2.1.1(c)-(e) - Relative mass and mass spectrometry

The carbon-12 relative mass scale

Relative masses compare particle masses with a fixed standard. For atomic masses, the standard is one twelfth of the mass of one atom of carbon-12, written as ¹²C. Carbon-12 is assigned a relative mass of exactly 12, so one twelfth of the mass of a ¹²C atom is the reference unit.

This is about one isotope only. For example, ³⁵Cl and ³⁷Cl have different relative isotopic masses because their atoms contain different numbers of neutrons.

The word weighted matters. Most elements have more than one isotope, and the isotopes do not usually occur in equal amounts. $A_r$ therefore depends on both the relative isotopic masses and the relative abundances of the isotopes.

Worked example: why $A_r$ is a weighted mean

Imagine a sample of chlorine atoms made from 75 atoms of ³⁵Cl and 25 atoms of ³⁷Cl. The total relative mass of these 100 atoms is:

The mean mass of one atom in the sample is:

So the relative atomic mass of this sample of chlorine is 35.5. No chlorine atom has mass 35.5; the value is the weighted mean of many atoms.

Reading isotope data from a mass spectrum

Mass spectrometry can be used to determine the relative isotopic masses and relative abundances of isotopes. For this lesson, you do not need to know how the mass spectrometer works. You only need to interpret the data it provides.

The simplified isotope mass spectra in this lesson use singly charged positive ions only. That means the charge is 1+, so the mass-to-charge ratio, m/z, has the same numerical value as the relative isotopic mass.

[DIAGRAM: asset_name: Lesson 2.1.1c-e: Relative Mass and Mass Spectrometry - diagram 01; asset_slug: 02_01_01b_relative_mass_and_mass_spectrometry__diagram_01; recommended_method: drawn_chem; description: A clean 16:9 mass spectrum for chlorine with m/z on the x-axis, relative abundance on the y-axis, bars at 35 and 37 labelled 35Cl+ and 37Cl+, and a note that for singly charged ions m/z equals relative isotopic mass.]

To read a simple isotope spectrum:

1. Use the x-axis positions to identify the relative isotopic masses.
2. Use the peak heights or given percentages to identify the relative abundances.
3. Combine mass and abundance using a weighted mean.

For the chlorine example, the peak at m/z 35 represents ³⁵Cl⁺ ions and the peak at m/z 37 represents ³⁷Cl⁺ ions. These are atomic-ion peaks. A molecular-ion peak belongs to a whole molecule and must not be mistaken for a new atomic isotope. For example, a singly charged molecule containing two chlorine-35 atoms has mass-to-charge ratio 70, rather than indicating a chlorine-70 atom. Molecular spectra are developed later in the course.

The taller m/z 35 peak shows that ³⁵Cl is more abundant in the sample.

Worked example: reading spectrum information

A spectrum of an element has two peaks:

• m/z 10, relative abundance 20
• m/z 11, relative abundance 80

Because the ions are singly charged, the isotopic masses are 10 and 11. The relative abundances are in the ratio 20:80, so the isotope at m/z 11 is four times as abundant as the isotope at m/z 10.

Calculating relative atomic mass

The calculation is a weighted mean. Multiply each isotope mass by its abundance, add those values, then divide by the total abundance.

If the abundances are percentages, the total abundance is 100:

Worked example 1: percentages

Bromine has two isotopes in this sample:

Substitute into the percentage form:

To three significant figures, this is:

This answer is close to 80 because the two isotopes are almost equally abundant and lie either side of 80.

Worked example 2: relative peak heights

A mass spectrum of boron has these simplified relative abundances:

Do not divide by 100 here, because the abundances are not percentages. Divide by their total:

To three significant figures:

Working backwards from a mean

If an element has isotopic masses 10 and 11 and relative atomic mass 10.8, let the fraction of mass-10 atoms be $x$. The mass-11 fraction is $1-x$, so $10x+11(1-x)=10.8$. Rearranging gives $x=0.20$: 20% mass-10 and 80% mass-11. A mean nearer 11 therefore corresponds to more mass-11 atoms; heavier does not automatically mean more abundant.

Calculation habits

Show your substitution clearly, keep one or two extra figures during working, then round the final answer to the requested precision. Relative atomic mass has no unit because it is a ratio on the carbon-12 scale.

Using $M_r$ and relative formula mass

Relative mass language depends on the type of substance.

For simple molecules, use the term relative molecular mass, $M_r$. Calculate it by adding the relative atomic masses of all the atoms in one molecule.

Worked example: relative molecular mass of carbon dioxide

Carbon dioxide is a simple molecular substance with formula CO₂.

Using $A_r$ values C = 12.0 and O = 16.0:

For compounds with giant structures, use the term relative formula mass. This includes ionic compounds such as MgBr₂ and giant covalent structures. Calculate it in the same arithmetic way: add the relative atomic masses in the formula.

Worked example: relative formula mass of magnesium bromide

Magnesium bromide, MgBr₂, is ionic, so use relative formula mass rather than relative molecular mass.

Using $A_r$ values Mg = 24.3 and Br = 79.9:

Quick Recap

Use a weighted mean, divide by the total abundance, and compare with one twelfth of a carbon-12 atom. Relative masses are ratios and have no units.