Relative Masses and the Mole

AS · 7 min

Atoms are far too small to weigh or count one at a time, so chemists compare their masses on a relative scale and count them in large, fixed bundles called moles. Every calculation in AS Chemistry, from reacting masses and titrations to equilibrium constants and enthalpy changes, rests on the definitions and the one equation in this note. The definitions of relative atomic mass, relative isotopic mass and the mole are examined word for word, so learn them exactly.

The unified atomic mass unit

The mass of a hydrogen atom is about $1.67 \times 10^{-24}\ \text{g}$ . Numbers like this are unhelpful, so masses of atoms are measured relative to a standard: the carbon-12 atom.

Definition

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

On this scale a $\ce{^{12}C}$ atom has a mass of exactly $12\ u$ , and a proton or a neutron has a mass of very nearly $1\ u$ . One unified atomic mass unit is $1.66 \times 10^{-24}\ \text{g}$ , but you will not need that value.

Four relative masses

Each relative mass compares the mass of a particle with the unified atomic mass unit. Because it is a ratio of two masses, a relative mass has no units.

Definition

Relative isotopic mass is the mass of an atom of an isotope compared with one twelfth of the mass of an atom of carbon-12.
Relative atomic mass, $A_r$ , is the weighted mean mass of the atoms of an element compared with one twelfth of the mass of an atom of carbon-12.
Relative molecular mass, $M_r$ , is the weighted mean mass of a molecule compared with one twelfth of the mass of an atom of carbon-12.
Relative formula mass is the weighted mean mass of one formula unit of a compound compared with one twelfth of the mass of an atom of carbon-12. It is used for ionic compounds, which are made of ions rather than molecules.

The phrase weighted mean matters. Most elements are mixtures of isotopes. Chlorine is about three-quarters $\ce{^{35}Cl}$ and one-quarter $\ce{^{37}Cl}$ , so the average mass of a chlorine atom is closer to 35 than to 37. A weighted mean takes account of how common each isotope is.

Key result

A_r = \frac{\sum (\text{relative isotopic mass} \times \text{abundance})}{\sum \text{abundance}}

If the abundances are percentages, $\sum \text{abundance} = 100$ .

Relative molecular and formula masses

Add up the relative atomic masses of every atom in the formula. Use the values in the Periodic Table in the data booklet, which are given to one decimal place.

substance	working	$M_r$
$\ce{H2O}$	$2(1.0) + 16.0$	18.0
$\ce{CO2}$	$12.0 + 2(16.0)$	44.0
$\ce{Ca(OH)2}$	$40.1 + 2(16.0 + 1.0)$	74.1
$\ce{CuSO4.5H2O}$	$63.5 + 32.1 + 4(16.0) + 5(18.0)$	249.6

Watch out

In $\ce{Ca(OH)2}$ the subscript 2 multiplies everything in the bracket. In $\ce{CuSO4.5H2O}$ the 5 multiplies the whole water molecule. Missing a bracket or the water of crystallisation is the commonest source of a wrong $M_r$ .

Relative atomic mass from isotopic abundances

A sample of chlorine contains 75.8% $\ce{^{35}Cl}$ and 24.2% $\ce{^{37}Cl}$ . Calculate the relative atomic mass of chlorine to three significant figures.

Solution

A_r = \frac{(35 \times 75.8) + (37 \times 24.2)}{100} = \frac{2653 + 895.4}{100} = 35.484 = 35.5 \text{ (3 s.f.)}

Three isotopes

The mass spectrum of magnesium shows peaks at $m/e$ 24, 25 and 26 with relative abundances 78.99%, 10.00% and 11.01%. Calculate $A_r$ of magnesium to four significant figures.

Solution

A_r = \frac{(24 \times 78.99) + (25 \times 10.00) + (26 \times 11.01)}{100} = \frac{1895.76 + 250.0 + 286.26}{100} = 24.32

Working backwards to an abundance

Boron has two isotopes, $\ce{^{10}B}$ and $\ce{^{11}B}$ . Its relative atomic mass is 10.8. Calculate the percentage abundance of each isotope.

Solution

Let the percentage of $\ce{^{10}B}$ be $x$ . Then the percentage of $\ce{^{11}B}$ is $100 - x$ .

\frac{10x + 11(100 - x)}{100} = 10.8 \quad\Rightarrow\quad 1100 - x = 1080 \quad\Rightarrow\quad x = 20

So boron is 20% $\ce{^{10}B}$ and 80% $\ce{^{11}B}$ .

The mole and the Avogadro constant

A mole is a counting unit, like a dozen, but much larger. It is chosen so that one mole of atoms of an element has a mass in grams equal to its relative atomic mass.

Definition

One mole of a substance is the amount of that substance that contains $6.022 \times 10^{23}$ particles. This number is the Avogadro constant, $L$ (also written $N_A$ ): $L = 6.022 \times 10^{23}\ \text{mol}^{-1}$ .

The particles must always be specified: atoms, molecules, ions, electrons or formula units. One mole of water molecules contains $6.022 \times 10^{23}$ molecules, but three times as many atoms.

The molar mass, $M$ , is the mass of one mole of a substance. It is numerically equal to $A_r$ or $M_r$ , with units of $\text{g mol}^{-1}$ . The molar mass of water is $18.0\ \text{g mol}^{-1}$ .

Key result

n = \frac{m}{M} \qquad\qquad N = n \times L

$n$ = amount in moles (mol), $m$ = mass (g), $M$ = molar mass ( $\text{g mol}^{-1}$ ), $N$ = number of particles, $L = 6.022 \times 10^{23}\ \text{mol}^{-1}$ .

Method

Moving between mass, moles and particles

Mass to moles: divide by the molar mass.
Moles to particles: multiply by $L$ .
If the question asks about atoms or ions within a formula, multiply by the number of those atoms or ions in one formula unit.
Keep at least one extra significant figure through the working; round only the final answer.

Counting molecules and atoms

Calculate (a) the number of molecules and (b) the number of hydrogen atoms in 4.50 g of water.

Solution

$n(\ce{H2O}) = \dfrac{4.50}{18.0} = 0.250\ \text{mol}$

(a) Molecules $= 0.250 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$

(b) Each molecule contains two H atoms: $2 \times 1.5055 \times 10^{23} = 3.01 \times 10^{23}$

The mass of a single atom

Calculate the mass, in grams, of one atom of iron.

Solution

One mole of iron atoms, $6.022 \times 10^{23}$ atoms, has a mass of $55.8\ \text{g}$ .

\text{mass of one atom} = \frac{55.8}{6.022 \times 10^{23}} = 9.27 \times 10^{-23}\ \text{g}

Exam-style: ions in an ionic compound

Calculate the total number of ions in $2.84\ \text{g}$ of sodium sulfate, $\ce{Na2SO4}$ .

Solution

$M(\ce{Na2SO4}) = 2(23.0) + 32.1 + 4(16.0) = 142.1\ \text{g mol}^{-1}$

$n = \dfrac{2.84}{142.1} = 0.01999\ \text{mol}$

Each formula unit contains three ions: $2\ \ce{Na+}$ and $1\ \ce{SO4^2-}$ .

N(\text{ions}) = 3 \times 0.01999 \times 6.022 \times 10^{23} = 3.61 \times 10^{22}

Exam tip

Definitions: "weighted mean (average) mass", "of an atom" (or "of the atoms of an element"), "compared with one twelfth of the mass of an atom of carbon-12". Missing one twelfth or carbon-12 loses the mark.
Relative masses have no units. Molar masses have units of $\text{g mol}^{-1}$ . Do not write " $A_r = 35.5\ \text{g}$ ".
In calculations, quote answers to the number of significant figures the data justify, usually three. Never round intermediate values to one or two figures.

Summary

$1\ u$ is one twelfth of the mass of a $\ce{^{12}C}$ atom.
Relative isotopic mass: one isotope. $A_r$ : weighted mean of all the isotopes. $M_r$ : molecules. Relative formula mass: ionic formula units.
$A_r = \sum(\text{mass} \times \text{abundance}) / \sum \text{abundance}$ .
A mole contains $6.022 \times 10^{23}$ particles (the Avogadro constant, $L$ ).
$n = m / M$ and $N = nL$ . Always state which particle you are counting.

Practice

Question

Define relative isotopic mass and relative atomic mass.
Gallium consists of 60.1% $\ce{^{69}Ga}$ and 39.9% $\ce{^{71}Ga}$ . Calculate $A_r$ of gallium to three significant figures.
Silicon contains 92.23% $\ce{^{28}Si}$ , 4.67% $\ce{^{29}Si}$ and 3.10% $\ce{^{30}Si}$ . Calculate $A_r$ of silicon to three significant figures.
Rubidium has two isotopes, $\ce{^{85}Rb}$ and $\ce{^{87}Rb}$ , and $A_r = 85.5$ . Find the percentage of $\ce{^{85}Rb}$ .
For $0.250\ \text{mol}$ of carbon dioxide, calculate the mass, the number of molecules and the number of oxygen atoms.
Calculate the mass of $1.00 \times 10^{22}$ molecules of methane, $\ce{CH4}$ .
Calculate the number of hydroxide ions in $3.70\ \text{g}$ of calcium hydroxide, $\ce{Ca(OH)2}$ .
One atom of element X has a mass of $3.82 \times 10^{-23}\ \text{g}$ . Calculate $A_r$ of X and identify it.
Calculate the relative formula mass of hydrated copper(II) sulfate, $\ce{CuSO4.5H2O}$ , and the mass of $0.0200\ \text{mol}$ of it.
Neon consists of $\ce{^{20}Ne}$ (90.48%), $\ce{^{21}Ne}$ (0.27%) and $\ce{^{22}Ne}$ (9.25%). Calculate $A_r$ to four significant figures, and explain why the $A_r$ of most elements is not a whole number even though each isotope has a relative isotopic mass very close to a whole number.

Answers

Relative isotopic mass: the mass of an atom of an isotope compared with one twelfth of the mass of an atom of carbon-12. Relative atomic mass: the weighted mean mass of the atoms of an element compared with one twelfth of the mass of an atom of carbon-12.
$A_r = \dfrac{69(60.1) + 71(39.9)}{100} = \dfrac{4146.9 + 2832.9}{100} = 69.798 = 69.8$
$A_r = \dfrac{28(92.23) + 29(4.67) + 30(3.10)}{100} = \dfrac{2582.44 + 135.43 + 93.0}{100} = 28.1$
$85x + 87(100 - x) = 8550 \Rightarrow 8700 - 2x = 8550 \Rightarrow x = 75$ . So 75% $\ce{^{85}Rb}$ .
Mass $= 0.250 \times 44.0 = 11.0\ \text{g}$ . Molecules $= 0.250 \times 6.022 \times 10^{23} = 1.51 \times 10^{23}$ . Oxygen atoms $= 2 \times 1.51 \times 10^{23} = 3.01 \times 10^{23}$ .
$n = \dfrac{1.00 \times 10^{22}}{6.022 \times 10^{23}} = 0.01661\ \text{mol}$ ; mass $= 0.01661 \times 16.0 = 0.266\ \text{g}$ .
$M = 74.1\ \text{g mol}^{-1}$ ; $n = \dfrac{3.70}{74.1} = 0.04993\ \text{mol}$ . Each formula unit has two $\ce{OH-}$ : $2 \times 0.04993 \times 6.022 \times 10^{23} = 6.01 \times 10^{22}$ ions.
Mass of one mole $= 3.82 \times 10^{-23} \times 6.022 \times 10^{23} = 23.0\ \text{g}$ , so $A_r = 23.0$ : X is sodium.
$M_r = 63.5 + 32.1 + 64.0 + 90.0 = 249.6$ . Mass $= 0.0200 \times 249.6 = 4.99\ \text{g}$ .
$A_r = \dfrac{20(90.48) + 21(0.27) + 22(9.25)}{100} = \dfrac{1809.6 + 5.67 + 203.5}{100} = 20.19$ . $A_r$ is a weighted mean of the masses of all the naturally occurring isotopes; because an element is usually a mixture of isotopes with different masses, the mean lies between the whole-number isotopic masses.

The unified atomic mass unit

Four relative masses

Relative molecular and formula masses

The mole and the Avogadro constant

Practice

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