Analytical Techniques (A Level Only) (CIE A Level Chemistry)

Verdigris is a green pigment that contains both copper(II) carbonate, CuCO₃, and copper(II) hydroxide, Cu(OH)₂, in varying amounts.

Both copper compounds react with dilute hydrochloric acid.

CuCO₃ (s) + 2HCl (aq) → CuCl₂ (aq) + CO₂ (g) + H₂O (l)

Cu(OH)₂ (s) + 2HCl (aq) → CuCl₂ (aq) + 2H₂O (l)

You are to plan an experiment to determine the percentage of copper(II) carbonate in a sample of verdigris. Your method should involve the reaction of verdigris with excess dilute hydrochloric acid.

You are provided with the following:

• 0.494 g of verdigris
• 10.0 mol dm^–3 hydrochloric acid, HCl (aq)
• commonly available laboratory reagents and equipment.

You may assume that any other material present in verdigris is unaffected by heating and is not acidic or basic.

 [1]

[3]

[2]

[2]

volume of 0.500 mol dm^–3 HCl = .................................................... cm³ [2]

Azurite is a blue copper-containing mineral. The copper compound in azurite has the formula Cu₃(CO₃)₂(OH)₂. This copper compound reacts with sulfuric acid according to the equation.

Cu₃(CO₃)₂(OH)₂ (s) + 3H₂SO₄ (aq) → 3CuSO₄ (aq) + 2CO₂ (g) + 4H₂O (l)

A student carries out a series of titrations on 1.50 g samples of solid azurite using 0.400 mol dm^–3 sulfuric acid.

Assume that any other material present in azurite does not react with sulfuric acid. Some titration data is given in Table 1.1.

Table 1.1

The indictor for the titration is bromophenol blue. Bromophenol blue is blue at pH 4.6 and yellow at pH 3.0.

[1]

[1]

percentage by mass of Cu₃(CO₃)₂(OH)₂  in the sample of azurite = ..........................% [3]

Activated charcoal is a form of carbon with a very high surface area. It can be used to remove impurities from mixtures. It does this by a process called adsorption, where particles of the impurity bond (adsorb) to the activated charcoal surface.

A student wants to determine the ability of activated charcoal to adsorb a blue dye (the impurity) from aqueous solution.

The equation that links the mass of activated charcoal with the amount of blue dye adsorbed is shown.

log  = A + b log [X]

D = difference in concentration of dye (in g dm^–3) before and after adsorption
m = mass of activated charcoal (in g)
[X] = final concentration of dye (in g dm^–3) after adsorption
A and b are constants

The student uses the following procedure to investigate the ability of activated charcoal to adsorb a blue dye from an aqueous solution.

• Place a 50.0 cm³ sample of a 25.00 g dm^–3 solution of blue dye in a conical flask.
• Add a weighed mass of activated charcoal to the flask.
• Stir the contents of the flask for three minutes and then leave for one hour.
• Filter the mixture.
• Determine the final concentration of the blue dye, [X].
• Repeat the procedure using different masses of activated charcoal.

The procedure is carried out. The final concentrations of blue dye, [X], are shown in Table 2.1.

Table 2.1

mass of activated charcoal, m / g	initial concentration of blue dye / g dm–3	final concentration of blue dye, [X] / g dm–3	difference in concentration of blue dye, D / g dm–3		log	log [X]
0.20	25.00	0.96		120.20	2.08
0.25	25.00	0.69		97.24	1.99
0.30	25.00	0.60		81.33	1.91
0.35	25.00	0.41		70.26	1.85
0.40	25.00	0.33		61.68	1.79
0.45	25.00	0.27		54.96	1.74
0.50	25.00	0.23		49.54	1.69
0.55	25.00	0.20		45.09	1.65
0.60	25.00	0.17		41.38	1.62

[2]

[1]

[2]

Plot a graph on the grid to show the relationship between log  and log [X].

Use a cross (×) to plot each data point. Draw the straight line of best fit.

Circle the most anomalous point on the graph.

Suggest why this anomaly may have happened during the experimental procedure.

[2]

A = ........................................................... [1]

Column chromatography is a variation of thin-layer chromatography. Column chromatography and gas chromatography work by the same principles of components travelling through the column at different rates. 

State the difference between the mobile phases involved in column chromatography and gas-liquid chromatography.

Three compounds, A, B and C, of similar volatility, are mixed together. The mixture is then analysed in a gas chromatograph.

The gas chromatogram produced is shown in Fig. 3.1.

Fig. 3.1 

State which compound, A, B or C, has the greatest affinity for the solid phase. Explain your reasoning. 

Use the gas chromatogram shown in Fig 3.1 to identify the most abundant compound in the mixture. Explain your answer.

An oil tanker is travelling through the English Channel. The tanker has a slight leak which is not large enough to result in an oil slick but some oil is noticed on a beach.

Suggest how gas chromatography could be used to identify the tanker as the source of crude oil pollution on the beach. 

A mixture is analysed using thin-layer chromatography. 

Draw the labelled apparatus that would be used to identify the number of compounds in the mixture. 

A student runs a thin-layer chromatography experiment and plans to determine the compounds from their R_f values.

Describe the steps that the student needs to perform to determine the identity of the compounds. 

The student’s results are shown in Fig. 4.1.

Fig. 4.1

For their measurements, the student locates the centre of each spot by estimating its rough position by eye.

Suggest an improved method to locate the centre of each spot. 

The student runs another TLC practical on a mixture of benzaldehyde and benzyl alcohol using 7:3 pentane / diethyl ether as a solvent. 

The student's chromatogram is shown in Fig. 4.2.

Fig. 4.2

Calculate the R_f values for both compounds in the chromatogram. 

Explain why the maximum R_f value of a compound cannot exceed 1.

During the production of an NMR spectrum, tetramethylsilane (TMS) is mixed with the sample.

State the number of peaks in the C-13 NMR spectrum of 1,3-dichlorobenzene.

Compound A contains the elements carbon, hydrogen, oxygen and nitrogen only. Compound A contains a benzene ring.

Part of the mass spectrum of A is shown in Fig. 5.2.

Methyl cinnamate, C₁₀H₁₀O₂, is a white crystalline solid used in the perfume industry.

The proton NMR spectrum of methyl cinnamate in the solvent CDCl₃ is shown in Fig. 6.1.

Fig. 6.1

The structure of methyl cinnamate is shown in Fig. 6.2.

Fig. 6.2

The data in Table 6.1 should be used in answering this question.

Identify the proton environment that gives rise to the peak at a chemical shift of 3.8 ppm in Fig. 6.1. Explain your answer.

Proton NMR spectroscopy can be used to distinguish between isomers of C₆H₁₂O₂.

Draw the two esters with formula C₆H₁₂O₂ that each have only two peaks, both singlets, in their ¹H NMR spectra. The relative peak areas are 3:1 for both esters.

The proton NMR spectrum of another isomer of C₆H₁₂O₂ is shown in Fig. 6.3.

Four isomers of C₆H₁₂O₂, A, B, C and D, are shown in Fig. 6.4. 

Fig. 6.4

The C-13 NMR spectrum of one of the four isomers of C₆H₁₂O₂ is shown in Fig. 6.5.

Fig. 6.5

The data in Table 6.3 should be used in answering this question.

Identify which of the four isomers, A, B, C or D of C₆H₁₂O₂ produced the C-13 NMR spectrum shown in Fig 6.5.

Hybridisation of the carbon atom	Environment of carbon atom	Example	Chemical shift range δ/ppm
sp3	alkyl	CH3–, CH2–, –CH<, >C<	0 – 50
sp3	next to alkene / arene	–C–C=C, –C–Ar	25 – 50
sp3	next to carbonyl / carboxyl	C–COR, C–O2R	30 – 65
sp3	next to halogen	C–X	30 – 60
sp3	next to oxygen	C–O	50 – 70
sp2	alkene or arene	>C=C<,	110 – 160
sp2	carboxyl	R−COOH, R−COOR	160 – 185
sp2	carbonyl	R−CHO, R−CO−R	190 – 220
sp	nitrile	R−C≡N	100 – 125

Ethane-1,2-diol, C₂H₆O₂, can be distinguished from ethanedioic acid, C₂H₂O₄, by a number of analytic techniques including MS, IR and NMR

Fig. 7.1 (spectrum A) and Fig. 7.2 (spectrum B) show the mass spectra of ethane-1,2-diol and ethanedioic acid.

Complete Table 7.1 to suggest which compound is responsible for each spectrum? Explain your answer.

The IR spectra of ethane-1,2-diol, C₂H₆O₂, and ethanedioic acid dihydrate, C₂H₂O₄.2H₂O, are shown in Fig. 7.3 (Spectrum C) and Fig. 7.4 (Spectrum D).

The data in Table 7.3 should be used in answering this question.

Complete Table 7.2 to suggest which compound is responsible for each spectrum? Explain your answer.

The proton NMR spectrum of ethane-1,2-diol is shown in Fig. 7.5.

Describe and explain the splitting patterns of the spectrum.

Suggest the number of proton NMR peaks and splitting pattern for ethanedioic acid.

Compound P is a naturally occurring chemical found in strawberries, apples and Parmesan cheese.

The percentage by mass is carbon 58.82%, hydrogen 9.80% and oxygen 31.38%.

The mass spectrum of compound P is recorded in Fig. 1.1.

Fig. 1.1

Determine the molecular formula of compound P. Show your working.

Table 1.1 shows the results of qualitative tests performed on compound P. 

Table 1.1

Analyse the potential functional groups in compound P. Explain your answers.

The carbon-13 (¹³C) NMR spectrum of compound P is shown in Fig. 1.2. 

Fig. 1.2

Table 1.2 

Hybridisation of the carbon atom	Environment of carbon atom	Example	Chemical shift range δ / ppm
sp3	alkyl	CH3–, CH2–, –CH<, >C<	0 – 50
sp3	next to alkene / arene	–C–C=C, –C–Ar	25 – 50
sp3	next to carbonyl / carboxyl	C–COR, C–O2R	30 – 65
sp3	next to halogen	C–X	30 – 60
sp3	next to oxygen	C–O	50 – 70
sp2	alkene or arene	>C=C<,	110 – 160
sp2	carboxyl	R−COOH, R−COOR	160 – 185
sp2	carbonyl	R−CHO, R−CO−R	190 – 220
sp	nitrile	R−C≡N	100 – 125

Identify the functional group(s) present in compound P using your answer in (b) and information from Fig. 1.2 and Table 1.2. Explain your answer.

The high-resolution proton NMR spectrum of compound P was recorded as shown in Fig. 1.3.

Fig. 1.3

Table 1.3

Suggest the structure of compound P using your answers to (a), (b) and (c) and information from Fig. 1.3 and Table 1.3. Explain your answer.

Three hydrocarbons, L, M and N, have the molecular formula C₈H₁₀. Information about the number of peaks seen in the carbon-13 (¹³C) NMR spectrum of the three isomers is shown in Table 2.1.

Table 2.1

Complete Table 2.1 to give details of the proton NMR spectra for isomers L, M and N.

Table 2.1

Explain which of the three isomers, L, M or N has the highest melting point.

Gas chromatography is often connected to mass spectroscopy in order to analyse each compound as it exits the gas chromatography column. This combined technique is called GC-MS.

Fig. 3.1

Two of the compounds in the organic mixture are hydrocarbons with similar retention times.

The mass spectra of the two compounds revealed that they have the same molecular formula but with subtle differences indicating structural isomerism is present.

The mass spectrum of one of the isomers is shown in Fig. 3.2.

Both isomers are tested with bromine water, which remains orange.

Fig. 3.2

There is a small, additional peak on the far right of the mass spectrum. 

One other isomer present in the original mixture was 2,3-dimethylbutane.

Dinitrobenzene has the molecular formula C₆H₄N₂O₄. 

Draw the isomers of dinitrobenzene and name the type of isomerism.

State the factors that determine the distance travelled by a spot in thin-layer chromatography.

Fig. 4.1 shows the chromatogram with the spot for 1,4-dinitrobenzene.

Fig. 4.1

Draw the expected position of the spot for 1,2-dinitrobenzene. Explain your answer.

Explain what the student could do to reverse the relative positions of the 1,2-dinitrobenzene and 1,4-dinitrobenzene spots.