Relative masses of atoms and molecules

1. Overview

Atoms are far too light to weigh one at a time, so chemists compare their masses on a relative scale. This section defines the unit of that scale, the unified atomic mass unit, and the four relative masses built on it: relative isotopic mass, relative atomic mass, relative molecular mass and relative formula mass. Most mole calculations in the course use these values, and the definitions are often asked for in full.

Key Definitions

• Unified atomic mass unit (u): one twelfth of the mass of one atom of carbon-12; it is about 1.66 × 10⁻²⁷ kg, or 1.66 × 10⁻²⁴ g.
• Isotopes: atoms of the same element, with the same number of protons but different numbers of neutrons, and so with different masses.
• Abundance (of an isotope): the proportion of the atoms of an element that are that isotope, usually given as a percentage.
• Relative isotopic mass: the mass of one atom of a particular isotope compared with one twelfth of the mass of one atom of carbon-12; it is the mass of that atom in unified atomic mass units, written without the unit.
• Relative atomic mass (Ar): the weighted mean mass of an atom of an element, taking account of the abundance of each isotope, compared with one twelfth of the mass of one atom of carbon-12.
• Relative molecular mass (Mr): the weighted mean mass of a molecule of a substance compared with one twelfth of the mass of one atom of carbon-12; it is the sum of the relative atomic masses of all the atoms in the molecular formula.
• Relative formula mass: the weighted mean mass of one formula unit of a substance compared with one twelfth of the mass of one atom of carbon-12; it is the term used for ionic and giant covalent substances, which are not made of molecules.
• Formula unit: the group of ions or atoms shown by the formula of a substance with a giant structure, for example one Na⁺ ion and one Cl⁻ ion in NaCl, or one Si atom and two O atoms in SiO₂.

Content

Why a relative scale?

A hydrogen atom has a mass of about 1.67 × 10⁻²⁴ g. Numbers this small are awkward, so chemists compare every atom with one standard instead. The standard is an atom of carbon-12, the isotope of carbon whose nucleus has 6 protons and 6 neutrons.

The unified atomic mass unit

The unified atomic mass unit, symbol u, is one twelfth of the mass of one atom of carbon-12.

• A carbon-12 atom therefore has a mass of exactly 12 u. This is true by definition, not by measurement.
• Every other atom's mass is measured against it, and none comes out as an exact whole number of u.
• 1 u is about 1.66 × 10⁻²⁷ kg, or 1.66 × 10⁻²⁴ g.

You can get this value from the Avogadro constant, $L$, in the Data Booklet. One mole of carbon-12 atoms is $L$ atoms and has a mass of 12 g (very nearly exactly), so one atom has a mass of $12/L$ g and

$$1\ \text{u} = \frac{1}{12} \times \frac{12\ \text{g}}{L} = \frac{1\ \text{g}}{6.022 \times 10^{23}} = 1.66 \times 10^{-24}\ \text{g}$$

The unit is called unified because in 1961 it replaced two older scales, both based on oxygen. Older books may write it as amu, and in biochemistry it is called the dalton (Da).

Relative masses have no units

Each relative mass in this section is a mass divided by 1 u:

$$\text{relative mass} = \frac{\text{mass of the particle}}{\tfrac{1}{12} \times \text{mass of one carbon-12 atom}} = \frac{\text{mass of the particle}}{1\ \text{u}}$$

When the particles are not all identical, the top line is their weighted mean mass. The units cancel, so every relative mass is a pure number: write $M_{\text{r}}$ = 18.0 for water, not 18.0 g.

Relative mass	Symbol	What is compared with 1 u
Relative isotopic mass	also $A_{\text{r}}$ (no symbol of its own)	one atom of one isotope
Relative atomic mass	$A_{\text{r}}$	the weighted mean mass of an atom of an element
Relative molecular mass	$M_{\text{r}}$	the weighted mean mass of a molecule
Relative formula mass	often also $M_{\text{r}}$	the weighted mean mass of one formula unit

Relative isotopic mass

Isotopes of an element have the same number of protons but different numbers of neutrons, so their atoms have different masses. The relative isotopic mass is the mass of one atom of a particular isotope compared with one twelfth of the mass of a carbon-12 atom. Put simply, it is the mass of that atom in u, without the unit.

All atoms of one isotope have the same mass, so there is no averaging in this definition.

Relative isotopic mass has no symbol of its own. Where a symbol is needed, $A_{\text{r}}$ is used for it too, so always say in words which of the two you mean.

A relative isotopic mass is close to the isotope's mass number (its number of protons plus neutrons), but only carbon-12 gives an exact whole number:

Isotope	Mass number	Relative isotopic mass
hydrogen-1	1	1.008
carbon-12	12	12 (exactly)
carbon-13	13	13.003
oxygen-16	16	15.995
chlorine-35	35	34.969
iodine-127	127	126.904

The gaps arise because a proton and a neutron each have a mass slightly above 1 u, the electrons add a little more, and a nucleus has slightly less mass than its protons and neutrons would have apart. You do not need to explain this. Use the values a question gives, and use mass numbers only when nothing more precise is given.

Relative atomic mass: a weighted mean

Most elements are a mixture of isotopes. The relative atomic mass, $A_{\text{r}}$, is the weighted mean mass of an atom of the element compared with one twelfth of the mass of a carbon-12 atom.

Weighted means each isotope counts in proportion to its abundance, the share of the element's atoms that are that isotope. A simple mean would count every isotope equally, which is wrong whenever the abundances differ. Imagine an element with two isotopes of relative isotopic masses 10.0 and 12.0:

• With equal numbers of each, $A_{\text{r}}$ = 11.0, exactly halfway.
• With 90% of the lighter isotope, $A_{\text{r}} = \frac{(10.0 \times 90) + (12.0 \times 10)}{100} = 10.2$.

$A_{\text{r}}$ always lies between the lightest and the heaviest relative isotopic masses. For an element with two isotopes, it lies nearer the more abundant one. Three consequences follow.

1. No single atom need have a mass equal to $A_{\text{r}}$. Bromine's $A_{\text{r}}$ is 79.9, but every bromine atom is either bromine-79 or bromine-81, in roughly equal numbers.
2. An element with only one natural isotope has $A_{\text{r}}$ equal to that isotope's relative isotopic mass. Natural iodine is all iodine-127, so its $A_{\text{r}}$ is 126.9, not 127. Fluorine, sodium, aluminium and phosphorus are other examples.
3. The standard is the isotope carbon-12, not the element carbon. Natural carbon contains about 1.1% carbon-13, so $A_{\text{r}}$(C) is 12.01 (12.0 in the Data Booklet), slightly more than 12.

Calculating relative atomic mass from isotope data

$$A_{\text{r}} = \frac{\sum (\text{relative isotopic mass} \times \text{abundance})}{\sum \text{abundance}}$$

1. Multiply each relative isotopic mass by its abundance.
2. Add the products.
3. Divide by the total abundance. This is 100 for percentages. For ratios or relative abundances (such as peak heights in a mass spectrum), it is the sum of the values given. If a table lists isotopes of more than one element, use only those of the element asked about.
4. Round only at the end, to the decimal places or significant figures the question asks for.

Worked example 1: the relative atomic mass of silicon

Question: Natural silicon is a mixture of three isotopes. Calculate the relative atomic mass of silicon to two decimal places.

Isotope	Relative isotopic mass	Abundance / %
silicon-28	27.98	92.23
silicon-29	28.98	4.68
silicon-30	29.97	3.09

Step 1: check the total abundance. 92.23 + 4.68 + 3.09 = 100.00, so divide by 100.

Step 2: multiply, add and divide. $$A_{\text{r}} = \frac{(27.98 \times 92.23) + (28.98 \times 4.68) + (29.97 \times 3.09)}{100}$$ $$A_{\text{r}} = \frac{2580.5954 + 135.6264 + 92.6073}{100} = \frac{2808.8291}{100} = 28.088$$

Answer: $A_{\text{r}}$(Si) = 28.09 (2 d.p.), which agrees with 28.1 in the Data Booklet.

The answer sits close to 27.98 because more than nine atoms in ten are silicon-28.

Worked example 2: finding abundances from a relative atomic mass

Question: A sample of boron has $A_{\text{r}}$ = 10.811. It contains only boron-10 (relative isotopic mass 10.013) and boron-11 (relative isotopic mass 11.009). Calculate the percentage abundance of each isotope.

Step 1: choose one unknown. Let the abundance of boron-10 be $x$%. The rest, $(100 - x)$%, is boron-11.

Step 2: write the weighted mean. $$10.811 = \frac{10.013x + 11.009(100 - x)}{100}$$

Step 3: solve. Multiply both sides by 100 and expand the bracket: $$1081.1 = 10.013x + 1100.9 - 11.009x$$ $$0.996x = 1100.9 - 1081.1 = 19.8$$ $$x = \frac{19.8}{0.996} = 19.88$$

Answer: boron-10 is 19.9% and boron-11 is 80.1% (3 s.f.).

Check: $\frac{(10.013 \times 19.9) + (11.009 \times 80.1)}{100} = 10.811$. The answer comes from a small difference (1100.9 − 1081.1), so keep every digit of the data until the end.

Relative molecular mass and relative formula mass

A molecule is a group of atoms held together by covalent bonds. The relative molecular mass, $M_{\text{r}}$, is the weighted mean mass of a molecule compared with one twelfth of the mass of a carbon-12 atom. In practice, add the $A_{\text{r}}$ values of every atom in the molecular formula. For water, $M_{\text{r}} = (2 \times 1.0) + 16.0 = 18.0$.

Ionic compounds and giant covalent substances have no molecules. Their formula shows only the ratio of ions or atoms, called a formula unit. The relative formula mass is the weighted mean mass of one formula unit compared with one twelfth of the mass of a carbon-12 atom. You calculate it in exactly the same way.

Substance	Structure	Term to use	Value
H₂O	simple molecular	relative molecular mass	18.0
Cl₂	simple molecular (an element)	relative molecular mass	71.0
C₂H₅OH	simple molecular	relative molecular mass	46.0
NaCl	giant ionic	relative formula mass	58.5
MgCl₂	giant ionic	relative formula mass	95.3
SiO₂	giant covalent	relative formula mass	60.1
Cu	giant metallic (an element)	relative atomic mass	63.5

Points to note:

• The symbol. $M_{\text{r}}$ is often used for relative formula mass too. In words, though, NaCl has a relative formula mass, never a relative molecular mass.
• Ions. An electron has a mass of only about 0.0005 u, so gaining or losing a few electrons does not change a relative mass at one decimal place. NaCl has the same relative formula mass whether you count Na⁺ and Cl⁻ ions or Na and Cl atoms.
• Water of crystallisation. In a hydrated salt, the water is part of the formula unit, so it must be included. The dot in CuSO₄·5H₂O means "together with": add 5 × 18.0 for the water.
• Data Booklet values. Use the $A_{\text{r}}$ values exactly as printed, to one decimal place (Mg 24.3, Cl 35.5, K 39.1), unless a question gives its own, more precise value. A sum of one-decimal-place values is also given to one decimal place.

Worked example 3: the relative formula mass of a hydrated salt

Question: Hydrated iron(III) nitrate has the formula Fe(NO₃)₃·9H₂O. State which term describes its relative mass, and calculate it using Data Booklet values.

Step 1: choose the term. The compound is ionic: a lattice of Fe³⁺ and NO₃⁻ ions with water of crystallisation. There are no separate molecules of Fe(NO₃)₃·9H₂O, so the term is relative formula mass.

Step 2: count each kind of atom. The 3 after the bracket multiplies N and O₃, so there are 3 N and 9 O in the nitrate ions.

Part of the formula	Working	Contribution
Fe	1 × 55.8	55.8
N	3 × 14.0	42.0
O (in the nitrate ions)	9 × 16.0	144.0
9H₂O	9 × [(2 × 1.0) + 16.0]	162.0
Total		403.8

Answer: the relative formula mass of Fe(NO₃)₃·9H₂O is 403.8 (no units). Leaving out the water would give 241.8, the value for the anhydrous salt.

From relative masses to real masses

Since 1 u = 1.66 × 10⁻²⁴ g, you can move between a relative mass and the actual mass of one particle:

$$\text{mass of one particle (g)} = \text{relative mass} \times 1.66 \times 10^{-24}\ \text{g}$$

This link is why relative masses are so useful. $L$ particles, each of relative mass $M_{\text{r}}$, have a total mass of $M_{\text{r}} \times L \times \frac{1}{L}$ g, which is $M_{\text{r}}$ grams. So the molar mass of a substance, in g mol⁻¹, has the same number as its relative mass: water has $M_{\text{r}}$ = 18.0 and a molar mass of 18.0 g mol⁻¹. Only the molar mass has a unit.

Worked example 4: identifying a molecule from its mass

Question: The mean mass of one molecule of a gaseous hydrocarbon with the general formula CₙH₂ₙ is 9.30 × 10⁻²³ g. Use $L$ = 6.022 × 10²³ mol⁻¹ to find its relative molecular mass and its molecular formula.

Step 1: the mass of 1 u in grams. $$1\ \text{u} = \frac{1\ \text{g}}{6.022 \times 10^{23}} = 1.6606 \times 10^{-24}\ \text{g}$$

Step 2: divide by 1 u. $$M_{\text{r}} = \frac{9.30 \times 10^{-23}\ \text{g}}{1.6606 \times 10^{-24}\ \text{g}} = 56.0$$

Step 3: find $n$. For CₙH₂ₙ, $M_{\text{r}} = 12.0n + (2 \times 1.0)n = 14.0n$, so $$n = \frac{56.0}{14.0} = 4.00$$

Answer: $M_{\text{r}}$ = 56.0 (3 s.f.), so the molecular formula is C₄H₈.

The question says "mean" mass because about 4 in every 100 of these molecules contain a carbon-13 atom and are slightly heavier than the rest.

Key Equations

Unified atomic mass unit ($L$ = Avogadro constant, 6.022 × 10²³ mol⁻¹): $$1\ \text{u} = \tfrac{1}{12} \times \text{mass of one carbon-12 atom} \approx \frac{1\ \text{g}}{L} = 1.66 \times 10^{-24}\ \text{g} = 1.66 \times 10^{-27}\ \text{kg}$$

Any relative mass (no units; use the weighted mean mass when the particles differ): $$\text{relative mass} = \frac{\text{mass of the particle}}{1\ \text{u}}$$

Relative atomic mass from isotopes (abundances as percentages, ratios or relative abundances): $$A_{\text{r}} = \frac{\sum (\text{relative isotopic mass} \times \text{abundance})}{\sum \text{abundance}}$$

Two isotopes ($m_1$ and $m_2$ = relative isotopic masses; $x$ = percentage abundance of the isotope of mass $m_1$): $$A_{\text{r}} = \frac{m_1 x + m_2 (100 - x)}{100}$$

Relative molecular or formula mass (add the $A_{\text{r}}$ of every atom in the formula, including any water of crystallisation): $$M_{\text{r}} = \sum A_{\text{r}}$$

Molar mass ($M$ in g mol⁻¹, numerically equal to the relative mass): $$M = M_{\text{r}}\ \text{g mol}^{-1}$$

Common Mistakes to Avoid

• Mistake: defining $A_{\text{r}}$ as "the average mass of an element" with no standard, or with "carbon" as the standard. Fix: give both parts: the weighted mean mass of an atom of the element, compared with one twelfth of the mass of one atom of carbon-12. $A_{\text{r}}$(Si) = 28.09 means a silicon atom, on average, is 28.09 times as heavy as one twelfth of a carbon-12 atom.
• Mistake: comparing with the whole carbon-12 atom. Fix: the standard is one twelfth of it. Against a whole carbon-12 atom, sodium would come out at 23.0 ÷ 12 = 1.92.
• Mistake: calling relative isotopic mass "the average mass of an isotope", or swapping it with $A_{\text{r}}$. Fix: relative isotopic mass is for one atom of one isotope, with no averaging. Silicon-28 has a relative isotopic mass of 27.98; silicon, the element, has $A_{\text{r}}$ = 28.09.
• Mistake: taking a simple mean of the isotopic masses. Fix: weight each by its abundance. In Worked example 1, the simple mean (27.98 + 28.98 + 29.97) ÷ 3 = 28.98, almost a whole unit above the correct 28.09.
• Mistake: dividing by 100 when the abundances are not percentages. Fix: divide by their total. Isotopes of mass 50 and 52 in the ratio 3 : 1 give $A_{\text{r}} = \frac{(50 \times 3) + (52 \times 1)}{4} = 50.5$, not 2.02.
• Mistake: giving a relative mass a unit. Fix: relative masses are ratios. $M_{\text{r}}$(C₄H₈) = 56.0; only the molar mass, 56.0 g mol⁻¹, has a unit.
• Mistake: writing "relative molecular mass" for an ionic or giant covalent substance. Fix: these have no molecules. SiO₂ and NaCl have relative formula masses (60.1 and 58.5), even when a question uses the symbol $M_{\text{r}}$.
• Mistake: rounding Data Booklet values, or leaving out water of crystallisation. Fix: K₂SO₄ is (2 × 39.1) + 32.1 + (4 × 16.0) = 174.3, not 174.1 from K = 39. Fe(NO₃)₃·9H₂O is 403.8, not 241.8.

Exam Tips

• Define each relative mass in two parts: what is compared ("the weighted mean mass of an atom of an element", "the mass of one atom of an isotope", "the weighted mean mass of a molecule") and the standard ("one twelfth of the mass of one atom of carbon-12"). Write "carbon-12", never just "carbon".
• When a definition must be in terms of the unified atomic mass unit, write "compared with the unified atomic mass unit, which is one twelfth of the mass of one atom of carbon-12". That covers both forms in one sentence.
• Calculate $A_{\text{r}}$ by writing the whole weighted-mean expression with numbers before you work it out. Round only at the end, to the decimal places asked for (often 2 d.p.), or else to as many significant figures as the data justify.
• Explain why $A_{\text{r}}$ is not a whole number by naming both causes: it is a weighted mean over isotopes of different masses, and relative isotopic masses are not exact whole numbers (except carbon-12).
• Show an $M_{\text{r}}$ as a sum, part by part, using Data Booklet values to one decimal place. Give the answer without units, and check you have answered the quantity asked for: the $M_{\text{r}}$ of the compound and the $A_{\text{r}}$ of one element in it are different answers.