Mass Spectrometer Uses mass spectrometer Presence of isotopes and its abundance Relative atomic mass of an element Relative Molecular mass of a molecule Structure of organic compound Distinguish between structural isomers CH 3 CH 2 CH 2 OH OH | CH 3 CHCH 3 CH 3 | CH 3 C-CH 3 | CH 3 CO 2 structural formula Organic structure determination
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IB Chemistry on Mass Spectrometry and Isotopes for Option A HL.
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Mass Spectrometer
Uses mass spectrometer
Presence of isotopes and its abundance
Relative atomic mass of an element
Relative Molecular mass of a molecule
Structure of organic compound
Distinguish between structural isomers
CH3CH2CH2OH OH | CH3CHCH3
CH3 | CH3C-CH3
| CH3
CO2
structural
formula
Organic structure determination
Mass Spectrometer
Parts of Mass Spectrometer
Sample injection
Vaporization Chamber • Sample heat to vapour state
Ionization Chamber • Molecule bombard with electrons form positive ions
Accelerator Chamber • M+ ions accelerated by Electric field
Deflector • M+ ions deflected by magnetic field
Detector • Convert abundance of M+ ions to electrical current. • M+ ions neutralize by electrons (more e needed - higher current – higher intensity of peak) • Intensity of peak show -relative abundance of ions
Sample X bombarded by electron • Form positive M+ ion • Accelerated (Electric Field) • Deflected (Magnetic Field) and Detected X + e- → X+ + 2e-
Deflection depend: • mass/charge (m/z) ratio: (m/z) ratio LOW↓- Deflection HIGH ↑
37CI+
35CI+
35CI2+
2
3 4
1 5 Detector
• Convert abundance of M+ ions to electrical current. • M+ ion neutralize by electrons (more e needed - higher current – higher intensity of peak) • Intensity of peak show -relative abundance of ions
Weighted average mass- due to presence of isotopes
Relative Isotopic Mass, (Ar) of an element: •Relative isotopic mass = Average mass of one atom of element 1/12 x mass of one carbon-12 • Relative isotopic mass, carbon = 12.01
Video on Isotopes
RAM = 12.01
Relative Abundance 98.9% 1.07%
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Why RAM is not a whole number?
Relative Isotopic Mass: = (Mass 12C x % Abundance) + (Mass 13C x % Abundance) = (12 x 98.9/100) + (13 x 1.07/100) = 12.01
Video on weighted average Weighted average calculation
24 Mg – (100/127.2) x 100% - 78.6% 25 Mg – (12.8/127.2) x 100% - 10.0% 26 Mg – (14.4/127.2) x 100% - 11.3%
Relative Isotopic Mass: = (Mass 24 Mg x % Ab) + (Mass 25 Mg x % Ab) + (Mass 26M g x % Ab) = (24 x 78.6/100) + (25 x 10.0/100) + (26 x 11.3/100) = 24.30
Relative Abundance % Abundance
Pb - 4 Isotopes
204Pb – (0.2/10) x 100% - 2% 206Pb – (2.4/10) x 100% - 24% 207Pb – (2.2/10) x 100% - 22% 208Pb – (5.2/10) x 100% - 52%
Relative Isotopic Mass = (Mass 204Pb x % Ab) + (Mass 206Pb x % Ab) + (Mass 207Pb x % Ab) + (Mass 208Pb x % Ab) = (204 x 2/100) + (206 x 24/100) + (207 x 22/100) + (208 x 52/100) = 207.20
Relative Isotopic Mass = (204Pb x % Ab) + (206Pb x % Ab) + (207Pb x % Ab) + (208Pb x % Ab) = (204 x 2/100) + (206 x 24/100) + (207 x 22/100) + (208 x 52/100) = 207.20
Normal Mass Spectrometer Vs High Resolution Mass spectrometer
Normal Mass Spectrometer
• Molecular formula/weight
by adding all relative atomic mass • RMM for molecule = Sum of all RAM • RMM O2 = 16 + 16 = 32 • RMM N2H4 = (14 x 2) + (1 x 4) = 32 • RMM CH3OH = (12 + 3 + 16 + 1) = 32 • Molecular ion peak -O2, N2H4, CH3OH - SAME = 32
RAM, O = 16 RAM, N = 14 RAM, H = 1 RAM, C = 12
High Resolution Mass Spectrometer Measure to RMM to 4/5 decimal places
• Molecular formula/weight
by adding all relative atomic mass • RMM for molecule = Sum of all RAM • RMM O2 = 15.9949 + 15.9949 = 31.9898 • RMM N2H4 = (14.0031 x 2) + (1.0078 x 4) = 32.0375 • RMM CH3OH = (12.0000 )+ (3 x 1.0078) + 15.9949 = 32.0262 • Molecular ion peak- O2, N2H4, CH3OH is the NOT the same
RAM, O = 15.9949 RAM, N = 14.0031 RAM, H = 1.0078 RAM, C = 12.0000 Vs
Mass spectrometer used to investigate isotopic composition of elements. Thallium has two isotopes shown below. 1) State symbol of two singly charged ions form. 2) State which ion will follow path marked X on diagram. Lighter -> DEFLECTED MORE 3) Doubly charged ions form. Suggest reason whether they would be deflected less than or more than ions at X and Y. DEFLECTED MORE. Cause deflection depends on m/z ratio. •Low Mass + High charge -> m/z ratio is low -> deflected more.
Naturally occuring boron has 2 isotopes, shown below. RAM of boron is 10.81.
% abundance x% (100 – x)% Determine percentage abundance of these isotopes. Answer: Let % abundance be x. 19% 81%
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203 205
81 81
TI TI 203 205
81 81
X =
203
81
B 10 B 11
Relative Isotopic Mass: = (Mass 10B x % Ab) + (Mass 11B x % Ab) = (10 x x/100) + (11 x (100 – x)/100) = 10.81 X = 19%
A sample of germanium is analysed in mass spec. The first and last processes are vaporization and detection. 1) State the names of other three processes in order in which they occur Answer: Ionization -> Acceleration -> Deflection 2) For each of the processes named in a (i), outline how the process occur Ionization -> Sample bombarded with high energy/high speed electrons Acceleration -> Cations (+ve charged ions) accelerated by an electric field Deflection -> Cations deflected by a magnetic field 3) Sample of germanium found to have following composition i)Define relative atomic mass. Average / weighted masses of all isotopes of an element. ii) Calculate RAM of sample, giving answer to two decimal places. 19%
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Relative Isotopic Mass = (Mass 70Ge x % Ab) + (Mass 72Ge x % Ab) + (Mass 74Ge x % Ab) + (Mass 76Ge x % Ab) = (70 x 22.60/100) + (72 x 25.45/100) + (74 x 36.73/100) + (76 x 15.22/100) = 72.89
The following shows a mass spectrometer. 1)Identify the parts labelled A, B and C.
2)State and explain which one of the following will undergo greatest deflection. Answer : Greatest deflection -> lowest mass + highest charged -> m/z -> lowest 3) Mass spectrum for an element shown below: i) Explain why there is more than one peak. Existence of isotopes ii) Calculate the relative atomic mass of the element.
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Relative Isotopic Mass
= (Mass 24 Y x % Ab) + (Mass 25 Y x % Ab) + (Mass 26 Y x % Ab) = (24 x 79/100) + (25 x 10/100) + (26 x 11/100) = 24.32
• electron gun • ionisation chamber • ionizer
• Electric field • Charged plates • Potential difference
Vaporized magnesium is introduced into mass spec. One of the ions that reaches detector shown below. 1)Identify the number of protons, neutron and electrons Answer : 12 protons, 13 neutrons, 11 electrons
2) State how this ion is accelerated in mass spectrometer. Using a strong electric field/strong opposite charged plate/potential difference 3) The ion is also detected by changing the magnetic field. Deduce and explain by reference to m/z values of these two ions of magnesium, which of the ions and is detected using a stronger magnetic field. Answer: - due to lower charge -> m/z is higher -> deflected less -> needs a stronger magnetic field to deflect.
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Cations (+ve) accelerated by (-ve) plates
Mg+ 25
12
Mg2+ Mg+ 25 25
Mg2+ 25
Mg+ 25
Smallest deflection – high mass, low charged Mg+ 25
Strong magnet/magnetic field to deflect it to bottom
Rubidium contains two stable isotopes shown below. RAM for rubidium is 85.47 1) Calculate % of each isotope in rubidium. Answer : Let % abundance be x %.
% Abundance x% (100 – x)%
1) 76.5% 23.5%
2) Vaporized sample is ionized and accelerated in a mass spec. How the use of magnetic field and detector enables the percentage of two isotopes to be determined.
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Rb 85
Rb 87
Rb Rb 85 87
Relative Isotopic Mass: = (Mass 85Rb x % Ab) + (Mass 87Rb x % Ab) = (85 x x/100) + (87 x (100 – x)/100) = 85.47 X = 76.5%
Rb
Rb Rb 85 87
Detector • Convert abundance M+ ions to electrical current. • M+ ions neutralize by electrons (more e needed - higher current – higher intensity of peak) •Ratio of intensity peaks show ratio of ions in sample •Ratio of height of peaks due to 85Rb : 87Rb –> 76.5 : 23.5
Magnetic field/Deflector • M+ ions deflected by magnetic field