Qualitative analysis (mass spectrometry)

Class 12 Chemistry, unit 17, book pages 267–274

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Mass spectrometry turns atoms and molecules into positive ions, sorts the ions by their mass-to-charge ratio ( $m / z$ ) and records how many ions of each kind arrive at a detector. From the pattern of peaks we can count the isotopes of an element and calculate its relative atomic mass, read the molecular mass of a compound from its molecular ion peak, find the number of carbon atoms from the M+1 peak, detect chlorine and bromine from the M+2 peak, and suggest the structure of a molecule from the fragments it breaks into.

Learning outcomes (6)

Analyse mass spectra in terms of $m / z$ (written $m / e$ in the book) values and isotopic abundances.
Calculate the relative atomic mass of an element given the relative abundances of its isotopes or its mass spectrum.
Deduce the molecular mass of an organic molecule from the molecular ion peak in the mass spectrum.
Suggest the identity of molecules by simple fragmentation in a given mass spectrum.
Deduce the number of C atoms $n$ in a compound using the M+1 peak and the formula: number of carbon atoms $= \frac{relative intensity of M+1 peak}{0.01107 \times relative intensity of M ^{+} peak}$ .
Deduce the presence of chlorine and bromine atoms in a compound using the M+2 peak in the mass spectrum.

17Mass spectrometry and the mass spectrum

You met the mass spectrometer in Grade XI. It is an instrument that finds the masses of atoms and molecules and how common each kind is. It cannot weigh a single particle on a balance. Instead it turns the particles into positive ions, separates the ions according to their mass, and counts how many ions of each mass reach a detector.

Mass spectrometry: An analytical method in which a sample is changed into gaseous positive ions, the ions are separated according to their mass-to-charge ratio, and the relative number of ions of each kind is recorded.
Mass-to-charge ratio, $m / z$: The mass $m$ of an ion (in atomic mass units, amu) divided by the number of charges $z$ on it. It has no unit. The SLO box writes it as $m / e$ . Almost every ion formed has a charge of $+ 1$ , so the $m / z$ value is simply the mass of the ion.
Relative abundance (relative intensity): The height of a peak. It shows how many ions of that $m / z$ reach the detector, either as a percentage of all the ions or compared with the tallest peak, which is given the value 100.
Mass spectrum: A bar graph of relative abundance ( $y$ -axis) against $m / z$ ( $x$ -axis). Each kind of ion appears as one thin vertical line, called a peak.

A mass spectrum answers several questions at once. The book lists these uses:

the number of isotopes of an element and how common each one is (section 17.1);
the relative atomic mass of an element (section 17.2);
the relative molecular mass of a compound, from the molecular ion peak (section 17.3);
the structure of an organic molecule, from the pieces (fragments) it breaks into (section 17.4);
the number of carbon atoms (section 17.5) and the presence of chlorine or bromine (section 17.6);
the mechanism of a reaction, by finding the masses of the fragment cations and intermediate cations formed during the reaction.

Grade XI recap, 19.5.1The mass spectrometer: instrumentation and working

The book says the instrument was described in Grade XI, and section 19.5.1 summarises its working again. It is included here because the 2026 paper (Curriculum 2006) asked for it as a 7-mark question, and because it explains where every peak in this unit comes from. A magnetic sector mass spectrometer has five main parts, and the whole path of the ions is kept under a high vacuum so that the ions do not collide with air molecules.

A magnetic sector mass spectrometer. Only ions of one

m / z

follow the curve that ends at the detector; lighter ions are bent more and heavier ions less.

Working, stage by stage

1. Vaporisation. A very small sample (a few micrograms) is introduced and changed into vapour by heating at low pressure.
2. Ionisation. In the ionisation chamber a beam of fast electrons from an electron gun (a heated filament) strikes the molecules. Each collision knocks one electron out of a molecule and leaves a positive ion. This is called electron impact, or electron ionisation (EI).
$M + e X^{-} M X^{+∙} + 2 e X^{-}$
Many of these ions then break into smaller fragment ions (section 17.4).
3. Acceleration. The positive ions are attracted towards negatively charged accelerating plates at a high voltage $V$ . They pass through slits and form a narrow, fast beam. All ions with the same charge gain the same kinetic energy.
$z e V = \frac{1}{2} m v^{2}$
4. Deflection. The beam enters a strong magnetic field $B$ at right angles. The field bends each ion into a circular path. Light ions (small $m / z$ ) are bent more; heavy ions (large $m / z$ ) are bent less.
$r = \frac{m v}{z e B}$
5. Detection and recording. Only ions of one particular $m / z$ follow the curve that ends at the detector (ion collector). Each ion that arrives gives a tiny electric current, proportional to the number of ions. The magnetic field is changed slowly so that ions of each $m / z$ reach the detector in turn. The current is amplified and recorded (by a computer) as a peak whose height is the relative abundance.

Derivation The $m / z$ that reaches the detector

Aim. Find how the $m / z$ of the ions reaching the detector depends on the magnetic field $B$ , the radius $r$ of the path and the accelerating voltage $V$ .

Assumptions.

Each ion has charge $z e$ and is almost at rest before it is accelerated.
The magnetic field is uniform and at right angles to the velocity of the ion.

The electric field does work $z e V$ on the ion. This work becomes the kinetic energy of the ion.
$z e V = \frac{1}{2} m v^{2} (1)$
In the magnetic field the magnetic force $z e v B$ provides the centripetal force $m v^{2} / r$ needed for the circular path.
$z e v B = \frac{m v ^{2}}{r} ⟹ v = \frac{z e B r}{m} (2)$
Put $v$ from (2) into (1) to remove the speed, which is hard to measure.
$z e V = \frac{1}{2} m (\frac{z e B r}{m})^{2} = \frac{z ^{2} e ^{2} B ^{2} r ^{2}}{2 m}$
Rearrange to get the mass-to-charge ratio.
$\frac{m}{z e} = \frac{B ^{2} r ^{2}}{2 V}$

Result

\frac{m}{z e} = \frac{B ^{2} r ^{2}}{2 V}

The detector is at a fixed radius $r$ , so the $m / z$ that reaches it depends only on $B^{2} / V$ . Increasing $B$ slowly (magnetic scanning) brings heavier and heavier ions onto the detector, one $m / z$ value at a time. Older textbooks write the same result as $m / e = H^{2} r^{2} /2 E$ , with $H$ for the magnetic field and $E$ for the accelerating voltage.

17.1Number of isotopes and their relative abundances

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, so they have different masses. In a mass spectrometer the atoms of an element are ionised, and ions of different mass arrive at the detector separately. So the spectrum of an element shows one peak for each isotope.

The number of peaks gives the number of isotopes. Chlorine gives two peaks, at $m / z = 35$ and $m / z = 37$ , so natural chlorine is a mixture of two isotopes, $X_{35} X_{2}^{235} Cl$ and $X_{37} X_{2}^{237} Cl$ . The book adds that three peaks in the hydrogen spectrum show its three isotopes: protium ( $X_{1} X_{2}^{21} H$ ), deuterium ( $X_{2} X_{2}^{22} H$ ) and tritium ( $X_{3} X_{2}^{23} H$ ).
The position of each peak gives the mass number of that isotope, because $z = 1$ .
The height of each peak gives the relative abundance of that isotope, usually as a percentage.

Isotope data used in this unit (the boron, chlorine and bromine values are those printed in the book).
Element	Isotopes ( $m / z$ )	Relative abundance (%)
Hydrogen	$X_{1} X_{2}^{21} H$ (1), $X_{2} X_{2}^{22} H$ (2), $X_{3} X_{2}^{23} H$ (3)	about 99.99, about 0.01, trace
Boron	$X_{10} X_{2}^{210} B$ (10), $X_{11} X_{2}^{211} B$ (11)	19.90, 80.10
Carbon	$X_{12} X_{2}^{212} C$ (12), $X_{13} X_{2}^{213} C$ (13)	98.9, 1.1
Chlorine	$X_{35} X_{2}^{235} Cl$ (35), $X_{37} X_{2}^{237} Cl$ (37)	75.77, 24.23
Bromine	$X_{79} X_{2}^{279} Br$ (79), $X_{81} X_{2}^{281} Br$ (81)	50.50, 49.50

17.2Calculating the relative atomic massOften examined

Most elements are mixtures of isotopes, so the mass used for an element is an average. It is a weighted average: each isotope counts according to how common it is. The relative atomic mass $A_{r}$ of an element is the weighted average mass of its atoms compared with one-twelfth of the mass of a carbon-12 atom. The mass spectrum supplies both pieces of data: the mass of each isotope (the $m / z$ of its peak) and its abundance (the height of its peak).

A_{r} = \frac{\sum ( mass number \times percentage abundance )}{100}

The sum runs over all the isotopes, that is, over all the peaks of the element.

If the heights are not percentages (for example, when the tallest peak is scaled to 100), divide by the total of all the heights instead of by 100:

A_{r} = \frac{\sum ( m \times h )}{\sum h}

$m$ : mass number ( $m / z$ value) of each peak; $h$ : its height (relative abundance).

100

m / z

abundance

19.90% (^{10} B)

80.10% (^{11} B)

Fig 17.1: mass spectrum of boron. Two peaks show two isotopes; the heights give their percentage abundances.

Relative atomic mass of boron (book example, Fig 17.1)

The mass spectrum of boron (

Z = 5

) shows a peak at

m / z = 10

with relative abundance

19.90%

and a peak at

m / z = 11

with relative abundance

80.10%

. Calculate the average atomic mass of boron.

Two peaks mean two isotopes, $X_{10} X_{2}^{210} B$ and $X_{11} X_{2}^{211} B$ . Write the formula with both.
$A_{r} = \frac{( 10 \times 19.90 ) + ( 11 \times 80.10 )}{100}$
Multiply each mass number by its abundance and add the products.
$(10 \times 19.90) + (11 \times 80.10) = 199.0 + 881.1 = 1080.1$
Divide by 100, because the abundances are percentages.
$A_{r} = \frac{1080.1}{100} = 10.801$

A_{r} (B) = 10.801 \approx 10.80 amu

Concept assessment exercise 17.1: identify the element

Using the given spectrum (Fig 17.2), calculate the average atomic mass of the element and find out to which element this spectrum belongs. You can take the help of the periodic table.

m / z

abundance

51.45%

11.22%

17.15%

17.38%

2.80%

Fig 17.2: mass spectrum of an unknown element (concept assessment exercise 17.1).

Read each peak: the $m / z$ value is the mass number and the height is the percentage abundance. The five abundances add up to $100.00%$ , so the total is divided by 100.

$m / z$	Abundance (%)	$m / z \times$ abundance
90	51.45	4630.50
91	11.22	1021.02
92	17.15	1577.80
94	17.38	1633.72
96	2.80	268.80
Total	100.00	9131.84

A_{r} = \frac{9131.84}{100} = 91.32

In the periodic table the element with relative atomic mass close to 91.3 is zirconium, $Zr$ (atomic number 40, $A_{r} = 91.22$ ). Natural zirconium does have five isotopes with mass numbers 90, 91, 92, 94 and 96, which confirms the answer. The value 91.32 is slightly higher than 91.22 because whole mass numbers were used instead of exact isotopic masses.

A_{r} \approx 91.3; the element is zirconium, Zr

Heights scaled to the tallest peak (extra practice)

The mass spectrum of magnesium shows three peaks at

m / z = 24

25

and

26

with relative heights

100

12.7

and

13.9

. Calculate the relative atomic mass of magnesium.

The heights do not add up to 100, so find their total and divide by it.
$\sum h = 100 + 12.7 + 13.9 = 126.6$
Multiply each mass number by its height and add.
$\sum (m \times h) = 24 (100) + 25 (12.7) + 26 (13.9) = 2400 + 317.5 + 361.4 = 3078.9$
Divide.
$A_{r} = \frac{3078.9}{126.6} = 24.32$

A_{r} (Mg) \approx 24.3

17.3Molecular mass from the molecular ion peakOften examined

When a molecule M of an organic compound is hit by a fast electron, it loses one electron and becomes a positive ion. This ion still contains every atom of the molecule, and the lost electron has almost no mass, so the ion has the same mass as the molecule. It is called the molecular ion. It has lost one electron from a pair, so it has an unpaired electron and is written \ce{M^{+\bullet}} (a radical cation), or simply \ce{M+}.

\ce{M + e- -> M^{+\bullet} + 2e-}

Molecular ion, \ce{M+}: The positive ion formed when a whole molecule loses one electron. Its mass equals the relative molecular mass of the compound.
Molecular ion peak (\ce{M+} peak): The peak produced by the molecular ion. It is the peak with the largest m/z value, after ignoring the small peaks just beyond it (M+1, M+2) that are caused by heavier isotopes.
Base peak: The tallest peak in a mass spectrum. It is caused by the most abundant ion, which is usually the most stable ion. Its height is set to 100 and the heights of all other peaks are given relative to it.
M+1 peak: A small peak one unit above the molecular ion peak, caused by molecules that contain one \ce{^{13}C} atom in place of a \ce{^{12}C} atom.
M+2 peak: A peak two units above the molecular ion peak. It is tiny when it comes from two \ce{^{13}C} atoms, but large when the molecule contains \ce{^{37}Cl} or \ce{^{81}Br}.

The molecular ion peak gives the molecular mass. The molecular ion carries a charge of +1, so m/z=m/1=m. The m/z value of the molecular ion peak is therefore equal to the relative molecular mass M_r of the compound. For example, the molecular ion peak of hexane is at m/z=86, and M_r(\ce{C6H14})=6(12)+14(1)=86.

How to find the molecular ion peak

Go to the right-hand (high m/z) end of the spectrum.
Find the last peak of reasonable height. One or two very small peaks just after it are the M+1 and M+2 isotope peaks; they are not the molecular ion.
Read its m/z value. This is the relative molecular mass of the compound.

Example: the mass spectrum of ethene (Fig 17.3)

ethene, \ce{C2H4}, M_r=28

Fig 17.3: mass spectrum of ethene, \ce{C2H4} (M_r=28). The molecular ion peak at 28 is also the base peak; the tiny peak at 29 is the M+1 peak.

The molecular ion peak is at m/z=28, equal to the molecular mass of ethene, 2(12)+4(1)=28. The small peak at m/z=29 is the M+1 peak: carbon contains about 1.1\% of \ce{^{13}C}, so a few ethene molecules contain one \ce{^{13}C} atom. The book also mentions the M+2 peak at m/z=30, from molecules in which both carbon atoms are \ce{^{13}C}; it is much smaller than the M+1 and M peaks. All the other peaks come from fragment ions formed in the spectrometer.

Peaks in the mass spectrum of ethene.
m/z	Ion	How it forms
28	\ce{[C2H4]+}	molecular ion (M); here also the base peak
29	\ce{[C2H4]+} containing one \ce{^{13}C}	M+1 peak
30	\ce{[C2H4]+} containing two \ce{^{13}C}	M+2 peak (tiny)
27	\ce{[C2H3]+}	M−1: loss of one H atom
26	\ce{[C2H2]+}	M−2: loss of two H atoms
25, 24	\ce{[C2H]+}, \ce{[C2]+}	loss of three and of four H atoms
14, 13, 12	\ce{[CH2]+}, \ce{[CH]+}, \ce{[C]+}	the carbon to carbon double bond has broken

Molecular ion peak and base peak compared

Molecular ion peak compared with base peak.
Point	Molecular ion peak (\ce{M+})	Base peak
Caused by	the whole molecule after it loses one electron	the most abundant (usually the most stable) ion
Position	at the highest m/z, apart from the small M+1 and M+2 peaks	anywhere in the spectrum
Height	any height; can be small or even missing	always the tallest peak, given the value 100
Information	the relative molecular mass of the compound	the most stable fragment; the reference for all other heights
Example: n-pentane	m/z=72	m/z=43 (\ce{C3H7+})

17.4Identifying a molecule from its fragmentation patternOften examined

The electrons in the ionisation chamber carry far more energy than is needed to remove one electron. The extra energy makes the bonds of the molecular ion vibrate strongly, and some of them break. This breaking up is called fragmentation. When the molecular ion breaks, one piece keeps the positive charge (a cation) and the other piece is a neutral free radical.

\ce{M^{+\bullet} -> X+ + Y^{\bullet}}

\ce{X+}: fragment cation, which is detected; \ce{Y^{\bullet}}: neutral radical, which is not detected.

The fragment cations give the smaller peaks of the spectrum. The whole set of fragment peaks is called the fragmentation pattern. It works like a fingerprint: it shows which groups of atoms are present and how they are joined.

Rules of fragmentation

Only cations are detected. Radicals and neutral molecules have no charge, so they are not accelerated or deflected and never reach the detector.
The weakest bond breaks most easily. Which bond breaks depends on the functional group. In alkanes it is usually a C–C bond.
The most stable fragments form most often and give the tallest peaks. For carbocations the order of stability is tertiary > secondary > primary > methyl.
Some ions rearrange. A few stable ions form by the rearrangement of unstable ions, so fragments come both from electron impact and from chemical rearrangements.
The mass lost tells you what was lost. A peak at M-15 means a \ce{CH3} group was lost; M-17 means \ce{OH}; M-18 means \ce{H2O}; M-29 means \ce{C2H5} (or \ce{CHO}).

Common fragment ions and what they suggest.
m/z	Fragment ion	Usually shows
15	\ce{CH3+} (methyl cation)	a \ce{CH3} group
14	\ce{CH2+}	\ce{CH2} groups in a chain
17	\ce{OH+}	an alcohol or a carboxylic acid
29	\ce{CH3CH2+} (ethyl cation) or \ce{CHO+}	an ethyl group, or an aldehyde
31	\ce{CH2OH+}	a primary alcohol (\ce{-CH2OH} group)
43	\ce{CH3CH2CH2+} (propyl cation) or \ce{CH3CO+}	a propyl group, or a methyl ketone
57	\ce{C4H9+} (butyl cation)	a butyl group
77	\ce{C6H5+} (phenyl cation)	a benzene ring

Worked example: the mass spectrum of n-pentane (Fig 17.4)

n-pentane, \ce{C5H12}, M_r=72

Fig 17.4: mass spectrum of n-pentane, \ce{C5H12} (M_r=72). Molecular ion peak at 72, base peak at 43.

The molecular ion peak is at m/z=72, with a tiny M+1 peak at 73 and an M−1 peak at 71. The base peak is at m/z=43. The fragmentation (Fig 17.5 of the book) is shown below. In each step the charge stays on one piece and the other piece leaves as a radical.

Fragmentation of n-pentane

Ionisation. An electron knocks one electron out of the molecule, giving the molecular ion.
\ce{CH3CH2CH2CH2CH3 + e- -> [CH3CH2CH2CH2CH3]^{+\bullet} + 2e-}
m/z=72: molecular ion peak
The C2–C3 bond breaks, with the charge on the three-carbon piece. The propyl cation forms and an ethyl radical is lost.
\ce{[C5H12]^{+\bullet} -> CH3CH2CH2+ + CH3CH2^{\bullet}}
m/z=43: base peak
The same bond breaks, with the charge on the two-carbon piece. The ethyl cation forms and a propyl radical is lost.
\ce{[C5H12]^{+\bullet} -> CH3CH2+ + CH3CH2CH2^{\bullet}}
m/z=29
The C4–C5 bond (at the end of the chain) breaks. Loss of a methyl radical gives the butyl cation.
\ce{[C5H12]^{+\bullet} -> CH3CH2CH2CH2+ + CH3^{\bullet}}
m/z=57, that is M-15
The end bond breaks with the charge on the small piece. The methyl cation forms.
\ce{[C5H12]^{+\bullet} -> CH3+ + CH3CH2CH2CH2^{\bullet}}
m/z=15
A fragment loses hydrogen. The propyl cation loses two hydrogen atoms (as \ce{H2}) to give the propenyl (allyl) cation, which the book calls the propylene cation.
\ce{CH3CH2CH2+ -> CH2=CHCH2+ + H2}
m/z=41

**Why m/z=43 is the base peak.** When the C2–C3 bond breaks, the charge can stay on either piece, giving the propyl cation (43) or the ethyl cation (29); the book says both are possible. In the real spectrum the peak at 43 is about four times taller. The larger propyl group spreads (stabilises) the positive charge better, and the \ce{C3H7+} ion can rearrange into the more stable secondary cation \ce{(CH3)2CH+}. More stable ions form more often and survive to reach the detector, so they give taller peaks.

Main peaks in the mass spectrum of n-pentane.
m/z	Ion	Origin
73	M+1	molecules with one \ce{^{13}C} atom
72	\ce{[C5H12]+}	molecular ion, M
71	\ce{[C5H11]+}	M−1: loss of an H atom
57	\ce{[C4H9]+}	loss of \ce{CH3} (M-15)
43	\ce{[C3H7]+}	loss of \ce{C2H5} (M-29); base peak
42	\ce{[C3H6]+}	formed by rearrangement
41	\ce{[C3H5]+}	\ce{C3H7+} minus \ce{H2}
39	\ce{[C3H3]+}	\ce{C3H5+} minus \ce{H2}
29	\ce{[C2H5]+}	loss of \ce{C3H7} (M-43)
27	\ce{[C2H3]+}	\ce{C2H5+} minus \ce{H2}
15	\ce{[CH3]+}	methyl cation

Alcohols

For alcohols the book notes three useful peaks: M-1, from loss of a hydrogen atom; M-17, from loss of the \ce{OH} group; and a peak at m/z=17 when the \ce{OH} piece keeps the positive charge (\ce{OH+}). Primary alcohols also give a strong peak at m/z=31 (\ce{CH2OH+}), formed when the C–C bond next to the oxygen atom breaks. Short question v works through the spectrum of ethanol.

Using fragments to tell isomers apart

Isomers have the same molecular ion peak but can break up differently. n-Pentane and 2,2-dimethylpropane both have M_r=72. In 2,2-dimethylpropane, loss of a methyl radical gives the very stable tertiary butyl cation \ce{(CH3)3C+} at m/z=57. This is its base peak, and its molecular ion peak is hardly visible because the molecular ion breaks up so easily. n-Pentane has its base peak at 43 and a clear molecular ion peak at 72.

n-pentane: base peak 43

2,2-dimethylpropane: base peak 57

Identify a hydrocarbon from its spectrum

A hydrocarbon gives its molecular ion peak at m/z=44 and fragment peaks at m/z=29 and m/z=15. Suggest its structure.

The molecular ion peak gives M_r=44. Try the alkane formula \mathrm{C}_n\mathrm{H}_{2n+2}, whose mass is 12n+(2n+2)=14n+2.
14n+2=44\;\Longrightarrow\;n=3\;\Longrightarrow\;\ce{C3H8}
Explain the peak at 29: 44-29=15, so a methyl radical was lost and the ethyl cation remains.
\ce{[CH3CH2CH3]^{+\bullet} -> CH3CH2+ + CH3^{\bullet}}\qquad m/z=29
Explain the peak at 15: the same bond breaks with the charge on the methyl piece.
\ce{[CH3CH2CH3]^{+\bullet} -> CH3+ + CH3CH2^{\bullet}}\qquad m/z=15

\ce{CH3CH2CH3}\ \text{(propane)}

17.5Number of carbon atoms from the M+1 peakOften examined

Natural carbon is about 98.9\% \ce{^{12}C} and 1.1\% \ce{^{13}C}. A molecule that happens to contain one \ce{^{13}C} atom is one unit heavier than the normal molecular ion, so it appears at M+1. The more carbon atoms a molecule has, the greater the chance that one of them is \ce{^{13}C}. So the height of the M+1 peak, compared with the M peak, grows with the number of carbon atoms.

Derivation Formula for the number of carbon atoms

Aim. Relate the heights of the M and M+1 peaks to the number of carbon atoms n in the molecule.

Assumptions.

Each carbon atom has a 1.1\% chance of being \ce{^{13}C}, independent of the other carbon atoms.
The small contributions of other heavy isotopes (such as \ce{^2H}, \ce{^{15}N} and \ce{^{17}O}) to the M+1 peak are ignored.

A molecule with one carbon atom: for every 100 molecular ions at M, about 1.1 contain \ce{^{13}C} and appear at M+1.
\dfrac{I_{M+1}}{I_{M}}=\dfrac{1.1}{100}\qquad(n=1)
A molecule with n carbon atoms: any one of the n atoms can be the \ce{^{13}C} atom, so the chance is about n times larger.
\dfrac{I_{M+1}}{I_{M}}=\dfrac{1.1\,n}{100}
Make n the subject.
n=\dfrac{100\times I_{M+1}}{1.1\times I_{M}}
The SLO box writes the same formula with the more exact abundance 1.107\%, which is 0.01107 as a fraction.
n=\dfrac{I_{M+1}}{0.01107\times I_{M}}

Resultn=\dfrac{100\times\text{abundance of M+1 peak}}{1.1\times\text{abundance of M}^{+}\text{ peak}}

The ratio of the two peak heights counts the carbon atoms. Round the answer to the nearest whole number, because a molecule cannot contain part of an atom.

Number of carbon atoms (book example)

Determine the molecular formula of a molecule with a molecular ion peak of relative abundance 27.32\% and an M+1 peak with a relative intensity of 2.10\%.

Given

I_{M} = 27.32\% molecular ion peak
I_{M+1} = 2.10\% M+1 peak

To find

number of carbon atoms n

Write the formula.
n=\dfrac{100\times I_{M+1}}{1.1\times I_{M}}
Substitute the two relative intensities. Both are on the same scale, so the percentages cancel.
n=\dfrac{100\times2.10}{1.1\times27.32}=\dfrac{210}{30.05}
Evaluate and round to a whole number.
n=6.99\approx7

n=7

Carbon number and molecular formula together

A hydrocarbon shows its molecular ion peak at m/z=58 with relative intensity 12.0, and an M+1 peak with relative intensity 0.53. Find the number of carbon atoms and the molecular formula.

Find the number of carbon atoms from the heights of the M and M+1 peaks.
n=\dfrac{100\times0.53}{1.1\times12.0}=\dfrac{53}{13.2}=4.0\approx4
Find the mass of 4 carbon atoms and the mass left over for hydrogen.
4\times12=48;\qquad 58-48=10\;\Longrightarrow\;10\ \text{H atoms}
Write the formula. \ce{C4H10} fits \mathrm{C}_n\mathrm{H}_{2n+2}, so it is an alkane (butane or methylpropane).
\ce{C4H10}

n=4,\qquad \ce{C4H10}

17.6Detecting chlorine and bromine: the M+2 peakOften examined

Chlorine and bromine each have two isotopes that are both common, and in each case the two isotopes differ by 2 mass units. A compound containing one Cl or Br atom therefore gives two molecular ion peaks 2 units apart: the M peak (with the lighter isotope) and the M+2 peak (with the heavier isotope). The ratio of their heights shows which halogen is present.

Halogen	Isotopes (abundance)	Height ratio M : M+2	What the spectrum shows
Chlorine	\ce{^{35}Cl} (75.77%), \ce{^{37}Cl} (24.23%)	about 3:1	an M+2 peak about one-third the height of the M peak
Bromine	\ce{^{79}Br} (50.50%), \ce{^{81}Br} (49.50%)	about 1:1	M and M+2 peaks of almost equal height

Chlorine: the spectrum of chloromethane (Fig 17.6)

In the spectrum of chlorine atoms, the peak at m/z=35 is the larger one because \ce{^{35}Cl} is the more abundant isotope (75.77\%). A second peak of one-third its height appears at m/z=37 (the M+2 peak) from \ce{^{37}Cl} (24.23\%). The height of this second peak confirms the presence of chlorine.

chloromethane, \ce{CH3Cl}

Fig 17.6: mass spectrum of chloromethane, \ce{CH3Cl}. The M+2 peak at 52 is one-third the height of the M peak at 50: one Cl atom.

The molecular ion peak of chloromethane is at m/z=50, the mass of \ce{CH3^{35}Cl} (12+3+35=50). The peak at m/z=52 comes from \ce{CH3^{37}Cl} and is one-third as tall, which shows one chlorine atom. The tall peak at m/z=15 is \ce{CH3+}, formed when the C–Cl bond breaks.

Bromine: the spectrum of bromoethane (Fig 17.7)

The two bromine isotopes, \ce{^{79}Br} and \ce{^{81}Br}, have abundances of 50.50\% and 49.50\%. So bromine gives a peak at m/z=79 and another of almost the same height at m/z=81 (the M+2 peak).

bromoethane, \ce{CH3CH2Br}, M_r=109

Fig 17.7: mass spectrum of bromoethane, \ce{CH3CH2Br}. Molecular ion peaks at 108 and 110 of almost equal height: one Br atom.

Bromoethane gives a double molecular ion peak at m/z=108 (\ce{C2H5^{79}Br+}) and m/z=110 (\ce{C2H5^{81}Br+}). The two are almost equal; the peak at 110 is slightly smaller because \ce{^{81}Br} is slightly less abundant. The relative molecular mass printed as 109 uses the average A_r(\ce{Br})=79.9. The base peak at m/z=29 is \ce{C2H5+}, formed by loss of a bromine atom, and the pairs at 79 and 81 (\ce{Br+}) and at 93 and 95 (\ce{CH2Br+}) show the same 1:1 pattern.

Isotope patterns of halogen compounds. The last three rows go beyond the book but follow from the same abundances.
Halogen atoms in the molecule	Peaks	Height ratio
one Cl	M, M+2	3 : 1
one Br	M, M+2	1 : 1
two Cl	M, M+2, M+4	9 : 6 : 1
two Br	M, M+2, M+4	1 : 2 : 1
one Cl and one Br	M, M+2, M+4	3 : 4 : 1

Deduce the halogen

A compound shows two molecular ion peaks at m/z=78 and m/z=80, with heights in the ratio 3:1. What can you deduce?

Two peaks 2 units apart in the ratio 3:1 is the chlorine pattern, so the molecule contains one Cl atom.
The peak at 78 contains \ce{^{35}Cl}. Subtract its mass to find the mass of the rest of the molecule.
78-35=43
A mass of 43 fits \ce{C3H7} (3\times12+7=43).
\ce{C3H7Cl}\quad(1\text{-chloropropane or }2\text{-chloropropane})

\ce{C3H7Cl}

Formula sheet

Relative atomic mass (percentage abundances)

A_r=\dfrac{\begin{gathered}\sum(\text{mass number}\\{}\times\text{percentage abundance})\end{gathered}}{100}

the sum runs over all the isotopes (peaks) of the element

no unit (average atomic mass in amu)

Relative atomic mass (heights on any scale)

A_r=\dfrac{\sum(m\times h)}{\sum h}

m: mass number (m/z) of a peak; h: its height

no unit

Molecular mass from the molecular ion peak

M_r=\left(\dfrac{m}{z}\right)_{\text{molecular ion}}\qquad(z=1)

the m/z value of the molecular ion peak

no unit

Number of carbon atoms

n=\dfrac{100\times I_{M+1}}{1.1\times I_{M}}=\dfrac{I_{M+1}}{0.01107\times I_{M}}

I_{M+1}: relative intensity of the M+1 peak; I_M: relative intensity of the molecular ion peak

no unit; round n to a whole number

Mass of a fragment ion

m/z\ (\text{fragment})=M_r-\text{mass of the neutral piece lost}

for example M-15 (loss of \ce{CH3}), M-17 (loss of \ce{OH}), M-29 (loss of \ce{C2H5})

no unit

Chlorine and bromine isotope patterns

\ce{Cl}:\ I_M:I_{M+2}\approx3:1\qquad\ce{Br}:\ I_M:I_{M+2}\approx1:1

for one halogen atom in the molecule; I is the height of a peak

Unit: ratio

Magnetic sector mass spectrometer

\dfrac{m}{ze}=\dfrac{B^{2}r^{2}}{2V}

B: magnetic field; r: radius of the path of the ion; V: accelerating voltage; ze: charge of the ion

Unit: m/ze in \mathrm{kg\,C^{-1}}

Key terms

Mass spectrometry: A technique that changes a sample into gaseous positive ions, separates them by mass-to-charge ratio and records their relative abundances.
Mass spectrum: A graph of relative abundance against m/z, in which each kind of ion appears as a vertical line (peak).
m/z (or m/e) value: The mass of an ion divided by its charge. For singly charged ions it equals the mass of the ion.
Relative abundance (relative intensity): The height of a peak, showing how many ions of that m/z reached the detector.
Isotopes: Atoms of the same element with the same number of protons but different numbers of neutrons, and so different masses.
Relative atomic mass: The weighted average mass of the atoms of an element (from its isotopes and their abundances), compared with one-twelfth of the mass of a carbon-12 atom.
Electron impact (electron ionisation): Ionisation of molecules by a beam of high-energy electrons, which knocks out one electron: \ce{M + e- -> M^{+\bullet} + 2e-}.
Molecular ion: The positive ion (\ce{M+}) formed when a whole molecule loses one electron; its mass equals the relative molecular mass.
Molecular ion peak: The peak due to the molecular ion; the peak of highest m/z apart from the small M+1 and M+2 isotope peaks. It gives M_r.
Base peak: The tallest peak in a mass spectrum, due to the most abundant (most stable) ion; its height is set to 100.
Fragmentation: The breaking of the molecular ion into a smaller cation and a neutral radical inside the mass spectrometer.
Fragment ion: A positive ion formed when the molecular ion breaks up; only these charged pieces are detected.
Fragmentation pattern: The set of fragment peaks in a mass spectrum, characteristic of a compound and used to work out its structure.
M+1 peak: A small peak one unit above the molecular ion peak, due to molecules containing one \ce{^{13}C} atom; its height gives the number of carbon atoms.
M+2 peak: A peak two units above the molecular ion peak; it is large (one-third of M for Cl, equal to M for Br) when chlorine or bromine is present.
Radical: A neutral atom or group with an unpaired electron, such as \ce{CH3^{\bullet}}; it is not detected in the mass spectrometer.
Detector (ion collector): The part of the mass spectrometer that produces an electric current proportional to the number of ions reaching it.

Multiple choice questionsp. 271–272

Which is not the field of mass spectrometry?

AMolecular mass
BRelative abundance of isotopes
CConcentration of molecules in a sampleAnswer
DMolecular structure elucidation

Why

This unit uses mass spectrometry for three kinds of information: the molecular mass (from the molecular ion peak), the relative abundance of isotopes (from peak heights) and structure elucidation (from fragmentation). Measuring the concentration of molecules in a sample is not one of the uses taught here, so (C) is the answer. With careful calibration a mass spectrometer can also be used quantitatively (Unit 19), but that is not the field described in this unit.

Average atomic mass of an element can be calculated by using:

AIR spectroscopy
BUV-visible spectroscopy
CMass spectrometryAnswer
DNuclear magnetic resonance spectroscopy

Why

Only a mass spectrum gives both the mass of each isotope (m/z of a peak) and its relative abundance (height of the peak), which are needed in A_r=\sum(\text{mass number}\times\text{abundance})/100. IR, UV-visible and NMR spectroscopy give information about bonds, conjugation and hydrogen environments, not about isotope masses.

Note. The option letters are misprinted in the book (a, b, b, c, with 'b.' and 'd.' repeated inside the options); they are relabelled A to D here.

iii

Molecular mass of a compound cannot be deduced by:

AMolecular ion peak
BBase peakAnswer
CM+1 peak
DM+2 peak

Why

The base peak is just the tallest peak, caused by the most stable ion, which is usually a fragment. For n-pentane the base peak is at 43 while M_r=72. The molecular ion peak gives M_r directly, and the M+1 and M+2 peaks lie at fixed positions one and two units above it, so they also point to M_r.

Why is m/z value taken equivalent to mass?

ACharge is zero
BCharge of ions is equal to +1Answer
CCharge of ions is equal to -1
DMass is very high as compared to charge

Why

Almost every ion formed in the mass spectrometer has lost just one electron, so z=+1 and m/z=m/1=m. Particles with zero charge (A) are not detected at all, and the instrument separates positive ions, not negative ones (C).

Note. The book prints 'irons' in option (c); it means ions.

Fragments in mass spectrum are formed by the:

AImpact of electrons
BImpact of protons
CChemical rearrangements
DBoth A and CAnswer

Why

The high-energy electrons break the molecular ion into fragments (A), and some stable ions are also formed by the rearrangement of unstable ions (C). Protons are not used in electron impact ionisation, so (B) is wrong.

Bromine can be recognised by:

AMolecular ion peak
BM+1 peak
CM+2 peakAnswer
DM−1 peak

Why

Bromine has two isotopes, \ce{^{79}Br} and \ce{^{81}Br}, in almost equal amounts, so a bromine compound gives an M+2 peak of almost the same height as the M peak. The M+1 peak is due to \ce{^{13}C}, and the M−1 peak is due to loss of an H atom.

Short answer questionsp. 272

Answer each part with a definition or statement first, then the reason, then an equation or example.

i3 marks

How can you calculate average atomic mass of an element using relative abundance of all isotopes?

Solution

The average (relative) atomic mass is the weighted average of the masses of all the isotopes of the element. In the mass spectrum, the m/z of each peak gives the mass number of an isotope and the height of the peak gives its relative abundance. Multiply each mass number by its percentage abundance, add the products and divide by 100:

A_r=\dfrac{\begin{gathered}\sum(\text{mass number}\\{}\times\text{percentage abundance})\end{gathered}}{100}

For example, chlorine has \ce{^{35}Cl} (75.77\%) and \ce{^{37}Cl} (24.23\%):

A_r=\dfrac{(35\times75.77)+(37\times24.23)}{100}=\dfrac{2651.95+896.51}{100}=35.48

If the heights are not percentages, divide by the sum of all the heights instead of by 100.

Note. The book prints 'relative abundance is of all isotopes'; the stray 'is' is dropped.

ii2026 Annual I, Set B, Q.2 (xii)3 marks

How the number of carbon atoms in a molecule can be calculated by using intensities of the peaks in mass spectrum?

Solution

Carbon contains about 1.1\% of the heavier isotope \ce{^{13}C}. Molecules that contain one \ce{^{13}C} atom give a small peak one unit above the molecular ion peak, called the M+1 peak. Each carbon atom adds about 1.1\% of the M peak height to the M+1 peak, so the ratio of the two heights gives the number of carbon atoms n:

n=\dfrac{100\times\text{relative intensity of M+1 peak}}{1.1\times\text{relative intensity of M}^{+}\text{ peak}}

For example, if the M peak has relative intensity 27.32 and the M+1 peak 2.10, then n=\dfrac{100\times2.10}{1.1\times27.32}=6.99\approx7 carbon atoms. The answer is always rounded to the nearest whole number.

iii3 marks

How can you identify the presence of chlorine and bromine in the molecule using mass spectrum?

Solution

Chlorine and bromine each have two common isotopes that differ by 2 mass units. So a compound containing one of them shows a molecular ion peak (M) and a second large peak at M+2. The ratio of the two heights identifies the halogen:

Chlorine: \ce{^{35}Cl} (75.77\%) and \ce{^{37}Cl} (24.23\%) give M and M+2 peaks in the ratio about 3:1. Example: chloromethane shows peaks at m/z=50 and 52, the second one-third as tall.
Bromine: \ce{^{79}Br} (50.50\%) and \ce{^{81}Br} (49.50\%) give M and M+2 peaks of almost equal height (ratio about 1:1). Example: bromoethane shows peaks at m/z=108 and 110 of nearly the same height.

Compounds without these halogens show only a tiny M+2 peak (from two \ce{^{13}C} atoms), so a large M+2 peak is the sign of Cl or Br.

iv2026 Annual I, Set B, Q.2 (xi) OR3 marks

What is the role of molecular ion peak in the determination of molecular mass of a substance?

Solution

When a molecule is bombarded with high-energy electrons it loses one electron and forms the molecular ion, \ce{M+}, which contains all the atoms of the molecule:

\ce{M + e- -> M^{+\bullet} + 2e-}

The mass of the lost electron is negligible, so the molecular ion has the same mass as the molecule. It carries a charge of +1, so its m/z value equals its mass. The molecular ion peak is the peak of highest m/z (ignoring the small M+1 and M+2 isotope peaks), and its m/z value gives the relative molecular mass directly. For example, hexane shows its molecular ion peak at m/z=86, so M_r(\ce{C6H14})=86.

Observe following mass spectrum of an organic compound “g” carefully and answer the given questions.

MS of the compound “g” (book figure redrawn).

Solution

(a) Write few reactions that help to understand the fragmentation pattern of the sample molecule “g” whose MS is given.

From part (d), g is ethanol, \ce{CH3CH2OH} (M_r=46). Its main fragmentation reactions are:

Formation of the molecular ion by electron impact.
\ce{CH3CH2OH + e- -> [CH3CH2OH]^{+\bullet} + 2e-}
m/z=46 (M)
The C–C bond next to the oxygen breaks; the charge stays on the oxygen-containing piece.
\ce{[CH3CH2OH]^{+\bullet} -> CH2OH+ + CH3^{\bullet}}
m/z=31 (base peak)
Loss of a hydrogen atom.
\ce{[CH3CH2OH]^{+\bullet} -> [C2H5O]+ + H^{\bullet}}
m/z=45 (M-1)
Loss of the OH group.
\ce{[CH3CH2OH]^{+\bullet} -> CH3CH2+ + ^{\bullet}OH}
m/z=29 (M-17)
The C–C bond breaks with the charge on the methyl piece.
\ce{[CH3CH2OH]^{+\bullet} -> CH3+ + ^{\bullet}CH2OH}
m/z=15
The ethyl cation loses a hydrogen molecule.
\ce{CH3CH2+ -> CH2=CH+ + H2}
m/z=27

(b) Why is the peak at m/z 47 not the molecular ion peak?

The peak at m/z=47 is the M+1 peak, not the molecular ion peak. It is caused by the few ethanol molecules that contain one \ce{^{13}C} atom (about 1.1\% for each carbon atom), so it is very small, only about 2\% of the height of the peak at 46. The molecular ion peak is the peak at m/z=46, which matches the molecular mass of \ce{C2H5OH}: 2(12)+6(1)+16=46. The molecular ion peak is the last peak of reasonable height, not the tiny isotope peak after it.

The base peak is the tallest peak in a mass spectrum. It is caused by the most abundant ion, which is usually the most stable fragment ion. Its height is set to 100 and the heights of all the other peaks are measured relative to it. In the spectrum of g the base peak is at m/z=31, due to the \ce{CH2OH+} ion. This ion is stable because a lone pair on the oxygen atom shares the positive charge (\ce{CH2=OH+}).

(d) What is the structural formula of the molecule g?

The molecular ion peak at m/z=46 gives M_r=46.
The peak at 31 (\ce{CH2OH+}) shows a \ce{-CH2OH} group; the peak at 29 (\ce{C2H5+}, M-17) shows loss of \ce{OH}; the peak at 15 shows a \ce{CH3} group; the M-1 peak at 45 is typical of an alcohol.
Put the pieces together: \ce{CH3} (15) and \ce{CH2OH} (31) add up to 46.
\ce{CH3-CH2-OH}

ethanol

Its isomer dimethyl ether, \ce{CH3OCH3}, also has M_r=46, but it has no \ce{-CH2OH} or \ce{OH} group. The strong peak at 31 and the loss of OH (M-17) therefore point to ethanol.

\ce{CH3CH2OH}\ \text{(ethanol)}

Long answer questionsp. 272

i7 marks

Analyse in detail, the information obtained by the fragmentation pattern in mass spectrum, used for the structural elucidation of a molecule.

Solution

Fragmentation

In the ionisation chamber, high-energy electrons knock one electron out of the molecule and form the molecular ion \ce{M+}. The electrons carry far more energy than needed, so many molecular ions break into smaller pieces. This is fragmentation. In each break one piece keeps the positive charge and the other leaves as a neutral radical; only the cations reach the detector and give peaks.

\ce{M + e- -> M^{+\bullet} + 2e-}\qquad\qquad \ce{M^{+\bullet} -> X+ + Y^{\bullet}}

The weakest bonds break most easily, and the most stable cations (tertiary > secondary > primary) form most often and give the tallest peaks. So the pattern of fragment peaks is characteristic of the molecule, like a fingerprint.

Information obtained from the pattern

Molecular mass. The molecular ion peak (highest m/z, apart from the isotope peaks) gives M_r, because z=+1.
Number of carbon atoms. The M+1 peak, caused by \ce{^{13}C}, gives n=\dfrac{100\times I_{M+1}}{1.1\times I_{M}}. Together with M_r this suggests the molecular formula.
Chlorine or bromine. A large M+2 peak in the ratio 3:1 shows one Cl atom; in the ratio 1:1 it shows one Br atom.
Groups lost from the molecule. The gap between the M peak and a fragment peak shows the neutral piece lost: M-15 (\ce{CH3}), M-17 (\ce{OH}), M-18 (\ce{H2O}), M-29 (\ce{C2H5} or \ce{CHO}), M-45 (\ce{COOH}).
Groups present in the molecule. Common fragment ions reveal groups: 15 (\ce{CH3+}), 29 (\ce{C2H5+}), 31 (\ce{CH2OH+}, a primary alcohol), 43 (\ce{C3H7+} or \ce{CH3CO+}), 77 (\ce{C6H5+}, a benzene ring).
Arrangement of the carbon chain. The base peak shows the most stable fragment. A tertiary cation such as \ce{(CH3)3C+} (57) as the base peak shows branching; a ladder of peaks 14 units apart (15, 29, 43, 57) shows a straight chain.
Identification of unknowns. The whole pattern can be matched with the spectra of known compounds, and it can tell apart structural isomers that have the same M_r.

Example: n-pentane

n-Pentane (M_r=72) gives its molecular ion peak at m/z=72. Breaking its C–C bonds gives the main fragments:

The C2–C3 bond breaks, with the charge on the larger piece.
\ce{[C5H12]^{+\bullet} -> CH3CH2CH2+ + CH3CH2^{\bullet}}
m/z=43, the base peak
The same bond breaks, with the charge on the smaller piece.
\ce{[C5H12]^{+\bullet} -> CH3CH2+ + CH3CH2CH2^{\bullet}}
m/z=29
A methyl radical is lost from the end of the chain.
\ce{[C5H12]^{+\bullet} -> C4H9+ + CH3^{\bullet}}
m/z=57
The end bond breaks with the charge on the methyl piece.
\ce{[C5H12]^{+\bullet} -> CH3+ + C4H9^{\bullet}}
m/z=15

The peaks at 15, 29, 43 and 57 differ by 14 (\ce{CH2}), so the five carbon atoms form an unbranched chain, \ce{CH3CH2CH2CH2CH3}. Its isomer 2,2-dimethylpropane gives its base peak at 57 (the stable tertiary cation) and almost no molecular ion peak, so the two isomers are easily told apart.

Limitations

The fragmentation pattern cannot distinguish optical and geometric isomers, and hydrocarbons that give similar ions are hard to tell apart. So mass spectra are usually used together with other methods such as IR and NMR spectroscopy and chromatography.

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 141 mark

The molecular ion peak gives information about:

ANumber of atoms
BColour of compound
CMolecular mass of compoundAnswer
DFunctional group only

Why

The molecular ion is the whole molecule minus one electron and carries a charge of +1, so the m/z value of its peak equals the relative molecular mass of the compound. It says nothing about colour, and functional groups are found from the fragment peaks and from other kinds of spectra.

2026 Annual I, Set B, Q.2 (xi) OR1+1+1 = 3 marks

What is molecular ion peak? State its symbol and write its importance.

Solution

Definition (1 mark)

The molecular ion peak is the peak in a mass spectrum produced by the molecular ion, the ion formed when a molecule loses one electron. It appears at the highest m/z value in the spectrum, apart from the small M+1 and M+2 isotope peaks.

Symbol (1 mark)

It is written \ce{M+} (more exactly \ce{M^{+\bullet}}, a radical cation), formed by electron impact: \ce{M + e- -> M^{+\bullet} + 2e-}.

Importance (1 mark)

Its m/z value equals the relative molecular mass of the compound, because the charge on the ion is +1. For example, hexane gives \ce{M+} at m/z=86, so M_r=86. It is also the reference point for the M+1 peak (number of carbon atoms), the M+2 peak (chlorine or bromine) and the mass losses that identify fragments.

2026 Annual I, Set B, Q.2 (xii)3 marks

A mass spectrum shows \mathrm{M}^{+} peak =100\% and (\mathrm{M}+1) peak =5.5\%. Calculate the number of carbon atoms in the molecule.

Solution

Given

I_{M} = 100\% height of the molecular ion peak
I_{M+1} = 5.5\% height of the M+1 peak

To find

number of carbon atoms n

Formula (1 mark). Each carbon atom contributes about 1.1\% of the M height to the M+1 peak (from \ce{^{13}C}).
n=\dfrac{100\times I_{M+1}}{1.1\times I_{M}}
Substitution (1 mark).
n=\dfrac{100\times5.5}{1.1\times100}=\dfrac{550}{110}
Answer (1 mark).
n=5

The molecule contains 5 carbon atoms. With the SLO form of the formula, n=\dfrac{5.5}{0.01107\times100}=4.97\approx5, the same answer.

n=5

2026 Annual I, Set D, MCQ 151 mark

Identify the \dfrac{m}{z} value of \ce{CH3CH2CH2+}:

A33
B23
C43Answer
D37

Why

\ce{CH3CH2CH2+} contains 3 carbon and 7 hydrogen atoms: 3(12)+7(1)=43. Its charge is +1, so m/z=43. It is the propyl cation, the base peak in the spectrum of n-pentane.

2026 Annual I, Set D, MCQ 161 mark

The base peak in a mass spectrum corresponds to the:

AMolecular ion
BHydrogen bonding
CHeaviest fragment
DMost stable ionAnswer

Why

The base peak is the tallest peak, caused by the most abundant ion, and the most abundant ion is the most stable one (for n-pentane, \ce{C3H7+} at 43). The molecular ion (A) is the base peak only in a few cases such as ethene, and the heaviest ion (C) is usually the molecular ion, which is often small.

2026 Annual I, Set D, Q.2 (xii)1.5+1.5 = 3 marks

Differentiate between molecular ion peak and base peak in mass spectrometery.

Solution

Molecular ion peak (1.5 marks)

The peak produced by the molecular ion \ce{M+}, formed when the whole molecule loses one electron (\ce{M + e- -> M^{+\bullet} + 2e-}). It lies at the highest m/z (apart from the small M+1 and M+2 peaks) and its m/z value gives the relative molecular mass. It may be small, or even missing, if the molecular ion breaks up easily.

Base peak (1.5 marks)

The tallest peak in the spectrum, produced by the most abundant, most stable ion. Its height is set to 100 and all other peaks are measured relative to it. It can be at any m/z and is usually a fragment ion.

Point	Molecular ion peak	Base peak
Ion	whole molecule minus one electron	most stable (most abundant) ion
Position	highest m/z	anywhere
Height	any; may be small	always 100 (tallest)
Tells us	M_r of the compound	the most stable fragment
n-pentane	m/z=72	m/z=43 (\ce{C3H7+})

Note. The paper prints 'spectrometery'; it means spectrometry.

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 17, old syllabus1 mark

Which method would be preferred for trace element detection?

AGravimetric analysis
BAcid-base titration
CPrecipitation
DSpectroscopic methodsAnswer

Why

Trace elements are present at the level of parts per million or less. Spectroscopic methods (such as atomic absorption spectroscopy and mass spectrometry) can detect such tiny amounts. Gravimetric analysis, titration and precipitation need weighable masses or measurable volumes, so they are not sensitive enough.

2026 Annual I, Old syllabus (Curriculum 2006), Q.3, old syllabus2+5 = 7 marks

What is mass spectrometery? Briefly explain its instrumentation and working.

Solution

Mass spectrometry (2 marks)

Mass spectrometry is an analytical technique in which a sample is changed into gaseous positive ions, the ions are separated according to their mass-to-charge ratio (m/z), and the relative abundance of each kind of ion is recorded as a mass spectrum. It is used to find the isotopes of an element and their abundances, relative atomic and molecular masses, and the structures of organic compounds.

Instrumentation (labelled diagram)

Magnetic sector mass spectrometer.

Working (5 marks)

Vaporisation. A very small sample is introduced and vaporised by heating at low pressure. The whole apparatus is kept under high vacuum so that ions do not collide with air molecules.
Ionisation chamber. A beam of high-energy electrons from a heated filament (electron gun) knocks an electron out of each molecule, forming positive ions. Some of these ions break into fragment ions.
\ce{M + e- -> M^{+\bullet} + 2e-}
Accelerating plates. Negatively charged plates at high voltage V attract the positive ions and, through slits, form a narrow beam of fast ions with the same kinetic energy.
zeV=\tfrac{1}{2}mv^{2}
Magnetic analyser. A strong magnetic field bends the ions into circular paths. Ions of small m/z are deflected more and ions of large m/z less, so ions are separated by their m/z.
\dfrac{m}{ze}=\dfrac{B^{2}r^{2}}{2V}
Detector (ion collector). By slowly changing the magnetic field, ions of each m/z in turn are brought onto the detector. Each ion produces a small current proportional to the number of ions arriving.
Recorder. The current is amplified and recorded by a computer as a peak. The position of a peak gives the m/z value and its height gives the relative abundance. The complete record is the mass spectrum.

For example, the mass spectrum of boron shows two peaks, at m/z=10 (19.90\%) and m/z=11 (80.10\%), from which A_r(\ce{B})=10.80.

Note. The paper prints 'spectrometery'; it means spectrometry.

Exam focus

What is the molecular ion peak? Give its symbol and importance. (short)
Asked in 2026 (Set B Q.2 (xi) OR); the idea behind every molecular mass question.
Go to answer
Differentiate between the molecular ion peak and the base peak. (short)
Asked in 2026 (Set D Q.2 (xii)) and behind two 2026 MCQs.
Go to answer
Calculate the number of carbon atoms from the heights of the M and M+1 peaks. (numerical)
A 3-mark numerical in 2026 (Set B Q.2 (xii)); one formula, quick marks.
Go to answer
Calculate the relative atomic mass of an element from its mass spectrum. (numerical)
A direct SLO; the book's boron example and the zirconium concept assessment exercise.
Go to answer
How are chlorine and bromine identified from the M+2 peak? (short)
Book short question iii and MCQ vi; the 3 : 1 and 1 : 1 ratios are easy marks.
Go to answer
Write the fragmentation of n-pentane and give the m/z of each fragment. (short)
The book's worked example; the m/z of \ce{CH3CH2CH2+} was a 2026 MCQ (Set D MCQ 15).
Go to answer
Interpret the mass spectrum of an unknown compound (compound g, ethanol). (short)
Book short question v tests every idea of the unit on one spectrum.
Go to answer
Information obtained from the fragmentation pattern for structure elucidation. (long)
The only long question in the book's exercise.
Go to answer
Explain the instrumentation and working of a mass spectrometer. (long)
Asked as a 7-mark question in the 2026 Curriculum 2006 paper (Q.3).
Go to answer

17Mass spectrometry and the mass spectrum

Grade XI recap, 19.5.1The mass spectrometer: instrumentation and working

Working, stage by stage

Derivation The m/z that reaches the detector

17.1Number of isotopes and their relative abundances

17.2Calculating the relative atomic massOften examined

17.3Molecular mass from the molecular ion peakOften examined

How to find the molecular ion peak

Example: the mass spectrum of ethene (Fig 17.3)

Molecular ion peak and base peak compared

17.4Identifying a molecule from its fragmentation patternOften examined

Rules of fragmentation

Worked example: the mass spectrum of n-pentane (Fig 17.4)

Alcohols

Using fragments to tell isomers apart

17.5Number of carbon atoms from the M+1 peakOften examined

Derivation Formula for the number of carbon atoms

17.6Detecting chlorine and bromine: the M+2 peakOften examined

Chlorine: the spectrum of chloromethane (Fig 17.6)

Bromine: the spectrum of bromoethane (Fig 17.7)

Formula sheet

Key terms

Multiple choice questionsp. 271–272

Short answer questionsp. 272

Long answer questionsp. 272

Fragmentation

Information obtained from the pattern

Example: n-pentane

Limitations

FBISE 2026 questions

Definition (1 mark)

Symbol (1 mark)

Importance (1 mark)

Molecular ion peak (1.5 marks)

Base peak (1.5 marks)

Mass spectrometry (2 marks)

Instrumentation (labelled diagram)

Working (5 marks)

Exam focus

Derivation The $m / z$ that reaches the detector