Finding the molecular formula from a mass spectrum

Summary

This 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

00:00

🔬 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.

05:01

📊 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.

10:02

🧬 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.

15:05

🔍 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.

Keywords

💡Mass Spectrum

A mass spectrum is a plot of the intensity of ion signals as a function of the mass-to-charge ratio. It is a key tool in analytical chemistry for determining the elemental composition of a sample. In the video, the mass spectrum is central to understanding how molecules are ionized and fragmented, with the electron impact ionization being the method discussed.

💡Electron Impact Ionization

Electron impact ionization is a process where electrons are accelerated towards a molecule, causing it to lose an electron and become an ion. The video explains this process in the context of mass spectrometry, where a gas-phase molecule is ionized by colliding with high-energy electrons, leading to the formation of a molecular ion.

💡Molecular Ion

The molecular ion is the ion formed when a molecule is ionized without any loss of atoms. It is often the peak at the highest mass in a mass spectrum and is crucial for determining the molecular weight of a compound. The script describes the molecular ion as the starting point for interpreting mass spectra and determining the molecular formula.

💡Fragmentation

Fragmentation refers to the process where a molecule breaks apart into smaller pieces after ionization. The video script discusses how the original molecule often fragments in multiple pathways, and these fragments are what are observed as peaks in the mass spectrum, not the neutral pieces.

💡Isotope Peak

An isotope peak is a peak in a mass spectrum that corresponds to a molecule containing one or more heavy isotopes of an element. The script explains how the presence of carbon-13, for example, can create an M+1 peak, which is used to calculate the number of carbon atoms in a molecule.

💡Odd Electron Ion

An odd electron ion is a molecular ion that retains an unpaired electron after ionization, resulting in an odd number of electrons. The script mentions that these ions can be useful clues in determining the structure of the original molecule due to their presence in the mass spectrum.

💡Puzzle-Solving

The script uses the term 'puzzle-solving' to describe the process of interpreting a mass spectrum and determining the molecular structure. It suggests a stepwise approach: determining a molecular formula, drawing possible structures, predicting fragmentation products, and matching these predictions with the observed spectrum.

💡Double Bond Equivalent

Double bond equivalent is a concept used to assess the plausibility of a proposed molecular formula by calculating the number of rings and double bonds it implies. The script mentions this concept as a quick test for the plausibility of a molecular formula derived from a mass spectrum.

💡High-Resolution Mass Spectrometer

A high-resolution mass spectrometer is an instrument that can measure the mass of ions with very high precision, typically to the nearest 0.001 atomic mass units. The script suggests that such instruments can provide more accurate data for determining the molecular formula of a compound.

💡Quadrupole Instrument

A quadrupole instrument is a type of mass spectrometer that uses a set of four rods to filter ions based on their mass-to-charge ratio. The script mentions the use of quadrupole instruments in conjunction with special calibration processes to achieve high accuracy in mass measurements.

💡NIST Mass Spectral Library

The NIST Mass Spectral Library is a comprehensive collection of mass spectra data provided by the National Institute of Standards and Technology. The script mentions that the spectra used in the lesson are taken from this library, indicating its importance in mass spectrometry education and research.

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