Organic Chemistry · Structure Determination: Mass Spectrometry and Infrared Spectroscopy

Mass Spectrometry of Small Molecules: Magnetic-Sector Instruments

9 min read

On this page 8 sections

In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools

In 30 seconds

Mass spectrometry answers one question about an unknown molecule: what is its mass, and what pieces does it break into? The instrument vaporizes the sample, blasts the molecules with high-energy electrons to create ions, accelerates them to a known kinetic energy, and sorts them by their mass-to-charge ratio, written m/z. A magnetic-sector instrument does the sorting by bending the ion beam through a magnetic field: heavier ions curve less, lighter ions curve more, and the detector records how many ions of each m/z arrive.

The output — a mass spectrum — is a bar graph of relative abundance versus m/z. The peak at the highest m/z is usually the molecular ion M^+•, whose mass gives the molecular weight of the compound. The other peaks are fragments, and their masses are fingerprints of the structure. Mass spectrometry is the first tool in the structure-determination arsenal because it gives the molecular formula information that spectroscopy alone cannot.

Why this matters

Before infrared or NMR spectroscopy can be interpreted, you need to know what molecule you are holding — and mass spectrometry provides that answer. A single spectrum gives the molecular mass to several decimal places, reveals chlorine or bromine from their isotope patterns, and hints at functional groups from characteristic fragment losses. In pharmaceutical analysis, forensic toxicology, and environmental monitoring, mass spectrometers identify trace compounds. For this course, the magnetic-sector design matters because its physics — kinetic energy from an accelerating voltage and circular motion in a magnetic field — explains why the spectrum is plotted against m/z rather than plain mass, and why doubly charged ions appear at half their formula mass.

The college version

Core Concepts

From molecule to ion: the electron-impact ion source

The sample is vaporized and admitted as a dilute gas into a low-pressure chamber. A beam of electrons (typically 70 eV kinetic energy) strikes the neutral molecules, knocking an electron out and producing a :

$M + e - ⟶ M + • + 2e -$

The molecular ion M^+• has essentially the same mass as the neutral molecule (the lost electron weighs about 9.109 × 10^-31 kg, far below measurement precision) and carries one positive charge. It is the source of the peak at the highest m/z.

Because 70 eV far exceeds a typical bond energy (about 3–4 eV), the molecular ion usually fragments before measurement, producing additional peaks. The most abundant ion in the spectrum is the , set to 100% relative abundance; all other peaks are scaled to it.

Acceleration: giving every ion the same kinetic energy

Once formed, the ions are pulled out of the source by an electric field. An ion of charge q = ze (where z is the number of charges and e = 1.602 × 10^-19 C is the elementary charge) accelerated through a potential V gains kinetic energy:

$12 mv 2 = zeV$

Two consequences follow. First, all singly charged ions passing through the same voltage emerge with the same kinetic energy — so a light ion moves faster than a heavy one. Second, kinetic energy in electronvolts equals the product zV: an ion accelerated through 4000 V carries 4000 eV per charge.

The magnetic analyzer: sorting ions by m/z

The fast ion beam enters a magnetic field B perpendicular to its travel. The field exerts a force F = qvB that bends the ion into a circular arc of radius r; for the ion to stay on that arc, the magnetic force must equal the centripetal force:

$qvB = mv 2 r$

Combining this with the kinetic-energy equation gives the central result of the magnetic sector:

$m z = eB 2 r 2 2V$

In words: for fixed accelerating voltage V and analyzer radius r, the m/z reaching the detector depends only on the field strength B. The instrument scans B upward; as the field grows, heavier ions come into focus on the detector slit. Each current spike, plotted against the m/z that produced it, is the mass spectrum.

Exact masses and isotopes

The m/z scale records exact (monoisotopic) masses, not rounded atomic weights: ¹²C = 12.0000, ¹H = 1.00783, ¹⁶O = 15.9949, ¹⁴N = 14.0031, ³⁵Cl = 34.9689, ³⁷Cl = 36.9659, ⁷⁹Br = 78.9183, ⁸¹Br = 80.9163 (daltons, Da). Because chlorine and bromine have two abundant isotopes (Cl ≈ 3:1, Br ≈ 1:1), their compounds show tell-tale peak pairs two mass units apart — an instant fingerprint for those halogens.

Common Confusions

Do not confuse	With	Difference
Molecular ion peak	Base peak	Molecular ion = intact molecule minus one electron (highest m/z); base peak = most abundant ion (often a fragment)
m/z	Plain mass	m/z divides by charge; a doubly charged ion of mass 100 appears at m/z = 50
Exact mass	Nominal/rounded mass	Exact uses isotope masses (e.g., ¹²C = 12.0000); nominal rounds to integers
Mass of the molecule	Mass of the ion	The lost electron's mass is negligible, so M^+•'s m/z ≈ molecular mass
Heavier ions curve more	Lighter ions curve more	In a magnetic field, lighter ions are deflected more; heavier ions need a stronger field to reach the detector
Electron impact always gives M⁺•	A visible M⁺•	Fragmentation can make the molecular ion peak weak or invisible — a common exam trap
Isotope peaks (M+1, M+2)	Impurities	¹³C, ³⁷Cl, ⁸¹Br naturally produce small peaks above M; not contamination

Eli explains

The same idea, in plain words

Explain it like I’m 10

A magnetic-sector mass spectrometer is like a marble race with a twist: you roll all the marbles down the same ramp so they all have the same "push," then send them through a sideways wind (the magnet). Light marbles get blown way off course; heavy marbles barely curve. By making the wind stronger and stronger, you can catch each weight one at a time at the finish line. Every marble that arrives makes a line on a chart, and the chart tells you how heavy the molecule is and what pieces it broke into.

Worked example

Example 1: Which m/z reaches the detector for given instrument settings?

A magnetic-sector instrument has an analyzer radius r = 0.15 m and accelerating voltage V = 4000 V. What singly charged ion mass (in daltons) is focused when the field is B = 0.50 T?

Formula first: combine the centripetal and kinetic-energy relations to get the ion mass in kilograms:

$m = zeB 2 r 2 2V$

Substitute (z = 1, e = 1.602 × 10^-19 C, B = 0.50 T, r = 0.15 m, V = 4000 V):

$m = (1)(1.602 \times 10 -19 C)(0.50 T) 2 (0.15 m) 2 2(4000 V)$

$m = (1.602 \times 10 -19)(0.25)(0.0225) 8000 kg = 9.011 \times 10 -22 8000 kg = 1.126 \times 10 -25 kg$

Convert to daltons (1 Da = 1.6605 × 10^-27 kg):

$m = 1.126 \times 10 -25 kg 1.6605 \times 10 -27 kg / Da = 67.8 Da$

Answer: an ion with m/z ≈ 68 (e.g., C₅H₈⁺ or C₄H₄O⁺) is focused at this field. Raising B would focus heavier ions.

Example 2: How fast is a molecular ion moving?

The molecular ion of benzene, C₆H₆^+• (m/z = 78), is accelerated through 4000 V. Find its speed.

Formula first:

$12 mv 2 = zeV \Rightarrow v = \sqrt 2zeV m$

Substitute: m = 78 Da × 1.6605 × 10^-27 kg/Da = 1.295 × 10^-25 kg; z = 1; V = 4000 V; e = 1.602 × 10^-19 C.

$v = \sqrt 2(1)(1.602 \times 10 -19 C)(4000 V) 1.295 \times 10 -25 kg = \sqrt 1.282 \times 10 -15 J 1.295 \times 10 -25 kg$

$v = \sqrt 9.90 \times 109 m 2 / s 2 = 9.95 \times 104 m/s$

Answer: ≈ 1.0 × 10⁵ m/s — about 100 km/s. A lighter ion (smaller m) at the same voltage would be moving even faster, which is exactly why the magnetic field separates them.

Example 3: Distinguishing formulas with exact mass

A compound's molecular ion appears at m/z = 44.0262. Is it C₂H₄O or CO₂?

Formula first: exact mass = ∑(exact mass of isotope × count).

Substitute:

$C 2 H 4 O: 2(12.0000) + 4(1.00783) + 15.9949 = 24.0000 + 4.0313 + 15.9949 = 44.0262$

$CO 2 : 12.0000 + 2(15.9949) = 12.0000 + 31.9898 = 43.9898$

Answer: m/z = 44.0262 matches C₂H₄O exactly. The two formulas share a nominal mass of 44 but differ by 0.0364 Da — a difference a high-resolution sector instrument detects easily. This is why exact masses matter.

Key takeaways

The molecular ion M^+• is a radical cation; its m/z equals the molecular mass (for z = 1).
Sector sorting rule: m/z = eB²r²/(2V) — scan B with V and r fixed; ions arrive in order of increasing mass.
All ions with the same charge share the same kinetic energy zeV after acceleration, so v is larger for lighter ions.
The base peak is the most abundant ion (100%); it is not necessarily the molecular ion.
70 eV electron impact gives the molecular ion and fragments; the molecular ion is often weak or absent for fragile molecules.
Exact masses (not rounded atomic weights) are what appear on the m/z axis; this lets you distinguish formulas that have the same nominal mass.
Chlorine (3:1) and bromine (1:1) isotope patterns produce M and M+2 peaks — two mass units apart.
Doubly charged ions (z = 2) appear at half the mass of the singly charged ion of the same molecule.

Check yourself

6 review questions from the chapter. Try each one, then open the answer.

What particle is the molecular ion, and what is its charge?
Show answer
M^+• is a radical cation: the neutral molecule minus one electron, carrying one positive charge and one unpaired electron.
Why do all singly charged ions have the same kinetic energy after acceleration through the same voltage?
Show answer
Kinetic energy is zeV, which depends only on charge z and voltage V — not on mass. Lighter ions therefore move faster, heavier ions slower.
In a magnetic sector, what happens to the focused m/z as the magnetic field is increased?
Show answer
From m/z = eB²r²/(2V), increasing B focuses heavier ions — the spectrum is scanned from low mass to high mass.
What is the base peak, and is it always the molecular ion?
Show answer
The most abundant ion in the spectrum, normalized to 100%. It is often a fragment ion, not the molecular ion.
A molecule contains one bromine atom. Describe its molecular ion pattern.
Show answer
Two peaks of roughly equal height two mass units apart (M and M+2) because ⁷⁹Br and ⁸¹Br occur in nearly 1:1 natural abundance.
Why can high-resolution mass spectrometry distinguish C₂H₄O from CO₂?
Show answer
Their exact masses differ (44.0262 vs 43.9898 Da) even though both round to 44; a high-resolution instrument resolves the 0.0364 Da difference.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Mass-to-charge ratio (m/z): Mass of an ion divided by its number of charges
Molecular ion (M+•): The ion formed by removing one electron from the neutral molecule
Radical cation: A positive ion with an unpaired electron
Electron impact (EI): Ionization by a beam of ~70 eV electrons
Base peak: The most abundant ion in the spectrum, normalized to 100%
Accelerating voltage (V): Potential that gives ions kinetic energy zeV
Magnetic field (B): Field that bends the ion beam into a circular arc
Exact (monoisotopic) mass: Mass calculated from the most abundant isotope of each element
Nominal mass: Mass rounded to whole numbers
Resolution: Ability to separate peaks of similar m/z (m/Δm)

Educational content only. It is not medical, legal or professional advice. Found an error? Tell us.