI only
BI and II only
CI and III only
DI, II and III
A1.
B. 1%mv = 1g 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?
A1 g L–1 NaCl
B1% m/v NaCl
C1000 ppm NaCl
D0.017 M NaCl
A2.
B. All concentrations must be expressed in the same units, say mole L–1 in order to make comparison.
In A,1 gL–1 NaCl=mole L–1 = 0.017M
In B,1%m/v NaCl = 1g in 100mL = 10g L–1 = mole L–1 = 0.17M
In C,1000ppm = 1000g per 1 000 000g = 1 g per 1000g = 1 g L–1 = 0.017M
The concentration in alternatives A, C and D are all equivalent to 0.017M. 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 0.3M sodium iodide solution?
A25 mL
B75 mL
C150 mL
D300 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:
= 0.075 L = 75 mL
Q4.
A 20 cm3 sample of which solution contains the greatest number of chloride ions?
A0.2 M NaCl solution
B0.1 M KCl solution
C0.2 M CaCl2 solution
D0.1M AlCl3 solution
A4.
C. nCl–1 = 2 × 0.2 × 0.020 = 0.008mol
In A,nCl–1 = 0.2 × 0.020 = 0.004 mol
In B,nCl–1 = 0.1 × 0.020 = 0.002 mol
In D,nCl–1 = 3 × 0.1 × 0.020 = 0.006 mol
Q5.
A conjugate acid–base pair for this reaction is:
HSO3–(aq) + OH–(aq) SO32–(aq) + H2O(l)
AHSO3– and SO32–
BHSO3– and OH–
COH– and SO32–
DSO32– 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 L
B L
C L
D2 × 0.056 × 22.4 × 0.1 L
A6.
B.nCO2= mol
nKOH = 2 nCO2
=mol B.
VKOH = 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?
ADichromate ions have been reduced.
BChloride ions have been oxidised.
CHydrogen ions have been reduced.
DSulfate ions have not been oxidised or reduced.
A7.
C. The oxidation state of H has not changed.
Q8.
In gas–liquid 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:
Amake the flow of gas through the column easier
Breduce the overall mass of the column
Cincrease the surface area of the liquid phase available
Dmake 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:
Athe identity of a chemical
Bthe concentration of a chemical
Cthe amount of a chemical in the sample
Dall 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:
Apeak area
Bpeak height
Cretention time
DRf 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?
A23 g/mL
B27 g/mL
CCannot be determined as peak area is the only acceptable measure of concentration.
DCannot 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:
Aelectrons near the nucleus of the sodium atom absorbing energy
Bexcited electrons returning to lower energy levels and releasing energy
Csodium atoms losing an electron to form sodium ions
Dsodium 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:
Aa prism or grating to separate the light
Ba light that is passed through the sample
Cemission of light by the sample
Da 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:
Ievaporate the solvent
IIconvert metal ions to atoms
IIIexcite the metal atoms
AI
BI and II
CI, II and III
DII 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:
Agas chromatography
Bhigh pressure liquid chromatography
Ccolorimetry
Datomic 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:
AC–C compared to C=C
BC–Cl compared to C–Br
CC–O compared to C–H
DC=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: CH3–COOandO–CH3.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.
Iacceleration of particles in an electric field
IIvaporisation of the sample
IIIdeflection of particles in a magnetic field
IVbombardment of the sample with high energy electrons
Vproduction of an electric current when charged particles enter the detector
The correct sequence of events is:
AIV, II, I, III, V
BII, I, IV, III, V
CIV, I, III, II, V
DII, 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?
Agas chromatography
Bhigh performance liquid chromatography
Cnuclear magnetic resonance spectroscopy
Datomic 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)= = 0.1.05 mol