A2.2 Chemistry

• Mass spectrometry


  An analytical technique in which a sample can be analysed or identified from its mass spectrum

  
• Process of mass spectrometry

  1. Vaporised/atomised

  2. Ionised

  3. Accelerated

  4. Deflected

  5. Detected

  
• Mass spectrometry


  • All ions have a single positive charge and their mass determines how far it is deflected
  • Only ions with a mass suitable for the magnetic field being applied will reach the detector
  • By varying the magnetic field all ions and their abundances can be detected

  
• Features of a mass spectrum

  • Vertical (y) axis = relative abundance in comparison with base peak, which is 100
  • Horizontal (x) axis = m/z (mass to charge ratio. All charges are +1
  • Base peak – peak of greatest abundance
  • Molecular ion peak (M+) – last major peak in the spectrum with the highest m/z value
  • Fragmentation pattern is the pattern of peaks below the M+ peak
  • M+1 peak is a small peak above the M+ peak due to the presence of one carbon-13

  
• Mass spectrometry of elements and compounds

  1. Multiply the m/z value by the relative abundance

  2. Add all these values together

  3. Divide by the total number of relative abundances

  
• High-resolution mass spectrometry
• Ethyl Ethanoate - CH3COOCH2CH3

  • Molecular ion peak = 88; CH3COOCH2CH3+
  • M+1 peak = 89 due to the presence of one 13C in the molecular ion
  • The base peak is the most stable ion at 43. Species responsible is CH3CO+
  • The pattern for the m/z values below the molecular ion peak at 88 is caused by the molecule fragmenting during vaporisation and ionisation. The pattern is called the fragmentation pattern and is unique to a particular molecule.
• Fragmentation pattern - Ethanol (CH3CH2OH)


  • Base peak is the most stable m/z = 31
  • Molecular ion peak is C2H5OH+
  • CH3+ = 15, CH2+ = 14, OH+ = 17 (m/z values)
  • CH3CH2+ = 29, CH2OH+ = 31 (base peak)
  • (M-1)+ peak is when a compound loses one hydrogen
  • C-H bonds are usually the last to be broken
• (M+1)+ peak


  • Caused by 13C and this occurs in 1.1% of all carbon atoms
  • The relative abundance of the (M+1)+ peak as a percentage of the M+ peak gives a measure of the number of carbon atoms in the molecule
• Nuclear magnetic resonance spectroscopy

  • Process where a compound is analysed for hydrogen (1H) atoms
  • The sample is compared with a standard tetramethylsilane, Si(CH3)4 (TMS)

  
• 1H NMR spectroscopy


  • Nucleus of a 1H atom is a single proton so 1H NMR spectroscopy can be called proton NMR
  • Chemically equivalent hydrogen atoms are all in the same environment and appear as the same chemical shift on an NMR spectrum
  • A hydrogen atom in an organic compound has its nucleus exposed. The proton in the hydrogen atom is spinning creating a magnetic field/moment
  • The further bonded electrons are from the proton (Attached to a more electronegative molecule; N or O) the more de-shielded the proton

  
• De-shielded proton


  A proton is described as de-shielded when the electrons in the bond are further away from it
• TMS


  • The hydrogen atoms in TMS are less de-shielded than any hydrogen atoms in an organic compound
  • Si is more electronegative than carbon causing maximum shielding
  • The 12 hydrogen atoms attached to the carbons are chemically equivalent, so provide a string signal for comparison
  • Miscible with solvents used in NMR samples
  • Will not react with any compounds in the sample
• Features of an NMR spectrum

  • Horizontal axis is chemical shift, which runs from right to left. Measured in parts per million (ppm)
  • Vertical axis = signal intensity
  • A peak at δ=0 is caused by TMS
  • Number of peaks is the number of different types of chemical environments of hydrogen atoms
  • Area under each peak is equivalent to the ratio of the number of hydrogen atoms in each chemical environment. An integration trace can be given to indicate the ratio of hydrogen atoms
  • Spin-spin splitting pattern is caused by hydrogen atoms bonded to adjacent carbon atoms – n+1 rule
  • Low resolution NMR does not show the splitting pattern where as high resolution NMR does
  • Use data leaflet to relate chemical shift to the atom hydrogen atoms are bonded to

  
• Ethyl Ethanoate - CH3COOCH2CH3


  • δ=0 is due to chemically equivalent hydrogen atoms in TMS
  • 3 chemical environments = 3 peaks
  • Integration – 3:3:2
  • δ1.3 = CH3, with CH2 adjacent
  • δ2.0 = CH3 with not hydrogen atoms bonded to adjacent carbons
  • δ4.1 = three hydrogen atoms bonded to adjacent carbon atoms; CH2 with adjacent CH3
  • δ4.1 = ester group

  
• Iodine-thiosulfate titrations


  • Used to determine the concentration of an oxidising agent
  • An oxidising agent is an electron acceptor (it is reduced)
• Conditions for titrations


  1. A known volume of solution of the oxidising agent is placed in a conical flask

  2. The mixture is acidified using sulfuric acid

  3. Solid excess of potassium iodide is added

  4. Iodine solution in the conical flask is titrated with standard sodium thiosulfate solution

  5. Brown colour of the mixture fades to yellow, when the straw yellow colour is reached a few drops of starch indicator are added. Titration is continued until the blue-black colour produced by the starch colourless
• Reduction of IO3-


  2IO3- + 12H+ + 10e- I2 + 6H2O
• Oxidation of iodide ions, I-

  2I- I2 + 2e- (x5 to balance electrons)
• Combine equations


  2IO3- + 12H+ + 10I- 6I2 + 6H2O
• One mole of IO3- produces 3 moles of I2, when reacting with an excess of iodide ions in the presence of H+ ions
• IO3- : I2 = 1:3
• Reduction of H2O2


  H2O2 + 2H+ + 2e- H2O
• Oxidation of iodide ions, I-

  2I- I2 + 2e-
• Combine equations


  H2O2 + 2H+ + 2I- I2 + 2H2O
• One mole of H2O2 produces one mole of I2, when reacting with iodide ions in the presence of H+
• IO3- : H2O2 = 1:1
• Iodine, I2, produced in the reactions reacts with thiosulfate ions from the thiosulfate

  I2 + 2S2O32- 2I- + S4O62-
• Use titration calculations and ratios to determine the concentration of the oxidising agent
• Potassium manganite (VII) titrations

  • Potassium permanganate = KMnO4 and a purple solid
  • Manganate (VII) ions = MnO4-
  • Purple KMnO4 is reduced to pink manganese (II) ions (virtually colourless)
  • Used to determine the concentration of reducing agents
  • A reducing agent is an electron donor (it is oxidised)
• Conditions for titration


  1. 25cm3 of a solution of reducing agent is placed in a conical flask

  2. The mixture is acidified with sulfuric acid (additions of H+ ions)

  3. The standard solution of KMnO4 is added from the burette until the solution changes from colourless to pink
• Reduction of MnO4-


  MnO4- + 8H+ + 5e- Mn2+ + 4H2O
• Oxidation of Fe2+


  Fe2+ Fe3+ + e- (x5 to balance electrons)
• Combine equations


  MnO4- + 8H+ + 5Fe2+ Mn2+ + 5Fe3+ + 4H2O
• MnO4- : Fe2+ = 1:5
• Reduction of MnO4-


  MnO4- + 8H+ + 5e- Mn2+ + 4H2O (x2)
• Oxidation of C2O42-


  C2O42- 2CO2 + 2e- (x5)
• Combine equations


  2MnO4- + 16H+ + 5C2O42- 2Mn2+ + 10CO2 + 8H2O
• MnO4- : C2O42- = 2:5
• Back titration


  • A back titration is a method where an excess amount of reagent is reacted with a sample. The amount of unreacted reagent is then determined by titration
  • Used to determine the purity of a group I metal or group II metal oxide/carbonate
  • Used when the substance under analysis is not soluble in water but will react with acid