Unit 1 - D Spectrometry and Spectroscopy

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Transcript Unit 1 - D Spectrometry and Spectroscopy

• Mass spectrometry
• is used to detect isotopes.
• mass spectrometer uses an ionizing beam of electrons to analyze
a sample of an element by turning atoms into ions.
• The individual ions are then sorted by mass.
• Since a sample of a single element can contain different
isotopes with different masses.
• a number of distinct ions of different masses can be identified
within the spectrum.
• data can lead to the identification of different isotopes and the
calculation of average atomic masses.
• Analysis of mass spectra
• A typical mass spectrum for chlorine is shown below:
• Like PES,
• Y-axis: relative intensity (indicates abundance of each isotope)
• X-axis: mass/charge ratio (equivalent to the mass of each isotope)
• Chlorine has two isotopes, with masses of 35 and 37, in a
3:1 ratio (relative peak height).
• When chlorine atoms pair together in chlorine molecules,
there are three possibilities for their mass;
• Cl2 70 (two Cl-35 atoms),
• Cl2 72 (one Cl-35 atom and one Cl-37 atom) and
• Cl2 74 (two Cl-37 atoms).
• Since Cl-35 is three times more prevalent, molecules with
masses that are made from Cl-35 atoms are more
abundant compared to Cl-37 atoms.
• Task 2D
• 1. Naturally occurring chlorine molecules, Cl2, have masses of 70,
72 and 74 amu as seen in the mass spectrum above. They occur in
the percentages 56.25%, 37.50% and 6.250% respectively. Use
this data to calculate the average atomic mass of chlorine atoms
and to find the relative abundance of 35-Cl and 37-Cl isotopes.
• 2. Sketch the mass spectrum that you might expect to observe if
bromine were passed through a mass spectrometer. The two
common isotopes of bromine are Br-79 and Br-81 that are known to
exist in an approx. 1:1 ratio and like chlorine, bromine is known to
form diatomic molecules.
• 3. The mass spectrum for titanium produces peaks according to the
following data. Use the data to calculate the relative atomic mass of
• 4. A typical mass spectrum for Mg contains three peaks at m/z
values of 24, 25 and 26 respectively.
• (a) What does the existence of three peaks suggest?
• (b) The relative intensities of the three peaks (24, 25 and 26) are
found to be 63, 8.1 and 9.1 respectively.
• (i) What do these data tells us about the isotope with m/z = 26?
• (ii) Calculate the relative atomic mass of magnesium.
Spectroscopy and the Beer-Lambert Law
• Spectroscopy - the study of the interaction of
electromagnetic radiation and matter.
• Absorption spectroscopy methods involve a sample being
exposed to photons of varying energy, and then
measuring the extent to which the sample absorbs or
reflects that energy.
• Depending on the magnitude of energy (E) used (which
depends upon the frequency (ν) of the photons according
to E = hν), we can collect data about substances.
• infrared (IR) spectroscopy
• When covalent bonds are exposed to infrared radiation they absorb
that energy and tend to bend, stretch and vibrate.
• The interaction with the IR is unique for each type of bond, so IR
spectroscopy can be used to distinguish between compounds that
have different types of covalent bond.
• Ultraviolet (UV) and visible spectroscopy (PES)
• As we saw with PES, UV and visible light tends to cause electronic
transitions within atoms, so can be used to gather information
about electronic configurations.
• Beer-Lambert Law
• The Beer-Lambert law is used to relate the concentrations of
colored solutions to the amount of visible light they absorb.
Beer-Lambert Law
• The amount of absorbance is calculated using the formula A = a b c
• Where, A = absorbance,
a = molar absorptivity (a constant that depends on the material tested)
b = path length (the length of the sample that the light passes through)
c = concentration.
• When absorbance measurements are made at a
• (1) fixed wavelength,
• (2) in a cell of constant path length,
• both a, and b are constant, and the absorbance, A, will be directly
proportional to c.
• If a solution of a compound obeys the Beer-Lambert law,
• a plot of absorbance (y-axis) versus concentration (x-axis) gives a straight line
with a slope of ab.
• The y-intercept is zero (the line will pass through the origin of the graph). One
can use the graph to read corresponding concentrations and absorption