Relative Atomic Mass Revisited

Introduction

Before we can weigh out moles of a substance, we need reliable masses for its atoms. You have already met relative atomic mass as a way of comparing atoms. Now we look at it again with more care, because every molar mass you will ever use is built from these numbers. Two ideas matter most: the carbon-12 standard, and the fact that most elements are mixtures of isotopes.

Core explanation

A relative scale. A single atom has a mass of around 10⁻²⁷ to 10⁻²⁵ kg, which is awkward to use. Instead, chemists compare atoms with a chosen standard. The standard is the carbon-12 atom, which is given a relative mass of exactly 12. So one-twelfth of the mass of a carbon-12 atom is the reference unit, called the unified atomic mass unit (u), about 1.66 × 10⁻²⁷ kg.

Definition. The relative atomic mass, Aᵣ, of an element is the weighted average mass of its atoms compared with one-twelfth of the mass of an atom of carbon-12. Because it is a ratio of two masses, Aᵣ has no units.

Isotopes cause non-whole numbers. Most elements exist as a mixture of isotopes. Chlorine, for example, is about 75% chlorine-35 and 25% chlorine-37. No chlorine atom has a mass of 35.5, but a large sample behaves as if every atom had that average mass. That is why the periodic table gives chlorine an Aᵣ of 35.5.

Why a weighted average? A simple average of 35 and 37 would be 36, but that ignores the fact that chlorine-35 is three times as common. A weighted average multiplies each isotope's mass by its abundance:

Aᵣ = (mass₁ × abundance₁ + mass₂ × abundance₂ + …) ÷ 100

For chlorine: (35 × 75 + 37 × 25) ÷ 100 = (2625 + 925) ÷ 100 = 35.5.

The result lies nearer the common isotope. The weighted average is always pulled towards the most abundant isotope. For boron (about 20% boron-10 and 80% boron-11), Aᵣ = (10 × 20 + 11 × 80) ÷ 100 = 10.8, much closer to 11.

Accuracy note. Using mass numbers such as 35 and 37 gives a good estimate. Precise isotope masses differ very slightly from whole numbers, which is why accurate tables show values such as 35.45 for chlorine.

Formulae

Aᵣ = Σ(isotope mass × percentage abundance) ÷ 100. Aᵣ has no units.

Step-by-step reasoning

1. List each isotope with its mass number and percentage abundance. 2. Multiply each mass number by its abundance. 3. Add the products together. 4. Divide by 100 (or by the total abundance if it is not 100). 5. Check that the answer lies between the lightest and heaviest isotope, nearer the most common one.

Visual explanation

Picture a mass spectrum for chlorine: a tall line at mass 35 and a line one-third as tall at mass 37. The Aᵣ sits between the lines, much nearer the tall one, like a see-saw balance point shifted towards the heavier child.

Real-world analogy

A class's average test mark is worked out by weighting each mark by how many students got it. If 30 students scored 35 and 10 scored 37, the class average is 35.5, not 36. Isotopes are averaged the same way.

Real-world example

Mass spectrometers in laboratories measure isotope abundances very precisely. Geologists and forensic scientists use tiny variations in abundance — for example in lead or oxygen isotopes — to trace where a rock, a food product or a drug sample originated.

Why?

Why was carbon-12 chosen? Earlier scales were based on hydrogen and later on oxygen, but chemists and physicists used slightly different oxygen scales. In 1961 both groups agreed on carbon-12: it is easy to measure precisely in mass spectrometers and gives almost the same numbers as the older chemical scale.

Common misconception

"Chlorine atoms have a mass of 35.5." No individual chlorine atom has this mass. Every atom is either chlorine-35 or chlorine-37; 35.5 is the average over a very large number of atoms.

Worked example

Question: Magnesium consists of 79% magnesium-24, 10% magnesium-25 and 11% magnesium-26. Calculate its relative atomic mass.

Reasoning: (24 × 79 + 25 × 10 + 26 × 11) ÷ 100 = (1896 + 250 + 286) ÷ 100 = 2432 ÷ 100 = 24.32. The answer lies close to 24, the most abundant isotope.

Answer: Aᵣ(Mg) = 24.3 (no units).

Quick check

1. Lithium is 7.5% lithium-6 and 92.5% lithium-7. Calculate Aᵣ. Answer: (6 × 7.5 + 7 × 92.5) ÷ 100 = 6.925, so Aᵣ = 6.9.

Exam focus

Learn the definition word for word: "weighted average mass of an atom … compared with one-twelfth of the mass of an atom of carbon-12". In calculations, show the multiplication for each isotope, divide by 100 and give the answer to the precision asked, without units.

Advanced insight

Isotope abundances vary slightly from place to place on Earth, so for some elements, such as hydrogen, carbon and oxygen, IUPAC now gives Aᵣ as a small range rather than a single value. For everyday calculations a single "conventional" value, such as 12.0 for carbon, is perfectly adequate.

Summary

Relative atomic mass compares the average mass of an element's atoms with one-twelfth of the mass of a carbon-12 atom, so it has no units. Because most elements are mixtures of isotopes, Aᵣ is a weighted average: multiply each isotope's mass by its abundance, add and divide by 100. The result lies nearest the most common isotope.

Practice questions

1. Define relative atomic mass. Answer: The weighted average mass of the atoms of an element compared with one-twelfth of the mass of an atom of carbon-12. 2. Boron is 20% boron-10 and 80% boron-11. Calculate Aᵣ(B). Answer: (10 × 20 + 11 × 80) ÷ 100 = 10.8. 3. Bromine is about 50% bromine-79 and 50% bromine-81. Estimate Aᵣ(Br) and explain why a simple average works here. Answer: (79 × 50 + 81 × 50) ÷ 100 = 80.0. The two isotopes are equally abundant, so the weighted average equals the simple average. 4. Explain why relative atomic mass has no units. Answer: It is a ratio of two masses (an atom's mass divided by one-twelfth of a carbon-12 atom's mass), so the units cancel.