Concentration and Gas Volumes

What this lesson covers in the specification

Incandio is aligned to this specification. It is not published by, endorsed by or affiliated with Pearson, and it reproduces none of Pearson's wording — the statement numbers are given so you can check every lesson against your own copy.

• 1.34 — Calculations involving amount of substance, volume and concentration in mol/dm³ (bold C statement — Paper 2 only)
• 1.35 — Calculations involving gas volumes and the molar volume (24 dm³ at rtp) (bold C statement — Paper 2 only)

Specification section 1(e) Chemical formulae, equations and calculations — also taught in The Mole, Reacting Masses and Percentage Yield, Empirical and Molecular Formulae, Ionic Bonding.

1 · Understand it

No exam language yet. The only question this section answers is: do I actually understand what is happening?

Everything so far has been weighed. But a great deal of chemistry happens in solutions and in gases, and neither is convenient to put on a balance — you cannot weigh out 0.1 mol of hydrochloric acid, and weighing a gas requires knowing the mass of the container to a precision nobody has. What you CAN measure easily in both cases is a VOLUME, so this page is about converting volumes into moles.

For a solution, the link is CONCENTRATION: the number of moles dissolved in each cubic decimetre of solution, measured in mol/dm³. So moles = concentration × volume in dm³. A cubic decimetre is a litre, which is 1000 cm³ — and since burettes and pipettes are marked in cm³, dividing by 1000 is the single most common step in the whole of this topic.

Think of it like the strength of squash rather than the size of the glass

Two glasses of orange squash can be the same size and taste completely different, and one can be twice the size of the other and taste identical. Volume and strength are separate things, and how much orange you have actually drunk depends on BOTH — a big glass of weak squash and a small glass of strong squash can deliver the same amount. Concentration is the strength, volume is the size of the glass, and moles is how much orange there actually is. That is the whole equation, and it also explains why a question can hand you a large volume of a dilute acid and a small volume of a concentrated one and expect you to notice they contain the same number of moles.

Moles in a solution, and a concentration found by working backwards

(a) Calculate the moles of HCl in 25.0 cm³ of 0.100 mol/dm³ hydrochloric acid. (b) 0.0500 mol of sodium hydroxide is dissolved to make 250 cm³ of solution. Find its concentration.

1. (a) Convert the volume: 25.0 cm³ ÷ 1000 = 0.0250 dm³.
2. moles = concentration × volume = 0.100 × 0.0250 = 0.00250 mol.
3. (b) Convert the volume: 250 cm³ ÷ 1000 = 0.250 dm³.
4. concentration = moles ÷ volume = 0.0500 ÷ 0.250.
5. = 0.200 mol/dm³.

Answer: (a) 0.00250 mol. (b) 0.200 mol/dm³ — with the cm³-to-dm³ conversion done first both times.

Now gases, where something surprising happens. The volume a gas occupies barely depends on what gas it is. A mole of hydrogen, a mole of carbon dioxide and a mole of any other gas all occupy very nearly the SAME volume at the same temperature and pressure — about 24 dm³ at room temperature and pressure. That figure is the MOLAR VOLUME, and it means volume = moles × 24.

This looks like it ought to be impossible, because a carbon dioxide molecule is twenty-two times heavier than a hydrogen molecule. The resolution is that in a gas the particles are so far apart that the particles themselves take up almost none of the space — what you are measuring is overwhelmingly the gaps between them. Since the gaps are set by the temperature and pressure rather than by what the particles are, equal numbers of particles occupy equal volumes whatever they happen to be. That is Avogadro's hypothesis, and it took chemistry fifty years to accept it.

A gas volume from a reacting mass

What volume of carbon dioxide, measured at room temperature and pressure, is produced when 10.0 g of calcium carbonate decomposes completely? CaCO₃ → CaO + CO₂. (Relative formula mass of CaCO₃ = 100.)

1. moles of CaCO₃ = 10.0 ÷ 100 = 0.100 mol.
2. The ratio of CaCO₃ to CO₂ is 1 : 1, so 0.100 mol of CO₂ is formed.
3. volume = moles × 24 = 0.100 × 24.
4. = 2.4 dm³.
5. Notice that the relative formula mass of CO₂ was never needed. Once you have the moles, the molar volume does not care what the gas is.

Answer: 2.4 dm³ — and the identity of the gas was irrelevant after the mole step.

Which equation to use, decided by what the question gives you

1. A MASS in grams: use moles = mass ÷ relative formula mass.
2. A VOLUME AND A CONCENTRATION of a solution: use moles = concentration × volume in dm³.
3. A VOLUME OF A GAS: use moles = volume in dm³ ÷ 24.
4. Whichever route you came in by, you are now in moles — and every question crosses to the other side through the equation's ratio.
5. Then convert out again by whichever of the three the answer is wanted in.

That is the shape worth holding on to. There are three doors into moles and the same three doors out, and the balanced equation is the corridor between them. A question that gives a mass and asks for a gas volume looks new, and is simply one door in and a different door out.

One caution about the molar volume. The 24 dm³ figure applies at ROOM TEMPERATURE AND PRESSURE and nowhere else — heating a gas expands it and compressing it shrinks it, so the number changes. Questions on this specification will say 'at rtp' when they intend you to use it, and if they do not give those conditions the molar volume is not the route they want.

2 · Grade 9 Notes

A different job from the section above. You have already understood it; this is the precise set of things to LEARN — definitions to reproduce word for word, processes in order, equations with units, and the answers that score full marks.

moles = concentration × volume

concentration in moles per cubic decimetre (mol/dm³), volume in cubic decimetres (dm³)

Units: mol, mol/dm³, dm³. Rearranged: concentration = moles ÷ volume. Convert cm³ to dm³ by DIVIDING BY 1000.

volume of gas = moles × 24

volume in cubic decimetres (dm³), 24 dm³ is the molar volume at room temperature and pressure

Units: dm³, mol. Rearranged: moles = volume ÷ 24. Applies at rtp only, and is the same for every gas.

Learn this definition · Molar volume

The volume occupied by one mole of any gas at a stated temperature and pressure — 24 dm³ at room temperature and pressure. It is the same for every gas, because the particles occupy almost none of the volume.

Choosing the route into moles

1. Given a MASS in grams: moles = mass ÷ relative formula mass.
2. Given a SOLUTION's volume and concentration: moles = concentration × volume in dm³.
3. Given a GAS volume at rtp: moles = volume in dm³ ÷ 24.
4. Cross to the other substance using the ratio in the balanced equation.
5. Convert out by whichever of the three the answer is wanted in.

The units and conversions

mol/dm³

Concentration — the moles dissolved in each cubic decimetre of solution.

dm³

A cubic decimetre, the same as a litre and the same as 1000 cm³.

cm³ → dm³

DIVIDE BY 1000. Burettes and pipettes are marked in cm³, so this is the commonest step in the topic.

rtp

Room temperature and pressure — the conditions at which the molar volume is 24 dm³.

Model answer [3 marks]

Calculate the number of moles of solute in 50.0 cm³ of a 0.200 mol/dm³ solution. [3]

The volume must first be converted to cubic decimetres: 50.0 ÷ 1000 = 0.0500 dm³. Then moles = concentration × volume = 0.200 × 0.0500 = 0.0100 mol.

Model answer [3 marks]

Calculate the volume of hydrogen produced at rtp when 0.20 mol of magnesium reacts completely with excess hydrochloric acid: Mg + 2HCl → MgCl₂ + H₂. [3]

The ratio of Mg to H₂ in the equation is 1 : 1, so 0.20 mol of hydrogen is produced. The volume of a gas at rtp is the number of moles multiplied by the molar volume, so volume = 0.20 × 24 = 4.8 dm³.

Where the marks go

• FORGETTING to divide cm³ by 1000 — the single commonest error in this topic
• Using the molar volume when the question has not stated rtp
• Looking up the relative formula mass of a gas when the molar volume makes it unnecessary
• Multiplying by 1000 instead of dividing, giving an answer a million times out
• Using the ratio from an unbalanced equation

Not this: A mole of a heavy gas such as carbon dioxide takes up more room than a mole of hydrogen.

This: They occupy the SAME volume, about 24 dm³ at rtp. In a gas the particles are so far apart that the particles themselves take up almost none of the space — the volume is overwhelmingly the gaps, and the gaps depend on temperature and pressure rather than on what the particles are.

Mark-losing trap. cm³ to dm³ means DIVIDING BY 1000. This one conversion costs more marks than anything else here.

Mark-losing trap. The molar volume is 24 dm³ for EVERY gas — you never need a gas's relative formula mass to use it.

Mark-losing trap. 24 dm³ applies at rtp only. Different conditions, different molar volume.

3 · Prove it — the five questions

The five questions climb Grade 6 → Grade 7 → Grade 8 → Grade 9 → Grade 9 challenge, and are marked inside Incandio on your own device, by rule, with an authored diagnosis of the mistake you actually made. The mark schemes stay in the app so that the practice is worth doing; the questions themselves are here.

1. Grade 6 · Convert [1 mark] — A pipette delivers 25.0 cm³ of solution. What is this volume in dm³?
2. Grade 7 · Calculate [2 marks] — Calculate the volume occupied at rtp by 0.50 mol of oxygen gas.
3. Grade 8 · Calculate [3 marks] — Calculate the concentration, in mol/dm³, of a solution containing 0.0400 mol of solute in 200 cm³ of solution.
4. Grade 9 · Calculate [5 marks] — Select every statement that belongs in a full-mark calculation of the volume of carbon dioxide produced at rtp when 25.0 g of calcium carbonate decomposes completely. CaCO₃ → CaO + CO₂. (Relative formula mass of CaCO₃ = 100.)
5. 9+ · Analyse [6 marks] — A student says: 'One mole of any gas occupies 24 dm³, so one mole of water must occupy 24 dm³ too — and since a mole of water is only 18 g, that means 18 g of water fills a large bucket.' Select every statement that belongs in a full-mark analysis.

The people behind this science

Two ways into the same idea — the one who explained why every gas takes up the same room, and the one who first measured what happens to a gas's volume. Inside Incandio each of them answers knowing exactly which lesson you have just finished.

Amedeo Avogadro — the one who explained why every gas takes up the same room

The molar volume on this page is a direct consequence of his 1811 hypothesis: equal volumes of gases, at the same temperature and pressure, contain equal numbers of particles. Stated that way it sounds almost too simple to matter, and it is the reason a single number of 24 dm³ works for hydrogen and carbon dioxide alike despite one being twenty-two times heavier. He is worth asking here rather than about formulae because the gas case is where the hypothesis is most visibly strange, and he is honest that it required accepting something his contemporaries would not — that the elementary gases exist as pairs of atoms rather than single ones.

• “Why does a heavy gas not take up more room than a light one?”
• “What does equal volumes containing equal numbers really mean?”
• “How can the space be mostly gaps?”
• “Why did the elementary gases have to exist in pairs?”
• “What could chemists not calculate before your hypothesis?”

Robert Boyle — the one who first measured what happens to a gas's volume

The molar volume comes with a condition attached — 24 dm³ at room temperature and pressure, and a different figure otherwise — and Boyle is the person who established that a gas's volume depends on the pressure upon it at all. Trapping air in the short arm of a J-shaped tube and pouring in mercury, he found that the volume falls as the pressure rises, in inverse proportion. He is also the right figure for the picture behind this page, since his air-pump work is what established that a gas is mostly empty space, and he published his failures alongside his successes so that anyone could repeat the measurement.

• “What happens to a gas when you squeeze it?”
• “How did you measure that with a J-shaped tube?”
• “Why does a gas's volume depend on temperature and pressure at all?”
• “What did the air pump show you about what a gas is?”
• “Why publish the experiments that did not work?”

Carry on through the course

• Previous lesson: Empirical and Molecular Formulae
• Next lesson: Ionic Bonding
• The Three States of Matter — What the particles are doing in a solid, a liquid and a gas — and what really changes when ice melts.
• Pure Substances, Mixtures and How to Separate Them — Why a pure substance melts at one temperature, and which separation technique to reach for when.
• Inside the Atom — Protons, neutrons and electrons — where they are, what they weigh, what they carry, and how isotopes give an element a decimal mass.
• Electronic Configurations and the Periodic Table — Why the table has the shape it has: the group number IS the outer-shell electron count, and everything follows from that.
• Solubility and Solubility Curves — Why a solution can be full, why hot water holds more, and how one curve tells you exactly how many crystals you will get back.
• All of Chemistry · Chemistry key terms · required practicals · equations · Incandio Science · The 4CH1 specification