18
SPEKTROMETRI MASSA SPEKTROMETRI MASSA membom molekul-molekul fasa uap dengan membom molekul-molekul fasa uap dengan energi energi elektron yang sangat tinggi elektron yang sangat tinggi ion ion ion-ion dipisahkan menurut perbandingan ion-ion dipisahkan menurut perbandingan massa massa terhadap muatan (m/z atau m/e) terhadap muatan (m/z atau m/e) ion-ion yang mempunyai perbandingan m/z ion-ion yang mempunyai perbandingan m/z khusus khusus dideteksi oleh suatu alat yang dideteksi oleh suatu alat yang dapat menghitung dapat menghitung Σ Σ ion-ion yang ion-ion yang menabraknya, dibaca oleh detektor dan menabraknya, dibaca oleh detektor dan outputnya diperkuat o/ampli dan dibaca outputnya diperkuat o/ampli dan dibaca o/rekorder o/rekorder spektrum massa: spektrum massa: grafik yang memplot limpahan ion relatif grafik yang memplot limpahan ion relatif (ordinat) dan massa (absis, tepatnya: m/z (ordinat) dan massa (absis, tepatnya: m/z

Spektrometri Massa

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Page 1: Spektrometri Massa

SPEKTROMETRI MASSASPEKTROMETRI MASSAmembom molekul-molekul fasa uap dengan membom molekul-molekul fasa uap dengan

energi energi elektron yang sangat tinggi elektron yang sangat tinggi ionion

ion-ion dipisahkan menurut perbandingan ion-ion dipisahkan menurut perbandingan massa massa terhadap muatan (mz atau me)terhadap muatan (mz atau me)

ion-ion yang mempunyai perbandingan mz ion-ion yang mempunyai perbandingan mz khusus khusus dideteksi oleh suatu alat yang dapat dideteksi oleh suatu alat yang dapat menghitung menghitung ΣΣ ion-ion yang menabraknya dibaca ion-ion yang menabraknya dibaca oleh detektor dan oleh detektor dan outputnya diperkuat oampli outputnya diperkuat oampli dan dibaca orekorderdan dibaca orekorder

spektrum massaspektrum massa

grafik yang memplot limpahan ion relatif (ordinat) grafik yang memplot limpahan ion relatif (ordinat) dan massa (absis tepatnya mz dan massa (absis tepatnya mz perbandingan massa dengan perbandingan massa dengan ΣΣ muatan suatu ion) muatan suatu ion) Gambar 1(benzamida ) Gambar 1(benzamida )

Puncak ion + pada mz 121 molekul yg Puncak ion + pada mz 121 molekul yg utuh(M)-1 elektron yang dilepaskan oleh utuh(M)-1 elektron yang dilepaskan oleh tabrakan sinar tabrakan sinar ion molekulion molekul M M++ seri seri fragmen ionfragmen ion ReaksiReaksi

C6H5-C-NH2

O

+ e C6H5-C-NH2

O -NH2

C6H5-C =O

-CO

mz 105

mz 77

M mz 121

-C6H5

CNH2

O

mz 44

SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA

inlet system

ion source

mass analyzer

detector

signal processor

readout

vacuum system

sample

10-6 - 10-8 torr

Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)

100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak

Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip

Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli

Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum

mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui

isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum

Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam

H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom

normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas

puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang

paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak

terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 2: Spektrometri Massa

Puncak ion + pada mz 121 molekul yg Puncak ion + pada mz 121 molekul yg utuh(M)-1 elektron yang dilepaskan oleh utuh(M)-1 elektron yang dilepaskan oleh tabrakan sinar tabrakan sinar ion molekulion molekul M M++ seri seri fragmen ionfragmen ion ReaksiReaksi

C6H5-C-NH2

O

+ e C6H5-C-NH2

O -NH2

C6H5-C =O

-CO

mz 105

mz 77

M mz 121

-C6H5

CNH2

O

mz 44

SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA

inlet system

ion source

mass analyzer

detector

signal processor

readout

vacuum system

sample

10-6 - 10-8 torr

Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)

100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak

Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip

Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli

Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum

mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui

isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum

Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam

H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom

normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas

puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang

paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak

terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 3: Spektrometri Massa

SUSUNAN ALAT SPEKTROMETER MASSASUSUNAN ALAT SPEKTROMETER MASSA

inlet system

ion source

mass analyzer

detector

signal processor

readout

vacuum system

sample

10-6 - 10-8 torr

Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)

100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak

Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip

Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli

Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum

mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui

isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum

Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam

H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom

normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas

puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang

paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak

terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 4: Spektrometri Massa

Yang paling tinggi intensitas puncak dasar Yang paling tinggi intensitas puncak dasar (base peak)(base peak)

100 (tinggi x factor sensitivitas) dari 100 (tinggi x factor sensitivitas) dari puncak2 lain termasuk puncak ion puncak2 lain termasuk puncak ion molekul sebagai base peakmolekul sebagai base peak

Ion molekul MIon molekul M++ sinar electron sinar electron mengubah molekul2 menjadi ion positipmengubah molekul2 menjadi ion positip

Nilai mz ion hanya kehilangan satu Nilai mz ion hanya kehilangan satu electron berat molekuler dari molekul electron berat molekuler dari molekul yang asliyang asli

Untuk mengidentifikasi puncak ion Untuk mengidentifikasi puncak ion molekul dalam spektrum molekul dalam spektrum

mampu menggunakan mampu menggunakan spektrumspektrum untuk untuk menentukan berat molekul suatu menentukan berat molekul suatu komponen yang tidak diketahui komponen yang tidak diketahui

isotop-isotop tampak dalam spektrumisotop-isotop tampak dalam spektrum

Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam

H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom

normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas

puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang

paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak

terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 5: Spektrometri Massa

Molekul-molekul di alam tidak hanya sebagai spesies Molekul-molekul di alam tidak hanya sebagai spesies isotop murni semua atom mempunyai isotop-isotop isotop murni semua atom mempunyai isotop-isotop yang menjadi sifat khas kelimpahan alamyang menjadi sifat khas kelimpahan alam

H H 11H ~ 002 H sebagai isotop H ~ 002 H sebagai isotop 22HH C C 1212C ~ 11 C sebagai isotop C ~ 11 C sebagai isotop 1313CC Isotop 1 atau 2 unit massa gt massa atom Isotop 1 atau 2 unit massa gt massa atom

normalnormal M+ M+1 dan M+2M+ M+1 dan M+2 Pada gambar 1 Puncak M+1 ~ 8 dari intensitas Pada gambar 1 Puncak M+1 ~ 8 dari intensitas

puncak ion molekul puncak ion molekul M+1 M+1 1313C C 22H H 1515N dan N dan 1717OO Tabel limpahan relatip isotop terhadap isotop yang Tabel limpahan relatip isotop terhadap isotop yang

paling melimpahpaling melimpah M+2 M+2 1818O kelimpahannya sangat rendah tidak O kelimpahannya sangat rendah tidak

terdeteksiterdeteksi Jika hanya C H N O F P dan IJika hanya C H N O F P dan I

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 6: Spektrometri Massa

(M+1) asymp 11 x sum atom C + 036 x sum atom N(M+1) asymp 11 x sum atom C + 036 x sum atom N (11 x sum atom C)(11 x sum atom C)22

(M+2) asymp + 020 x sum (M+2) asymp + 020 x sum atom O atom O 200 200

Jika puncak2 isotop ini cukup intens diukur Jika puncak2 isotop ini cukup intens diukur secara tepat perhitungan di atas dapat secara tepat perhitungan di atas dapat berguna untuk menentukan rumus molekulberguna untuk menentukan rumus molekul

Jika S dan Si ada M+2 lebih intensJika S dan Si ada M+2 lebih intens 1 atom S 1 atom S 3434S ~ 44 pada puncak M+2S ~ 44 pada puncak M+2 1 atom Si 1 atom Si 3030Si ~ 335 pada puncak Si ~ 335 pada puncak

M+2M+2

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 7: Spektrometri Massa

Mengenal puncak ion molekul (M)Mengenal puncak ion molekul (M) EI (Electron Impact)EI (Electron Impact) Problem Puncak sangat lemah atau tidak Problem Puncak sangat lemah atau tidak

nampaknampak Puncak ion molekul dan bukan Puncak ion molekul dan bukan

puncak fragmentasi atau tidak murnipuncak fragmentasi atau tidak murni CI (Chemical Ionization) puncak yang CI (Chemical Ionization) puncak yang

kuat pada M+1 dan sedikit fragmentasikuat pada M+1 dan sedikit fragmentasi ldquoNitrogen rulerdquo molekul dengan ldquoNitrogen rulerdquo molekul dengan

berat molekul genap tidak mengandung N berat molekul genap tidak mengandung N atau nitrogen genap dan berat molekul atau nitrogen genap dan berat molekul ganjil mempunyai suatu atom nitrogen ganjil mempunyai suatu atom nitrogen ganjilganjil

C H O N S dan X C H O N S dan X PP B Si Ar dan B Si Ar dan alkali tanahalkali tanah

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 8: Spektrometri Massa

Fragmentasi ikatan tunggal Fragmentasi ikatan tunggal fragmen ion no ganjil dari suatu ion fragmen ion no ganjil dari suatu ion molekul no genap dan fragmen ion no molekul no genap dan fragmen ion no genap dari suatu ion molekul no ganjil genap dari suatu ion molekul no ganjil fragmen ion mengandung nitrogen fragmen ion mengandung nitrogen (jika ada) dari ion molekul(jika ada) dari ion molekul

Intensitas puncak ion molekul Intensitas puncak ion molekul bergantung pada kestabilan ion molekulbergantung pada kestabilan ion molekul

IIon on mmolekul yang paling stabil sistem olekul yang paling stabil sistem aromatik murniaromatik murni

Jika ada substituen p i m akan Jika ada substituen p i m akan kurang intens dan puncak fragmen kurang intens dan puncak fragmen akan lebih intensakan lebih intens

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 9: Spektrometri Massa

Ion molekul yang sering tidak terdeteksi Ion molekul yang sering tidak terdeteksi alkohol alifatik nitrit nitrat senyawa alkohol alifatik nitrit nitrat senyawa nitro nitril dan senyawa-senyawa yang nitro nitril dan senyawa-senyawa yang banyak cabangnyabanyak cabangnya

AdanyaAdanya Puncak M-15 (kurang ndashCH3) puncak M-18 Puncak M-15 (kurang ndashCH3) puncak M-18

(kurang H2O) M-31 (kurang -OCH3 dari (kurang H2O) M-31 (kurang -OCH3 dari metil ester) konfirmasi dari suatu metil ester) konfirmasi dari suatu puncak ion molekulpuncak ion molekul

Umum M-1 kadang-kadang M-2 (kurang Umum M-1 kadang-kadang M-2 (kurang H2 karena fragmentasi atau termolisis) M-H2 karena fragmentasi atau termolisis) M-3 (dari alkohol)3 (dari alkohol)

(M-3) ndash (M-14) mungkin ada (M-3) ndash (M-14) mungkin ada kontaminan atau diduga puncak im kontaminan atau diduga puncak im adalah puncak ion fragmenadalah puncak ion fragmen

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 10: Spektrometri Massa

(M-19)-(M-25) tidak mungkin (kecuali (M-19)-(M-25) tidak mungkin (kecuali ndashF(19) atau -20(HF) dari senyawa2 flourinndashF(19) atau -20(HF) dari senyawa2 flourin

(M-16 O) (M-17OH) M-18 H2O) (M-16 O) (M-17OH) M-18 H2O) mungkin jika ada 1 atom O dalam molekulmungkin jika ada 1 atom O dalam molekul

PENENTUAN RUMUS MOLEKULPENENTUAN RUMUS MOLEKUL index of hydrogen deficiency (unsaturation index of hydrogen deficiency (unsaturation

index) the number π bonds andor index) the number π bonds andor rings a molecule containsrings a molecule contains

= CarbonsndashHydrogens2ndashhalogens2 = CarbonsndashHydrogens2ndashhalogens2 +nitrogens2 +1+nitrogens2 +1

Thus the compound C7H7NO= 7-Thus the compound C7H7NO= 7-35+05+1=535+05+1=5

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 11: Spektrometri Massa

Generalized molecular formulaGeneralized molecular formulaαIβIIγIIIδIV = IV ndash 12I + 12III + 1 αIβIIγIIIδIV = IV ndash 12I + 12III + 1

wherewhereα is H D or halogen (ie any α is H D or halogen (ie any

monovalent atom) monovalent atom) β is O S or any other bivalent atomβ is O S or any other bivalent atomγ is N P or any other trivalent atomγ is N P or any other trivalent atomδ is C Si or any other tetravalen δ is C Si or any other tetravalen

atomatomCompare C6H6 and C6H14 the index Compare C6H6 and C6H14 the index

is 4 for the former and 0 for the latteris 4 for the former and 0 for the latter

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 12: Spektrometri Massa

the rule of thirteenthe rule of thirteen High resolution mass spectrometry provides High resolution mass spectrometry provides

molecular mass information from which the molecular mass information from which the user can determine the exact molecular user can determine the exact molecular formula directlyformula directly

However when it is not available it is often However when it is not available it is often useful to be able to generate all the useful to be able to generate all the possible molecular formulas for a given possible molecular formulas for a given mass mass

A useful method for generating possible A useful method for generating possible molecular formulas for a given molecular molecular formulas for a given molecular mass mass

THE RULE OF THIRTEENTHE RULE OF THIRTEEN

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 13: Spektrometri Massa

a base formula which contains only C a base formula which contains only C and H M13 = and H M13 = nn + + rr1313

The base formula thus becomes The base formula thus becomes CCnnHHn+rn+r

the index of hydrogen deficiency the index of hydrogen deficiency (unsaturated index) U which (unsaturated index) U which corresponds to the preceding formula corresponds to the preceding formula is calculated easily by applying the is calculated easily by applying the relationship U = (relationship U = (n-r+n-r+2)22)2

Of course you can also calculate the Of course you can also calculate the index of hydrogen deficiency using the index of hydrogen deficiency using the preceding methodpreceding method

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 14: Spektrometri Massa

if we wish to derive a formula which if we wish to derive a formula which includes other atoms besides C and H then includes other atoms besides C and H then we must substract the mass of combination we must substract the mass of combination of carbons and hydrogens that equals the of carbons and hydrogens that equals the masses of the other atoms being included in masses of the other atoms being included in the formulathe formula

For example if we wish to convert the base For example if we wish to convert the base formula to a new formula containing one O formula to a new formula containing one O atom then we subtract one C and four atom then we subtract one C and four hydrogens at the same time that we add hydrogens at the same time that we add one O atom Both changes involve a one O atom Both changes involve a molecular mass equivalent of 16 molecular mass equivalent of 16 (O=CH4=16) (O=CH4=16)

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 15: Spektrometri Massa

Table below includes a number of CH Table below includes a number of CH equivalents for replacement of C and H in equivalents for replacement of C and H in the base formula by the most common the base formula by the most common elements likely to occur in an organic elements likely to occur in an organic compoundcompound

CarbonHydrogen equivalents for some CarbonHydrogen equivalents for some common elements common elements

TABELTABEL

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 16: Spektrometri Massa

To comprehend how the Rule of Thirteen To comprehend how the Rule of Thirteen might be applied consider an unknown might be applied consider an unknown subtance with a molecular mass of 94 subtance with a molecular mass of 94 amu Application of the formula provides amu Application of the formula provides

9413 = 7 + 3139413 = 7 + 313 According to the formula According to the formula n =n = 7 and 7 and r =r = 3 3 The base formula must be C7H10The base formula must be C7H10 The index of hydrogen deficiency is U = The index of hydrogen deficiency is U =

(7-3+2)2=3(7-3+2)2=3 A substance which fits this formula must A substance which fits this formula must

contain some combination of three rings or contain some combination of three rings or multiple bondsmultiple bonds

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 17: Spektrometri Massa

A possible structure might be A possible structure might be

If we were interested in a substance If we were interested in a substance which had the same molecular mass but which had the same molecular mass but which contained one O atom the which contained one O atom the molecular formula would become molecular formula would become C6H6O This formula is determined C6H6O This formula is determined according to the following schemeaccording to the following scheme

1 base formula = C7H10 U=31 base formula = C7H10 U=32 add +O2 add +O3 subtract -CH43 subtract -CH44 change the value of U ΔU = 1 (table)4 change the value of U ΔU = 1 (table)

CH3

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH

Page 18: Spektrometri Massa

5 new formula = C6H6O5 new formula = C6H6O6 new index of hydrogen deficiency U = 6 new index of hydrogen deficiency U =

44A possible substance which fits these A possible substance which fits these

data is data is

The new formula C6H6O U = 4The new formula C6H6O U = 4There are additional possible molecular There are additional possible molecular

formulas which conform to a molecular formulas which conform to a molecular mass of 94 amumass of 94 amu

C5H2O U = 5 C6H8N U=3frac12 C5H2S C5H2O U = 5 C6H8N U=3frac12 C5H2S U=5 CH3Br U=0U=5 CH3Br U=0

OH