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Chemistry (IAL)

Amount of Substance

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Edexcel IAL Chemistry · Topic 1.2

Amount of Substance

Everything in chemistry — mass, volume, concentration, gas behaviour, yield — comes down to counting particles in moles, and using ratios from balanced equations to convert between them.

Quick Summary

  • The mole is chemistry's "counting unit" — it links mass, particles, gas volume, and solution concentration all together.
  • Concentration can be measured as mass concentration (g dm⁻³) or molar concentration (mol dm⁻³) — and at very low levels, in parts per million (ppm).
  • Reacting mass calculations use the mole ratio from a balanced equation to find unknown masses of reactants or products.
  • You can even work backwards — using masses from an experiment to figure out the balanced equation itself.
  • Reacting volume calculations apply to gases: at r.t.p. one mole of any gas occupies 24.0 dm³; at s.t.p. it's 22.4 dm³.
  • The ideal gas equation PV = nRT handles gases at any temperature and pressure, not just standard conditions.
  • Percentage yield compares what you actually got to the maximum theoretically possible.
  • Atom economy tells you how much of your reactant mass actually ends up in the product you want (vs. wasted as by-products).

1. Concentration Calculations

The Mole — Your Universal Counting Unit

Atoms and molecules are absurdly small and absurdly numerous — you can't count them one by one. So chemists invented the mole: a "chemist's dozen" that lets us talk about huge numbers of particles using ordinary numbers we can calculate with.

Because "amount of substance" is really a blanket term for almost any calculation involving moles, there are several routes into finding n (moles) depending on what information you're given:

The Four Core Mole Equations
n = mass (m) ÷ molar mass (M)
n = concentration × volume
n = number of particles ÷ Avogadro's constant
n = PV ÷ RT  (ideal gas equation)

Think of these as four different "doors" into the same room. Whichever data the question hands you (mass, concentration+volume, particle count, or gas conditions), there's a formula waiting to convert it into moles.

Mass Concentration vs Molar Concentration

"Concentration" just means how much stuff is packed into a given volume of solution — but there are two ways to express "how much stuff":

  • Mass concentration — measured in g dm⁻³ (or g cm⁻³, kg m⁻³, mg/ml). This is literally grams of solute per volume of solution — no chemistry needed, just division.
  • Molar concentration — measured in mol dm⁻³. This tells you how many moles of solute are packed into the solution, which is far more chemically useful because moles react in fixed ratios, not grams.
Mass Concentration
mass concentration (g dm⁻³) = mass of solute (g) ÷ volume of solution (dm³)
Molar Concentration (two-step)
Step 1: n = mass ÷ molar mass
Step 2: molar concentration (mol dm⁻³) = n ÷ volume (dm³)

Molar concentration always needs this two-step journey: mass → moles → concentration. Never skip straight from mass to concentration without converting to moles first!

Watch the units!
Volumes are almost always given in cm³ but the formulas need dm³. Remember: divide cm³ by 1000 to get dm³. This single slip (forgetting to convert) is one of the most common lost marks in this topic.
Worked Example — Mass Concentration

Q: What is the mass concentration when 6.34 g of sodium chloride is dissolved into a 0.250 dm³ solution?

  1. mass concentration = mass ÷ volume
  2. = 6.34 ÷ 0.250
  3. = 25.4 g dm⁻³ (to 3 s.f.)
Worked Example — Molar Concentration

Q: What is the molar concentration when 6.34 g of sodium chloride is dissolved into a 250 cm³ solution?

  1. Convert volume: 250 cm³ ÷ 1000 = 0.250 dm³
  2. Molar mass of NaCl = 23.0 + 35.5 = 58.5 g mol⁻¹
  3. n(NaCl) = 6.34 ÷ 58.5 = 0.1084 mol
  4. molar concentration = n ÷ volume = 0.1084 ÷ 0.250
  5. = 0.434 mol dm⁻³
Practice Question 1

A sodium carbonate (Na₂CO₃) solution has a molar concentration of 1.25 mol dm⁻³. What volume of solution is made when 250 g of sodium carbonate is used? (M: Na = 23.0, C = 12.0, O = 16.0)

Practice Question 2

The molar concentration of a sodium bromide (NaBr) solution is 0.250 mol dm⁻³. What is the mass of sodium bromide in 500 cm³ of this solution? (M: Na = 23.0, Br = 79.9)

Parts Per Million (ppm)

Sometimes a concentration is so tiny that grams-per-litre becomes an awkward, fiddly number (like 0.000003 g dm⁻³). Instead, chemists switch to parts per million — perfect for describing pollutants in water or air, where the "stuff" is a whisker of the total.

Definition of 1 ppm (by mass)
1 ppm = 1 mg dissolved in 1 dm³ of water

Since 1 dm³ of water weighs 1 kg, this is the same as saying: 1 mg dissolved in 1 kg — i.e. a ratio of 1 part in a million (1 in 10⁶). That's literally where the name comes from.

Worked Example

Q: The concentration of chlorine in a swimming pool should be between 1 and 3 ppm. Calculate the maximum mass, in kg, of chlorine that should be present in an Olympic pool of size 2.5 million litres.

  1. Total mass in mg (using the max, 3 ppm; 1 litre = 1 dm³): 3 × 2.5×10⁶ = 7.5×10⁶ mg
  2. Convert mg to kg (1 mg = 10⁻⁶ kg): 7.5×10⁶ × 10⁻⁶ = 7.5 kg
Gases use ppmv (by volume), not mass
Atmospheric pollutant concentrations are measured by volume, not mass — written as ppmv. 1 ppmv means 1 cm³ of a gas in 1,000,000 cm³ (or 1000 dm³) of air.
Gas Concentration Formula
concentration (ppm) = (volume of gas × 1,000,000) ÷ volume of air

Both volumes just need to be in the same units — cm³ and cm³, or dm³ and dm³. Convert if they don't match!

Practice Question 3

A volume of 152 cm³ of ozone is found in 112 dm³ of air. Calculate the concentration in ppm.

2. Reacting Mass Calculations

Using Balanced Equations to Find Unknown Masses

This is the workhorse skill of the whole chapter. The logic is always a three-step relay race:

  • Step 1 — Mass → Moles: convert the mass you're given into moles, using n = mass ÷ molar mass.
  • Step 2 — Moles → Moles: use the ratio in the balanced equation to convert moles of the substance you have into moles of the substance you want.
  • Step 3 — Moles → Mass: convert those moles back into a mass using mass = moles × Mr.

The balanced equation is the "exchange rate" — it tells you exactly how many moles of one substance correspond to how many moles of another. Get the ratio wrong, and everything downstream is wrong too, so always balance the equation first.

The universal roadmap
Mass A → Moles A → Moles B (using ratio) → Mass B. Memorise this four-stage journey and almost every reacting-mass question becomes mechanical.
Worked Example 1

Q: Calculate the maximum mass of magnesium oxide produced by completely burning 7.5 g of magnesium in oxygen.

  1. Balanced equation: 2Mg(s) + O₂(g) → 2MgO(s)
  2. Molar masses: Mg = 24.3, MgO = 40.3 g mol⁻¹
  3. n(Mg) = 7.5 ÷ 24.3 = 0.3086 mol
  4. Ratio Mg : MgO is 2 : 2 = 1 : 1, so n(MgO) = 0.3086 mol
  5. mass(MgO) = 0.3086 × 40.3 = 12.4 g
Worked Example 2 — Larger-Scale (Tonnes)

Q: Calculate the mass of aluminium, in tonnes, produced from 51 tonnes of aluminium oxide. 2Al₂O₃ → 4Al + 3O₂

  1. 51 tonnes = 51,000,000 g (× 10⁶)
  2. n(Al₂O₃) = 51,000,000 ÷ 102.0 = 500,000 mol
  3. Ratio Al₂O₃ : Al is 2 : 4 = 1 : 2, so n(Al) = 500,000 × 2 = 1,000,000 mol
  4. mass(Al) = 1,000,000 × 27.0 = 27,000,000 g
  5. Convert back to tonnes: 27,000,000 ÷ 10⁶ = 27 tonnes

As long as you're consistent, it doesn't matter whether you work in grams, tonnes, or any other mass unit — reacting masses are always in proportion to the balanced equation.

Practice Question 4

Using 2Mg(s) + O₂(g) → 2MgO(s), calculate the mass of oxygen needed to completely react with 7.5 g of magnesium. (M: Mg = 24.3, O₂ = 32.0)

Working Backwards: Finding the Balanced Equation from Masses

Sometimes an exam flips the whole process on its head: instead of giving you a balanced equation and asking for a mass, it gives you the masses of everything involved and asks you to find the balanced equation itself.

The trick is simple: convert every mass to moles, then divide all the mole values by the smallest one. Whatever whole numbers pop out are your balancing coefficients.

Worked Example

Q: A student reacts 1.2 g of carbon with 16.2 g of zinc oxide. The products are 4.4 g of carbon dioxide and 13.0 g of zinc. Determine the balanced equation.

  1. Unbalanced skeleton: C + ZnO → Zn + CO₂
  2. Convert each mass to moles:
    n(C) = 1.2 ÷ 12.0 = 0.1 mol  |  n(ZnO) = 16.2 ÷ 81.4 = 0.2 mol
    n(Zn) = 13.0 ÷ 65.4 = 0.2 mol  |  n(CO₂) = 4.4 ÷ 44.0 = 0.1 mol
  3. Divide all by the smallest value (0.1):
    C = 1, ZnO = 2, Zn = 2, CO₂ = 1
  4. Balanced equation: C + 2ZnO → 2Zn + CO₂
Common mistake
Students often forget to divide by the smallest mole value — dividing by the wrong one gives fractional or backwards ratios. Always scan all your mole values first, spot the smallest, then divide everything (including that smallest one) by it.

3. Reacting Volume Calculations

Molar Gas Volume

Here's a genuinely beautiful fact about gases: one mole of ANY gas occupies the same volume, at a given temperature and pressure — it doesn't matter if it's helium or carbon dioxide, big molecule or small. This works because gas volume is really about the space between particles (which is huge compared to the particles themselves), not the particles' own size.

ConditionTemperaturePressureMolar Gas Volume
Room temperature & pressure (r.t.p.)293 K / 20°C101.3 kPa24.0 dm³ mol⁻¹
Standard temperature & pressure (s.t.p.)273 K / 0°C101.3 kPa22.4 dm³ mol⁻¹
Moles of Gas from Volume
n = volume of gas (dm³) ÷ molar gas volume (dm³)
Worked Example

Q: Calculate the number of moles present in 4.5 dm³ of carbon dioxide at room temperature and pressure.

  1. n = 4.5 ÷ 24 = 0.1875 mol
Worked Example — Combining with Reacting Mass Logic

Q: Calculate the volume of gas produced when 1.50 g of sodium reacts with water at s.t.p.
2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g)

  1. n(Na) = 1.5 ÷ 23.0 = 0.0652 mol
  2. Ratio Na : H₂ is 2 : 1, so n(H₂) = 0.0652 ÷ 2 = 0.0326 mol
  3. Volume = n × molar gas volume = 0.0326 × 22.4 = 0.730 dm³
Practice Question 5

Calculate the total volume of gas produced when 6.50 dm³ of propane combusts completely: C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)

The Ideal Gas Equation

The molar gas volume (24.0 or 22.4 dm³) only works at those two specific "standard" conditions. But what if a question gives you some random temperature and pressure — say, 220 kPa and 21°C? That's where the ideal gas equation comes in: a single formula that works at any temperature and pressure.

It's built on the kinetic theory of gases, which assumes:

  • Gas molecules move fast and randomly
  • Molecules have (almost) no volume of their own
  • There's no attraction or repulsion between molecules (no intermolecular forces)
  • Collisions are perfectly elastic (no kinetic energy lost)
  • Temperature reflects the average kinetic energy of the molecules

Gases that obey this perfectly are called ideal gases. Real gases only approximate this — but they get close enough that the equation works well in practice.

Ideal Gas Equation
PV = nRT

P = pressure in pascals (Pa)  |  V = volume in  |  n = moles  |  R = gas constant = 8.31 J K⁻¹ mol⁻¹  |  T = temperature in kelvin (K)

Units are EVERYTHING here
This is the #1 place marks get lost. Before plugging into PV = nRT, always convert:
  • kPa → Pa: × 1000
  • cm³ → m³: ÷ 1,000,000 (or dm³ → m³: ÷ 1000)
  • °C → K: + 273
Worked Example — Finding Volume

Q: Calculate the volume occupied by 0.781 mol of oxygen at a pressure of 220 kPa and a temperature of 21°C.

  1. Rearrange: V = nRT ÷ P
  2. P = 220,000 Pa  |  T = 21 + 273 = 294 K  |  n = 0.781 mol
  3. V = (0.781 × 8.31 × 294) ÷ 220,000 = 0.00867 m³
  4. Convert to dm³ (× 1000): 8.67 dm³
Worked Example — Finding Molar Mass

Q: A flask of volume 1000 cm³ contains 6.39 g of a gas at 300 kPa and 23°C. Calculate the relative molecular mass of the gas.

  1. Rearrange: n = PV ÷ RT
  2. P = 300,000 Pa  |  V = 1000 cm³ = 0.001 m³  |  T = 23 + 273 = 296 K
  3. n = (300,000 × 0.001) ÷ (8.31 × 296) = 0.12 mol
  4. molar mass = mass ÷ n = 6.39 ÷ 0.12 = 53.25 g mol⁻¹
Practice Question 6

A gas sample occupies 500 cm³ at a pressure of 150 kPa and a temperature of 25°C. Calculate the number of moles of gas present.

4. Calculations of Product: Yield & Atom Economy

Percentage Yield

In theory, a reacting mass calculation tells you the maximum possible amount of product you could ever get. In real lab life, you almost never get that much — because of side reactions, incomplete reactions, or product lost during filtering and purification.

Percentage yield measures how close your real-world result got to that theoretical maximum.

Percentage Yield
% yield = (actual yield ÷ theoretical yield) × 100

Actual yield = what you really measured in the lab. Theoretical yield = the maximum calculated using a reacting mass calculation (as in Section 2).

Worked Example

Q: 6.5 g of zinc was added to excess copper(II) sulfate solution. The copper produced was filtered, washed, and dried, giving 4.8 g. Calculate the percentage yield.
Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)

  1. n(Zn) = 6.5 ÷ 65.4 = 0.10 mol
  2. Ratio Zn : Cu is 1 : 1, so max n(Cu) = 0.10 mol
  3. Theoretical mass of Cu = 0.10 × 63.55 = 6.4 g (theoretical yield)
  4. % yield = (4.8 ÷ 6.4) × 100 = 75%
Can yield ever exceed 100%?
Calculated yield can occasionally come out above 100% — but this always signals an experimental error, not a real result. Common causes: the product wasn't fully dry when weighed, impurities added extra mass, or the filter paper's mass was accidentally included.
Practice Question 7

A reaction between 10.0 g of calcium carbonate and excess hydrochloric acid was expected to produce 4.40 g of CO₂ (theoretical yield). Only 3.74 g was actually collected. Calculate the percentage yield.

Atom Economy

Atom economy asks a different question than percentage yield. It doesn't care how efficiently the reaction was carried out in the lab — it asks: of all the atoms that go INTO the reaction, what fraction end up in the product you actually wanted? The rest becomes waste (by-products).

This matters hugely for green chemistry and industrial cost — a reaction can have a perfect 100% yield but a terrible atom economy if most of the reactant mass ends up as unwanted by-product.

Atom Economy
atom economy = (Mr of desired product ÷ sum of Mr of ALL reactants) × 100

Unlike percentage yield, atom economy is calculated purely from the balanced equation — no experimental data needed.

Addition reactions = automatic 100%
In an addition reaction (where everything combines into one single product, nothing left over), the atom economy is always 100% — because every single atom you put in ends up in the product. Example: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.
Worked Example — Qualitative

Q: Ethanol can be made by hydration of ethene (C₂H₄ + H₂O → C₂H₅OH) or by substitution of bromoethane (C₂H₅Br + NaOH → C₂H₅OH + NaBr). Which has the higher atom economy?

A: Hydration of ethene has a higher atom economy (100%) because all reactant atoms end up in the ethanol product. The bromoethane route produces NaBr as an unwanted by-product, wasting atoms.

Worked Example — Quantitative

Q: The blast furnace reduces iron(III) oxide with carbon monoxide: Fe₂O₃ + 3CO → 2Fe + 3CO₂. Calculate the atom economy, assuming iron is the desired product. (Mr: Fe₂O₃ = 159.6, CO = 28.0, Fe = 55.8)

  1. atom economy = (2 × 55.8) ÷ [159.6 + (3 × 28.0)] × 100
  2. = 111.6 ÷ 243.6 × 100
  3. = 45.9%
Practice Question 8

Ammonia is made industrially by: N₂ + 3H₂ → 2NH₃. Calculate the atom economy for this reaction. (Mr: N₂ = 28.0, H₂ = 2.0, NH₃ = 17.0)

What to Memorise

Term / FormulaMeaning
n = m ÷ MMoles = mass ÷ molar mass
n = c × VMoles = concentration × volume (V in dm³)
mass conc. = mass ÷ volumeUnits: g dm⁻³
1 ppm1 mg per dm³ of water; or 1 part in 10⁶
ppm (gas) = (Vgas × 10⁶) ÷ VairConcentration by volume, "ppmv"
r.t.p.293 K (20°C), 101.3 kPa → molar gas volume = 24.0 dm³
s.t.p.273 K (0°C), 101.3 kPa → molar gas volume = 22.4 dm³
n = V(gas) ÷ molar gas volumeMoles of gas from volume, at standard conditions
PV = nRTIdeal gas equation. P in Pa, V in m³, R = 8.31 J K⁻¹ mol⁻¹, T in K
K = °C + 273Converting Celsius to Kelvin
% yield = (actual ÷ theoretical) × 100How close to the maximum possible amount you got
atom economy = (Mr desired product ÷ Σ Mr reactants) × 100What fraction of reactant mass ends up as wanted product

Concepts Checklist

Exam Tips & Common Mistakes

Unit conversions are where most marks disappear
Examiners deliberately give data in "awkward" units (cm³, kPa, °C) to test whether you convert correctly. Before touching a formula, write out every value in the units the formula actually needs. This single habit prevents the majority of errors in this topic.
Always write the balanced equation first
For any reacting mass or reacting volume question, write out (and check!) the balanced equation before doing any arithmetic. The mole ratio it gives you is the entire basis of the calculation — get it wrong, and every subsequent number is wrong too.
Match your significant figures to the question
If a question gives data to 2 significant figures, your final answer should usually be to 2 s.f. too — not blindly copying your calculator display. Examiners specifically check this in mark schemes.
Don't confuse percentage yield with atom economy
These sound similar but measure completely different things:
  • % yield — compares your actual lab result to the theoretical maximum. Needs experimental data.
  • Atom economy — compares the desired product's mass to all reactant mass, straight from the equation. No experiment needed — it's a property of the reaction itself.
A reaction can have 100% yield but low atom economy (efficient process, wasteful reaction) — or the reverse. Exam questions love testing this distinction.
PV = nRT: the units checklist
Before every ideal gas calculation, run through this checklist:
  • Pressure in pascals? (×1000 if given in kPa)
  • Volume in ? (÷1000 from dm³, or ÷1,000,000 from cm³)
  • Temperature in kelvin? (+273 from °C)
Skipping just one of these three conversions will give an answer that's out by a factor of 1000 or more — always sanity-check whether your final number is a "sensible" size for the situation.
Gas volume ratios = mole ratios (shortcut!)
When both the reactant and product are gases, you can skip the "convert to moles" step entirely — the ratio of gas volumes equals the mole ratio directly, since equal moles of any gas occupy equal volume under the same conditions. This can save serious time in reacting volume questions.
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  • 4. Calculations of Product: Yield & Atom Economy
  • Exam Tips & Common Mistakes
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