Modern analytical techniques IIEdexcel A-Level Chemistry: Topic test

20 questions, 54 marks

Edexcel A-Level Chemistry

Modern analytical techniques II topic test

Total 54 marks

Name

Class

Date

1
A colourless gas gives a molecular ion peak at m/z 44.0262 in a high-resolution mass spectrum. Accurate relative atomic masses: H = 1.0078, C = 12.0000, N = 14.0031, O = 15.9949.
(a)
Which molecular formula is consistent with the molecular ion peak?
[1 mark]
AC₃H₈
BC₂H₄O
CCO₂
DN₂O
(b)
Why can a high-resolution mass spectrum distinguish compounds that have the same nominal relative molecular mass?
[1 mark]
AThey have different numbers of electrons
BTheir nominal masses differ in the fourth decimal place
CThe molecular ion is always lost in the spectrometer
DThe accurate relative atomic masses of the elements are not whole numbers, so different formulae have slightly different accurate masses
(c)
Calculate the accurate relative molecular mass of CH₄N₂ and decide whether it could be the gas.
[2 marks]
Total for question 1: 4 marks
2
The three isomers of dichlorobenzene, C₆H₄Cl₂, are 1,2-dichlorobenzene, 1,3-dichlorobenzene and 1,4-dichlorobenzene. An analyst distinguishes them using ¹³C NMR spectroscopy.
(a)
How many peaks are there in the ¹³C NMR spectrum of 1,4-dichlorobenzene?
[1 mark]
A2
B3
C4
D6
(b)
Which compound gives four peaks in its ¹³C NMR spectrum?
[1 mark]
A1,2-Dichlorobenzene
B1,4-Dichlorobenzene
C1,3-Dichlorobenzene
DHexachlorobenzene, C₆Cl₆
(c)
The ¹³C NMR spectrum of 1,2-dichlorobenzene has three peaks. Explain why there are only three peaks.
[2 marks]
Total for question 2: 4 marks
3
1-Bromopropane, CH₃CH₂CH₂Br, is analysed by high-resolution ¹H NMR spectroscopy, and its spectrum is compared with that of its isomer 2-bromopropane, (CH₃)₂CHBr.
(a)
Predict the number of peaks, the ratio of the peak areas and the splitting pattern of each peak in the ¹H NMR spectrum of 1-bromopropane.
[3 marks]
(b)
Explain how the ¹H NMR spectrum of 2-bromopropane differs from that of 1-bromopropane.
[4 marks]
Total for question 3: 7 marks
4
Compound J has a molecular ion peak at m/z 88.0522 in a high-resolution mass spectrum. Its ¹³C NMR spectrum has three peaks, at δ 19, 34 and 183. Its ¹H NMR spectrum has three peaks: δ 1.2 (doublet, area 6), δ 2.6 (septet, area 1) and δ 11.4 (singlet, area 1). A forensic laboratory later finds J in a mixture with its isomer butanoic acid. Accurate relative atomic masses: H = 1.0078, C = 12.0000, O = 15.9949.
(a)
J is either C₄H₈O₂ or C₅H₁₂O. Use all of the data to decide which, and deduce the structure of J.
[6 marks]
(b)
Explain how GC–MS can show that J, and not butanoic acid, is present in the mixture, and why mass spectrometry alone would not be enough.
[6 marks]
Total for question 4: 12 marks
5
A laboratory measures the caffeine content of energy drinks using high-performance liquid chromatography (HPLC). A pure sample of caffeine has a retention time of 3.2 minutes under the conditions used.
(a)
What is the mobile phase in HPLC?
[1 mark]
AA solid packed into the column
BA liquid solvent pumped through the column at high pressure
CAn inert gas
DThe dissolved sample
(b)
Another compound in the drink is held more strongly by the stationary phase than caffeine. What is its retention time under the same conditions?
[1 mark]
AShorter than 3.2 minutes
BExactly 3.2 minutes
CIt cannot be measured
DLonger than 3.2 minutes
(c)
Explain how the chromatogram of an energy drink can be used to show that it contains caffeine.
[2 marks]
Total for question 5: 4 marks
6
A student separates the pigments in a leaf extract by thin-layer chromatography. After development, the solvent front is 7.5 cm above the baseline. Three spots, X, Y and Z, are found at 1.5 cm, 4.5 cm and 6.0 cm above the baseline.
(a)
What is the Rf value of spot Y?
[1 mark]
A1.7
B0.75
C0.60
D0.40
(b)
Which statement about the three pigments is correct?
[1 mark]
AX is held most strongly by the stationary phase
BZ is held most strongly by the stationary phase
CX is the most soluble in the solvent
DAll three have the same Rf value
(c)
Explain, in terms of the mobile phase and the stationary phase, why the pigments separate.
[2 marks]
Total for question 6: 4 marks
7
Methylbenzene, C₆H₅CH₃, is a solvent used in a school laboratory. A student records both its ¹³C NMR spectrum and its high-resolution ¹H NMR spectrum.
(a)
The ¹³C NMR spectrum of methylbenzene has five peaks. Explain why there are five peaks when the molecule has seven carbon atoms.
[3 marks]
(b)
Predict the ¹H NMR spectrum of methylbenzene, giving the number of peaks, the ratio of the peak areas, the approximate chemical shifts and any splitting.
[4 marks]
Total for question 7: 7 marks
8
Compound R has a molecular ion peak at m/z 72.0573 in a high-resolution mass spectrum. Possible molecular formulae are C₃H₄O₂, C₃H₈N₂ and C₄H₈O. R gives an orange precipitate with 2,4-dinitrophenylhydrazine. Its ¹³C NMR spectrum has four peaks, one at δ 209. Its ¹H NMR spectrum has three peaks: δ 1.0 (triplet, area 3), δ 2.1 (singlet, area 3) and δ 2.4 (quartet, area 2). Accurate relative atomic masses: H = 1.0078, C = 12.0000, N = 14.0031, O = 15.9949.
(a)
Use all of the data to deduce the molecular formula and structure of R.
[6 marks]
(b)
Butanal, CH₃CH₂CH₂CHO, is an isomer of R. Explain how the ¹³C NMR and ¹H NMR spectra would allow an analyst to tell R from butanal.
[6 marks]
Total for question 8: 12 marks

End of questions

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).