Chapter 5: Mole Concept and Stoichiometry
On a September morning in 1804, a young French chemist climbed into the basket of a hydrogen balloon in Paris and cut it loose. Joseph Louis Gay-Lussac went up alone, through the clouds, to about seven thousand metres — higher than any human being had been before.
He was not there for the view. He carried empty glass flasks, sealed and emptied of air. At the top of the flight he opened them, closed them again, and brought the trapped air back down to be weighed and analysed. The air that high up, he found, had the same composition as the air in the street below.
That habit — measure the gas, then measure it again — stayed with him. Back in the laboratory, working with gases that react, he began to notice something odd about the volumes. Two litres of hydrogen and one litre of oxygen gave two litres of water vapour. One volume of nitrogen with three volumes of hydrogen gave two volumes of ammonia. Never an awkward ratio like 1.37 : 2.6. Always small whole numbers.
Gay-Lussac could report the pattern, but he could not say why gases should behave so obligingly. The answer came a few years later from Amedeo Avogadro in Turin, in one bold sentence about equal volumes holding equal numbers of molecules. Between them they opened a door that chemistry still walks through: measure a volume of gas, and you can count its particles.
Diagram
Fig. 5.0-1 — Measuring Gases, From The Balloon To The Bench Left panel: a hydrogen balloon high above a cloud layer with a lone observer in the basket holding a stoppered glass flask, altitude marked as about 7000 m. Right panel: three labelled gas jars on a bench showing 1 volume of nitrogen plus 3 volumes of hydrogen giving 2 volumes of ammonia, all at the same temperature and pressure, with a caption strip reading 1 : 3 : 2.
5.1 Gas Laws Revisited
A gas has no shape and no fixed volume of its own. Squeeze it, warm it, or move it to a taller container, and it fills whatever space it is given. That is why a chemist who wants to talk about a quantity of gas must always say three things about it: how much space it occupies, how hard it pushes on the walls, and how hot it is.
You met these three quantities in Class IX. Volume \(V\) is the space the gas fills, pressure \(P\) is the force it exerts on each unit area of the container wall, and temperature \(T\) measures how fast its particles are moving. The gas laws are the rules that connect them — statements of how a fixed mass of gas responds when one of the three is changed.
The important thing about the gas laws is that they apply to every gas. Hydrogen, chlorine, ammonia and carbon dioxide are chemically nothing like one another, yet all four shrink by the same fraction when the pressure on them is doubled. That common behaviour is the foundation on which this whole chapter is built.
5.1.1 Boyle’s Law — Pressure And Volume
Take a plastic syringe, seal the nozzle with your finger, and push the plunger in. The trapped air takes up less room, and you can feel it pushing back harder as you go. Squeeze it into half the space and it pushes back twice as hard. This is the relation drawn in Fig. 5.1-1.
Diagram
Fig. 5.1-1 — Volume Falls As Pressure Rises Left: two drawings of the same sealed syringe, the first with the plunger out and the label ‘volume V, pressure P’, the second with the plunger pushed halfway in and the label ‘volume V/2, pressure 2P’; show the trapped gas particles more closely packed in the second. Right: two graphs side by side for a fixed mass of gas at constant temperature — a smooth falling curve of volume against pressure, and a straight line through the origin of volume against 1/pressure.
Robert Boyle set this pattern down as Boyle’s law. Notice the two conditions built into the statement — the mass of gas must not change, and the temperature must be held constant. Leave either one out and the law is no longer true.
Written as a relation, this is
\[V \propto \frac{1}{P} \quad \text{(constant mass, constant } T\text{)}\]
\[P V = k \qquad \text{or} \qquad P_1V_1 = P_2V_2\]
where \(P_1\) and \(V_1\) are the pressure and volume before the change, \(P_2\) and \(V_2\) the pressure and volume after it, and \(k\) is a constant for that sample at that temperature. Since \(P_1V_1\) and \(P_2V_2\) must come out equal, the two pressures may be measured in any unit you like, provided both are in the same unit — and the same is true of the two volumes.
5.1.2 Charles’s Law — Temperature And Volume
Now hold the pressure steady and warm the gas instead. A balloon left in the sun swells; the same balloon carried into a cold room shrinks a little. The gas particles move faster when heated, strike the walls harder and more often, and push the walls outward until the volume has grown enough to restore the balance.
Jacques Charles measured this carefully, and found that the volume does not rise in step with the Celsius temperature but with the temperature counted from a far lower starting point — what we call the absolute or kelvin temperature, described fully in Section 5.1.5. For now, take it that a kelvin temperature is simply the Celsius reading plus 273.
In symbols,
\[V \propto T \quad \text{(constant mass, constant } P\text{)}\]
\[\frac{V}{T} = k \qquad \text{or} \qquad \frac{V_1}{T_1} = \frac{V_2}{T_2}\]
with \(T_1\) and \(T_2\) in kelvin. This is the one place where the unit really does matter. Doubling the temperature from \(300\ \text{K}\) to \(600\ \text{K}\) doubles the volume; doubling it from \(27^\circ\text{C}\) to \(54^\circ\text{C}\) does nothing of the kind, because the Celsius scale does not start at the true zero of temperature.
5.1.3 The Combined Gas Equation
In the laboratory, pressure and temperature rarely stay put together. A gas collected over water on a warm afternoon may be measured on a cold morning at a different atmospheric pressure. What is needed is one relation covering both changes at once.
Boyle’s law says \(V\) falls as \(P\) rises; Charles’s law says \(V\) rises as \(T\) rises. Putting the two together for a fixed mass of dry gas:
\[V \propto \frac{T}{P} \qquad \text{so} \qquad \frac{PV}{T} = k\]
If such a gas goes from pressure \(P_1\), volume \(V_1\) and temperature \(T_1\) to pressure \(P_2\), volume \(V_2\) and temperature \(T_2\), then
\[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]
This is the gas equation, and it is the workhorse of every calculation in which a measured volume of gas has to be shifted to some other set of conditions. Five of the six quantities are usually known; the sixth is what you are after. Both temperatures must be in kelvin, and the pressures must share a unit, as must the volumes.
5.1.4 Standard Temperature And Pressure
Because the volume of a gas changes so sharply with temperature and pressure, a volume quoted on its own tells you almost nothing. Ten litres of oxygen on a Jaipur rooftop in June and ten litres on a Shimla morning in January are not the same amount of oxygen at all. Chemists therefore agreed on one set of conditions to which every gas volume can be referred.
These agreed conditions are called standard temperature and pressure, written S.T.P.
A volume ‘at S.T.P.’ is therefore a volume that has been corrected, using the gas equation, to those conditions. Two such volumes can be compared directly, and — as you will see in Section 5.9 — a volume at S.T.P. can even be turned straight into a number of particles.
5.1.5 The Kelvin Scale Of Temperature
Return to Charles’s law and follow it downwards. A fixed mass of gas loses \(1/273\) of its volume at \(0^\circ\text{C}\) for every degree it is cooled. Cool it by \(100^\circ\text{C}\) and it has lost \(100/273\) of that volume. Cool it by \(273^\circ\text{C}\) and, on paper, nothing is left. Fig. 5.1-2 shows what the measurements look like when they are plotted and the line is carried back.
Diagram
Fig. 5.1-2 — Volume Of A Gas Plotted Against Temperature A straight-line graph of volume (vertical axis) against temperature in degrees Celsius (horizontal axis) for a fixed mass of gas at constant pressure; the measured portion drawn as a solid line and its backward extension as a dashed line meeting the temperature axis at −273 °C; a second scale printed below the first showing 0 K at −273 °C, 273 K at 0 °C and 373 K at 100 °C.
No real gas ever reaches that point — every gas liquefies well before it. But the line from every gas, whatever its identity, points back to the same temperature, \(-273^\circ\text{C}\). That temperature is absolute zero, the lowest temperature there can be, and a scale that begins there is called the absolute scale or kelvin scale.
To convert, add 273 to a Celsius reading:
\[T\ (\text{in K}) = t\ (\text{in }^\circ\text{C}) + 273\]
So room temperature of \(27^\circ\text{C}\) is \(300\ \text{K}\), the boiling point of water is \(373\ \text{K}\), and standard temperature is \(273\ \text{K}\). The reason the gas laws insist on kelvin is now plain: on this scale, and only on this scale, doubling the number really does double the quantity that drives the gas outward.
5.2 Reaction Stoichiometry
Up to now, a chemical equation has been a statement of what turns into what. Zinc and dilute sulphuric acid give zinc sulphate and hydrogen — that is a qualitative account, and for many purposes it is enough.
But a chemist filling a cylinder, or a manufacturer buying limestone by the tonne, needs more than that. They need to know how much: how much acid to buy, how much zinc will be used up, how much hydrogen will come off. Those questions are answered by stoichiometry.
When these quantitative relationships are read off from a balanced chemical equation, the study is called reaction stoichiometry. The equation is the recipe; stoichiometry is the reading of the quantities in it.
Everything that follows in this chapter is preparation for that reading. To use an equation quantitatively you must be able to move between the volume of a gas, its mass, and the number of particles it contains. The rest of the chapter builds those three bridges one at a time, and Section 5.14 puts them to work on equations.
5.3 Gay-Lussac’s Law Of Combining Volumes
Gay-Lussac began where any careful worker would: with the reaction he could measure most easily. Hydrogen and oxygen combine to form water, and when the water is kept as vapour, every substance in the reaction is a gas whose volume can be read off directly.
What he found was that one litre of oxygen always took exactly two litres of hydrogen — no more and no less. The two gases combined in the ratio \(1 : 2\) by volume. When he went on to other pairs of reacting gases, the same kind of thing kept happening: the volumes stood to one another in small whole-number ratios, never in awkward fractions.
He generalised these observations into the law of combining volumes of gases.
Two conditions inside that statement do the real work. First, the volumes must all be measured at the same temperature and pressure — otherwise, as Section 5.1 showed, you are not comparing like with like. Second, the law applies to gases only.
That second restriction has a practical consequence you will use in every calculation. A solid or a liquid takes up so little room compared with the same amount of matter as a gas that its volume is treated as zero. If water is formed and then cooled so that it condenses, the water contributes nothing to the volume of the mixture. If the reaction is kept at \(100^\circ\text{C}\) so that the water stays as steam, it counts in full.
5.3.1 Illustrating The Law
Two reactions show the pattern clearly.
Formation of ammonia. Experiment shows that one volume of nitrogen combines with three volumes of hydrogen to give two volumes of ammonia, all measured at the same temperature and pressure.
nitrogen + hydrogen → ammonia
\[\mathrm{N_2} + 3\,\mathrm{H_2} \rightarrow 2\,\mathrm{NH_3}\]
The ratio is \(1 : 3 : 2\) — as simple as it could be.
Formation of carbon dioxide. Two volumes of carbon monoxide burnt in one volume of oxygen give two volumes of carbon dioxide at the same temperature and pressure. Fig. 5.3-1 shows the three volumes side by side.
carbon monoxide + oxygen → carbon dioxide
\[2\,\mathrm{CO} + \mathrm{O_2} \rightarrow 2\,\mathrm{CO_2}\]
Here the ratio is \(2 : 1 : 2\).
Diagram
Fig. 5.3-1 — Combining Volumes In The Burning Of Carbon Monoxide Three identical gas syringes drawn in a row at the same temperature and pressure: the first labelled 2 volumes of carbon monoxide, the second labelled 1 volume of oxygen, an arrow leading to the third labelled 2 volumes of carbon dioxide; below the row print the ratio 2 : 1 : 2.
Look again at the coefficients of the two balanced equations — \(1 : 3 : 2\) and \(2 : 1 : 2\). They are exactly the volume ratios. For gases, and only for gases, the numbers in front of the formulae can be read straight off as volumes. That single fact is what makes the calculations in the next subsection so short.
5.3.2 Calculations Based On Combining Volumes
Every problem of this kind is worked the same way. Write the balanced equation. Write the volume ratio underneath it, taking each coefficient as a volume and putting ‘nil’ under any solid or liquid. Then scale the ratio up or down to the volumes you were given.
Sometimes the two gases are not supplied in the ratio the equation demands. One of them then runs out first and the reaction stops, leaving the other partly unused.
The chlorine in that example was used up completely, and once it was gone no more hydrogen could react, however much of it was left. A reactant that behaves in this way is called the limiting reagent — the reactant which is completely used up, and which therefore fixes how much product can form.
When a fuel is burnt in air rather than in pure oxygen, remember that only about one-fifth of air is oxygen. The other four-fifths pass through the reaction untouched and are still there at the end.
Problems of this kind are set in a eudiometer — a graduated tube, closed at one end, in which a gas mixture is measured, sparked to make it react, and then measured again. The change in volume is the evidence. When several gases in the tube burn at once, write a separate equation for each and add up the oxygen at the end. Fig. 5.3-2 shows the tube.
Diagram
Fig. 5.3-2 — A Eudiometer Tube A thick-walled graduated glass tube closed at the upper end and standing inverted in a trough of mercury, with a volume scale marked along its length and two platinum wires sealed through the glass near the top for passing an electric spark; label the trapped gas column, the mercury level, the scale and the spark terminals.
5.4 Avogadro’s Law
Gay-Lussac had a pattern but no explanation. Why should reacting gases care about volume at all? In 1811 Amedeo Avogadro, working in Turin, proposed an answer so simple that most chemists of his day refused to take it seriously.
His suggestion was about counting. If equal volumes of every gas, measured under the same conditions, hold the same number of particles, then a ratio of volumes is really a ratio of particles in disguise — and particles, being whole things, naturally combine in whole numbers.
Read that carefully, because it says something a beginner rarely expects. One litre of hydrogen contains the same number of molecules as one litre of oxygen, or one litre of chlorine, or one litre of any other gas — provided all of them are measured at the same temperature and pressure. The molecules differ enormously in mass, but not in number. Fig. 5.4-1 makes the point.
Diagram
Fig. 5.4-1 — Equal Volumes, Equal Numbers Three identical flasks of the same volume drawn side by side, labelled hydrogen, oxygen and chlorine, each marked ‘same temperature, same pressure’; inside each flask draw the same number of molecules (say eight), shown as small paired spheres of different sizes and shades so that the molecules clearly differ in size and mass but not in number; below, the caption strip ‘same volume → same number of molecules’.
Avogadro went further. He argued that the smallest particle of a gaseous element is the molecule and not the atom, and that such a molecule may contain one, two or more atoms. A molecule of nitrogen is made of two nitrogen atoms; a molecule of ammonia is made of one nitrogen atom and three hydrogen atoms.
5.4.1 Atoms And Molecules Distinguished
Avogadro’s argument only works if atoms and molecules are kept firmly apart, so he drew the distinction sharply.
The phrase ‘may or may not exist independently’ is the whole difference. An atom of hydrogen cannot be kept on a shelf; left to itself it immediately joins another to form a molecule. An atom of helium, on the other hand, is perfectly content alone. Both are atoms; only one of them is also a molecule.
5.4.2 Atomicity Of Elementary Gases
Since a molecule of an element may hold one atom or several, we need a word for how many.
Elements are grouped by their atomicity as shown in Table 5.1.
A molecule built from atoms of the same kind — \(\mathrm{P_4}\) or \(\mathrm{O_3}\) — is called homoatomic. One built from atoms of different kinds, such as \(\mathrm{HCl}\) or \(\mathrm{NH_3}\), is called heteroatomic. Note that atomicity is used for elements; the number of atoms in a compound molecule is simply read from its formula.
5.4.3 Avogadro’s Law Explains Gay-Lussac’s Law
Now put the two laws together. Substances react in simple ratios by number of molecules, because molecules are whole objects and cannot be halved in a reaction. Avogadro’s law says that number of molecules is proportional to volume. Therefore the volumes must also be in a simple ratio — which is exactly what Gay-Lussac found by measurement.
Take the formation of hydrogen chloride. By Gay-Lussac’s law the volumes are \(1 : 1 : 2\); by Avogadro’s law those volumes hold \(n\), \(n\) and \(2n\) molecules.
hydrogen + chlorine → hydrogen chloride
\[\mathrm{H_2} + \mathrm{Cl_2} \rightarrow 2\,\mathrm{HCl}\] \[\text{1 vol} \qquad \text{1 vol} \qquad \text{2 vols}\] \[n \text{ molecules} \qquad n \text{ molecules} \qquad 2n \text{ molecules}\]
The volume ratio and the molecule ratio are the same ratio. Gay-Lussac’s law is therefore not a separate rule of nature at all; it is Avogadro’s law seen from the outside, through a measuring cylinder.
5.4.4 Avogadro’s Law Predicts Atomicity
The same reasoning can be pushed one step further, to work out how many atoms a molecule of an elementary gas contains. This is one of the neatest arguments in the chapter, and it turns entirely on one fact: an atom cannot be split in a chemical reaction.
Hydrogen and chlorine. The volumes are \(1 : 1 : 2\), so by Avogadro’s law
1 molecule of \(\mathrm{H_2}\) + 1 molecule of \(\mathrm{Cl_2}\) → 2 molecules of \(\mathrm{HCl}\)
Divide throughout by two:
\(\tfrac{1}{2}\) molecule of hydrogen + \(\tfrac{1}{2}\) molecule of chlorine → 1 molecule of hydrogen chloride
Half a molecule of hydrogen has ended up inside one molecule of hydrogen chloride. Since atoms are indivisible, that half-molecule must contain at least one whole atom. So a full molecule of hydrogen must contain at least two atoms — and the same for chlorine. Both gases are diatomic.
Nitrogen and oxygen. Nitrogen and oxygen combine to form nitric oxide in the volume ratio \(1 : 1 : 2\).
nitrogen + oxygen → nitric oxide
\[\mathrm{N_2} + \mathrm{O_2} \rightarrow 2\,\mathrm{NO}\] \[\text{1 vol} \qquad \text{1 vol} \qquad \text{2 vols}\]
By Avogadro’s law, 1 molecule of nitrogen and 1 molecule of oxygen give 2 molecules of nitric oxide. Halving again, half a molecule of nitrogen and half a molecule of oxygen give one molecule of nitric oxide. As an atom is indivisible, half a molecule of nitrogen must be at least one atom of nitrogen. So a molecule of nitrogen, and likewise of oxygen, contains two atoms each.
5.4.5 Avogadro’s Law And The Molecular Formula
A molecular formula is the chemical formula that gives the actual number of atoms of each element present in one molecule of a compound. Avogadro’s law lets us find it from volume measurements alone.
It is found by experiment that 2 volumes of ammonia decompose to give 1 volume of nitrogen and 3 volumes of hydrogen.
ammonia → nitrogen + hydrogen
\[\text{2 volumes} \qquad \text{1 volume} \qquad \text{3 volumes}\] \[\text{2 molecules} \qquad \text{1 molecule} \qquad \text{3 molecules}\]
Nitrogen and hydrogen are both diatomic, as Section 5.4.4 established. So the 1 molecule of nitrogen is 2 atoms of nitrogen, and the 3 molecules of hydrogen are \(3 \times 2 = 6\) atoms of hydrogen:
\[\text{2 molecules of ammonia} \rightarrow \text{2 atoms of nitrogen} + \text{6 atoms of hydrogen}\]
Dividing by two, one molecule of ammonia contains 1 atom of nitrogen and 3 atoms of hydrogen. Its molecular formula is therefore \(\mathrm{NH_3}\) — a formula arrived at without ever weighing anything.
5.4.6 Calculations Based On Avogadro’s Law
Problems here ask you to compare numbers of molecules. The rule to hold on to is short: at the same temperature and pressure, more volume means proportionally more molecules, whatever the gas.
5.5 Relative Atomic Mass
Atoms are far too small to be seen, and far too light to be placed on any balance. Indirect methods from physics, however, have made it possible to find the absolute mass of nearly every kind of atom. The mass of a hydrogen atom is \(1.6735 \times 10^{-24}\ \text{g}\), and that of an oxygen atom is \(26.565 \times 10^{-24}\ \text{g}\).
Numbers like these are correct but useless in daily work. Nobody wants to multiply \(10^{-24}\) through every calculation, and a gram is absurdly large as a unit for a single atom. The sensible thing is to stop measuring atoms against the gram altogether, and to measure them against one another instead — choosing one atom as the standard and asking how many times heavier each other atom is.
In the beginning hydrogen was the obvious choice, since it is the lightest element of all. Later, in 1961, carbon-12 was finally selected as the standard, because adopting it disturbed the values already in use less than any other choice would have done. The masses obtained on this scale are the relative atomic masses (RAM), also called atomic weights.
Written as a ratio,
\[\text{relative atomic mass} = \frac{\text{mass of 1 atom of the element}}{\frac{1}{12} \times \text{mass of one C-12 atom}}\]
Because it is a ratio of two masses, a relative atomic mass has no unit of its own. Fig. 5.5-1 shows what the ratio means for oxygen.
Diagram
Fig. 5.5-1 — Weighing An Atom Against The Standard A simple beam balance in equilibrium: on the left pan a single labelled oxygen atom, on the right pan sixteen identical small blocks each labelled ‘1/12 of a carbon-12 atom’; beneath the balance the line ‘relative atomic mass of oxygen = 16’.
The quantity \(\frac{1}{12}\) of the mass of a carbon-12 atom is itself given a name — it is one atomic mass unit, written a.m.u. Atomic masses are therefore quoted in a.m.u.: hydrogen 1, carbon 12, nitrogen 14, oxygen 16, sulphur 32.
One small point saves confusion later. When an atom gains or loses electrons to become an ion, its mass is taken as unchanged. An electron is so much lighter than the protons and neutrons in the nucleus that its gain or loss can be ignored beside the total mass of the atom.
5.5.1 Fractional Atomic Masses
If atoms are built of whole protons and whole neutrons, you would expect every relative atomic mass to come out as a whole number. Most of them nearly do — but not quite, and chlorine is famously \(35.5\).
The reason is that most natural elements are not made of one kind of atom at all. They are a mixture, of constant composition, of two or more isotopes — atoms of the same element that have the same atomic number but different mass numbers, because they contain different numbers of neutrons. The relative atomic mass of such an element is the weighted average of the relative atomic masses of its natural isotopes.
Chlorine occurs as a mixture of two isotopes, of masses 35 and 37, in the ratio \(3 : 1\). In every four atoms of chlorine, three are of mass 35 and one is of mass 37. The average is therefore not halfway between them but much closer to 35:
\[\text{average relative atomic mass} = \frac{35 \times 3 + 37 \times 1}{4} = 35.5\]
This is why a bottle labelled chlorine has an atomic mass of \(35.5\) although no single chlorine atom weighs \(35.5\) a.m.u. The value describes the mixture, not the individual atom.
5.6 Relative Molecular Mass
A molecule is a group of atoms held together, so its mass is simply the masses of those atoms added up. Weighed against the same standard as before, that total is the relative molecular mass (RMM), also called the molecular weight.
Finding it needs no new apparatus and no new idea — only addition. Add together the relative atomic masses of all the atoms present in one molecule, counting each atom as many times as the formula shows.
For sulphuric acid, \(\mathrm{H_2SO_4}\), one molecule holds two hydrogen atoms, one sulphur atom and four oxygen atoms:
\[\text{RMM of }\mathrm{H_2SO_4} = 2 \times 1 + 32 + 4 \times 16 = 98\]
So a molecule of sulphuric acid is 98 times as heavy as \(\frac{1}{12}\) the mass of a carbon-12 atom. We say its molecular mass is 98 a.m.u.
Some formulae carry a dot followed by a number of water molecules, as in \(\mathrm{CuSO_4\!\cdot\!5H_2O}\). That water is water of crystallisation — water built into the crystal itself — and it is part of the substance, so its mass must be counted in.
5.7 Gram Atomic Mass And Gram Atom
A relative atomic mass is a comparison, not a weight. It tells you that an oxygen atom is 16 times as heavy as one atomic mass unit, but it does not put anything on a balance. To work in the laboratory we need a bridge from that pure number to a mass we can actually weigh out.
The bridge is disarmingly simple: take the same number, and call it grams.
The atomic mass of oxygen is 16 a.m.u., so its gram atomic mass is \(16\ \text{g}\). For sodium the atomic mass is 23 a.m.u., so the gram atomic mass is \(23\ \text{g}\); for helium it is \(4\ \text{g}\).
A quantity of the element that weighs exactly this much has its own name.
So \(16\ \text{g}\) of oxygen is one gram atom of oxygen. In the same way,
1 gram atom of sodium \(= 23\ \text{g}\) of sodium
1 gram atom of helium \(= 4\ \text{g}\) of helium
Notice what has quietly happened here. Two different elements, weighed out as one gram atom each, have very different masses — \(23\ \text{g}\) against \(4\ \text{g}\) — but they stand in exactly the same ratio as the masses of their single atoms. Sodium atoms are \(23/4\) times as heavy as helium atoms, and these two heaps are in that same ratio. The two heaps must therefore hold the same number of atoms. Section 5.9 gives that number a name.
5.8 Gram Molecular Mass And Gram Molecule
What was done for atoms in Section 5.7 is now done for molecules, in exactly the same way and for exactly the same reason.
A sample of the substance whose mass is equal to its gram molecular mass is called one gram molecule of that substance.
The molecular mass of water, \(\mathrm{H_2O}\), is \(18.00\) a.m.u., so its gram molecular mass is \(18.00\ \text{g}\). The relative molecular mass of ammonia, \(\mathrm{NH_3}\), is \(17.00\), so its gram molecular mass is \(17.00\ \text{g}\). Weigh out \(18.00\ \text{g}\) of water and you have one gram molecule of water; weigh out \(17.00\ \text{g}\) of ammonia and you have one gram molecule of ammonia.
These two amounts look nothing alike — one fills a small beaker, the other is a jar of gas — yet the argument at the end of Section 5.7 applies here too. Because the masses are in the same ratio as the masses of the individual molecules, the two samples contain the same number of molecules. Everything in this chapter now points towards that number, and the next section names it.
5.9 The Mole Concept
You cannot weigh out a single atom, and you cannot count out atoms one by one either — there are far too many of them in even a speck of dust. What a chemist can do is weigh a heap and know, from the argument at the end of Section 5.8, that a certain mass always holds a certain number of particles. What is needed is a name for that number, and a name for that heap.
Shopkeepers solved this problem long ago. A dozen is a collection of 12 objects; a gross is a collection of 144. In the same spirit, a mole is a collection of \(6.022 \times 10^{23}\) particles — atoms, or molecules, or ions.
The two ways of putting it agree, because 12 grams of carbon-12 happens to contain \(6.022 \times 10^{23}\) atoms. When the kind of particle is not specifically mentioned, one mole is taken to mean \(6.02 \times 10^{23}\) molecules.
5.9.1 Avogadro’s Number
The number itself is named after the man whose law made counting by volume possible.
It is worth pausing over how large \(6.022 \times 10^{23}\) is. Written out, it is 602 200 000 000 000 000 000 000. This is not a number chosen for convenience; it is the number that had to be chosen, so that the gram atomic mass of every element would hold the same count of atoms.
5.9.2 Mole Of Atoms
One mole of atoms contains \(6.022 \times 10^{23}\) atoms, and has a mass equal to the gram atomic mass of that element.
So one mole of oxygen atoms contains \(6.022 \times 10^{23}\) atoms of oxygen and weighs \(16\ \text{g}\). One mole of sodium atoms contains the same number of atoms and weighs \(23\ \text{g}\). The count is fixed; only the mass changes from element to element, and it changes in exactly the ratio of the atomic masses.
5.9.3 Mole Of Molecules
One mole of a molecular substance contains \(6.022 \times 10^{23}\) molecules, and its mass is equal to the gram molecular mass of that substance.
One mole of oxygen gas, \(\mathrm{O_2}\), therefore contains \(6.022 \times 10^{23}\) molecules and weighs \(32\ \text{g}\). One mole of water contains \(6.022 \times 10^{23}\) molecules and weighs \(18\ \text{g}\).
Notice the trap sitting in those two statements about oxygen. One mole of oxygen atoms weighs \(16\ \text{g}\); one mole of oxygen molecules weighs \(32\ \text{g}\), since each molecule holds two atoms. The number of particles is the same in both; the particles are not.
5.9.4 Mole And Molar Volume
For gases there is a third way of measuring out one mole, and it is the easiest of the three. It has been found that one mole of any gaseous substance occupies \(22.4\ \text{dm}^3\) (litres), that is \(22400\ \text{cm}^3\), at S.T.P. This volume is known as the molar volume.
This follows straight from Avogadro’s law. Equal volumes of all gases at the same temperature and pressure hold equal numbers of molecules; so if \(6.022 \times 10^{23}\) molecules of one gas fill \(22.4\ \text{dm}^3\) at S.T.P., the same number of molecules of any other gas must fill the same volume.
Three quite different measurements now describe the same amount of matter:
one gram molecular mass of a substance \(=\) 1 mole \(=\) \(6.02 \times 10^{23}\) molecules \(=\) \(22.4\ \text{L}\) at S.T.P. if the substance is a gas.
Remember also that \(1\ \text{litre} = 1\ \text{dm}^3 = 1000\ \text{cm}^3 = 1000\ \text{mL}\). Table 5.2 sets out these relationships for three familiar substances.
| Substance | Moles | Molar mass (g) | Volume at S.T.P. (dm³) | Number of molecules | Number of atoms or ions |
|---|---|---|---|---|---|
| Sulphur dioxide, \(\mathrm{SO_2}\) | 1 | 64 | 22.4 | \(6.022 \times 10^{23}\) | S: \(6.022 \times 10^{23}\); O: \(2 \times 6.022 \times 10^{23}\) |
| Ammonia, \(\mathrm{NH_3}\) | 1 | 17 | 22.4 | \(6.022 \times 10^{23}\) | N: \(6.022 \times 10^{23}\); H: \(3 \times 6.022 \times 10^{23}\) |
| Calcium chloride, \(\mathrm{CaCl_2}\) | 1 | 111 | — | \(6.022 \times 10^{23}\) | \(\mathrm{Ca^{2+}}\): \(6.022 \times 10^{23}\); \(\mathrm{Cl^{-}}\): \(2 \times 6.022 \times 10^{23}\) |
Table 5.2 — One mole of three substances compared
The dash in the last row is not an oversight. Calcium chloride is a solid at room temperature, so the molar volume of a gas does not apply to it.
Diagram
Fig. 5.9-1 — One Mole, Three Ways Of Measuring It Three items on a laboratory bench, each labelled ‘one mole = 6.022 × 10²³ particles’: a watch glass of black powder labelled 12 g carbon, a small measuring cylinder of liquid labelled 18 g water, and a large gas jar labelled 32 g oxygen gas, 22.4 dm³ at S.T.P.
5.9.5 Calculations Based On The Mole
Every mole calculation is a journey between four quantities: the mass of the sample, the number of moles, the number of particles, and — for a gas at S.T.P. — the volume. The number of moles sits in the middle, and every journey passes through it. FC 5.1 shows the routes.
Flowchart
FC 5.1 — Converting Between Mass, Moles, Particles And Volume Mass in grams → divide by molar mass → number of moles → multiply by 6.022 × 10²³ → number of particles; branch from number of moles → multiply by 22.4 dm³ → volume of gas at S.T.P.; each arrow can be reversed by using the opposite operation
Written as relations, the two that do most of the work are
\[n = \frac{m}{M} \qquad \text{and} \qquad N = n \times N_\mathrm{A}\]
where \(n\) is the number of moles, \(m\) the mass of the sample in grams, \(M\) the molar mass in \(\text{g mol}^{-1}\), \(N\) the number of particles and \(N_\mathrm{A}\) Avogadro’s number. Table 5.3 collects the whole set for reference.
| Quantity wanted | Relation |
|---|---|
| Moles of an element (gram atoms) | mass of the element ÷ atomic mass |
| Moles of a compound | mass of the compound ÷ molecular mass |
| Mass of one atom | atomic mass ÷ \(6.022 \times 10^{23}\) |
| Mass of one molecule | molecular mass ÷ \(6.022 \times 10^{23}\) |
| Number of molecules | number of moles × \(6.022 \times 10^{23}\) |
| Number of atoms | number of moles × \(6.022 \times 10^{23}\) |
| Volume of a gas at S.T.P. | number of moles × \(22.4\ \text{dm}^3\) |
Table 5.3 — Relations used in mole calculations
5.10 Vapour Density And Relative Molecular Mass
Section 5.6 found relative molecular masses by adding up atomic masses — but that requires you to know the formula already. Suppose you have an unknown gas in a jar. Can you find how heavy its molecules are without knowing what it is?
You can, and the method is simply to weigh it against hydrogen, the lightest gas of all. Take a vessel of some fixed volume, fill it with the gas and weigh it; empty it, fill it with hydrogen under the same conditions and weigh it again. The ratio of the two masses is the relative vapour density of the gas.
\[\text{relative V.D.} = \frac{\text{mass of volume } v \text{ of the gas}}{\text{mass of volume } v \text{ of hydrogen, same conditions}}\]
Being a ratio of two masses, vapour density has no unit. Fig. 5.10-1 shows the comparison being made.
Diagram
Fig. 5.10-1 — Weighing A Gas Against Hydrogen A two-pan balance carrying two identical sealed flasks of the same volume, each marked ‘same temperature, same pressure’; the left flask labelled ‘unknown gas’ rests lower, the right flask labelled ‘hydrogen’ rides higher; an arrow to a line reading ‘vapour density = mass of gas ÷ mass of hydrogen’.
Now bring in Avogadro’s law. Equal volumes of the two gases, under the same conditions, contain equal numbers of molecules — so instead of comparing two volumes we may compare one molecule with one molecule:
\[\text{relative V.D.} = \frac{\text{mass of 1 molecule of the gas or vapour}}{\text{mass of 1 molecule of hydrogen}} = \frac{\text{mass of 1 molecule of the gas or vapour}}{\text{mass of 2 atoms of hydrogen}}\]
since a molecule of hydrogen contains two atoms. Multiplying both sides by 2,
\[2 \times \text{relative V.D.} = \frac{\text{mass of 1 molecule of the gas or vapour}}{\text{mass of 1 atom of hydrogen}}\]
and the quantity on the right is precisely the relative molecular mass.
For example, the relative vapour density of gaseous chlorine is \(35.5\), so its relative molecular mass is \(2 \times 35.5 = 71.0\).
5.10.1 Molar Volume From Vapour Density
The molar volume met in Section 5.9.4 was simply stated. Vapour density lets us see where the figure \(22.4\ \text{dm}^3\) actually comes from.
Direct weighing shows that \(1\ \text{dm}^3\) of hydrogen at S.T.P. weighs \(0.089\ \text{g}\). If some other gas has vapour density \(D\), then by the definition each of its volumes is \(D\) times as heavy as the same volume of hydrogen. So \(1\ \text{dm}^3\) of that gas weighs \(D \times 0.089\ \text{g}\) at S.T.P.
Turn the statement round. If \(D \times 0.089\ \text{g}\) of the gas occupies \(1\ \text{dm}^3\), then
\[D\ \text{g of the gas occupies } \frac{1}{0.089} = 11.2\ \text{dm}^3\]
and therefore \(2D\) grams occupy \(22.4\ \text{dm}^3\). But \(2D\) is the molecular mass. So the gram molecular mass of any gas occupies \(22.4\) litres at S.T.P., contains \(6.02 \times 10^{23}\) molecules, and is one mole of that gas. The three descriptions of one mole in Section 5.9.4 are now not three separate facts but one fact seen three ways.
5.10.2 Calculations Based On Vapour Density
When a gas is not measured at S.T.P., it must be brought there first. The gas equation of Section 5.1.3 does that job, and only then is the vapour density worked out.
5.11 Percentage Composition
A formula tells you which elements are present and in what ratio of atoms. It does not, at a glance, tell you how the mass is shared out. In washing soda, for instance, most of the mass turns out to be water, which you would never guess from the symbols alone.
The share of mass belonging to each element is the percentage composition of the compound.
It is found from the formula in one step. Work out the relative molecular mass, then ask what fraction of that total comes from each element:
\[\%\ \text{of an element} = \frac{\text{total mass of the element in one molecule}}{\text{gram molecular weight of the compound}} \times 100\]
The mass of an element ‘in one molecule’ means its atomic mass multiplied by the number of its atoms in the formula — two sodium atoms contribute \(2 \times 23 = 46\), not 23.
Fig. 5.11-1 shows how unevenly that mass is shared, even though there are more sodium atoms than carbon atoms in the formula.
Diagram
Fig. 5.11-1 — How The Mass Of Sodium Carbonate Is Shared A single pie chart divided into three labelled sectors — sodium 43.4 per cent, oxygen 45.3 per cent and carbon 11.3 per cent — with the caption strip ‘total relative molecular mass = 106’; use three clearly distinguishable tints and place the percentage inside each sector.
The same method works for water of crystallisation. Here the whole \(\mathrm{H_2O}\) group is treated as the unit whose mass share is wanted.
A mixture can be handled in the same way, provided you deal with each component separately and add the contributions at the end.
5.12 Empirical Formula Of A Compound
When a compound is analysed, what comes out of the laboratory is not a formula but a set of masses or percentages. Turning those numbers into a formula is the subject of this section and the next.
The first formula that can be obtained from analysis is the simplest one.
Hydrogen peroxide is a good example. Its actual formula is \(\mathrm{H_2O_2}\), but the ratio of hydrogen atoms to oxygen atoms in that molecule is \(1 : 1\), so its empirical formula is written \(\mathrm{HO}\). In the same way glucose, \(\mathrm{C_6H_{12}O_6}\), has an empirical formula of \(\mathrm{CH_2O}\), because the ratio of C, H and O atoms in a molecule of glucose is \(1 : 2 : 1\).
The empirical formula mass is the sum of the atomic masses of the elements present in the empirical formula. For hydrogen peroxide the empirical formula is \(\mathrm{HO}\), and its empirical formula mass is \(1 + 16 = 17\).
5.12.1 Determination Of The Empirical Formula
The empirical formula can be calculated from the percentage composition of the compound. The route is set out in FC 5.2 and then explained step by step.
Flowchart
FC 5.2 — Finding An Empirical Formula From Percentage Composition Percentage of each element → divide by its atomic mass → atomic ratio → divide every value by the smallest of them → simplest ratio → multiply throughout if any value is still fractional → whole-number ratio → write the symbols side by side with these numbers as subscripts; branch: percentage of water of crystallisation → divide by 18 instead of by an atomic mass → rejoin at ‘divide by the smallest’
- The percentage weight of each element is divided by its atomic weight. This gives the ratio of the number of atoms of that element in a molecule of the compound.
- If the molecule contains water of crystallisation, its percentage weight is divided by the molecular weight of water, that is 18. This gives the ratio of the number of water molecules in one molecule of the compound.
- To get a whole number ratio of atoms of different elements and water molecules, the numbers obtained in steps 1 and 2 are divided by the smallest number among them.
- The empirical formula is now written down by placing the symbols of the elements side by side, with the number of atoms of each as the subscript to the lower right of its symbol.
Two practical notes. If a number comes out as \(3.5\) or \(2.5\) after step 3, do not round it — multiply every figure by 2 (or by whatever small whole number is needed) so that all become whole. And in an organic compound, the percentage of carbon, oxygen or hydrogen is sometimes not given; it is then found by subtracting the sum of the other percentages from 100.
Sometimes the analysis is given as masses rather than percentages. Nothing changes — the mass of each element is divided by its atomic mass, exactly as before, and the number obtained is the number of gram atoms.
5.13 Molecular Formula Of A Compound
The empirical formula stops short of the truth. It tells you that glucose has one carbon for every two hydrogens and one oxygen, but not whether the molecule is \(\mathrm{CH_2O}\), \(\mathrm{C_2H_4O_2}\) or \(\mathrm{C_6H_{12}O_6}\). To go further we need the actual count of atoms in one molecule — the molecular formula.
Read a molecular formula slowly and it says a great deal. The molecular formula of blue vitriol is \(\mathrm{CuSO_4\!\cdot\!5H_2O}\), and it tells us that one molecule of blue vitriol is made of
- one atom of copper,
- one atom of sulphur,
- four atoms of oxygen,
- five molecules of water of crystallisation.
5.13.1 Determination Of The Molecular Formula
The molecular formula is always a whole-number multiple of the empirical formula, because both describe the same ratio of atoms — one at its simplest, the other at full size. All that is needed is the multiplier, and that comes from comparing the two masses.
- Calculate the empirical formula weight of the compound from its empirical formula.
- Divide its molecular weight by the empirical formula weight, which gives the number \(n\).
- Multiply the empirical formula by this number to get the molecular formula.
\[\text{molecular formula} = \text{empirical formula} \times n \qquad \text{where}\quad n = \frac{\text{molecular weight}}{\text{empirical formula weight}}\]
For hydrogen peroxide the empirical formula is \(\mathrm{HO}\) and its empirical formula weight is \(1 + 16 = 17\), while its molecular weight is 34. Hence \(n = 34/17 = 2\), and the molecular formula is twice the empirical formula, that is \(\mathrm{H_2O_2}\).
The value of \(n\) must come out as a whole number, since you cannot have a fraction of an atom. If experiment gives \(1.98\) or \(3.02\), round it to the nearest whole number — the small departure is experimental error.
Where does the molecular weight itself come from? Usually from a vapour density measurement, as in Section 5.10: molecular mass \(= 2 \times\) vapour density. Empirical formula from analysis, molecular mass from vapour density — between them the molecule is pinned down.
A very common form of this question asks for the water of crystallisation in a hydrated salt. The crystals are heated until all the water has been driven off, and the loss in mass is the water.
5.14 Calculations Based On Chemical Equations
Everything built up in this chapter now comes together. A balanced equation is more than a statement of what changes into what — it is a set of quantities waiting to be read.
Chemical equations give information about the moles and masses of the various reactants and products, and about the volumes of gaseous reactants and products measured at S.T.P.
5.14.1 Information Conveyed By A Chemical Equation
Consider the decomposition of potassium chlorate. Solid potassium chlorate, heated in the presence of manganese dioxide, gives solid potassium chloride and oxygen gas.
potassium chlorate → potassium chloride + oxygen
\[2\,\mathrm{KClO_3}\text{(s)} \xrightarrow{\ \mathrm{MnO_2},\ \Delta\ } 2\,\mathrm{KCl}\text{(s)} + 3\,\mathrm{O_2}\text{(g)}\]
That single line can be read in three different ways.
The molecular proportion of substances. Two molecules of solid potassium chlorate, on heating in the presence of manganese dioxide, give two molecules of solid potassium chloride and three molecules of oxygen gas.
The relative masses of substances. Replace each molecule by its gram molecular mass. Then \(2 \times 122.5 = 245\ \text{g}\) of potassium chlorate gives \(2 \times 74.5 = 149\ \text{g}\) of potassium chloride and \(3 \times 32 = 96\ \text{g}\) of oxygen.
The volumes of gaseous substances. Replace each mole of a gas by \(22.4\ \text{L}\) at S.T.P. Then \(3 \times 22.4 = 67.2\ \text{L}\) of oxygen at S.T.P. is evolved when \(245\ \text{g}\) of potassium chlorate is heated.
The routine for every problem in this section follows from those three readings, and is set out in FC 5.3.
Flowchart
FC 5.3 — Solving A Problem From A Chemical Equation Write a balanced equation for the reaction → write the molecular weight of each substance below its formula → for any gaseous substance also write the product of 22.4 L and the number of molecules below it → keep the same units of weight and volume throughout → set the known quantity against its own substance → scale to find the unknown quantity
5.14.2 Problems Based On Reacting Masses
Here every quantity in the question is a mass, and the equation is used purely as a ratio of masses.
5.14.3 Problems Based On Mass–Volume Relationships
In these problems one substance is weighed and another is measured as a gas. The bridge between the two is the molar volume, \(22.4\ \text{dm}^3\) at S.T.P., written under every gaseous formula.
A real sample is rarely pure, and a real reaction rarely gives quite the mass the equation promises. Comparing what was actually obtained with what the equation predicts is how purity is measured.
Two Harder Problems
The next two examples go a step beyond the rest. In the first, the gas is not measured at S.T.P., so it must be corrected before the equation can be used; in the second, purity enters at the start rather than at the end.
5.15 Summary
- The gas laws describe how a fixed mass of gas responds when its pressure, volume or temperature is changed, and they hold for every gas alike.
- Boyle’s law: the volume of a given mass of dry gas is inversely proportional to its pressure at constant temperature. \[P_1V_1 = P_2V_2\]
- Charles’s law: the volume of a given mass of dry gas is directly proportional to its absolute temperature at constant pressure. \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]
- Combining the two gives the gas equation. \[\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\]
- S.T.P. means \(0^\circ\text{C}\) or \(273\ \text{K}\), and 1 atmosphere, that is \(760\ \text{mm}\) or \(76\ \text{cm}\) of mercury.
- Absolute zero is \(0\ \text{K}\), equal to \(-273^\circ\text{C}\); a kelvin temperature is the Celsius reading plus 273, and gas laws are always worked in kelvin.
- Stoichiometry is the measurement of the quantitative relationships in a chemical reaction; read from a balanced equation, it is called reaction stoichiometry.
- Gay-Lussac’s law of combining volumes: when gases react, they do so in volumes bearing a simple ratio to one another and to the volume of the gaseous product, all volumes measured at the same temperature and pressure.
- The law applies only to gases; the volumes of solids and liquids are taken as zero, so water counts only when it is present as steam.
- For gases, the coefficients of a balanced equation may be read directly as volumes.
- The reactant used up completely is the limiting reagent, and it fixes how much product can form.
- Avogadro’s law: equal volumes of all gases, under similar conditions of temperature and pressure, contain the same number of molecules.
- An atom is the smallest particle of an element that can take part in a chemical reaction, and may or may not exist independently; a molecule is the smallest particle of an element or compound that can exist by itself.
- Atomicity is the number of atoms in one molecule of an element: monoatomic (inert gases), diatomic (\(\mathrm{H_2}\), \(\mathrm{O_2}\), \(\mathrm{N_2}\), \(\mathrm{Cl_2}\)), triatomic (\(\mathrm{O_3}\)), tetratomic (\(\mathrm{P_4}\)) and octatomic (\(\mathrm{S_8}\)).
- Molecules of one kind of atom are homoatomic; molecules of different kinds of atoms are heteroatomic.
- Avogadro’s law explains Gay-Lussac’s law, predicts the atomicity of hydrogen, chlorine, nitrogen and oxygen as two, and yields the molecular formula of a gaseous compound such as \(\mathrm{NH_3}\).
- Relative atomic mass is the number of times one atom of an element is heavier than \(\frac{1}{12}\) the mass of a carbon-12 atom; \(\frac{1}{12}\) of that mass is one atomic mass unit.
- Atomic masses are often fractional because most natural elements are mixtures of isotopes; the relative atomic mass is the weighted average, as in chlorine, \(\frac{35 \times 3 + 37 \times 1}{4} = 35.5\).
- Relative molecular mass is the number of times one molecule is heavier than \(\frac{1}{12}\) the mass of a carbon-12 atom, and is found by adding the atomic masses of all the atoms in the formula.
- The atomic mass expressed in grams is the gram atomic mass, and that quantity of the element is one gram atom; the molecular mass expressed in grams is the gram molecular mass or molar mass, and that quantity is one gram molecule.
- A mole is the amount of pure substance containing the same number of chemical units as there are atoms in exactly 12 grams of carbon-12.
- Avogadro’s number, \(N_\mathrm{A} = 6.022 \times 10^{23}\), is the number of elementary units in one mole of a substance.
- Number of moles and number of particles follow from \[n = \frac{m}{M} \qquad \text{and} \qquad N = n \times N_\mathrm{A}\]
- Molar volume is the volume occupied by one mole of a gas at S.T.P., equal to \(22.4\ \text{dm}^3\) or \(22\,400\ \text{cm}^3\).
- One gram molecular mass = 1 mole = \(6.02 \times 10^{23}\) molecules = \(22.4\ \text{L}\) at S.T.P. for a gas.
- Relative vapour density is the ratio of the masses of equal volumes of a gas and of hydrogen under the same conditions, and \[\text{molecular mass} = 2 \times \text{vapour density}\]
- Since \(1\ \text{dm}^3\) of hydrogen at S.T.P. weighs \(0.089\ \text{g}\), \(2D\) grams of any gas occupy \(22.4\ \text{dm}^3\) — which is why the gram molecular mass of every gas fills the same volume.
- Percentage composition is the percentage by weight of each element in a compound. \[\%\ \text{of an element} = \frac{\text{mass of the element in one molecule}}{\text{gram molecular weight}} \times 100\]
- The empirical formula gives the simplest whole-number ratio of atoms, and is found by dividing each percentage by the atomic mass (or by 18 for water of crystallisation), then dividing by the smallest value.
- The molecular formula gives the actual number of atoms in one molecule. \[\text{molecular formula} = \text{empirical formula} \times n, \qquad n = \frac{\text{molecular weight}}{\text{empirical formula weight}}\]
- A balanced equation gives the molecular proportions, the relative masses (using gram molecular masses) and the volumes of gaseous substances (using \(22.4\ \text{L}\) per mole at S.T.P.).
- Comparing the mass of product actually obtained with the mass the equation predicts gives the percentage purity of the sample used.
Flowchart
FC 5.4 — From Analysis To Molecular Formula Percentage composition or combining masses → divide by atomic masses → simplest whole-number ratio → empirical formula → empirical formula weight; separately, vapour density → multiply by 2 → molecular weight; the two branches meet → n = molecular weight ÷ empirical formula weight → multiply the empirical formula by n → molecular formula
5.16 Key Terms
- Absolute scale (kelvin scale) — the temperature scale that starts at absolute zero; a kelvin value equals the Celsius value plus 273.
- Absolute zero — the lowest temperature there can be, \(0\ \text{K}\) or \(-273^\circ\text{C}\).
- Atom — the smallest particle of an element that can take part in a chemical reaction; it may or may not exist independently.
- Atomic mass unit (a.m.u.) — one-twelfth of the mass of an atom of carbon-12, used as the unit for the mass of a single atom.
- Atomicity — the number of atoms present in one molecule of an element.
- Avogadro’s law — equal volumes of all gases, under similar conditions of temperature and pressure, contain the same number of molecules.
- Avogadro’s number (\(N_\mathrm{A}\)) — the number of elementary units in one mole of a substance, \(6.022 \times 10^{23}\).
- Boyle’s law — the volume of a given mass of dry gas is inversely proportional to its pressure at constant temperature.
- Charles’s law — the volume of a given mass of dry gas is directly proportional to its absolute temperature at constant pressure.
- Chemical equation — a balanced account of a chemical change, giving both what reacts and in what quantities.
- Empirical formula — the simplest formula, giving the simplest whole-number ratio of the atoms present in a molecule of a compound.
- Empirical formula mass — the sum of the atomic masses of the elements shown in the empirical formula.
- Eudiometer — a graduated tube in which a gas mixture is measured, sparked to react, and measured again.
- Gas equation — the relation \(\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}\), obtained by combining Boyle’s and Charles’s laws.
- Gas laws — the rules connecting the pressure, volume and temperature of a fixed mass of gas.
- Gay-Lussac’s law of combining volumes — reacting gases combine in volumes bearing a simple ratio to one another and to the gaseous product, all measured at the same temperature and pressure.
- Gram atom — the quantity of an element whose mass equals its gram atomic mass.
- Gram atomic mass — the atomic mass of an element expressed in grams.
- Gram molecular mass (molar mass) — the molecular mass of a substance expressed in grams.
- Gram molecule — a sample of a substance whose mass equals its gram molecular mass.
- Heteroatomic — describing a molecule built from atoms of different elements, such as \(\mathrm{HCl}\).
- Homoatomic — describing a molecule built from atoms of the same element, such as \(\mathrm{O_3}\).
- Isotopes — atoms of the same element having the same atomic number but different mass numbers.
- Limiting reagent — the reactant that is completely used up, and which therefore fixes the amount of product formed.
- Molar volume — the volume occupied by one mole of a gas at S.T.P., equal to \(22.4\ \text{dm}^3\).
- Mole — the amount of pure substance containing as many chemical units as there are atoms in exactly 12 grams of carbon-12.
- Molecular formula — the formula giving the actual number of atoms of each element in one molecule of a compound.
- Molecule — the smallest particle of an element or compound that can exist by itself.
- Percentage composition — the percentage by weight of each element present in a compound.
- Reaction stoichiometry — the quantitative relationships among substances as read from a balanced chemical equation.
- Relative atomic mass — the number of times one atom of an element is heavier than one-twelfth of the mass of a carbon-12 atom.
- Relative molecular mass — the number of times one molecule of a substance is heavier than one-twelfth of the mass of a carbon-12 atom.
- Relative vapour density — the ratio of the masses of equal volumes of a gas and of hydrogen under the same conditions; half the molecular mass.
- Standard temperature and pressure (S.T.P.) — \(0^\circ\text{C}\) (that is \(273\ \text{K}\)) and 1 atmosphere, or \(760\ \text{mm}\) of mercury.
- Stoichiometry — the measurement of the quantitative relationships in a chemical reaction.
5.17 Points to Ponder
Gay-Lussac’s law works for gases but not for solids or liquids. What is it about the way particles are arranged in a gas that makes volume such a reliable stand-in for number, and why does that arrangement break down in a liquid?
Avogadro’s proposal was ignored by most chemists for nearly fifty years. Looking at the argument in Section 5.4.4, what would a sceptic of the time have found hardest to accept — and what kind of evidence would have settled the matter?
Carbon-12 was chosen as the standard for atomic masses in 1961, replacing earlier standards. If a different element had been chosen instead, which of the quantities in this chapter would change their numerical values, and which would not change at all?
The molar volume \(22.4\ \text{dm}^3\) is the same for oxygen, ammonia and carbon dioxide, whose molecules differ greatly in size. How can molecules of such different sizes claim the same amount of room, and what does that suggest about how much of a gas is actually empty space?
Two compounds may share an empirical formula yet behave completely differently. What does this tell you about how much of a substance’s chemistry is decided by the ratio of its atoms, and how much by their number and arrangement?
A reaction carried out in a factory rarely gives the full mass the equation predicts. List as many chemical reasons as you can for the shortfall, and decide which of them would show up as impurity in the product rather than as a smaller yield.
5.18 Objective Questions
Multiple Choice
- The number of molecules present in \(8\ \text{g}\) of oxygen gas is: [1]
- \(1.5 \times 10^{23}\)
- \(3.0 \times 10^{23}\)
- \(6.02 \times 10^{23}\)
- \(1.2 \times 10^{24}\)
- Which of the following occupies the greatest volume at S.T.P.? [1]
- \(2\ \text{g}\) of hydrogen
- \(16\ \text{g}\) of oxygen
- \(14\ \text{g}\) of nitrogen
- \(22\ \text{g}\) of carbon dioxide
- A gas has a relative molecular mass of 71. Its vapour density is: [1]
- \(142\)
- \(71\)
- \(35.5\)
- \(17.75\)
- The empirical formula of benzene, \(\mathrm{C_6H_6}\), is: [1]
- \(\mathrm{CH}\)
- \(\mathrm{C_2H_2}\)
- \(\mathrm{C_3H_3}\)
- \(\mathrm{C_6H_6}\)
- Two litres of nitrogen and two litres of chlorine are taken at the same temperature and pressure. Which statement is correct? [1]
- Both contain the same number of molecules
- Both contain the same mass
- Nitrogen contains more molecules because it is lighter
- Chlorine contains more molecules because it is heavier
- Which conversion requires multiplication by \(22.4\)? [1]
- Moles of a gas into its volume at S.T.P.
- Mass of a gas into its number of moles
- Moles of a gas into number of molecules
- Vapour density into molecular mass
Assertion–Reason
For each of the following, choose the correct option from the codes given below. (1) Both A and R are true and R is the correct explanation of A. (2) Both A and R are true but R is not the correct explanation of A. (3) A is true but R is false. (4) A is false but R is true.
Assertion (A): Gas volumes in a chemical equation may be read straight off the coefficients. Reason (R): Equal volumes of all gases under the same conditions contain equal numbers of molecules. [1]
Assertion (A): One mole of oxygen atoms weighs \(32\ \text{g}\). Reason (R): The atomicity of oxygen is two. [1]
Assertion (A): The relative atomic mass of chlorine is \(35.5\). Reason (R): No single chlorine atom has a mass number of \(35.5\). [1]
Match The Following
- Match each item in Column A with the correct item in Column B. [1]
| Column A | Column B |
|---|---|
| a. \(22.4\ \text{dm}^3\) at S.T.P. | i. atomic mass unit |
| b. \(6.022 \times 10^{23}\) | ii. octatomic molecule |
| c. One-twelfth the mass of a carbon-12 atom | iii. molar volume |
| d. \(\mathrm{S_8}\) | iv. Avogadro’s number |
Fill In The Blanks
The molecular mass of a gas is ______ times its vapour density. [1]
A temperature of \(-273^\circ\text{C}\) is called ______. [1]
The reactant that is completely used up in a reaction is called the ______. [1]
True Or False
Gay-Lussac’s law of combining volumes applies to solids and liquids as well as to gases. [1]
One mole of water and one mole of carbon dioxide contain the same number of molecules. [1]
Name The Following
The volume occupied by one mole of any gas at S.T.P. [1]
Atoms of the same element having the same atomic number but different mass numbers. [1]
Give One Word Or Phrase For
The formula that gives the simplest whole-number ratio of atoms in a compound. [1]
The number of atoms present in one molecule of an element. [1]
Odd One Out, And Give A Reason
\(\mathrm{H_2}\), \(\mathrm{O_2}\), \(\mathrm{O_3}\), \(\mathrm{Cl_2}\) [1]
Helium, neon, argon, nitrogen [1]
5.19 Check Your Concepts
Define molar volume, and state its value at S.T.P. [2]
State Avogadro’s law. [2]
Give a reason why the relative atomic mass of an element is often not a whole number. [2]
Differentiate between an atom and a molecule, stating the basis of difference. [2]
Why must the temperature in a gas-law calculation always be converted to kelvin? [2]
Differentiate between empirical formula and molecular formula, stating the basis of difference, and illustrate with one example. [3]
Explain, with the help of an example, what is meant by a limiting reagent. [3]
What would happen to the volume of a fixed mass of dry gas if its pressure were halved and its absolute temperature doubled at the same time? Explain your reasoning. [3]
Explain why the volume of water formed in a reaction is taken as nil at room temperature but counted in full at \(100^\circ\text{C}\). [3]
Show how Avogadro’s law explains Gay-Lussac’s law of combining volumes, using the formation of hydrogen chloride as your example. [4]
Using the reaction between nitrogen and oxygen to form nitric oxide, explain how Avogadro’s law establishes that a molecule of nitrogen contains two atoms. [4]
State the three kinds of quantitative information that a balanced chemical equation conveys, and illustrate each from the decomposition of potassium chlorate. [5]
5.20 Equations, Conversions & Identification
Write the balanced equation for the complete combustion of ethane, \(\mathrm{C_2H_6}\), in oxygen. [2]
Complete and balance: \(\mathrm{Ca(NO_3)_2} \xrightarrow{\ \Delta\ }\) ______ + ______ + ______ [2]
Zinc is added to dilute hydrochloric acid. Name the gas evolved and state one test by which it may be identified. [2]
Limestone is strongly heated in a crucible. State what you would observe, and write the balanced equation for the change. [3]
How would you obtain dry crystals of copper sulphate, \(\mathrm{CuSO_4\!\cdot\!5H_2O}\), starting from anhydrous copper sulphate powder? State what colour change accompanies the conversion. [3]
A colourless gas turns lime water milky, is denser than air, and is produced when a metal carbonate is heated. Identify the gas and write the equation for its formation from calcium carbonate. [3]
Study Fig. 5.3-2. Name the piece of apparatus shown, state the purpose of the two wires sealed through the glass, and explain why mercury rather than water is used in the trough. [3]
Study Fig. 5.1-1. State which gas law the two graphs illustrate, and explain why one graph is a curve while the other is a straight line. [3]
5.21 Practice With Numericals
Calculate the number of moles, the number of molecules and the volume at S.T.P. of \(11\ \text{g}\) of carbon dioxide. [\(\mathrm{C} = 12\), \(\mathrm{O} = 16\)] [3]
Calculate the mass of \(5.6\ \text{dm}^3\) of ammonia gas measured at S.T.P. [\(\mathrm{N} = 14\), \(\mathrm{H} = 1\)] [3]
\(4.4\ \text{g}\) of a gas occupies \(2.24\ \text{dm}^3\) at S.T.P. Calculate its molar mass and its vapour density. [3]
\(200\ \text{cm}^3\) of a gas is collected at \(27^\circ\text{C}\) and \(750\ \text{mm}\) pressure. Calculate the volume it would occupy at S.T.P. [3]
\(60\ \text{cm}^3\) of hydrogen and \(40\ \text{cm}^3\) of chlorine are mixed and exploded. Calculate the composition by volume of the resulting mixture, and name the limiting reagent. [3]
Calculate the volume of oxygen required for the complete combustion of \(50\ \text{cm}^3\) of propane, and the volume of carbon dioxide formed, all volumes being measured under the same conditions. [3] \[\mathrm{C_3H_8} + 5\,\mathrm{O_2} \rightarrow 3\,\mathrm{CO_2} + 4\,\mathrm{H_2O}\]
Calculate the percentage of oxygen in potassium nitrate, \(\mathrm{KNO_3}\). [\(\mathrm{K} = 39\), \(\mathrm{N} = 14\), \(\mathrm{O} = 16\)] [3]
A compound contains \(26.7\ \%\) carbon, \(2.2\ \%\) hydrogen and \(71.1\ \%\) oxygen by mass, and its vapour density is 45. Find its empirical formula and its molecular formula. [4]
\(24.6\ \text{g}\) of hydrated magnesium sulphate, \(\mathrm{MgSO_4\!\cdot\!xH_2O}\), on strong heating left \(12\ \text{g}\) of the anhydrous salt. Find the value of \(x\). [\(\mathrm{Mg} = 24\), \(\mathrm{S} = 32\), \(\mathrm{O} = 16\), \(\mathrm{H} = 1\)] [4]
Calculate the mass of calcium carbonate required to produce \(5.6\ \text{dm}^3\) of carbon dioxide at S.T.P. when it reacts with dilute hydrochloric acid. [4] \[\mathrm{CaCO_3} + 2\,\mathrm{HCl} \rightarrow \mathrm{CaCl_2} + \mathrm{H_2O} + \mathrm{CO_2}\]
\(10\ \text{g}\) of an impure sample of zinc, treated with excess dilute sulphuric acid, gave \(2.24\ \text{dm}^3\) of hydrogen at S.T.P. Calculate the percentage purity of the zinc. [\(\mathrm{Zn} = 65\)] [4] \[\mathrm{Zn} + \mathrm{H_2SO_4} \rightarrow \mathrm{ZnSO_4} + \mathrm{H_2}\]