Ch 1 · Relative masses of atoms and molecules · AS & A Level Chemistry 9701

Topic 1 of 4 in Atoms, molecules and stoichiometry — AS & A Level Chemistry 9701 — 48 lessons.

Introduction

Because atoms are far too small to weigh individually, chemists compare atomic and molecular masses on a relative scale defined against a single fixed standard: the carbon-12 atom. This lesson defines that scale, distinguishes the different relative mass quantities used in chemistry, and shows how mass spectrometry data is used to calculate relative atomic mass.

Key ideas — The unified atomic mass unit

The unified atomic mass unit (u) is defined as exactly one twelfth of the mass of one atom of carbon-12 (^12C), measured in its ground state. This single, universally agreed standard allows the masses of all atoms, ions and molecules to be expressed as ratios on a consistent scale, without needing to know their absolute masses in kilograms.

🧑‍🏫 Teacher's explanation

Imagine you are trying to weigh tiny grains of sand. Using a standard kitchen scale would be impossible because the grains are too light. Instead, you decide to use a single, specific grain of sand as your 'standard unit'. You say, 'This grain is 1 unit.' Now, you can describe every other grain by saying it is 'twice as heavy' or 'half as heavy' as your standard grain. This is exactly what chemists do with atoms.

📖 Definitions

🔍 Worked example — Easy

Calculate the relative molecular mass (M_r) of water, H_2O, given A_r(H) = 1 and A_r(O) = 16.

🔍 Worked example — Standard

An element X has two isotopes: ^63X with an abundance of 70\% and ^65X with an abundance of 30\%. Calculate the relative atomic mass (A_r) of X.

🔍 Worked example — Exam-style

A mass spectrum shows two peaks for element Y. Peak 1 is at m/z = 35 with a relative intensity of 75\%. Peak 2 is at m/z = 37 with a relative intensity of 25\%. Calculate the A_r of Y.

⚠️ Common errors

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🎯 Exam question

Key ideas — Defining the relative mass quantities

All four quantities are ratios and therefore have no units.

🧑‍🏫 Teacher's explanation

Imagine you are trying to weigh very tiny objects, like grains of sand, using a giant industrial scale. The scale isn't sensitive enough to show the weight of one grain, so instead, you compare the weight of your sand to a standard 'reference weight' that everyone agrees on. In chemistry, atoms are too light to weigh directly on a normal scale, so we use the Carbon-12 atom as our 'standard reference weight'.

📖 Definitions

🔍 Worked example — Easy

Calculate the relative isotopic mass of an isotope of Boron that has a mass of 11.01 u.

🔍 Worked example — Standard

An element X has two isotopes. Isotope A has a mass of 35 u and an abundance of 75\%. Isotope B has a mass of 37 u and an abundance of 25\%. Calculate the relative atomic mass (A_r) of element X.

🔍 Worked example — Exam-style

Calculate the relative molecular mass (M_r) of Calcium Nitrate, ceCa(NO3)2, given the following relative atomic masses: ceCa = 40, ceN = 14, ceO = 16.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

(a) Calculate the relative atomic mass (A_r) of this element. [2] (b) Explain why the relative atomic mass is not an integer. [1] [3 marks]

Key ideas — Calculating relative atomic mass from isotopic abundances

Where an element exists as a mixture of isotopes, its relative atomic mass is the weighted mean of the relative isotopic masses, weighted by their percentage (or relative) abundance:

🧑‍🏫 Teacher's explanation

Imagine you are buying a bag of mixed fruit. If you have 90 apples weighing 100g each and 10 oranges weighing 200g each, the 'average' weight of a piece of fruit isn't just the middle point (150g); it is much closer to 100g because there are so many more apples. This is exactly how atoms behave.

📖 Definitions

🔍 Worked example — Easy

Calculate the relative atomic mass of Boron, which consists of 20% ^10 extB and 80% ^11 extB.

🔍 Worked example — Standard

An element X has two isotopes. Isotope 1 has a mass of 63 and an abundance of 69%. Isotope 2 has a mass of 65 and an abundance of 31%. Calculate the A_r of element X.

🔍 Worked example — Exam-style

A mass spectrum shows two peaks for element Y. The first peak is at m/z = 35 with a relative abundance of 75.6. The second peak is at m/z = 37 with a relative abundance of 24.4. Calculate the A_r of element Y.

⚠️ Common errors

🧠 Quick check 1

🧠 Quick check 2

🎯 Exam question

Key ideas — Interpreting a mass spectrum

A mass spectrum plots relative abundance (vertical axis) against mass-to-charge ratio, m/z (horizontal axis). For singly charged ions (z = 1), each peak's m/z value corresponds directly to the relative isotopic mass of one isotope present in the sample, and the height of each peak gives its relative abundance. To find relative atomic mass from a mass spectrum: read off the m/z value and relative peak height (abundance) of every peak, then apply the weighted-mean formula above, using the abundances as the weighting.

🧑‍🏫 Teacher's explanation

Imagine you are weighing a bag of mixed fruit. If you have 3 apples weighing 100g each and 1 orange weighing 200g, you can't just say the average weight is 150g. You have to account for how many of each fruit you have. A mass spectrum works exactly like this. The peaks on the graph tell you the 'weight' of each isotope (the m/z value) and how 'common' it is (the relative abundance).

📖 Definitions

🔍 Worked example — Easy

An element has two isotopes. Isotope A has an m/z of 35 and a relative abundance of 75%. Isotope B has an m/z of 37 and a relative abundance of 25%. Calculate the relative atomic mass (A_r).

🔍 Worked example — Standard

A mass spectrum shows two peaks for an element. Peak 1 is at m/z = 63 with a relative abundance of 69.1%. Peak 2 is at m/z = 65 with a relative abundance of 30.9%. Calculate the A_r.

🔍 Worked example — Exam-style

A mass spectrum of a compound shows a molecular ion peak (M^+) at m/z = 44. The peaks for the isotopes of the elements in the molecule are as follows: Element X has isotopes with m/z 12 (abundance 99%) and 13 (abundance 1%). Element Y has isotopes with m/z 1 (abundance 99.9%) and 2 (abundance 0.1%). Element Z has isotopes with m/z 16 (abundance 99.9%) and 17 (abundance 0.1%). Calculate the relative molecular mass (M_r) of the compound based on the molecular ion peak.

Ch 1 · Relative masses of atoms and molecules · AS & A Level Chemistry 9701

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