Finding the molecular formula from a mass spectrum
Summary
TLDRThis educational video script delves into the interpretation of mass spectra generated by electron impact ionization, focusing on organic molecules. It introduces the concept of molecular ions and their fragments, highlighting the significance of the parent ion and the use of isotope patterns to deduce molecular formulas. The script guides viewers through the process of determining molecular structures by analyzing mass spectra, employing strategies such as calculating carbon numbers and considering the presence of elements like chlorine, bromine, and sulfur. It emphasizes the art of puzzle-solving in mass spectrometry and the importance of flexibility and high-resolution data in accurate molecular formula determination.
Takeaways
- 🔬 The video introduces the process of interpreting mass spectra, focusing on those produced by electron impact ionization.
- 🌟 The spectra used in the lesson are sourced from the NIST mass spectral library with permission.
- 💡 In the ionization chamber, a heated tungsten filament emits electrons that collide with gas-phase molecules, knocking out an electron and creating a positive ion.
- 🚀 The collision is often violent enough to cause the molecule to fragment, with the positive charge and unpaired electron typically ending up on different fragments.
- 🔎 The mass spectra will only show peaks from charged fragments, not neutral ones, making it crucial to observe the charged fragments to interpret the spectrum.
- 🔍 The molecular ion is a key peak in the spectrum, often at the high mass end, and can help in determining the molecular weight of the original molecule.
- 🧩 The main objective in interpreting a mass spectrum is to determine the molecular structure, which involves a step-by-step approach of determining a molecular formula, predicting fragmentation products, and finding evidence for these in the spectrum.
- 🌐 The number of carbon atoms in a molecule can be inferred from the intensity of the M+1 peak, which is related to the presence of carbon-13 isotopes.
- 🌀 The presence of other elements like chlorine, bromine, and sulfur can affect the mass spectrum, showing distinct patterns due to their isotopes.
- 🌡 High accuracy mass data from high-resolution mass spectrometers can provide more precise information for determining molecular formulas and will be discussed in future lessons.
Q & A
What is the primary focus of the video script?
-The video script focuses on interpreting mass spectra, specifically those produced by electron impact ionization, and discusses strategies for determining molecular structures from these spectra.
What is the role of the heated tungsten filament in the ionization chamber?
-The heated tungsten filament in the ionization chamber emits electrons that, when accelerated by an electric field, collide with gas phase molecules, causing ionization and the production of positive ions.
Why are the collisions between electrons and molecules often violent enough to cause the molecules to fragment?
-The collisions are violent because the electrons are accelerated to a high kinetic energy of 70 electron volts, which is much greater than the ionization potential of organic molecules, leading to fragmentation after ionization.
What is a molecular ion and why is it significant in mass spectrometry?
-A molecular ion is an ion that retains the original composition of the molecule after losing an electron. It is significant because its peak in the mass spectrum, often at the high mass end, provides information about the molecular weight of the original molecule.
How can the presence of a single bond breaking in a molecule lead to the formation of charged fragments?
-When a single bond breaks, the positive charge and the unpaired electron typically separate, resulting in two charged fragments, one with the positive charge and the other with the unpaired electron.
What is the general approach chemists follow to interpret a mass spectrum and determine molecular structure?
-The general approach involves determining a reasonable molecular formula first, then drawing possible structures for that formula, predicting fragmentation products for each candidate structure, and finally looking for evidence of these predicted fragments in the mass spectrum to decide which structure fits best.
How can the presence of carbon-13 isotopes affect the mass spectrum of an organic molecule?
-The presence of carbon-13 isotopes can create a satellite peak at M plus 1 (where M is the mass of the molecular ion) in the mass spectrum. The relative intensity of this peak compared to the molecular ion can be used to estimate the number of carbon atoms in the molecule.
What is the significance of an odd molecular weight in determining the number of nitrogen atoms in a molecule?
-An odd molecular weight indicates an odd number of nitrogen atoms because nitrogen contributes one amu to the molecular weight and has an odd atomic number.
How can the presence of chlorine, bromine, or sulfur affect the mass spectrum and the calculation of the number of carbon atoms?
-The presence of these elements, which have isotopes one or two atomic mass units higher than their most abundant forms, can contribute to the intensity at M plus 1 or M plus 2, affecting the calculation of the number of carbon atoms. Their contributions must be accounted for or corrected before calculating the carbon number.
What is the concept of 'double bond equivalence' and how is it used in determining a plausible molecular formula?
-The concept of 'double bond equivalence' is used to estimate the number of rings and double bonds in a molecule based on its molecular formula. It is calculated by taking the number of carbon atoms, subtracting half the number of hydrogen atoms, adding half the number of nitrogen atoms, and adding one. If the result is an integer, it suggests a plausible molecular formula.
How can high-resolution mass spectrometry provide more accurate data for determining molecular formulas?
-High-resolution mass spectrometers or quadrupole instruments using a special calibration process can provide mass assignments for ions and fragments accurate to the nearest 0.001 atomic mass units, which can help in more accurately determining the molecular formula.
Outlines
🔬 Electron Impact Ionization and Mass Spectrum Interpretation
This paragraph introduces the process of electron impact ionization in mass spectrometry, where electrons with high kinetic energy collide with gas-phase molecules, causing ionization and fragmentation. The focus is on interpreting mass spectra from the NIST mass spectral library. It explains the production of positive ions and the resulting molecular ion peak, which is crucial for determining molecular structure. The paragraph outlines a general approach to interpreting mass spectra, which includes determining a molecular formula, predicting fragmentation products, and matching these predictions with observed fragments. The importance of the molecular ion peak and the presence of odd electron ions as clues for structure determination are highlighted.
📊 Isotope Patterns and Calculating Carbon Atoms in Mass Spectra
This paragraph discusses the use of isotope patterns in mass spectra to infer the number of carbon atoms in a molecule. It explains the concept of the M plus 1 peak, which is indicative of the presence of carbon-13 isotopes, and provides a method to calculate the number of carbon atoms based on the relative intensity of this peak. The paragraph also addresses the presence of other elements with isotopes that can affect the M plus 1 peak, such as chlorine and bromine, and how to account for these when determining the number of carbon atoms. It also introduces the distinctive isotope patterns of chlorine and bromine, which can be used to identify these elements in a molecule.
🧬 Molecular Formula Deduction Using Mass Spectra and Isotope Patterns
The paragraph delves into the process of deducing molecular formulas from mass spectra by identifying the molecular ion and using the intensities of associated peaks to calculate the number of carbon atoms. It discusses the use of the molecular ion's mass and the presence of nitrogen, as indicated by odd or even molecular weights, to further refine the molecular formula. The paragraph also introduces the concept of double bond equivalence to assess the plausibility of a proposed molecular formula. It provides examples of how to apply these principles to interpret mass spectra and generate reasonable molecular formulas for different compounds.
🔍 Advanced Techniques for Molecular Formula Determination in Mass Spectrometry
This paragraph discusses advanced techniques for determining molecular formulas with greater accuracy in mass spectrometry. It highlights the use of high-resolution mass data and the importance of considering contributions from elements like sulfur to the M plus 2 peak. The paragraph explains how to adjust for these contributions when calculating the number of carbon atoms and provides an example of how to deduce the molecular formula of diethyl sulfide from its mass spectrum. It concludes with a note on the iterative and flexible nature of interpreting mass spectra, emphasizing the need for high accuracy and the use of additional data sources for more precise molecular formula determination.
Mindmap
Keywords
💡Mass Spectrum
💡Electron Impact Ionization
💡Molecular Ion
💡Fragmentation
💡Isotope Peak
💡Odd Electron Ion
💡Puzzle-Solving
💡Double Bond Equivalent
💡High-Resolution Mass Spectrometer
💡Quadrupole Instrument
💡NIST Mass Spectral Library
Highlights
Introduction to interpreting mass spectra, focusing on spectra produced by electron impact ionization.
Use of NIST mass spectral library with permission for teaching purposes.
Explanation of the ionization process involving a heated tungsten filament and an electrical plate.
Description of how gas phase molecules become positive ions through electron collisions.
Discussion on the fragmentation of molecules post-ionization and the formation of charged fragments.
Importance of observing the parent ion and its odd electron count for structural determination.
General approach to interpreting a mass spectrum: determining molecular formula, drawing structures, predicting fragments, and matching to the spectrum.
Identification of the molecular ion peak as the most useful in a mass spectrum.
Method to determine the molecular weight of an organic molecule from the molecular ion peak.
Use of isotope patterns to infer the number of carbon atoms in a molecule.
Explanation of the significance of the M plus 1 peak in mass spectra for carbon atom count.
Impact of other elements with isotopes on the M plus 1 peak and the need for correction in carbon atom calculations.
Unique isotope patterns of chlorine, bromine, and sulfur in mass spectra and their implications for molecular structure.
The role of nitrogen in determining odd or even molecular weights in mass spectra.
Application of the principles of odd and even molecular weights to deduce the presence of nitrogen atoms.
Use of high accuracy mass data from high-resolution mass spectrometers for precise molecular formula determination.
Practical examples demonstrating the process of deducing molecular formulas from mass spectra.
Importance of flexibility and cross-referencing information when interpreting mass spectra.
Transcripts
this video provides an introduction to
interpreting mass spectrum we will focus
on spectra produced by electron impact
ionization in this lesson spectra used
in this lesson we're taken from the NIST
mass spectral librarian were used with
permission the vacuum chamber where
ionization takes place contains a heated
tungsten filament below this wire is an
electrical plate a voltage of 70
electron volts is applied between the
filament and the plate electrons escape
the filament and speed toward the plate
with a kinetic energy of a 70 electron
volts an energy that is far greater than
the ionization potential of organic
molecules so if we introduce a gas phase
molecule into the chamber so that it
drifts into the beam a collision with
one of these electrons will knock out an
electron from the original molecule and
produce a positive ion this collision is
usually so violent that even after
losing the electron the molecule is left
reeling and often breaks apart if a
single bond breaks the positive charge
remains on one piece and the unpaired
electron ends up on the other usually
the molecule fragments in multiple
pathways occasionally both charged and
unpaired electron remain on the same
piece in order for this to happen
two or more bonds must break in the
process this will be very important to
us later since all of our mass sorting
strategies involve manipulating the
charged fragments we never are able to
observe the neutral pieces directly our
mass spectra will only show Peaks from
these charged fragments if we are lucky
we will still be able to observe a
signal from the original molecule or
parent ion we will occasionally refer to
the fact that the parent ion has an odd
number of electron other odd electron
ions that appear in this spectrum will
be especially useful clues to
determining the structure of the
original molecule we'll come back to
this idea in a later lesson usually the
main objective in interpreting a mass
spectrum is determining molecular
structure getting from the spectrum to a
structure amounts to solving a puzzle
puzzle-solving is an art people have
their own strategies but most chemists
follow this general approach first
determine a reasonable molecular formula
then draw possible structures for that
formula then predict what fragmentation
products will be produced for each
candidate structure finally look for
evidence for the predicted fragments in
order to decide which structure fits
best
this lesson will concentrate on the
first step let's take a look at a few
simple mass spectra and attempt to
interpret them the single most useful
peak in the spectrum is associated with
the molecular ion let's assume that the
peak at 26 mass units is the molecular
ion the molecular ion is not always the
tallest peak in the spectrum however we
know that it must be at the high mass
end of the scale we're going to assume
that we're working with organic
molecules in this video what organic
molecule has a molecular weight of 26
pause the video a moment and when you
have an idea go on
most chemists would reason this way we
know that an organic molecule has at
least one carbon atom carbon has an
atomic mass of 12 so one carbon atom
would account for 12 units so 26 minus
12 leaves 14 nitrogen weighs 14 atomic
mass units but CN is not a realistic
molecule let's try two carbons they
would have come for 24 atomic mass units
leaving two AMU for something else it's
hard to account for two atomic mass
units in any way other than two atoms of
hydrogen we have a molecular formula of
c2h2
acetylene
let's try another let's assume that the
molecular ion appears at 128 what do you
think this molecule is
when the mass gets bigger the problem
gets much tougher but what if I told you
this molecule has 10 carbon atoms then
what would you say is a reasonable
molecular formula
if we have 10 carbon atoms then we can
account for 120 amu that leaves eight
mass units to account for if we assume
now that we are not dealing with an
organometallic compound then we can
infer that eight implies eight hydrogen
atoms so we have C 10 H 8 which is the
formula for naphthalene I hope that you
see that knowing the number of carbon
atoms is a very valuable clue in
generating a reasonable molecular
formula let me show you how to find the
carbon number when we began looking at
the spectrum I asserted that the peak at
128 was associated with molecular I we
might think that the mercury ion ought
to be the heaviest ion in spectrum since
everything else is a fragment of that
parent ion we ought to appear at the
high end of the mass scale but if we
look closely at the spectrum we see a
small peak just to the right at 129 amu
for ease of reference let's refer to the
mass of the molecular ion as M and the
peak to the right as M plus 1 this
satellite peak is associated with
versions of the parent molecule that
contained one heavy isotope of carbon
here is a table of the most common
elements found in organic molecules to
these elements the lightest isotope is
the most abundant of the common isotopes
for that particular element the table is
telling us that for every 100 atoms of
carbon that have a mass of 12
there are approximately 1.1 atoms of
carbon 13 so what is the implication in
a spectrum of methane we would expect to
find a peak at n plus 1 that is 1% as
tall as the molecular ion in ethane or
any other molecule with two carbons we
would expect the heavy isotope peak to
be 2.2 percent as tall as the molecular
ion M a three carbon molecule would have
a M plus one peak that is 3.3 percent as
tall a four carbon molecule will have an
isotope peak that is four point four
percent is tall and so on we can say
then that the relative intensity of the
M plus one peak compared to the
molecular ion intensity times 100
percent
is equal to one point one percent times
the number of carbon atom or said
another way the number of carbon atoms
can be calculated from this ratio times
one hundred percent divided by 1.1
percent let's see how that applies to
the last example here we have a cluster
of ions in two high mass end and the
relative intensities we divide the 10.9
by 100 and multiply by 100 percent over
one point one percent and we get
approximately ten carbon atoms real data
have some uncertainty associated with
them so we don't expect an exact integer
when the signals are weak the relative
uncertainty may be large and we have to
be prepared to be flexible in our
conclusions let's see what else we can
learn from this table notice that there
are a few other elements that have fe
isotopes that are one atomic mass unit
higher than their most abundant forms if
we expect that these elements are
present they will contribute to the
intensity at M plus one and we should be
ready to correct for that contribution
before calculating the number of carbon
atoms we will see an example of this
later
chlorine stands out from the other
elements in that it has a heavy isotope
two units higher than the most abundant
isotope at 35 it's also distinctive
because the heavy isotope is practically
one third is abundant as the light
isotope this leads to an unmistakable
pattern in the high mass cluster in
which a peak 1/3 is tall as the
molecular ion appears at M plus two with
two chlorine atoms per molecule we see a
pattern with a strong M plus two peak
and a significant peak at M plus four
their relative intensities also fed a
distinctive pattern roughly ten to six
to one here we see a molecule with three
chlorine atoms and we see a combination
of isotope Peaks which represents
molecules with one heavy chlorine two
heavy chlorines and three heavy chlorine
atoms this also follows a distinctive
pattern of intensities the intensity
pattern for the distribution of heavy
isotopes can be predicted from their
natural abundances there are several
websites that have free
calculator programs where you can
predict the distribution of isotopes for
a given molecular formula here is a good
link notice that bromine is also a
special case it's population is a nearly
1:1 mixture of isotopes with mass 79 and
mass 81 this fact also leads to an
unmistakable signature in the mass
spectrum here we see the molecular ion
at 156 units and a peak 2 units higher
with almost the same intensity with two
bromine atoms in a molecule the
probability is greatest for the molecule
to have one heavy atom and one light
bromine atom the pattern looks very much
like a triplet in an NMR spectrum the
tall pica 236 is actually the M plus 2
peak notice the doublet at 155 and 157
this is a strong clue that the fragment
at 155 contains one bromine atom a few
other elements have heavy isotopes that
are 2 units above the most abundant
isotope sulfur is a good example it has
a small but very significant
contribution to the M plus 2 peak oxygen
is a very important element in organic
compounds it has a very tiny
contribution to the M plus 2 peak this
signal may be useful when the molecular
ion is very intense usually it is only
suggestive nitrogen contributes weakly
to the M plus 1 peak although its
contribution is tiny it should be
subtracted out before calculating the
number of carbon atoms there is a very
useful concept that we can employ with
regard to nitrogen take a moment to look
at this table pause the video and see if
you can see a pattern in the data
off the compounds in the first column
have an even molecular weight and
contain no nitrogen atoms all the
compounds in the middle column contain
one nitrogen atom and exhibit an odd
molecular weight all of the compounds on
the right have an even number of
nitrogen's and once again exhibit an
even molecular weight so here is an
important principle an odd molecular
weight indicates an odd number of
nitrogen AB let's apply these ideas to a
few example spectra and generate
reasonable molecular formula for each
start by identifying the molecular ion
if it is present it should be in the
cluster of ions at the high mass end of
the scale let's assume that the tallest
peak in this cluster is the molecular
ion note that is not always the case
with the molecular ion at 68 the peak at
69 must be the n plus 1 ion let's use
their intensities to calculate the
number of carbon atoms we get 3.8 so we
assume 4 carbon atom this accounts for
48 atomic mass units leaving us with 20
we don't see any indication of chlorine
bromine sulfur or silicon from the n
plus 2 peak molecular ion has an even
mass so we can roll out nitrogen
fluorine has an atomic mass of 19 but if
an atom of fluorine were present that
would leave us with only one atomic mass
unit and consequently one hydrogen atom
let's assume that we have an oxygen atom
that leaves us four atomic mass units
that could be explained by four hydrogen
atoms so we have a molecular formula of
C 4 h4 o a quick test at the
plausibility of the molecular formula
can be made by calculating the number of
rings plus double bonds that this
formula implies you may have seen this
concept before sometimes it has been
called the number of double bond
equivalence we calculate the number of
double bond equivalence by taking the
number of carbon atoms subtracting half
the number of hydrogen atoms adding half
the number of nitrogen atoms plus one if
this is a plausible molecular formula
then this number should be an integer in
this case we get three a reasonable
number indeed this is suspect
of Furai will save the discussion of
fragmentation processes for another
lesson let's work a couple more examples
here we have the high mass cluster ions
in the table we might guess that the
tallest peak in that group will leave
them like it awry
if so the peaks at 73 and 74 must be the
result of heavy isotopes for this
particular formula the calculation for
carbon number gives us 3 so subtracting
36 atomic mass units from our molecular
weight gives us 36 atomic units
unaccounted for we see no evidence of
chlorine or bromine sulfur or silicon in
the M plus 2 peak we have an even
molecular weight so we would expect an
even number of nitrogen's since zeroed
also be an even number let's first try
the idea of no nitrogen's we could have
two oxygen atoms which would account for
32 atomic mass units leaving us four to
be explained by four hydrogen atoms
calculating the double bond equivalent
gives us two a reasonable number
croelick acid would be one structure
that we might draw that meets all of
these criteria let's go back to the
possibility of two nitrogen atoms we get
a difference of eight atomic mass units
which suggest eight hydrogen's and a
molecular formula of c3h8 and two once
again the double bond equivalence is an
integer so we have two possible formulas
that seem to fit the data for the high
mass cluster in order to decide which is
the better choice we will need to look
at the fragmentation pattern for each
structure that we can draw for these
two's formula that will take up in
another lesson let's try another once
again we look at the high mass cluster
we'll consider the possibility that the
tallest peak is the molecular ion using
the intensities for the n plus 1 peak
and the molecular ion we calculate 5
carbon atoms which accounts for all but
30 atomic mass units but in this case we
have a fairly strong n plus
to peak it's certainly not the result of
chlorine or bromine but it could
indicate the presence of sulfur if an
atom of sulfur were present then it
would contribute to the M plus 2
intensity a signal of 4.4% as intense as
the signal of the molecular ion 4 point
4 percent of 75.2 would give us a signal
of 3 point 3 units and that jives very
well with the intensity of the peak at
mass 92 also notice that a sulfur atom
will contribute to the M plus 1 peak so
if one sulfur atom is present we would
should adjust the intensity for the M
plus one ion before calculating the
number of carbon atoms you see that the
sulfur would contribute 0.62 the
intensity of the M plus one ion so we
correct the intensity at M plus 1 by
subtracting 0.6 from the value of 4.1 in
order to calculate the carbon number
this gives us a value of approximately 4
subtracting out the mass of 4 carbons
and one sulfur leaves us with 10 amu
which suggests the formula of c4h10 s
calculating rings plus double bonds
gives us a number of 0 a very good
number indeed this is the spectrum of
diethyl sulfide a take-home message here
is that there is some uncertainty in the
data that we use for calculating the
number of atoms of the various elements
in our molecular formula consequently
the process is somewhat similar to
solving a crossword puzzle we need to be
flexible with our choices and be ready
to give up on an earlier idea if it
conflicts with other information another
helpful approach to finding the best
molecular formula is the consideration
of high accuracy mass data that is mass
assignments for ions and fragments that
are accurate to the nearest 0.001 atomic
mass units that sort of data can be
provided by high resolution mass
spectrometers or quadrupole instruments
using a special calibration process we
will take up this topic
in another lesson
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