Mass-to-Charge Ratio and the Mass Spectrum

Introduction

A mass spectrum looks like a bar chart: a row of thin vertical lines of different heights standing on a horizontal axis. To read one you need to understand exactly what each axis means. The horizontal axis is labelled m/z , not "mass", and the vertical axis shows relative abundance , not a number of ions. Getting these two ideas straight is the foundation for every later skill, from finding relative molecular mass to spotting isotopes and interpreting fragments.

Core explanation

The m/z axis. The instrument separates ions according to how they respond to electric and magnetic fields, and that response depends on the ratio of mass to charge. The quantity plotted is m/z , where m is the relative mass of the ion (on the carbon-12 scale) and z is the number of charges it carries. Because m/z is a ratio of a relative mass to a charge number, it has no units .

Most ions are singly charged. In electron impact and electrospray of small molecules, nearly all ions carry a single positive charge, so z = 1 and the m/z value equals the relative mass of the ion. This is why chemists casually read an m/z of 46 as "a mass of 46".

Doubly charged ions. Occasionally an ion loses two electrons. An ion of mass 46 carrying two charges appears at m/z 23. Doubly charged peaks are usually small for simple molecules, but large biomolecules analysed by electrospray often carry many charges, giving a series of peaks at m/z values far below their true mass.

The relative abundance axis. The detector current for each m/z value measures how many ions of that type arrived. Absolute ion counts depend on sample size and instrument settings, so they are not compared directly. Instead the tallest peak — the base peak — is set to 100%, and every other peak is expressed as a percentage of it. A peak labelled 25 is one quarter as abundant as the base peak.

Line spectra. Because ion masses are close to whole numbers, and low-resolution instruments round them, the spectrum appears as a set of sharp lines at integer m/z values, not a continuous curve.

Two uses of the same plot.

- For an element , each peak is a different isotope. The m/z values give the isotope masses and the abundances give their proportions, from which the relative atomic mass Ar can be calculated. - For a compound , the peaks include the molecular ion and many fragment ions. The m/z values reveal the masses of pieces of the molecule.

Isotopic data for magnesium (a typical exam case):

m/z Relative abundance (%) --- --- 24 79.0 25 10.0 26 11.0

Formulae

m/z = relative mass of ion ÷ charge number. Ar = Σ(isotope mass × abundance) ÷ Σ(abundance). If abundances are given as percentages summing to 100, divide by 100.

Step-by-step reasoning

To calculate Ar from a spectrum of an element:

1. Read the m/z value and relative abundance of every peak. 2. Assume z = 1 unless told otherwise, so m/z equals isotope mass. 3. Multiply each mass by its abundance. 4. Add the products. 5. Divide by the total abundance (which may not be 100 if heights are relative to the base peak). 6. Round to a sensible number of significant figures.

Visual explanation

Sketch two perpendicular axes. Label the horizontal one m/z from 0 upwards and the vertical one relative abundance from 0 to 100%. Draw a tall line reaching 100 at m/z 24 and two short lines, of heights 10 and 11, at m/z 25 and 26. That simple picture is the mass spectrum of magnesium.

Real-world analogy

Think of a coin-sorting machine at a bank. Coins roll down a track and drop through slots according to size, and a counter records how many go into each slot. The slot label is like m/z, and the count in each slot, compared with the fullest slot, is like relative abundance.

Real-world example

Geologists date rocks by measuring isotope ratios such as ⁸⁷Sr/⁸⁶Sr or ²⁰⁶Pb/²³⁸U in mass spectrometers. The ratio of peak heights at the relevant m/z values reveals how much radioactive decay has taken place since the rock formed, giving ages of millions or billions of years.

Why?

Why is abundance quoted relative to the base peak rather than as raw ion counts? The number of ions depends on how much sample was injected and on the detector's sensitivity, both of which vary between runs. Ratios of peak heights stay the same, so normalising makes spectra comparable between instruments and laboratories.

Common misconception

"The tallest peak is always the molecular ion." The base peak is simply the most abundant ion. For many organic compounds it is a stable fragment, and the molecular ion may be quite small.

Worked example

Question: Use the magnesium data in the table to calculate the relative atomic mass of magnesium.

Reasoning: Ar = (24 × 79.0 + 25 × 10.0 + 26 × 11.0) ÷ 100 = (1896 + 250 + 286) ÷ 100 = 2432 ÷ 100.

Answer: Ar = 24.3.

Quick check

1. An ion of relative mass 32 carries a charge of 2+. At what m/z value will it appear on the spectrum? Answer: At m/z 16, because m/z = 32 ÷ 2.

Exam focus

Always state that m/z is mass-to-charge ratio and that for singly charged ions it equals the relative mass. In Ar calculations, check whether the abundances add up to 100; if they are heights relative to the base peak, divide by their actual sum.

Advanced insight

The m/z scale is defined relative to carbon-12, whose mass is exactly 12 u. Other nuclides have masses slightly different from whole numbers because of nuclear binding energy, so ¹⁶O is 15.9949 and ¹H is 1.0078. Low-resolution spectra hide these differences by rounding, but high-resolution instruments exploit them to identify molecular formulae.

Summary

A mass spectrum plots relative abundance against m/z, the mass-to-charge ratio of each ion. Most ions carry a single charge, so m/z usually equals the ion's relative mass. Peak heights are normalised so that the base peak is 100%. For an element the peaks give isotope masses and abundances, from which the relative atomic mass is calculated.

Practice questions

1. What quantity is plotted on the horizontal axis of a mass spectrum, and why does it have no unit? Answer: m/z, the mass-to-charge ratio; it is a relative mass divided by a charge number, both of which are unitless. 2. Chlorine gives peaks at m/z 35 (abundance 75.8%) and m/z 37 (abundance 24.2%). Calculate Ar for chlorine. Answer: (35 × 75.8 + 37 × 24.2) ÷ 100 = 35.5. 3. A spectrum shows peaks at m/z 20 (height 100) and m/z 22 (height 11). Assuming no other peaks, calculate Ar. Answer: (20 × 100 + 22 × 11) ÷ 111 = 2242 ÷ 111 = 20.2. 4. A peak appears at m/z 14 in the spectrum of nitrogen gas, N₂. Suggest two ions that could cause it. Answer: A nitrogen atom ion N⁺ (mass 14, charge 1) or a doubly charged molecular ion N₂²⁺ (mass 28, charge 2).