Sec 3 Chemistry — Mole Concept and Stoichiometry
Prepared by Miss Clarissa Ng · www.clartutors.com
Part A · Counting particles — the mole
1 What Is a Mole?
The SI unit for the amount of a substance is the mole. One mole of any substance contains 6.02 × 10²³ particles — a number known as Avogadro's constant.
A mole is simply a counting unit, the chemist's version of "a dozen". Particles here means whatever the substance is made of — atoms, molecules, ions or electrons.
number of moles (mol) = number of particles ÷ 6.02 × 10²³ particles mol⁻¹
Worked example — particles to moles
A beaker contains 1.45 × 10²⁴ water molecules. Calculate the number of moles of water.
n = (1.45 × 10²⁴) ÷ (6.02 × 10²³) = 2.41 mol (3 s.f.)
Exam habit: always write the unit mol after a number of moles. A bare number in a moles question loses the mark even when the arithmetic is right.
2 Relative Atomic, Molecular and Formula Mass
| Quantity | What it compares | How you get it |
|---|---|---|
| Relative atomic mass, Ar | The average mass of one atom of an element, compared to 1/12 of the mass of one carbon-12 atom | Read it straight off the Periodic Table |
| Relative molecular mass, Mr | The average mass of one molecule of a molecular substance, on the same scale | Add up the Ar of all atoms in the chemical formula |
| Relative formula mass, Mr | The average mass of one formula unit of an ionic compound, on the same scale | Add up the Ar of all atoms in the formula |
All three are ratios, so they have no unit. This is why you can add them up and why they never carry a "g" — the grams arrive later, in molar mass.
Worked example — an element with three isotopes
Magnesium occurs naturally as 79 % magnesium-24, 10 % magnesium-25 and 11 % magnesium-26. Find its relative atomic mass.
Ar = (79/100 × 24) + (10/100 × 25) + (11/100 × 26)
= 18.96 + 2.50 + 2.86 = 24.3 (3 s.f.)
= 18.96 + 2.50 + 2.86 = 24.3 (3 s.f.)
Sanity check: the answer must land between the lightest and heaviest isotope and lean towards whichever is most abundant. An answer outside that range means a slipped decimal.
3 Molar Mass — from Moles to Grams
The molar mass of a substance is the mass of one mole of it, in grams per mole (g mol⁻¹). Its value comes straight from the Ar or Mr, with the unit attached.
number of moles (mol) = mass of substance (g) ÷ molar mass of substance (g mol⁻¹)
| The substance is an… | Its molar mass comes from… |
|---|---|
| Element | the relative atomic mass, Ar — e.g. Na is 23 g mol⁻¹ |
| Molecular substance | the relative molecular mass, Mr — e.g. O₂ is 32 g mol⁻¹ |
| Ionic compound | the relative formula mass, Mr — e.g. NaCl is 58.5 g mol⁻¹ |
Worked example — grams to moles
Calculate the number of moles in 5.85 g of sodium chloride.
M(NaCl) = 23 + 35.5 = 58.5 g mol⁻¹ n = 5.85 ÷ 58.5 = 0.100 mol
Unit trap: convert first. If a mass is given in kg, multiply by 1000 before dividing by the molar mass.
Part B · Getting from numbers to a formula
4 Percentage by Mass of an Element
percentage by mass = (number of atoms of the element in the formula × Ar of the element) ÷ Mr of the compound × 100 %
This tells you the mass fraction of one element inside a compound — the calculation examiners like to hide inside a fertiliser or ore question.
Worked example — nitrogen in a fertiliser
Calculate the percentage by mass of nitrogen in ammonium nitrate, NH₄NO₃.
Mr = 14 + (4 × 1) + 14 + (3 × 16) = 80
percentage by mass of N = (2 × 14) ÷ 80 × 100 % = 35.0 %
percentage by mass of N = (2 × 14) ÷ 80 × 100 % = 35.0 %
Exam habit: count the atoms of that element in the whole formula first. In NH₄NO₃ there are two nitrogen atoms — missing the second one is the mistake this question is built around.
5 Empirical Formula from Reacting Masses
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. It is found from the masses that actually react.
Three steps, every time:
| Step | What you do |
|---|---|
| 1 | Find the mass of each element that reacted. If a product mass is given, subtract the starting mass to get the mass of the element that came from the air. |
| 2 | Convert each mass to moles (mass ÷ molar mass), then divide every value by the smallest one to get the mole ratio. |
| 3 | Write the ratio as the empirical formula. If a ratio comes out as a fraction such as 1.5, multiply every value until they are whole numbers. |
Worked example — an oxide of iron
An oxide of iron contains 2.24 g of iron and 0.96 g of oxygen. Determine its empirical formula.
| Fe | O | |
|---|---|---|
| Mass / g | 2.24 | 0.96 |
| Molar mass / g mol⁻¹ | 56 | 16 |
| Number of moles / mol | 2.24 ÷ 56 = 0.040 | 0.96 ÷ 16 = 0.060 |
| Divide by the smallest (0.040) | 1 | 1.5 |
| Multiply by 2 to make them whole | 2 | 3 |
∴ the empirical formula is Fe₂O₃. A ratio of 1 : 1.5 is not wrong — it just means you have not finished. Multiply both numbers by 2 and the formula appears.
6 Empirical Formula from Percentage by Mass
Same three steps, with one change: if you are given percentages instead of masses, take each percentage as the mass in 100 g of the compound. The numbers then work exactly as before.
Worked example — a compound of carbon, hydrogen and oxygen
A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen by mass. Determine its empirical formula.
| C | H | O | |
|---|---|---|---|
| Mass in 100 g / g | 40.0 | 6.7 | 53.3 |
| Molar mass / g mol⁻¹ | 12 | 1 | 16 |
| Number of moles / mol | 3.33 | 6.7 | 3.33 |
| Divide by the smallest (3.33) | 1 | 2 | 1 |
∴ the empirical formula is CH₂O. Whenever you are given percentages that add to 100 %, this is the fastest route: read them as grams and carry on.
7 Molecular Formula from the Empirical Formula
The molecular formula shows the actual number of atoms of each element in one molecule. It is always a whole-number multiple of the empirical formula: molecular formula = (empirical formula)n.
| Substance | Molecular formula | Empirical formula |
|---|---|---|
| Water | H₂O | H₂O — they are the same |
| Methane | CH₄ | CH₄ — they are the same |
| Ethanoic acid | C₂H₄O₂ | CH₂O — the molecular formula is twice the empirical one |
| Ethane | C₂H₆ | CH₃ — the molecular formula is twice the empirical one |
Worked example — finding n
A compound has the empirical formula CH₂O and a relative molecular mass of 60. Determine its molecular formula.
Let the molecular formula be (CH₂O)n. Mr of CH₂O = 12 + (2 × 1) + 16 = 30
30n = 60 → n = 60 ÷ 30 = 2
molecular formula = (CH₂O)₂ = C₂H₄O₂
30n = 60 → n = 60 ÷ 30 = 2
molecular formula = (CH₂O)₂ = C₂H₄O₂
Exam habit: n must be a whole number. If you get 1.5, either your Mr or your empirical formula is wrong — go back rather than rounding to 2.
Part C · Gases and solutions
8 Molar Volume of Gases
At room temperature and pressure (r.t.p.) — 25 °C and 1 atm — one mole of any gas occupies 24 dm³, which is 24 000 cm³. This is the molar volume of a gas.
n = volume of gas (dm³) ÷ 24 dm³ mol⁻¹ or n = volume of gas (cm³) ÷ 24 000 cm³ mol⁻¹
Worked example — moles to volume
Calculate the volume occupied by 0.500 mol of carbon dioxide at r.t.p.
V = n × 24 = 0.500 × 24 = 12.0 dm³ (which is 12 000 cm³)
The one condition that matters: the molar volume of 24 dm³ is only valid at r.t.p. If a question states different conditions, the 24 does not apply — and the same goes for the mole ratio of gases, which relies on equal volumes holding equal numbers of particles.
9 Concentration of Solutions
mass concentration (g dm⁻³) = mass of solute (g) ÷ volume of solution (dm³)
molar concentration (mol dm⁻³) = number of moles of solute (mol) ÷ volume of solution (dm³)
molar concentration (mol dm⁻³) = mass concentration (g dm⁻³) ÷ molar mass of solute (g mol⁻¹)
molar concentration (mol dm⁻³) = number of moles of solute (mol) ÷ volume of solution (dm³)
molar concentration (mol dm⁻³) = mass concentration (g dm⁻³) ÷ molar mass of solute (g mol⁻¹)
Worked example — a solution of sodium carbonate
5.30 g of sodium carbonate is dissolved in distilled water to make 250 cm³ of solution. Calculate the molar concentration.
250 cm³ = 250 ÷ 1000 = 0.250 dm³ M(Na₂CO₃) = (2 × 23) + 12 + (3 × 16) = 106 g mol⁻¹
n = 5.30 ÷ 106 = 0.0500 mol c = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³
n = 5.30 ÷ 106 = 0.0500 mol c = 0.0500 ÷ 0.250 = 0.200 mol dm⁻³
Exam habit: the volume must be in dm³ before you divide. Convert cm³ to dm³ by dividing by 1000 — this single step is the most commonly dropped mark in the whole topic.
Part D · Stoichiometry — using the equation
10 Volume Ratios of Gases
Stoichiometry is the relationship between the number of moles of reactants and the number of moles of products, taken from the balanced equation. The numbers written in front of the formulae are the stoichiometric coefficients, and they give you the mole ratio.
Avogadro's law: equal volumes of all gases, at the same temperature and pressure, contain the same number of particles. So for gases the mole ratio is also the volume ratio.
Worked example — burning methane
100 cm³ of methane is burnt completely in oxygen at r.t.p.
CH₄ (g) + 2O₂ (g) → CO₂ (g) + 2H₂O (l)
| Gas | CH₄ | O₂ | CO₂ |
|---|---|---|---|
| Mole ratio from the equation | 1 | 2 | 1 |
| Volume ratio (same, for gases) | 1 | 2 | 1 |
| Volume / cm³ | 100 | 200 | 100 |
Oxygen needed = 200 cm³; carbon dioxide produced = 100 cm³. The water is left out of the volume ratio because at r.t.p. it is a liquid, and Avogadro's law only applies to gases.
11 Limiting and Excess Reactants
The reactant that is completely used up is the limiting reactant. The reactant that is not completely used up is the excess reactant. The amount of product formed is always decided by the limiting reactant.
The method, in four steps: find the moles of each reactant → use the mole ratio to work out how much of one reactant is needed for the other → compare with what is available → the reactant that runs out is the limiting one, and only its moles go into the product calculation.
Worked example — hydrogen and oxygen
4.00 g of hydrogen is mixed with 16.0 g of oxygen and ignited.
2H₂ (g) + O₂ (g) → 2H₂O (l)
| H₂ | O₂ | |
|---|---|---|
| Moles available | 4.00 ÷ 2 = 2.00 mol | 16.0 ÷ 32 = 0.500 mol |
| Ratio needed from the equation | 2 parts | 1 part |
| Moles needed for 0.500 mol O₂ | 1.00 mol | 0.500 mol |
| Verdict | excess — 1.00 mol left over | limiting — all used up |
mass of water = moles of O₂ × 2 × molar mass of H₂O = 0.500 × 2 × 18 = 18.0 g
Exam habit: never add the two masses and call that the product. Work from the limiting reactant, state which one it is, and the marks for the explanation come with the answer.
12 Percentage Yield and Percentage Purity
percentage yield = (actual yield ÷ theoretical yield) × 100 %
percentage purity = (mass of pure substance in sample (g) ÷ mass of impure sample (g)) × 100 %
percentage purity = (mass of pure substance in sample (g) ÷ mass of impure sample (g)) × 100 %
The percentage yield of a reaction is usually less than 100 % — product is lost to side reactions, an incomplete reaction, or simply left behind on the apparatus during transfer.
Worked example — percentage yield
4.00 g of calcium carbonate is strongly heated and 2.00 g of calcium oxide is collected.
CaCO₃ (s) → CaO (s) + CO₂ (g)
| Step | Working |
|---|---|
| 1 · moles of CaCO₃ | M = 40 + 12 + (3 × 16) = 100 g mol⁻¹; n = 4.00 ÷ 100 = 0.0400 mol |
| 2 · theoretical moles of CaO | ratio 1 : 1 → 0.0400 mol of CaO should form |
| 3 · theoretical mass of CaO | 0.0400 × (40 + 16) = 2.24 g |
| 4 · percentage yield | (2.00 ÷ 2.24) × 100 % = 89.3 % (3 s.f.) |
Worked example — percentage purity
A 1.00 g sample of impure zinc reacts completely with excess dilute hydrochloric acid and gives 288 cm³ of hydrogen at r.t.p.
Zn (s) + 2HCl (aq) → ZnCl₂ (aq) + H₂ (g)
| Step | Working |
|---|---|
| 1 · moles of H₂ | 288 ÷ 24 000 = 0.0120 mol (using the molar volume at r.t.p.) |
| 2 · moles of Zn that reacted | ratio 1 : 1 → 0.0120 mol |
| 3 · mass of pure Zn | 0.0120 × 65 = 0.780 g |
| 4 · percentage purity | (0.780 ÷ 1.00) × 100 % = 78.0 % |
Yield and purity sound alike but answer different questions. Yield asks "how much of what I expected did I actually get?" Purity asks "how much of my starting material was the real thing?" Yield needs a theoretical mass from the equation; purity needs the mass of pure substance hidden inside a weighed sample.
13 Put It Together — Exam-Style Question
A 25.0 cm³ sample of dilute hydrochloric acid is titrated against 0.200 mol dm⁻³ sodium hydroxide solution. The equation is:
HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l)
[1](a) State the mole ratio of HCl to NaOH in this reaction.
[2](b) 20.0 cm³ of the sodium hydroxide solution was needed for complete reaction. Calculate the number of moles of NaOH used.
[2](c) Hence calculate the concentration of the hydrochloric acid in mol dm⁻³.
[2](d) Calculate the mass of sodium chloride that would be formed.
[3](e) A 2.00 g sample of impure calcium carbonate is added to excess dilute nitric acid and 432 cm³ of carbon dioxide is collected at r.t.p. Calculate the percentage purity of the calcium carbonate.
CaCO₃ (s) + 2HNO₃ (aq) → Ca(NO₃)₂ (aq) + CO₂ (g) + H₂O (l)
Model answers.
(a) 1 : 1 — one mole of HCl reacts with one mole of NaOH.
(b) 20.0 cm³ = 0.0200 dm³. n = c × V = 0.200 × 0.0200 = 0.00400 mol.
(c) Ratio 1 : 1, so n(HCl) = 0.00400 mol in 25.0 cm³ = 0.0250 dm³. c = 0.00400 ÷ 0.0250 = 0.160 mol dm⁻³.
(d) M(NaCl) = 23 + 35.5 = 58.5 g mol⁻¹. Ratio 1 : 1, so n(NaCl) = 0.00400 mol. mass = 0.00400 × 58.5 = 0.234 g.
(e) n(CO₂) = 432 ÷ 24 000 = 0.0180 mol. Ratio 1 : 1, so n(CaCO₃) = 0.0180 mol. M(CaCO₃) = 40 + 12 + (3 × 16) = 100 g mol⁻¹, so mass of pure CaCO₃ = 0.0180 × 100 = 1.80 g. percentage purity = (1.80 ÷ 2.00) × 100 % = 90.0 %.
(a) 1 : 1 — one mole of HCl reacts with one mole of NaOH.
(b) 20.0 cm³ = 0.0200 dm³. n = c × V = 0.200 × 0.0200 = 0.00400 mol.
(c) Ratio 1 : 1, so n(HCl) = 0.00400 mol in 25.0 cm³ = 0.0250 dm³. c = 0.00400 ÷ 0.0250 = 0.160 mol dm⁻³.
(d) M(NaCl) = 23 + 35.5 = 58.5 g mol⁻¹. Ratio 1 : 1, so n(NaCl) = 0.00400 mol. mass = 0.00400 × 58.5 = 0.234 g.
(e) n(CO₂) = 432 ÷ 24 000 = 0.0180 mol. Ratio 1 : 1, so n(CaCO₃) = 0.0180 mol. M(CaCO₃) = 40 + 12 + (3 × 16) = 100 g mol⁻¹, so mass of pure CaCO₃ = 0.0180 × 100 = 1.80 g. percentage purity = (1.80 ÷ 2.00) × 100 % = 90.0 %.
★ Chapter Concept Map
Mole Concept and Stoichiometry — always convert to moles first
Particlesn = N ÷ 6.02 × 10²³ · one mole is 6.02 × 10²³ particles (Avogadro's constant)
THE MOLE
Massn = m ÷ M · M in g mol⁻¹ comes from Ar or Mr, and Ar and Mr have no unit
Gas volumen = V ÷ 24 at r.t.p. (25 °C, 1 atm) · 24 dm³ = 24 000 cm³ for any gas
OR
Solutionn = c × V · convert cm³ to dm³ first — divide by 1000
Numbers into a formula: percentage by mass → empirical formula from masses or percentages (divide by the smallest, multiply up if you get a fraction) → molecular formula = (empirical formula)n, where n = Mr ÷ empirical mass
Mole ratiothe stoichiometric coefficients in the balanced equation give the mole ratio of reactants to products
GASES
Avogadro's lawequal volumes of gases at the same temperature and pressure hold equal numbers of particles — so mole ratio = volume ratio
Limiting reactantcompletely used up, and it decides the amount of product · work out what is NEEDED, then compare with what is AVAILABLE
SO
Excess reactantnot completely used up · never use its moles to find the product
Percentage yieldactual ÷ theoretical × 100 % · usually below 100 %
VERSUS
Percentage puritymass of pure substance ÷ mass of impure sample × 100 %