Download Ionization Energy

Ionization Energy
introduction
In this experiment, with help from our friends in the UK,
a mass spectrometer is used to investigate two aspects of
electron bombardment of gas phase atoms and molecules.
We first see how much energy is required to produce ions and
then examine what happens when this energy is exceeded.
The mass spectrometer produces ions from the atoms and
molecules by bombarding them with high energy electrons.
These electrons are produced by the filament shown in the
diagram below. The ions are then separated by accelerating them into a magnetic field in which they are deflected
to an extent dependent on their mass to charge ratio. To
prevent unwanted collisions, the instrument is evacuated
throughout the experiment to a pressure of about 1 x ­10-9
atm, and even the sample under study must be maintained
at this low pressure.
A basic equation of mass spectrometry is:
m
V
=
e H2 r2
This relates the mass to charge ratio of the ions to the accelerating voltage V and the magnetic field strength H. In
the spectrometer used in the video, which has a permanent
magnet and fixed slits, the mass selector control varies the
ion accelerating potential, which causes ions of chosen
m/e values to hit the detector. The small ion current thereby
produced in the detector can be amplified and displayed on
a meter or on a chart recorder by continuously scanning
a range of m/e values. The size of the ion current (or the
height of a chart peak) is proportional to the number of ions
of that m/e value.
Mass spectrometers often have controls for varying the
magnetic field. In this case, V is a constant in the above
equation, while H varies. The m/e value is still the important
quantity measured.
The energy of the electrons streaming off the heated filament
can be varied. Above a threshold value simple ionization
occurs, and at still higher energies ionization may be accompanied by break-up of molecules to produce fragment
ions. The ions produced are accelerated and deflected by
the magnetic field and some, with a particular radius of
curvature r, pass through three slits in metal plates and are
detected.
There are other methods of mass detection used. Wikipedia
has a nice section on the various types of mass detectors
in use.
Evacuated
Chamber
Ion Detector
Region of
Magnetic
Field
Ion Flight
Defined by
Mass and
Charge
Filament
Entry Port
Chart Recorder
1
I.
Ionization of Argon
A. Take down the data as it shows on the screen. (After data is given, put video on pause.)
Voltage/eV
Current/µa
__________ __________
__________ __________
__________ __________ __________ __________
__________ __________
__________ __________
__________
__________
__________ __________
Explain why the m/e ratio of 40 is used here? How would the machine be set if we were working
with neon?
___________________________________________________________________
1.0
B. After the video is finished, plot the data: voltage on x
axis, current on y axis.
0.9
C. Extrapolate the line down to the x axis, making the best
straight line you can through the greatest number of points.
Do not use the first few points, which will be curved away
from the majority of the points. The x intercept at 0 µamp
is the voltage you want.
____________ eV
____________ kJmol-1
(1eV = 96.5 kJmol-1 ; 1500 kJmol-1 = book value)
% difference =
=
graph value − 1500
1500
kJ
mol
0.7
Current in µa
Voltage =
0.8
kJ
mol x 100
0.6
0.5
0.4
0.3
0.2
%
0.1
0.0
15.0
16.0
17.0
Voltage in eV
2
18.0
For the rest of the data, the voltage was set to 70 eV. This is well in excess of atomic and molecular ionization energies, and
can even cause some second ionizations. When the m/e ratio is from a 1st ionization (e=1), then the m/e ratio equals the
molecular mass. If the m/e ratio is from a 2nd ionization (e=2) , then the m/e ratio is 1/2 the molecular mass.
II. Air in Mass Spectrometer
A. Take down the m/e ratios as given on the video, and then later, the peak heights & causing species.
m/e ratio
______________
______________
peak height
species
______________
______________
______________
______________
B. The N2/O2 mole ratio in air is 4 to 1. From the data above, calculate the peak height ratio of N2+ to O2+.
N2+ peak height /O2+ peak height =
______________
Note that on page 1, it is stated that: the size of the ion current (or the height of a chart peak) is proportional to the number
of ions of that m/e value. What should the peak height ratio be for the ions to make the ratio correspond with the abundance
of the molecules?
To properly interpret the observed peak heights, we would need to see the peak heights of atomic N and atomic O, formed
by the breakup of the N2 and O2 molecules. How do you predict what these peak heights will look? Note that the total
number of nitrogen atoms, whether in N2 or in N atoms, will be four times the total number of oxygen atoms, whether in
O2 or in O atoms. See the illustration below to help you visualize what is happening here.
After electron bombardment: All are charged;
50% of the O2's, 25% of the N2's break up
Before electron bombardment:
N N
N N
N N
O O
N
N N
N N
N
+
N N
+
+
+
N N
N
N
+
+
+
+
O
N N
N N
+
O
N N
+
+
O O
+
N N
+
N
N
N N
N N
O O
What do you think this data says about the relative bond strength of the O2 versus the N2 molecule?
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Name_________________________________________ Grade___________ Date ___________
III. H2O in Mass Spectrometer
A. Take down the m/e values as given, and as best you can, estimate the relative peak heights. (Video will have to
be on pause to estimate.) Take down the species as they are later given during this video segment.
m/e
peak height estimate
species
______________
______________
______________
______________
______________
______________
______________
______________
______________
______________
______________
______________
B. The peaks from the water vapor sample overlap the fragment ions from the oxygen and nitrogen present in the atmosphere. Thus, ions of mass to charge ratio 18, 17 and 16 can be identified as H2O+, HO+ and O+ from ionization of
the water molecules and loss of one or both hydrogen atoms. Air present will contribute fragment ions O+ and N+ at
m/e = 16 and 14 as a result of dissociation of O2 and N2 molecules. A small contribution to the peaks at m/e 16 and
14 will also come from the double ionization of the oxygen and nitrogen molecules to produce O22+ and N22+ which
will have m/e values of 32/2 and 28/2. What could be done to limit the peaks to singly ionized species rather than to
multiple ionized species? (Reread paragraph just before part II, if necessary.)
IV. CH3Cl in Mass Spectrometer
A. Take down the m/e values as given and estimate the peak heights as best you can. (Video will have to be on pause
to make an estimate.) Take down the species as they are later given during this video segment.
m/e
peak height estimate
________
_______
________
_______
species
________
________
B. The fractional value for the relative atomic mass of chlorine (35.46) results from the mixture of atoms of differing
mass in the naturally occurring element. These isotopes are 35Cl and 37Cl, which occur in the ratio of approximately
3:1. Chloromethane, like other compounds of chlorine, also shows a similar isotopic abundance ratio and this is seen
from its mass spectrum.
37Cl
to 35Cl? Explain.
Does the peak height ratio correspond to the isotope abundance ratio of
Do you think this machine shows enough accuracy to be used as an analytical tool? Why or why not?
V. Question:
Write out a description of the basic principle of operation of the mass spectrometer.
(Include ionizing energy and magnetic fields in your answer.)
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