ANDERSON JUNIOR COLLEGEPHYSICAL CHEM]STRYMASS SPECTROMETRY

Contents

(l) lntroduction(ll) Mass Spectrometer - lts Basic Functions and Principles(lll) Interpretation of Mass Spectra

(A) The Mass Spectra ol an Element(B) The Mass Spectra of a Molecule

(lV) Applications of Mass Spectrometry(A) Determine Molecular Formula of a Compound(B) Determine the Molecular Structure of a Compound

ObjectivesCandidates should be able to:

(a) analyse mass spectra in terms of isotopic abundances and molecularfragments[knowledge of the workings of the mass spectrometer is not required]

(b) calculate the relative atomic mass of an element given that the relativeabundances of its isotopes, or its mass spectrum

References

Longman A - Level Guides Chemistry (3' Edition); J.G.R. BriggsA - Level Chemistry (2"d Edition); E.N. RamsdenSpectroscopic ldentification ol Organic Compounds (5'n Edition); R.M. Silverstein, G.Clayton Bassler, Terence C. Morrill

http://masspec.scripps.edu

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INTRODUCTION

Mass spectrometry is an instrumental approach that can be used to determine:

the relative atomic mass of an elementthe relative isotopic masses and abundances of isotopesthe relative molecular masses andslructure of organic compounds

.t The instrument used is known as a mass spectrometer.It measures the mass-to-charge (m/e) ratio ol a chemical species and thisinformation is translated into a graphical plot known as a mass spectrum.

F.Y.f. : History of mass spectronelry @xtracted from http://masspec.scripps.edu)The history df mass spectrometry began with Sir J.J. Thomson of the CavendishLaboratory ol the University of Cambridge, whose studies on electrical discharges ingases led to the discovery ol the electron in 1897.

In lhe tirst decade of the 20tn century, Thomson went on to construct the lirst massspectrometer (then called a parabola spectrograph) tor the determination of mass-to-charge ratios of ions.

Thomson received the 1906 Nobel Prize in Physics "in recognition of the greal meritsof his theoretical and experimental investigations on the conduction of electricity bygases".

Thomson's prot6g6, Francis W. Aston ol the University of Cambridge, designed amass spectrometer in which ions were dispersed by mass and tocused by velocity -which improved the resolving power by an order of magnitude over the resolutionThomson had been able to achieve. Aston received the 1922 Nobel Prize inChemistry for isotopic studies carried out with this type of instrument.

Around 1920, professor of physics, A.J. Dempster of the University ol Chicagodeveloped a magnetic deflection instrument with direction locusing - a format lateradopted commercially and still in use today.

Dempster also developed the first electron impact source, which ionizes volatizedmolecules with a beam of electrons from a hot wire filament. Electron impact ion

i sources are still very widely used in modern mass spectrometer.:.................,,.......

* Mass spectrometers have become crucial for a wide range ol applications as theprocess is fast and accurate and only a very small sample (10'to 10-'' g), whichcan be in solid, liquid or gaseous state, is needed.

Examples of analysis of inorganic, organic and bio-organic chemicals. dating of geologic samples. drug testing and discoveryo process monitoring in the petroleum, chemical and pharmaceutical

industries. structural identification of unknown comoounds

(t)

(D(ii)(iiD(iv)

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(il)

I

MASS SPECTROMETER -

ITS BASIC FEATURES & PRINCIPLES

3) Acceleration Chamber.l A high voltage is applied between

olates X and Y

* Positive ions are accelerated byan electric field towards olate Y

4) Maonetic Fleld* Positive ions are deflected by a strong magnetic field

* Deflection is inversely proportional to mass/charge (m/e)ratio

- lons with large m/e are deflected less

- lons with small rn/e are deflected more

Species Charge. e mle DeflectionNa* -rt 23/l = 23 LessNa'- 'fz 23/2 = 11.5 More

d0t'.1

Fo

1) Vaporisation Chamber.:. Sample is vaporized into

gaseous aloms ormolecules

+ Gaseous sample isdirectly introduced intochamber

+-'.^ Lt"Uv - {ottl.

2) lonlsation Chamber* Heated filament produces high energy

free electrons which bombard samole

{. produces gaseous posilive ions whichare usually singly charged

Vacuum Dump

Maintains a vacuum insidemass spectrometer as airmolecules will blockmovement of the ions

5) Detector and recorder* Determines oosition and

number of ions on detector

* Data is then fed into arecorder or computer whichthen generates a massspectrum

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* Elements with isotopes will produce a mass spectrum with the number of peaks &jrcorrgsponding to thg number of isotopes present

I wltrr trrn rrurrr,nr ur ""o^" 'f Il l , l

.:. The following information can be obtained directly from the mass spectrum of "n

l#t'

element:

. the isotopes present

. the relative isotopic mass of each isotope

. the relative abundance of each isotope (also known as peak height)

* The relative atomic mass (A,) of the element can then be calculated from therelative isotopic mass of the isotopes and their relative abundances using thefollowing formulae:

(ilD INTERPRETATTON OF MASS SPECTRA(A) The Mass Spectra of an Element

r rA

* Elements that do not have any isotopei will produce a mass spectrum with 1 'l 'i "peak only I Il l . rzi tn It

&=Z (..lrlNc aur.n lara X flcbtrrr fri'?'r

-ars )Z (nla-tiv< atudrla )

/ ( l. asunlar@ \ rrlali ve ttoly;. -ost

)nr - tco

%a6undsnsg= ftar ila't\^l r 199Tclqt Paar $ eigh*

OR

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Example 1: Masspectrum of lead, Pb. Calculate the A of Pb.

Peak Heighl

m3 204 205 206 207 204 2@

N= (zo+v o,z) * t Ly 2.9) + (

'oz x ' t ,z) + lL"S y 5.a)

o.2 + 7.9 + 7.2+ 5.2

b?.>')-

,"4 l,

m/e= Relativelsotopic

Mass

lsotope lon RelativeAbundance

"/o Abundance

204,ot Pt '*Pt* 9.1 2!L * r* -- ,o1o

206. tob ol

l l t -'Pb* z+ 3I l ton. z+l-lo207

^'Pb ofb* 9 ,1 t* = 22' /"208

"tPb '*Pb* 5,L i l x rco = sz%

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Try thisl -

Mass spectrum of Neon, Ne. Calculate the A of Ne.

BelativeAbundance

19 20 21 22 23

( r -xrr t ) + ( t t x o 'z) + [1!x l l r ' )i , A, r15+ o.2 { l r .2

ll= 2a,>

rn/e= Relativelsotopic

Mass

lsotope lon RelativeAbundance

7o Abundance

20-Ne

hNcY I l l+ #** ' t6

' qs.1l

21 At.rNc ' t &* 0, r. 9t3- ^t* -- o. r b

22"'Nn "N.t i l1 1#*^ '* = & 13

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* The relative abundance of isotopes of an elema can also be calculated

Example 2

Naturally occuning chlorine contains 2 isotopes, scl and 37C1. Calculate thepercentage abundance of these 2 isotopes, given that the relative atomic mass ofchlorine is 35.5.

kt o/o eb"n Aanc. .+ " iCL tc r t ,

. ' , '1" adrlat'q 4 5?C X tr I too - r ' )

o1o

*=

35.6a5/

.5cL

3" c7

tlr" c , alo

%a\WnlanQ

alual er.cr

.(4

Tsetd .,

(ps- 45) ' lo,5"k I

too

Try this!

The element Rubidium, Rb, comprises 2 isotopes, ssRb and 87Rb. Calculate thepercentage abundance of these isotopes, given that the relative atomic mass ofrubidium is 85.6.

Lt 1o aL"&re r{ "tRb k , '1" .

", 1o Gvqrat4 a '{ {1c1, ;r ( r"t -r ) %

rla" r ,%%

alo,,a /avr c

6bo^l tvta

= 86L

-- +o

sAn' ttb

4*+

i s 1o'/o ,

' ( roo -*rYo-- zoX, f,

8 5.t

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" (B) Tha Mass Spectra of ltlolecules

s Recall : In ionization chamber, high energy eleclrons bombard tho moleculos andproduces po6itive ions

* Posiiive ions fomed may be

(1) Molecular or oarent ions

. 4"s 4-r"a). .ln r" rna!3rk! r.rrfir ry tcr s 4 .t.,,",! o,lo!^9 , F!4

-*,glq* "\3! .,p.|.r..a. .-M (5) + e-

--

Mt (1) + e- re-

a' E'g. ! t" t3) + @- "t cl ' r+ f ?) t o' + e-

.

Cat L &-r a+ flc b'54+ .fe volu! i\4 rll Ma$sp?ct'.rro / Nihr.h t, al" {re ru lallvc tdcetlAr

-at ' , fc'* 1,-)

. It the molecule (RMM = M) contains atoms that have heavier isotopes,then ions may be found at (M+1) or (M+2) m/e values

(2) Fraoment ions

. e.d dt asglnrt b.n.l| rrlr\t' ' d. cul.s 6i? Lolrn .

P h" *t6 ?o'eb' "E.g. ttL1, q\ H- (t) * ct+ t3l '\"b

- 6^ '-

'

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Example 3: Mass spectra of ethanol, CH3CHTOH (ueithout isotopes)

.z-g(l

m/e

rsv ' l

n

.lhabl

25 30

'( t ! .o) ++b.o

35 ,O 45

atr ,o) . r lb o

mle Structure of fragments B6nds Broken

46+

I crl5 cl t ou'1 ' ion'

45^lc = 45 =4t ' - \

(cu 3 LAy O)r

H t l It l

A-L'c-of \l l

hH

3lvl. ' 31 = +1o ' t l

( c*- o.x)Tr l l

H-1+L-o-n1l lH:f '

29nle = >4'- 4L' ta

( ca gca,\a

HHIl l l

A-c-. - l ]_o-r ,t l) . l t i

28-1" "

29 = +'- t '

Ccrl 5 Ct l )7( c4ocat\+

| 4t '\ l , . I l ' l . -l t - a

-L*o- l . l l H-L-cio ' t ' lt , { l -+.r ' .r r 'h I u \ , ,

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' Example 4: Mass spectrum of hydrogen chloride, HCI (with isotopes)

RelativeAbundance

Chlorine molecules comprise of sCl and 37Cl isotopes.

33 34 35 36 37 38 39

m/e Structure of fragments Bonds Broken

ulg = 16= t6- l

t6 cL*

H .4 . - CL

JO ( r t t , r ) '

JI

nle 37 -- 38 - l

3a cL+

l1at- l + - c7"

38 (n - ' }cz)t M& tuldr'iovr '

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{. Given the relative abundance of each isotope, the peak heights of parent andpossible fragment ions can be calculated and the mass spectrum sketched

Example 5:Calculate the relative abundance of all the ions that may be formed from chlorine, Cl2(with isotopes) in the mass spectrometer. Then use these data to sketch the massspectrum of chlorine.

Chlorine has 2 isotopesRelative abundance

*cla

ttclI

RelativoAbundance

lons m/e AbundanceRelative

abundance= Peak Heioht

tu c1-* t5

. l

4 3 LI 'J

t* cr ' 3a In r [+l

[ 'uct - t tct f ?o4)t-- i

?\ T - , r q

(5rr -

+rt)rot

(ttlt -

25627+t1

( i . t ) " (+-7)6

- l l 'b

1ia ct - t+ n)1 f r r ** * = +

I

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f.XT't?": ff ""mine,

Bre was analvsed in a mass spectrometer ang tllt- l9't9#1"-Yshows the ion paths in the spectrometer. Bromine consists of 2 Isotopes' '-ttr ano818r.

Each of the groups of lines A, B and C is due to one of the these ions, Br*, Brz* andBf'.

(D Match the ions to the conect group of linss'..iii) ldentify the ion which gives rise to each line and calculate the relative

abundance of each line in each group.

analyra.rgfi!ld

G) fttall '. Def h{q" u t"v P'g Url. DrlL t.[.y. : A

peP,+,o al +o '^ )c->9 >c

(lr) Lc+ .rlai1x o$t)btha. oI 4$ "R lortrr aun&"o ol llBY i!

14' + rr(r-*) _- 11.1

t t - 27 -- \9.1

x -- O,55

Atefta ^b{.!a&L 4

*&- --

Pr lqflv< abtdetia ,l , , E .

W l.-

( r - ' " )

o,5t .

t5 ( l - o,ss)

a4^ .

(tv)(A) Iletermlne Molecular Formula of a Compound* The m/e value of the parent ion obtained from the mass spectrurn can be used.to

determine the molecular formula of an unknown compound; given that theempirical formula is known

(B) Determine the Molecular Structure ot a Compound* Molecules generally rupture prelerentially at their weakest points

.:. The pattern of {ragmentation can hence supply detailed structural intormation

Example 7:Comp6und Xhas an empirical formula of CaHoO. Determine the molecular formula ofX given that the molecular ion appeared al mle = 92 in the mass spectrum.

rrb\. c.rl Q' lov. rf ral = 9Z'

lzMM "I CczALof , = q>.

n f> {rz o) r ' 6( t o) + rc.ol = 9l46a

-- q )-

n" ), l ruopc., tar S,vrr la cI * tJ ( c1x, o) , = C* H,, o. , .

Example 8rWhich compound, COz or NOz, would give the complete mass spectrum below?

Relativeintensity

*

I Aryn,.rr>ook** tal hh

N rO.te^ +ta+ k

M/e Possible lons14 r+ N+

16 r ' b+28

, ' , A), * t= C' ' O'44 , . .1 fbn. t>"tbn *re

, v L-

-t

i . }o o

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. Mass spectra of isomers of pentanol

. Isomers : same molecular formula but difrere'lrt stsucfural formula

. Differe,nt weak points in each isomer, hence different splitting pattemobtained, resulting in different mass spectra produced.

100908070@50403020l00

@

oaQ

o-

@

oie

-(HrO.nd CH:-CH2)

M -(H@ .nd CHr)

M-(H.rO+CHr)

M -Hro

-cHr

l-hrd MWeSIII

cHr(CH:ETCHToHl-cr----

. 2-Pcnt nol MW 88

cnrcn.cx" -l- cx - cn,

2-Mclu-2-but.nol MW aA

I i "*""'i:;*"Lu:------

100908070OU

4030

t0o

- l

M -(Hro.nd

CHz: CH2)

M-(Hro .ndcH2-gHtr

M -(Hro

M

o|e

1009080706050403020l00

FIGUR: 2.9. tsornsric Fntanols.

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M -H.rg

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