School Chemistry — Grades 1 to 12 · Grades 1–12

25The Mole and Molar Mass

At the end of the day, a bank does not count its coins one by one: it weighs them. If one coin has a mass of $7.5\,\mathrm{g}$ , a bag of coins with a mass of $7.5\,\mathrm{kg}$ holds a thousand coins, and the scale has done the counting. Chemists face the same problem with atoms, only far worse: a spoonful of water holds more molecules than there are grains of sand on all the beaches of a coast. They solve it the same way, by weighing, and the bag they use holds $602\,214\,076\,000\,000\,000\,000\,000$ entities. This chapter defines that bag, the mole, and shows how to go from a mass on a balance to a number of atoms or molecules.

You already know

The mass of an atom is concentrated in its nucleus; a proton and a neutron each have a mass of about $1.67 \times 10^{-27}\,\mathrm{kg}$ , so an atom of mass number $A$ has a mass of about $A \times 1.67 \times 10^{-27}\,\mathrm{kg}$ . Large and small numbers are written in scientific notation (Chapter 16). A chemical formula gives the number of atoms of each element in a molecule (Chapter 11).

A bank counts coins by weighing them. Chemists count atoms the same way.

25.1 Counting by weighing

Example 25.1 (A bag of coins)

A coin has a mass of $7.5\,\mathrm{g}$ . A bag of these coins has a mass of $1.80\,\mathrm{kg}$ , that is $1800\,\mathrm{g}$ , the mass of the bag itself left aside. It holds

\frac{1800\,\mathrm{g}}{7.5\,\mathrm{g}} = 240 \text{ coins.}

To count, the bank divides the total mass by the mass of one coin.

Example 25.2 (A spoonful of water)

A water molecule, $\ce{H2O}$ , has $2 + 16 = 18$ nucleons (the oxygen-16 and hydrogen-1 atoms make up nearly all natural water), so its mass is about $18 \times 1.67 \times 10^{-27}\,\mathrm{kg} = 3.0 \times 10^{-26}\,\mathrm{kg}$ , or $3.0 \times 10^{-23}\,\mathrm{g}$ . A spoonful of $5.0\,\mathrm{g}$ of water therefore holds

\frac{5.0\,\mathrm{g}}{3.0 \times 10^{-23}\,\mathrm{g}} \approx 1.7 \times 10^{23} \text{ molecules.}

The division works as for the coins, but the answer is a number no one can picture. Chemists need a package of atoms adapted to the samples of the laboratory.

Definition 25.3 (Amount of substance and mole)

The amount of substance $n$ of a sample counts the entities it contains (atoms, molecules or ions, stated each time). Its unit is the mole, symbol $\mathrm{mol}$ : one mole contains exactly $6.022\,140\,76 \times 10^{23}$ entities.

Remark 25.4 (A dozen, a ream, a mole)

The mole is a counting unit, like a dozen eggs (12) or a ream of paper (500 sheets), only very much larger. “Two moles of water molecules” means $2 \times 6.022\,140\,76 \times 10^{23} = 1.20 \times 10^{24}$ molecules, just as “two dozen eggs” means 24 eggs. Always say which entities are counted: one mole of oxygen molecules $\ce{O2}$ holds two moles of oxygen atoms.

History — Avogadro’s hypothesis, 1811

In 1811 the physicist Amedeo Avogadro proposed that equal volumes of different gases, at the same temperature and pressure, contain the same number of particles. He also understood that a gas such as oxygen is made of molecules of two atoms. His idea was neglected for nearly fifty years, until Stanislao Cannizzaro used it in 1860 to obtain a coherent set of atomic weights. The constant that counts the entities of a mole was later named after Avogadro; he never knew its value.

25.2 The Avogadro constant

Definition 25.5 (Avogadro constant)

The Avogadro constant $N_A$ is the number of entities per mole:

N_A = 6.022\,140\,76 \times 10^{23}\,\mathrm{mol}^{-1} .

Its value is exact, fixed by definition. In calculations it is rounded to $6.02 \times 10^{23}\,\mathrm{mol}^{-1}$ .

Proposition 25.6 (Number of entities and amount)

A sample holding $N$ entities contains the amount

n = \frac{N}{N_A}, \qquad\text{that is}\qquad N = n \times N_A .

Proof. Each mole contains $N_A$ entities, so $n$ moles contain $n \times N_A$ entities; dividing by $N_A$ gives $n$ . ∎

Example 25.7 (Molecules and moles)

A sample contains $1.5 \times 10^{24}$ molecules of carbon dioxide. Its amount of carbon dioxide is $n = 1.5 \times 10^{24} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 2.5\,\mathrm{mol}$ . Conversely, $0.20\,\mathrm{mol}$ of iron holds $N = 0.20\,\mathrm{mol} \times 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 1.2 \times 10^{23}$ atoms of iron.

Remark 25.8 (Where the number comes from)

For more than fifty years the mole was defined as the number of atoms in exactly $12\,\mathrm{g}$ of carbon-12, and $N_A$ had to be measured. Since 2019 the number itself has been fixed, which makes the mole a pure count. The old and new definitions agree to better than one part in a hundred million, so twelve grams of carbon-12 still contain one mole of atoms, as the next section checks.

25.3 Molar mass

Definition 25.9 (Molar mass)

The molar mass $M$ of a chemical species is the mass of one mole of its entities, in grams per mole ( $\mathrm{g}/\mathrm{mol}$ ). The molar mass of an element, taken over its natural mixture of isotopes, is its atomic molar mass; it is printed in the periodic table.

Example 25.10 (One mole of carbon atoms)

An atom of carbon-12 has a mass of about $12 \times 1.67 \times 10^{-27}\,\mathrm{kg} = 2.00 \times 10^{-26}\,\mathrm{kg}$ . One mole of them has a mass of

2.00 \times 10^{-26}\,\mathrm{kg} \times 6.02 \times 10^{23} = 1.20 \times 10^{-2}\,\mathrm{kg} = 12.0\,\mathrm{g}.

The mole has been chosen so that the molar mass of an atom, in grams per mole, is close to its mass number: about $1\,\mathrm{g}/\mathrm{mol}$ for hydrogen, $12\,\mathrm{g}/\mathrm{mol}$ for carbon, $16\,\mathrm{g}/\mathrm{mol}$ for oxygen.

element	symbol	$M$ ( $\mathrm{g}/\mathrm{mol}$ )	element	symbol	$M$ ( $\mathrm{g}/\mathrm{mol}$ )
hydrogen	$\ce{H}$	1.0	silicon	$\ce{Si}$	28.1
carbon	$\ce{C}$	12.0	sulfur	$\ce{S}$	32.1
nitrogen	$\ce{N}$	14.0	chlorine	$\ce{Cl}$	35.5
oxygen	$\ce{O}$	16.0	potassium	$\ce{K}$	39.1
sodium	$\ce{Na}$	23.0	calcium	$\ce{Ca}$	40.1
magnesium	$\ce{Mg}$	24.3	iron	$\ce{Fe}$	55.8
aluminium	$\ce{Al}$	27.0	copper	$\ce{Cu}$	63.5
			gold	$\ce{Au}$	197.0

Atomic molar masses of common elements, rounded to

0.1\,\mathrm{g}/\mathrm{mol}

. These are the values used in this book.

Remark 25.11 (Why chlorine is 35.5)

Natural chlorine is a mixture of two isotopes: about three quarters of its atoms are chlorine-35, one quarter chlorine-37 (Chapter 16). Its atomic molar mass is the average, weighted by these shares, which lies between 35 and 37: $35.5\,\mathrm{g}/\mathrm{mol}$ . That is why some atomic molar masses are far from whole numbers.

Proposition 25.12 (Molar mass of a molecule)

The molar mass of a molecule is the sum of the atomic molar masses of all its atoms, each counted as many times as it appears in the formula. The same rule gives the molar mass of an ionic solid from its formula unit.

Proof. One mole of molecules $\ce{A_xB_y}$ contains $x$ moles of atoms $\ce{A}$ and $y$ moles of atoms $\ce{B}$ , and nothing else; its mass is the sum of their masses, $x\,M(\ce{A}) + y\,M(\ce{B})$ . ∎

Example 25.13 (Three molar masses)

\begin{align*} M(\ce{H2O}) &= 2 \times 1.0 + 16.0 = 18.0\,\mathrm{g}/\mathrm{mol},\\ M(\ce{NaCl}) &= 23.0 + 35.5 = 58.5\,\mathrm{g}/\mathrm{mol},\\ M(\ce{C12H22O11}) &= 12 \times 12.0 + 22 \times 1.0 + 11 \times 16.0 = 342.0\,\mathrm{g}/\mathrm{mol}. \end{align*}

The last one is sucrose, table sugar.

25.4 From mass to amount and back

Proposition 25.14 (Mass and amount)

A sample of mass $m$ of a species of molar mass $M$ contains the amount

n = \frac{m}{M}, \qquad\text{that is}\qquad m = n \times M .

With $m$ in grams and $M$ in grams per mole, $n$ is in moles.

Proof. Each mole has a mass $M$ , so $n$ moles have a mass $n \times M$ ; this is the coin count of the bank, with the mole as the coin. ∎

Method 25.15 (From a mass to a number of entities)

Write the formula of the species and compute its molar mass $M$ from the table.
Convert the mass to grams if needed, then $n = m / M$ .
If the number of entities is wanted, $N = n \times N_A$ .
Check the order of magnitude: a few grams of a small molecule is of the order of a tenth of a mole, some $10^{22}$ to $10^{23}$ entities.

The amount n sits in the middle: the balance gives m, the molar mass leads to n, and the Avogadro constant leads on to the number of entities. — The amount $n$ sits in the middle: the balance gives $m$ , the molar mass leads to $n$ , and the Avogadro constant leads on to the number of entities.

Example 25.16 (Nine grams of water)

How many molecules are there in $9.0\,\mathrm{g}$ of water? $M(\ce{H2O}) = 18.0\,\mathrm{g}/\mathrm{mol}$ , so

n = \frac{9.0\,\mathrm{g}}{18.0\,\mathrm{g}/\mathrm{mol}} = 0.50\,\mathrm{mol}, \qquad N = 0.50\,\mathrm{mol} \times 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 3.0 \times 10^{23} \text{ molecules.}

Example 25.17 (A mole on the bench)

One mole of water is $18.0\,\mathrm{g}$ , about a tablespoon. One mole of salt, $58.5\,\mathrm{g}$ , is a small heap; one mole of sugar, $342.0\,\mathrm{g}$ , fills a mug; one mole of iron, $55.8\,\mathrm{g}$ , is a cube a little under $2\,\mathrm{cm}$ across. All four hold the same number of entities: the molecules of sugar are simply much heavier than those of water, and the atoms of iron are packed much more tightly.

One mole each of water, salt (sodium chloride), sugar (sucrose) and iron, drawn as cubes of their true volumes, all at the same scale. Each holds 6.02 × 1023 entities. — One mole each of water, salt (sodium chloride), sugar (sucrose) and iron, drawn as cubes of their true volumes, all at the same scale. Each holds $6.02 \times 10^{23}$ entities.

25.5 Gases: the molar volume

Definition 25.18 (Molar volume)

The molar volume $V_m$ of a gas is the volume occupied by one mole of that gas, at a given temperature and pressure. It is expressed in litres per mole ( $\mathrm{L}/\mathrm{mol}$ ).

Proposition 25.19 (All gases have the same molar volume)

At a given temperature and pressure, one mole of any gas occupies very nearly the same volume, whatever the gas. At $20\,{}^{\circ}\mathrm{C}$ and normal atmospheric pressure,

V_m = 24.1\,\mathrm{L}/\mathrm{mol};

at $0\,{}^{\circ}\mathrm{C}$ and the same pressure, $V_m = 22.4\,\mathrm{L}/\mathrm{mol}$ . The amount of a gas of volume $V$ is then $n = V / V_m$ .

Proof. Admitted here: this is Avogadro’s hypothesis of 1811, now a law of physics, and the value of $V_m$ follows from the gas laws studied in the physics book. ∎

Three balloons at 20\, C, each holding one mole of a different gas. The volumes are the same; the masses are not. — Three balloons at $20\,{}^{\circ}\mathrm{C}$ , each holding one mole of a different gas. The volumes are the same; the masses are not.

Example 25.20 (A balloon of carbon dioxide)

A balloon holds $2.4\,\mathrm{L}$ of carbon dioxide at $20\,{}^{\circ}\mathrm{C}$ . Its amount is $n = 2.4\,\mathrm{L} / 24.1\,\mathrm{L}/\mathrm{mol} = 0.10\,\mathrm{mol}$ , and its mass $m = 0.10\,\mathrm{mol} \times 44.0\,\mathrm{g}/\mathrm{mol} = 4.4\,\mathrm{g}$ . The same balloon filled with hydrogen would hold the same $0.10\,\mathrm{mol}$ , but only $0.20\,\mathrm{g}$ of gas.

Remark 25.21 (Heavy and light gases)

Since a litre of any gas holds the same amount, the mass of a litre of gas is proportional to its molar mass. Air, a mixture of about four fifths nitrogen ( $28.0\,\mathrm{g}/\mathrm{mol}$ ) and one fifth oxygen ( $32.0\,\mathrm{g}/\mathrm{mol}$ ), behaves like a gas of molar mass about $29\,\mathrm{g}/\mathrm{mol}$ . A gas of larger molar mass, such as carbon dioxide ( $44.0\,\mathrm{g}/\mathrm{mol}$ ), is denser than air and gathers at the bottom of a closed cellar; a gas of smaller molar mass, such as helium ( $4.0\,\mathrm{g}/\mathrm{mol}$ ), rises.

25.6 Exercises

Exercise 25.1 ★

Compute the molar masses of dioxygen $\ce{O2}$ , dinitrogen $\ce{N2}$ , methane $\ce{CH4}$ , ammonia $\ce{NH3}$ and carbon dioxide $\ce{CO2}$ .

Solution

Solution of Exercise 25.1.

$M(\ce{O2}) = 2 \times 16.0 = 32.0\,\mathrm{g}/\mathrm{mol}$ ; $M(\ce{N2}) = 2 \times 14.0 = 28.0\,\mathrm{g}/\mathrm{mol}$ ; $M(\ce{CH4}) = 12.0 + 4 \times 1.0 = 16.0\,\mathrm{g}/\mathrm{mol}$ ; $M(\ce{NH3}) = 14.0 + 3 \times 1.0 = 17.0\,\mathrm{g}/\mathrm{mol}$ ; $M(\ce{CO2}) = 12.0 + 2 \times 16.0 = 44.0\,\mathrm{g}/\mathrm{mol}$ .

Exercise 25.2 ★

What amount of substance is contained in $36.0\,\mathrm{g}$ of water? In $4.0\,\mathrm{g}$ of methane?

Solution

Solution of Exercise 25.2.

Water: $n = 36.0 / 18.0 = 2.00\,\mathrm{mol}$ . Methane: $n = 4.0 / 16.0 = 0.25\,\mathrm{mol}$ .

Exercise 25.3 ★

What is the mass of $0.25\,\mathrm{mol}$ of sodium chloride $\ce{NaCl}$ ? Of $2.0\,\mathrm{mol}$ of carbon dioxide?

Solution

Solution of Exercise 25.3.

$m(\ce{NaCl}) = 0.25 \times 58.5 = 14.6\,\mathrm{g}$ , about $15\,\mathrm{g}$ ; $m(\ce{CO2}) = 2.0 \times 44.0 = 88\,\mathrm{g}$ .

Exercise 25.4 ★

How many molecules are there in $0.50\,\mathrm{mol}$ of dioxygen? What amount of iron contains $3.0 \times 10^{24}$ atoms of iron?

Solution

Solution of Exercise 25.4.

$N = 0.50 \times 6.02 \times 10^{23} = 3.0 \times 10^{23}$ molecules of dioxygen. $n = 3.0 \times 10^{24} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 5.0\,\mathrm{mol}$ of iron.

Exercise 25.5 ★

At $20\,{}^{\circ}\mathrm{C}$ and normal atmospheric pressure, what volume does $0.50\,\mathrm{mol}$ of dinitrogen occupy? What amount of dioxygen is there in $12\,\mathrm{L}$ of that gas?

Solution

Solution of Exercise 25.5.

$V = n \times V_m = 0.50 \times 24.1 = 12\,\mathrm{L}$ . $n = V / V_m = 12 / 24.1 = 0.50\,\mathrm{mol}$ .

Exercise 25.6 ★★

A drop of water has a volume of $0.050\,\mathrm{mL}$ , and $1\,\mathrm{mL}$ of water has a mass of $1.0\,\mathrm{g}$ . How many water molecules does the drop contain?

Solution

Solution of Exercise 25.6.

The drop has a mass of $0.050\,\mathrm{g}$ , so $n = 0.050 / 18.0 = 2.8 \times 10^{-3}\,\mathrm{mol}$ and $N = 2.8 \times 10^{-3} \times 6.02 \times 10^{23} = 1.7 \times 10^{21}$ molecules.

Exercise 25.7 ★★

Which contains more atoms, $10\,\mathrm{g}$ of aluminium or $10\,\mathrm{g}$ of iron? Explain without computing first, then check by computing.

Solution

Solution of Exercise 25.7.

An aluminium atom ( $27.0\,\mathrm{g}/\mathrm{mol}$ ) is about half as heavy as an iron atom ( $55.8\,\mathrm{g}/\mathrm{mol}$ ), so the same mass of aluminium holds about twice as many atoms. Check: aluminium $10/27.0 = 0.370\,\mathrm{mol}$ , that is $2.23 \times 10^{23}$ atoms; iron $10/55.8 = 0.179\,\mathrm{mol}$ , that is $1.08 \times 10^{23}$ atoms. Aluminium wins, by a factor of about 2.1.

Exercise 25.8 ★★

At $20\,{}^{\circ}\mathrm{C}$ , compute the mass of one litre of carbon dioxide and of one litre of dinitrogen. Why does carbon dioxide collect at the bottom of a closed cellar where grape juice is fermenting?

Solution

Solution of Exercise 25.8.

One litre is $1/24.1 = 0.0415\,\mathrm{mol}$ of gas. Carbon dioxide: $0.0415 \times 44.0 = 1.83\,\mathrm{g}$ ; dinitrogen: $0.0415 \times 28.0 = 1.16\,\mathrm{g}$ . Carbon dioxide is about one and a half times denser than air, whose molar mass is close to that of dinitrogen. The fermenting juice gives off carbon dioxide, which sinks and builds up near the floor of a closed cellar, where it can suffocate anyone who goes in.

Exercise 25.9 ★★

Using the figure of the three boxes $m$ , $n$ , $N$ , give the mass of $1.0 \times 10^{22}$ molecules of glucose $\ce{C6H12O6}$ . Write down the two arrows followed.

Solution

Solution of Exercise 25.9.

$M(\ce{C6H12O6}) = 6 \times 12.0 + 12 \times 1.0 + 6 \times 16.0 = 180.0\,\mathrm{g}/\mathrm{mol}$ . Arrow $\div N_A$ : $n = 1.0 \times 10^{22} / 6.02 \times 10^{23} = 1.66 \times 10^{-2}\,\mathrm{mol}$ ; arrow $\times M$ : $m = 1.66 \times 10^{-2} \times 180.0 = 3.0\,\mathrm{g}$ .

Exercise 25.10 ★★

A lump of sugar (sucrose, $\ce{C12H22O11}$ ) has a mass of $5.0\,\mathrm{g}$ . Compute the amount of sucrose it contains, then the number of sucrose molecules, then the number of carbon atoms.

Solution

Solution of Exercise 25.10.

$M = 342.0\,\mathrm{g}/\mathrm{mol}$ , so $n = 5.0 / 342.0 = 1.46 \times 10^{-2}\,\mathrm{mol}$ ; $N = 1.46 \times 10^{-2} \times 6.02 \times 10^{23} = 8.8 \times 10^{21}$ molecules. Each has 12 carbon atoms: $12 \times 8.8 \times 10^{21} = 1.1 \times 10^{23}$ carbon atoms.

Exercise 25.11 ★★

Which contains the larger amount of molecules: $1.0\,\mathrm{kg}$ of water $\ce{H2O}$ or $1.0\,\mathrm{kg}$ of ethanol $\ce{C2H6O}$ ? By what factor?

Solution

Solution of Exercise 25.11.

Water: $1000 / 18.0 = 55.6\,\mathrm{mol}$ . Ethanol: $M = 2 \times 12.0 + 6 \times 1.0 + 16.0 = 46.0\,\mathrm{g}/\mathrm{mol}$ , so $1000 / 46.0 = 21.7\,\mathrm{mol}$ . Water has the larger amount, by a factor $55.6 / 21.7 \approx 2.6$ , which is simply $46.0 / 18.0$ .

Exercise 25.12 ★★★

A gold ring has a mass of $5.0\,\mathrm{g}$ ; three quarters of its mass is gold, the rest copper. How many atoms of gold does it contain? How many atoms of copper? Which metal has more atoms in the ring?

Solution

Solution of Exercise 25.12.

Gold: $3.75\,\mathrm{g}$ , $n = 3.75 / 197.0 = 1.90 \times 10^{-2}\,\mathrm{mol}$ , that is $1.15 \times 10^{22}$ atoms. Copper: $1.25\,\mathrm{g}$ , $n = 1.25 / 63.5 = 1.97 \times 10^{-2}\,\mathrm{mol}$ , that is $1.19 \times 10^{22}$ atoms. Surprisingly, the ring holds slightly more copper atoms than gold atoms: a gold atom is about three times heavier than a copper atom.

Exercise 25.13 ★★★

A classroom measures $8.0\,\mathrm{m}$ by $6.0\,\mathrm{m}$ by $3.0\,\mathrm{m}$ , and the air in it is at $20\,{}^{\circ}\mathrm{C}$ .

What amount of gas does it contain ( $1\,\mathrm{m}^{3} = 1000\,\mathrm{L}$ )? How many molecules?
Taking air as 78 molecules of dinitrogen, 21 of dioxygen and 1 of argon ( $40.0\,\mathrm{g}/\mathrm{mol}$ ) in every hundred, compute the mean molar mass of air.
What is the mass of the air in the room?

Solution

Solution of Exercise 25.13.

$V = 8.0 \times 6.0 \times 3.0 = 144\,\mathrm{m}^{3} = 1.44 \times 10^{5}\,\mathrm{L}$ ; $n = 1.44 \times 10^{5} / 24.1 = 5.98 \times 10^{3}\,\mathrm{mol}$ ; $N = 5.98 \times 10^{3} \times 6.02 \times 10^{23} = 3.6 \times 10^{27}$ molecules.
$M = (78 \times 28.0 + 21 \times 32.0 + 1 \times 40.0) / 100 = 2896 / 100 = 29.0\,\mathrm{g}/\mathrm{mol}$ .
$m = 5.98 \times 10^{3} \times 29.0 = 1.73 \times 10^{5}\,\mathrm{g}$ , about $173\,\mathrm{kg}$ : the air of a classroom weighs as much as two or three adults.

Exercise 25.14 ★★★

A sphere of pure silicon has a mass of exactly $1.000\,\mathrm{kg}$ .

How many silicon atoms does it contain?
Deduce the mass of one silicon atom.
Before 2019 the Avogadro constant was not fixed but measured. Explain how an independent count of the atoms of such a sphere, whose amount is known from its mass, would give a value of it.

Solution

Solution of Exercise 25.14.

$n = 1000 / 28.1 = 35.6\,\mathrm{mol}$ , so $N = 35.6 \times 6.02 \times 10^{23} = 2.14 \times 10^{25}$ atoms.
$m = 1.000\,\mathrm{kg} / 2.14 \times 10^{25} = 4.67 \times 10^{-26}\,\mathrm{kg}$ .
The mass of the sphere and the molar mass give its amount, $n = m / M$ . If the atoms $N$ are counted independently (from the volume of the sphere and the spacing of the atoms in the crystal), the ratio $N / n$ is the Avogadro constant.

Exercise 25.15 ★★★

Natural chlorine contains $75.8\,\%$ of chlorine-35 atoms, of molar mass very close to $35.0\,\mathrm{g}/\mathrm{mol}$ , and $24.2\,\%$ of chlorine-37 atoms, of molar mass very close to $37.0\,\mathrm{g}/\mathrm{mol}$ . Compute the atomic molar mass of natural chlorine, and compare it with the value of the table.

Solution

Solution of Exercise 25.15.

$M = 0.758 \times 35.0 + 0.242 \times 37.0 = 26.53 + 8.95 = 35.5\,\mathrm{g}/\mathrm{mol}$ : the value of the table.

25.7 Problem: Kelvin’s Glass of Water

Problem 25.1

Weekend problem — pour a glass of water into the sea, wait for it to mix through all the oceans, and fill a glass again: how many of the first molecules come back?

A famous thought experiment, often attributed to the physicist Lord Kelvin, goes like this. Mark every molecule in a glass of water, pour the glass into the sea, and wait until the marked molecules have spread evenly through all the oceans of the world. Then dip the glass into the sea again, anywhere. Will it bring back any marked molecules? The glass holds $250\,\mathrm{mL}$ , and $1\,\mathrm{mL}$ of water has a mass of $1.0\,\mathrm{g}$ . The oceans hold about $1.335 \times 10^{9}\,\mathrm{km}^{3}$ of water. Treat sea water as pure water for the counting.

Part I — Water in a glass.

Compute the molar mass of water.
What is the mass of the water in the glass?
Compute the amount of water molecules in the glass.
What amount of hydrogen atoms does the glass contain? Of oxygen atoms?

Part II — Molecules in the glass.

How many water molecules does the glass hold?
Compute the mass of one water molecule, in grams.
Check the answers to 5 and 6 by multiplying them: what should the product be?
Counting one molecule per second, day and night, how many years would it take to count the molecules of the glass? (A year lasts about $3.16 \times 10^{7}\,\mathrm{s}$ .)

Part III — The ocean.

Convert the volume of the oceans to cubic metres ( $1\,\mathrm{km}^{3} = 10^{9}\,\mathrm{m}^{3}$ ), then to litres.
Once the marked molecules have spread evenly, how many of them are there in each litre of ocean?
How many marked molecules does the second glass bring back?
Second method. Compute the amount of water in the oceans, taking $1.0\,\mathrm{kg}$ for the mass of one litre.
What fraction of all the ocean’s water molecules are marked?
Multiply this fraction by the number of molecules of the second glass, and compare with the answer to question 11.

Part IV — What it means.

How many glasses of $250\,\mathrm{mL}$ could be filled with the water of the oceans?
Compare this number with the number of molecules of one glass. Explain in one sentence why the second glass brings back so many marked molecules.
What volume of sea water must be drawn, on average, to bring back a single marked molecule? Compare with a drop of $0.05\,\mathrm{mL}$ .
The answer to question 11 has many digits. Why should it be given only as an order of magnitude? Give two reasons.
State the final answer: about how many molecules of the first glass come back in the second?

Solution

Solution of Problem 25.1.

1. $M(\ce{H2O}) = 2 \times 1.0 + 16.0 = 18.0\,\mathrm{g}/\mathrm{mol}$ .

2. $m = 250 \times 1.0\,\mathrm{g} = 250\,\mathrm{g}$ .

3. $n = 250 / 18.0 = 13.9\,\mathrm{mol}$ .

4. Each molecule has two hydrogen atoms and one oxygen atom: $27.8\,\mathrm{mol}$ of hydrogen atoms, $13.9\,\mathrm{mol}$ of oxygen atoms.

5. $N = 13.9 \times 6.02 \times 10^{23} = 8.36 \times 10^{24}$ molecules.

6. $m = 18.0\,\mathrm{g}/\mathrm{mol} / 6.02 \times 10^{23}\,\mathrm{mol}^{-1} = 2.99 \times 10^{-23}\,\mathrm{g}$ .

7. $8.36 \times 10^{24} \times 2.99 \times 10^{-23}\,\mathrm{g} = 250\,\mathrm{g}$ , the mass of the water in the glass, as it should be.

8. $8.36 \times 10^{24} / 3.16 \times 10^{7} = 2.6 \times 10^{17}$ years: a number of years with seventeen zeros, far beyond any possible count.

9. $1.335 \times 10^{9}\,\mathrm{km}^{3} \times 10^{9} = 1.335 \times 10^{18}\,\mathrm{m}^{3}$ , and with $1\,\mathrm{m}^{3} = 1000\,\mathrm{L}$ , $1.335 \times 10^{21}\,\mathrm{L}$ .

10. $8.36 \times 10^{24} / 1.335 \times 10^{21} = 6.26 \times 10^{3}$ marked molecules per litre.

11. $6.26 \times 10^{3} \times 0.250 = 1.57 \times 10^{3}$ marked molecules in the second glass.

12. Mass $1.335 \times 10^{21} \times 1.0\,\mathrm{kg} = 1.335 \times 10^{21}\,\mathrm{kg} = 1.335 \times 10^{24}\,\mathrm{g}$ ; $n = 1.335 \times 10^{24} / 18.0 = 7.42 \times 10^{22}\,\mathrm{mol}$ .

13. $13.9 / 7.42 \times 10^{22} = 1.87 \times 10^{-22}$ .

14. $1.87 \times 10^{-22} \times 8.36 \times 10^{24} = 1.57 \times 10^{3}$ : the same answer as question 11, as it must be.

15. $1.335 \times 10^{21} / 0.250 = 5.34 \times 10^{21}$ glasses.

16. $8.36 \times 10^{24} / 5.34 \times 10^{21} \approx 1570$ . A glass holds about 1600 times more molecules than the oceans hold glasses of water; spread evenly, the molecules of one glass leave about 1600 in every glassful of ocean.

17. $1 / 6.26 \times 10^{3} = 1.6 \times 10^{-4}\,\mathrm{L} = 0.16\,\mathrm{mL}$ , about three drops.

18. The ocean volume is an estimate known to three or four figures at best, and it was rounded; sea water is not pure water (it holds salts); the glass is not exactly $250\,\mathrm{mL}$ ; and the mixing through all the oceans would never be perfectly even. Only the order of magnitude can be trusted.

19. About 1600 molecules of the first glass come back in the second.

Terms defined in this chapter

See all 852 terms in the glossary