missballinger.files.wordpress.com · Web viewQ1. An extract from a plant was analysed using thin-layer chromatography with a non-polar solvent. The chromatogram obtained is shown

Worked solutions to textbook questions 1

Chapter 6 Chromatography

Q1.

An extract from a plant was analysed using thin-layer chromatography with a non-polar solvent. The chromatogram obtained is shown in Figure 6.4.

Figure 6.4Chromatogram.

Table 6.2 gives the Rf values of some chemicals commonly found in plants.Table 6.2 Rf values of some plant materials.

Chemical Rf

Xanthophyll 0.67-carotene 0.82Chlorophyll a 0.48Chlorophyll b 0.35Leutin 0.39Neoxanthin 0.27

a Measure and record the distance from the origin to the centre of each band, and the distance of the solvent front from the origin.

b Calculate the Rf value of each band.c Compare Rf values for the bands to the Rf values in Table 6.2 and name the

chemicals present in the extract.d If water had been used as the solvent, would the chromatogram be likely to have

been similar in appearance? Explain.

Heinemann Chemistry 2 (4th edition)Copyright © Pearson Education Australia (a division of Pearson Australia Group Pty Ltd) 2007


A1.

a, b, c It would be useful to set your answers out in a table.

Band a Distance from origin (mm)

b Rf c compound

Light green 20 0.33 Chlorophyll bDark green 27 0.45 Chlorophyll aOrange 40 0.67 XanthophyllYellow 50 0.83 β-caroteneSolvent front 60 – –

d The chromatogram would be likely to be different because separation of components depends on their solubility in the mobile phase (as well as strength of adsorption to the stationary phase). The polarity of the solvent used in TLC and paper chromatography will affect the Rf of the sample components. A polar solvent will dissolve polar samples readily; a non-polar solvent will dissolve non-polar samples readily.

E1.

Solids are not injected into the inlet port of a gas chromatograph. How might the chromatograms of the chips have been obtained?

AE1.

A sample of air from near the chips may have been injected into the inlet port of the gas chromatograph. Alternatively, the chips may have been mashed and mixed with a solvent that would cause the compounds responsible for the off-flavours to dissolve; this solution would then have been injected into the instrument.

E2.

What method was used to confirm that new peaks in the chromatogram of the off-flavoured chips could be attributed to the substances that caused the disagreeable flavour?

AE2.

The effluent from the column was smelt and compared with the smell of the off-flavoured chips.

E3.

What recommendations would you make to overcome the problem at the factory?

AE3.

To prevent further problems at the factory, regular inspections for the presence of soft rot on the stored potatoes should be conducted; potatoes should be stored under conditions that minimise the problem of soft rot.



Q2.

Australian wines are routinely tested for ethanol content. A quick and reliable method is by gas–liquid chromatography. The peak areas produced by a sample of wine and a number of standard solutions of ethanol are shown in Table 6.3.

Table 6.3 Peak areas from GLC analysis of a wine sample and standards.

% ethanol Relative ethanol peak areaWine ? 82 400Standard 1 4.00 31 200Standard 2 8.00 62 900Standard 3 12.0 94 200Standard 4 16.0 125 700

a Plot a calibration curve of concentration of ethanol against peak area.b Determine the percentage of ethanol in the wine sample.c Why is it necessary to measure the peak areas produced by a number of

standards?

A2.

a

b 10.4%c Techniques such as AA, GLC, HPLC, UV, etc. do not directly produce measures

of concentration. Standards must be used. More than one standard should be used and the unknown sample should lie between these standards. This is because a zero standard may be contaminated with trace amounts of the chemical being tested. Calibration graphs are often non-linear and multiple standards increase the chance of detecting incorrectly prepared solutions.



Chapter review

Q3.

Write a definition of each of the following terms: adsorption, desorption, mobile phase, stationary phase, eluent, retention time, carrier gas.

A3.

adsorption – The attraction of one substance to the surface of another.desorption – The breaking of the bonds between a substance and the surface to which the substance is adsorbed. mobile phase – The phase that moves over the stationary phase in chromatography.stationary phase – A solid, or a solid that is coated in a viscous liquid, used in chromatography. The components of a mixture undergo adsorption to this phase as they are carried along by the mobile phase.eluent – A liquid used as the mobile phase in chromatography.retention time – The time taken for a component to pass through a chromatography column.carrier gas – The gas used as the mobile phase in gas chromatography.

Q4.

There are several types of chromatography, including thin-layer and paper, gas and high performance liquid chromatography. What features are common to all kinds of chromatography?

A4.

All chromatographic techniques involve a stationary phase, mobile phase and separation of components.

Q5.

a Use the terms adsorbed and absorbed correctly in each of the sentences below.i Water was __________ by the towel as the wet swimmer dried himself.ii A thin layer of grease ___________ onto the cup when it was washed in the

dirty water.b Explain the difference between the two terms adsorbed and absorbed.

A5.

a Water was absorbed by the towel as the wet swimmer dried himself.A thin layer of grease adsorbed onto the cup when it was washed in the dirty water.

b Adsorb – Atoms or molecules accumulate and bond weakly to the surface of a solid or liquid.Absorb atoms or molecules are taken into the material.

Q6.

Phenacetin was once included in analgesic drugs, but is not used now because it causes liver damage. It is soluble in chloroform. A chemist wishes to analyse a brand of analgesic using thin-layer chromatography to determine whether it contains phenacetin. Outline the steps in the analysis. (Assume that a sample of pure phenacetin is available to the chemist.)



A6.

1 Dissolve a sample of pure phenacetin in a volume of chloroform. This is the standard solution.

2 Dissolve a tablet of the analgesic in chloroform. This is the sample solution.3 Place a small spot of the sample solution near the bottom of a thin-layer plate.

Place a spot of the standard solution next to it, at the same distance from the bottom of the plate.

4 When the spots are dry, place the plate in a container with a small volume of solvent, such as chloroform or another organic solvent. The lower edge of the plate, but not the spots, should be immersed.

5 Allow the solvent to rise until it almost reaches the top of the plate and remove the plate from the container.

6 Let the plate dry and examine it under ultraviolet light. If a spot from the sample appears at the same distance from the origin as the spot from the standard solution, the tablet is likely to contain phenacetin.

Q7.

A sample of brown dye from a lolly is placed at the origin on a strip of a chromatography plate. The solvent front moves 9.0 cm from the origin. A blue component of the dye moves 7.5 cm and a red component 5.2 cm in the same time. Calculate the Rf values of the two components.

A7.

Rf = distance dye has moved/distance solvent front has moved.Blue dye Rf = = 0.83

Red dye Rf = = 0.58

Q8.

Consider the diagram of thin-layer chromatography of three food colours in Figure 6.2.a Why must the level of the solvent be lower than the origin where spots of the

mixture are originally placed?b Why are Rf values always less than one?c How many different components have been used to make colour A?d Which components present in colours B and C are also in colour A? Explain.e Which component of colour A is most strongly adsorbed on the stationary phase?f Calculate the Rf values of each component of colour C.

A8.

a If the solvent were above the level of the origin, the compounds under test would dissolve and disperse throughout the solvent.

b Components in a mixture undergoing chromatography cannot move faster than the solvent that is carrying them over the stationary phase. Rf values must therefore be less than one.

c 3d B: blue; C: green. They can be identified on the basis of their colour and Rf values.e bluef 0.63; 0.13



Q9.

Using thin-layer chromatography showed that the black dye used in a brand of writing ink contained blue, red, orange and yellow components. The Rf values of these substances using ethanol as the solvent are 0.59, 0.32, 0.80 and 0.19, respectively.a How far apart would the blue and yellow components be after the solvent front

had moved 8.0 cm from the origin?b When the red component had travelled 6.0 cm from the origin, how far would the

orange component have travelled?c Sketch the chromatogram of the ink to scale after the solvent front had moved

15 cm from the origin.

A9.

a 3.2 cmb 15 cmc See diagram below.



Q10.

The amino acids present in a sample of fruit juice can be detected by thin-layer chromatography. Rf values of some amino acids using two separate solvents are given in Table 6.4.

Table 6.4

Amino acid Solvent 1 Rf Solvent 2 Rf

Lysine 0.12 0.55Leucine 0.58 0.82Proline 0.39 0.88Valine 0.40 0.742-Aminobutyric acid 0.28 0.58Threonine 0.21 0.49Hydroxyproline 0.21 0.67β-Phenylamine 0.50 0.86Isoleucine 0.57 0.81Alanine 0.24 0.55Serine 0.19 0.34Glutamic acid 0.25 0.33Glycine 0.20 0.40Arginine 0.13 0.60Taurine 0.12 0.33Tyrosine 0.38 0.62

To achieve better separation of the complex mixture of substances present in the juice, a ‘two-way’ chromatogram was prepared. The first step in this procedure was to run a chromatogram using Solvent 1. The results of this chromatogram are shown in Figure 6.17.

Figure 6.17

a Calculate the Rf value of each spot on the chromatogram.b Try to identify the amino acids responsible for each spot.



The paper was then turned around so that it lay at a right angle to the original and a second chromatogram was run using Solvent 2. Figure 6.18 shows the appearance of the paper after some time.c Use the table of Rf values to identify each component in the mixture.d What is the advantage of a two-way chromatogram?

Figure 6.18

A10.

a A: 0.59; B: 0.49; C: 0.38; D: 0.20; E: 0.13b A: leucine and/or isoleucine; B: β-phenylamine; C: proline and/or valine and/or

tyrosine; D: threonine and/or hydroxyproline and/or serine and/or glycine; E: lysine and/or arginine and/or taurine

c A: leucine and/or isoleucine; B: β-phenylamine; C: proline; D: serine; E: arginined A two-way chromatogram produces better separation of components of complex

mixtures, permitting easier isolation and identification.



Q11.

Read the following article from a 1990 British magazine and answer the questions.

Forensic expert slates Maguire analysis

A LEADING forensic scientist this week criticised the scientific evidence in the so-called Maguire case, in which seven people were convicted in 1976 for running an IRA bomb factory. The surviving defendants (one died in jail) have, since their release, sought to clear their names . . .

The forensic tests were the main evidence against the Maguire Seven. Government scientists told the 1976 court hearing that the seven had been ‘kneading’ the nitroglycerine [used in the bombing].

An 18-year-old trainee at the Royal Armament Research and Development Establishment (RARDE) found sizeable amounts of the explosive nitroglycerine in a thin-layer chromatography analysis of hand swabs from the accused. All seven were strongly positive. No other traces of explosives were found in the house . . .

a Nitroglycerine forms colourless solutions. How might it be made visible in thin-layer chromatography?

b In the mid 1970s, scientific policy was to use up all the sample when performing such tests. Suggest a reason for this.

c On what grounds might the forensic scientist be criticising the scientific evidence?

A11.

a Nitroglycerine can be seen by placing a chromatogram under an ultraviolet light since nitroglycerine absorbs light of this energy.

b If a very small amount of compound were available, all of it may need to be analysed to provide a detectable concentration.

c The relative inexperience of the person performing the analysis was criticised. Furthermore, since the entire sample was used up in performing the test, further confirmatory tests could not be undertaken.A court hearing in 1991 quashed the convictions upon hearing evidence that poor laboratory practice was involved in obtaining the forensic evidence. In particular, it was found that some testing materials were contaminated with traces of nitroglycerine.



Q12.

Refer to Figure 6.19.

Figure 6.19

a The following materials are commonly used in chromatography: water, ethanol, paraffin wax, glass sheets, paper strips, powdered alumina, nitrogen gas and hexane.Choose a suitable stationary phase and mobile phase from the list to use in the analysis of amino acids by thin-layer chromatography.

b Calculate the distance moved by the sample of leucine if the solvent front moves 13 cm from the origin.

c The separation of taurine and glycine on the chromatogram is not very great. State what would happen to the Rf values of the two amino acids if the separation was carried out for a longer period so the solvent front moved 10 cm instead of 5 cm from the origin. Explain your answer.

d Write a short paragraph explaining how leucine is separated from the other amino acids.

A12.

a Stationary phase = powdered alumina (also accept powdered alumina on glass sheet).Mobile phase = water or ethanol. The mobile phase should be a polar solvent in order to dissolve the polar sample.

b Rf leucine =

Rf leucine = = 0.58

Rf value is same regardless of the distance travelled by solvent front

0.58 =

Distance from origin by leucine = 13 cm ´ 0.58 = 7.5 cmc The Rf values would not change. The movement of the component and solvent

front both increase by the same proportion so there is no change in the Rf.



d As the mobile phase travels up the TLC plate it will absorb the amino acids and sweep them forward over the stationary phase. The amino acids are also attracted to the stationary phase and will tend to ‘stick’ or adsorb onto it. This adsorption and desorption process will happen many times. The amino acids have different solubilities and adsorption strengths so they undergo this process to different degrees, travel up the plate at different rates and separate.

Q13.

What are the advantages of HPLC for the analysis of drugs compared to an analysis technique based on column chromatography?

A13.

Compared with conventional column chromatography, HPLC is more sensitive, faster, resolves components better and it is able to detect colourless components readily.

Q14.

The organophosphorus insecticide parathion has been widely used in mosquito-prone areas. An empty drum of the insecticide was found close to a major reservoir. The EPA was asked to analyse the water to determine whether it was a threat to human health. Levels above 0.01 mg L–1 in water are a threat to human health. Parathion has an LD50 value (lethal dose to 50% of test animals exposed to this concentration) of 8 mg/kg.Parathion standards in water of 0 ppm, 10 ppm, 20 ppm and 30 ppm parathion were prepared and analysed by HPLC (Figure 6.20). An undiluted sample of the reservoir water was run on the column under the same conditions.

Figure 6.20

Chromatograms produced for suspected contaminated water sample and parathion standards.a Complete the table below.

Standard Peak height (cm)Standard 0 ppm parathionStandard 10.0 ppm parathionStandard 20.0 ppm parathionStandard 30.0 ppm parathionReservoir water

b Construct a calibration curve for the analysis of parathion.



c Determine the concentration of parathion in the water sample.d Is the reservoir water within the legal limits for safe drinking? Demonstrate your

answer with a calculation.e What volume of water would a laboratory mouse, mass 150 g, need to drink to

reach the LD50 dose?f Give possible reasons to explain the small peak observed with the 0 ppm standard

of parathion.

A14.

aStandard Peak height (cm)Standard 0 ppm parathion 0.2Standard 10 ppm parathion 2.5Standard 20 ppm parathion 5.2Standard 30 ppm parathion 7.4Suspected contaminated water 3.3

b

c Concentration of parathion in water sample = 13 ppmd 13 ppm º 13 mg/mL º 13 mg/L

No, this is over 1000 times the allowable concentration.e LD50 of 8 mg/kg º 8 mg/1000 g ´ 150 g = 1.2 mg. A mouse of 150 g has a 50%

chance of being killed by a 1.2 mg dose of parathion.

Volume of water which contains 1.2 mg parathion = L = 0.092 L = 92 mL

f A number of reasons could account for the small peak in the 0 ppm standard.· Water used for dilution of standards or glassware may have been contaminated

with parathion.· Small quantities of parathion may be leached off the column by solvent.· Small quantities of substances with similar retention times to parathion may be

present.



Q15.

In gas–liquid chromatography:a What is the usual mobile phase?b What is the stationary phase?c Why is the column packed with very fine particles?d Why is the injection port heated?

A15.

a nitrogenb high boiling liquid hydrocarbon or esterc Fine particles increase the surface area, which is coated with the stationary phase.

The greater the surface area of stationary phase, the more particles that can be adsorbed at any instant and the better the separation of components.

d To turn the sample components and solvent into a gas.

Q16.

Explain how gas chromatography can be used for:a qualitative analysisb quantitative analysis

A16.

a The time taken for a component to pass through the column of a GLC instrument is called the retention time, Rt. The Rt is characteristic of the component under the operating conditions in use and can be used to identify substances by comparing the Rt of unknown components with those of known substances.

b The relative amount of a component in a mixture may be determined by measuring the area under the peak of the component. This area is compared with the areas under peaks for samples containing the component in different known concentrations.

Q17.

A mixture of four alkanes, decane, heptane, hexane and octane, were separated by gas–liquid chromatography. The solubility of the hydrocarbons in the liquid phase on the column was directly proportional to the relative molecular mass of the alkane.

Figure 6.21Chromatogram of four hydrocarbons.

a Identity the hydrocarbons A, B, C and D.b Why does decreased solubility of a chemical in the stationary phase decrease the

retention time?c Why do the peaks become broader with increased retention time?d List the factors that can increase the retention time of a particular chemical in

GLC.



A17.

a A = hexane (lowest relative molecular mass); B = heptane; C = octane; D = decane (highest relative molecular mass).

b A compound which is relatively insoluble in the liquid stationary phase tends to spend little time dissolved in the liquid and therefore stationary, but is swept along the column by the gas and reaches the detector quickly. The retention time is therefore low.

c Compounds that are relatively insoluble in the stationary phase will tend to have all molecules swept along with very little ‘stationary’ time, so retention times of all molecules will be very similar, producing a narrow peak. More soluble compounds will have molecules that move in and out of solution in the stationary phase. More movements in and out of the stationary phase will increase the retention time in the column. The longer the time on the column, the greater the possible variation and the broader the peak.

d For example, longer column, stationary phase that has greater solubility with sample, lower the temperature, decreased flow rate of gas through column.

Q18.

The organic compounds dimethyl ether (CH3OCH3) and ethanol (CH3CH2OH) have the same molecular formula, C2H6O, and the same molar mass; however they have different structural formulas.A sample containing both ethanol and dimethyl ether is analysed by GLC using a flame ionisation detector.a Will the two compounds produce peaks with the same retention time? Explain

your answer.b If the sample has the same concentration of both chemicals, will the peaks

produced have the same area? Explain your answer.

A18.

a No. The retention time depends on solubility in the stationary phase. This depends on the functional groups on the compounds.

b No. The peak area depends on the sensitivity to the flame ionisation detector. This depends on the structure of the molecule and the ions that are formed in the FID cell. Consequently, if the concentration of more than one constituent of a sample is to be determined, separate calibration curves must be used for each compound.



Q19.

A herbal tea extract was analysed using HPLC. The chromatogram obtained is shown in Figure 6.22.

Figure 6.22

a Explain what information chemists can obtain from this chromatogram.b How many components are evident?c Briefly explain how the components are separated by the HPLC technique.

A19.

a Chemists can determine the number of components and the absorption of components, from which concentration can be determined.

b 4c Solid samples are dissolved in a suitable solvent. The liquid sample is injected

into the top of an HPLC column. The stationary and mobile phases are chosen to achieve a good separation of the components in the sample. The sample components alternately adsorb onto the stationary phase and then desorb into the solvent as they are swept forward. The time taken to exit the column increases if the component strongly absorbs onto the stationary phase and has a low solubility in the mobile phase.



Q20.

Various forms of heart disease including angina are very successfully treated using nitroglycerine patches. The small colourless plastic patches are stuck to the skin and slowly release nitroglycerine into the blood at a rate of 0.4 mg/h. The amount of nitroglycerine in the patches must be controlled carefully to prevent an overdose.A nitroglycerine patch was dissolved in approximately 20 mL ethanol and then diluted to 100 mL. 10.0 mL of this solution was diluted again to 100 mL with water. 20 mL volumes of the diluted sample and 20 mL of each of the prepared standards of nitroglycerine were injected onto a 1.5 m GLC column and analysed using a nitrogen-specific detector.

Nitroglycerine standards Peak area (mm2)Standard 5 mg/mL 7.2Standard 10 mg/mL 14.6Standard 15 mg/mL 22.0Diluted sample solution 10.8

a Construct a calibration curve and determine the concentration of nitroglycerine in the diluted sample.

b What mass of nitroglycerine was in the patch?c The patch has a total mass of 0.50 g. What is the percentage concentration (w/w)

of nitroglycerine in the patch?

A20.

a Concentration of nitroglycerine in diluted sample = 8.0 mg/mLb 100 mL of diluted sample solution 8.0 mg/mL ´ 100 mL = 800 mg

10 mL of stock solution contained 800 mg nitroglycerine100 mL of stock solution contained 8000 mg nitroglycerine º 8.0 mg nitroglycerine

c Percentage of nitroglycerine in patch = 8.0 mg/500 mg ´ 100% = 1.6% w/w



Q21.

Aspirin is a drug widely used in headache and cold preparations. Organic compounds such as aspirin in a formulation can be analysed by a variety of methods, for example paper chromatography, thin-layer chromatography, gas–liquid chromatography or high performance liquid chromatography. Final choice of a method of analysis involves consideration of the facilities available in the laboratory as well as problems inherent in each technique. A range of problems that can occur with some of these techniques is listed. Complete the table below using the letter codes to indicate the problem or problems is associated with each technique. Problems may be used more than once.A difficult or impossible to get quantitative dataB requires large amounts of solvent to operateC samples must be stable to heatD samples must be able to be dissolved in solventE expensive equipment neededF samples cannot have a high molecular mass

Technique Problem(s) (give letter)Paper chromatographyColumn chromatographyThin-layer chromatographyGas–liquid chromatographyHigh performance liquid chromatography

A21.

Technique Problem(s) (give letter)Paper chromatography A, DColumn chromatography A, B, DThin-layer chromatography A, DGas–liquid chromatography C, D, E, FHigh performance liquid chromatography B, D, E


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missballinger.files.wordpress.com · Web viewQ1. An extract from a plant was analysed using thin-layer chromatography with a non-polar solvent. The chromatogram obtained is shown