Topic 6 · The Mole Concept and Stoichiometry · spread 1 of 2

The Mole Concept and Stoichiometry

1

Relative Masses

Masses are compared with the carbon-12 standard: one unit is 112 of the mass of one mole of ¹²C atoms. Relative masses have no units.

TermDefinition
Relative isotopic massthe mass of one mole of an isotope relative to 112 the mass of one mole of ¹²C atoms
Relative atomic mass, Arthe average mass of one mole of atoms of an element relative to 112 the mass of one mole of ¹²C atoms
Relative molecular mass, Mrthe average mass of one mole of molecules relative to 112 the mass of one mole of ¹²C atoms
Relative formula mass, Mrthe average mass of one mole of formula units of a compound relative to 112 the mass of one mole of ¹²C atoms (ionic compounds)
Ar =Σ(% abundance × relative isotopic mass)100

Ar is rarely a whole number: it is a weighted average over the isotopes.

Worked exampleChlorine is 75.77% ³⁵Cl and 24.23% ³⁷Cl. Find Ar(Cl).

Ar = 75.77(35) + 24.23(37)100 = 35.5

Worked exampleA copper wire has Ar(Cu) = 63.54. Find the percentage abundances of ⁶³Cu and ⁶⁵Cu.

Let ⁶³Cu be x%:  63.54 = 63x + 65(100 − x)100 → x = 73.0, so ⁶³Cu = 73.0% and ⁶⁵Cu = 27.0%
6Topic 06 · The Mole Concept and Stoichiometry
2

The Mole

DefinitionMole: the amount of substance that contains 6.02 × 10²³ (the Avogadro constant, L) elementary entities. The entity must be stated: atoms, molecules, ions or electrons.
DefinitionMolar mass, M: the mass of one mole of the specified substance (g mol⁻¹), numerically equal to Ar or Mr.
DefinitionMolar gas volume, Vm: the volume occupied by one mole of a gas under given conditions of temperature and pressure: 24.0 dm³ mol⁻¹ at r.t.p. (20 °C, 1 atm), 22.7 dm³ mol⁻¹ at s.t.p. (0 °C, 1 bar).
DefinitionConcentration: the amount or mass of solute in a given volume of solution (mol dm⁻³ or g dm⁻³).
amountn / molmass / gm = n × M× M÷ Mno. of particlesN = n × L× L÷ Lgas volume / dm³V = n × Vₘ× Vₘ÷ Vₘconcentrationc = n ÷ V(dm³)÷ V× V

Mole ratio: in 1 mol of AmBn there are m mol of A and n mol of B, e.g. 1 mol CaCl₂ has 2 mol Cl⁻.

Note"1 mole of oxygen" is ambiguous: say 1 mol of O atoms or 1 mol of O₂ molecules. 1 dm³ = 1000 cm³.

2.1Particles and gas volume

Worked exampleA cylinder releases 16 dm³ of NH₃ gas at s.t.p. How many NH₃ molecules is this?

nNH₃ = 1622.7 = 0.7048 mol → N = 0.7048 × 6.02 × 10²³ = 4.24 × 10²³

2.2Preparing a solution

Weigh the solute, dissolve it in a little water, then make up to the mark in a volumetric flask and invert to mix.

Worked exampleA technician makes 500 cm³ of H₂SO₄ solution with [H⁺] = 2.00 mol dm⁻³. What mass of H₂SO₄ is needed?

nH⁺ = 2.00 × 0.500 = 1.000 mol. Each H₂SO₄ gives 2H⁺

nH₂SO₄ = 0.5000 mol → mass = 0.5000 × 98.1 = 49.1 g

Topic 06 · The Mole Concept and Stoichiometry7
3

Empirical and Molecular Formulae

DefinitionEmpirical formula: the formula showing the simplest whole number ratio of the atoms of each element in the substance.
DefinitionMolecular formula: the formula showing the actual number of atoms of the different elements in one molecule of the compound, a multiple of the empirical formula.
Answer structureFinding the formula: fill in the table
element 1element 2element 3
Mass / g (or %)givengivengiven
Amount / molmass ÷ Armass ÷ Armass ÷ Ar
Ratio (÷ smallest)whole numberwhole numberwhole number

Ratio → empirical formula → molecular formula = empirical formula × Mrempirical formula mass

3.1From % by mass

Worked exampleCompound X (Mr = 114) is 63.2% C, 12.2% H and 24.6% N by mass. Find its molecular formula.

CHN
Mass / g63.212.224.6
Amount / mol5.26712.201.757
Ratio (÷ 1.757)371

Take 100 g of X, so each percentage becomes a mass in grams.

Empirical C₃H₇N (M = 57.0), 114 ÷ 57.0 = 2 ∴ C₆H₁₄N₂

3.2From combustion data

Worked exampleA sugar, mannose (C, H and O only, Mr 180), is burnt completely in oxygen. 2.36 g of it gives 3.46 g of CO₂ and 1.42 g of H₂O. Find its molecular formula.

  1. All the C in the CO₂ came from the sample (12 g of C in every 44 g of CO₂): mass of C = 1244 × 3.46 = 0.9436 g
  2. All the H in the H₂O came from the sample (2 g of H in every 18 g of H₂O): mass of H = 218 × 1.42 = 0.1578 g
  3. O cannot be found this way, as some O in CO₂ and H₂O came from the O₂ burnt: mass of O = 2.36 − 0.9436 − 0.1578 = 1.259 g
sample+ O₂CO₂H₂Oall the Call the H
CHO
Mass / g0.94360.15781.259
Amount / mol0.078630.15780.07869
Ratio (÷ 0.07863)121

Empirical CH₂O (M = 30), 180 ÷ 30 = 6 ∴ C₆H₁₂O₆

8Topic 06 · The Mole Concept and Stoichiometry
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Stoichiometry, Limiting Reagent and Yield

  • A balanced equation gives the mole ratio of reactants and products: 2H₂ + O₂ → 2H₂O means 2 mol H₂ ≡ 1 mol O₂ ≡ 2 mol H₂O
  • The limiting reagent is used up first, it fixes the maximum amount of product (the theoretical yield)
percentage yield = actual mass (or amount) of producttheoretical mass (or amount) of product × 100%

Actual yield is lower: incomplete reaction, side reactions and losses during purification.

4.1Limiting reagent and yield

Worked exampleA student precipitates PbCl₂ by mixing 100 cm³ of 0.0600 mol dm⁻³ Pb(NO₃)₂ with 200 cm³ of 0.0500 mol dm⁻³ NaCl, and collects 1.1558 g of PbCl₂. Find the limiting reagent and the percentage yield.

  1. Pb(NO₃)₂ + 2NaCl → PbCl₂ + 2NaNO₃
  2. n(Pb(NO₃)₂) = 6.000 × 10⁻³ mol, nNaCl = 1.000 × 10⁻² mol
  3. 6.000 × 10⁻³ mol Pb(NO₃)₂ needs 1.200 × 10⁻² mol NaCl → NaCl is limiting
  4. nPbCl₂ = ½ × 1.000 × 10⁻² = 5.000 × 10⁻³ mol
  5. maximum mass = 5.000 × 10⁻³ × 278.1 = 1.391 g
  6. % yield = 1.15581.391 × 100 = 83.1%

4.2Reacting mass to gas volume

Worked exampleA 0.230 g piece of sodium is dropped into excess water. Find the volume of H₂ produced at s.t.p.

  1. 2Na + 2H₂O → 2NaOH + H₂. Water is in excess, so Na is limiting
  2. nNa = 0.230 ÷ 23.0 = 1.000 × 10⁻² mol → nH₂ = 5.000 × 10⁻³ mol
  3. V(H₂) = 5.000 × 10⁻³ × 22.7 = 0.114 dm³
Exam tipGive intermediate answers to 4 s.f. and final answers to 3 s.f. unless told otherwise, Ar and Mr to 1 d.p.
Topic 06 · The Mole Concept and Stoichiometry9
5

Gas Volumes and Combustion

5.1Gas volumes

  • Avogadro's law: at the same temperature and pressure, equal volumes of all gases contain the same number of particles → for gases, mole ratio = volume ratio
  • Only gases count: water at r.t.p. is a liquid, so its volume is ignored

5.2Combustion of hydrocarbons

CxHy(g) + (x + y4)O₂(g) → xCO₂(g) + y2H₂O(l)
O₂ addedCₓHᵧO₂ leftCO₂O₂ leftbeforeafter combustion, cooledafter excess NaOHcontractionCO₂ absorbed

NaOH(aq) absorbs CO₂: CO₂(g) + 2NaOH(aq) → Na₂CO₃(aq) + H₂O(l)

Worked exampleA chemist explodes 10 cm³ of a gaseous hydrocarbon CxHy with 110 cm³ of O₂ (excess). After cooling to r.t.p. the volume is 85 cm³, falling to 35 cm³ after passing through excess NaOH(aq). Find the formula of the hydrocarbon.

  1. CO₂ is absorbed by NaOH: V(CO₂) = 85 − 35 = 50 cm³ = 10x → x = 5
  2. Contraction on combustion = (10 + 110) − 85 = 35 cm³
  3. Contraction = V(CxHy) + V(O₂ reacted) − V(CO₂) = 10 + 10(x + y4) − 10x = 10 + 10y4 = 35 → y = 10
  4. ∴ the hydrocarbon is C₅H₁₀
6

Volumetric Analysis

TermDefinition
Standard solutiona solution whose concentration is accurately known
Titrantthe solution added from the burette
Analytethe fixed volume of solution in the conical flask
Equivalence pointthe point at which stoichiometric amounts of reactants have reacted, independent of the indicator used
End-pointthe point when the indicator changes colour
10Topic 06 · The Mole Concept and Stoichiometry
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Volumetric Analysis (continued)

6.1Choosing an acid–base indicator

TitrationIndicatorColour, acid → alkali
strong acid–strong basemethyl orange or phenolphthaleineither change below
strong acid–weak basemethyl orange (pH 3–5)red → yellow, orange at the end-point
weak acid–strong basephenolphthalein (pH 8–10)colourless → pale pink at the end-point
weak acid–weak baseno suitable indicator

A suitable indicator has a distinct colour change at the equivalence point.

6.2Dilution

c₁V₁ = c₂V₂

Amount of solute withdrawn = amount in the diluted solution.

6.3Back-titration

When A cannot be titrated directly (e.g. insoluble CaCO₃ or Mg(OH)₂)
  1. Add a known excess of B, so all of A reacts
  2. Titrate the unreacted B that is left over
  3. nB reacted with A = nB added − nB left over, then nA from the equation

Worked exampleAn antacid tablet contains impure Mg(OH)₂, which is insoluble, so it cannot be titrated directly. 1.00 g of it is added to 100 cm³ of 0.400 mol dm⁻³ HNO₃ (excess). The leftover HNO₃ needs 20.50 cm³ of 0.750 mol dm⁻³ KOH. Find the percentage of Mg(OH)₂.

all HNO₃ added = 0.04000 mol (known)reacted withthe Mg(OH)₂left overfound by difference0.04000 − 0.01538 = 0.02462 moltitrated with KOH0.01538 molMg(OH)₂ + 2HNO₃ → Mg(NO₃)₂ + 2H₂O
  1. nHNO₃ added = 0.400 × 0.100 = 0.04000 mol
  2. nHNO₃ left over = nKOH = 0.750 × 0.02050 = 0.01538 mol
  3. n(HNO₃ reacted with Mg(OH)₂) = 0.04000 − 0.01538 = 0.02462 mol → n(Mg(OH)₂) = ½ × 0.02462 = 0.01231 mol
  4. mass = 0.01231 × 58.3 = 0.7177 g → 71.8% Mg(OH)₂
Topic 06 · The Mole Concept and Stoichiometry11
7

Redox Reactions

OxidationReduction
gain of oxygenloss of oxygen
loss of hydrogengain of hydrogen
loss of electronsgain of electrons
increase in oxidation numberdecrease in oxidation number
Zn+Cu2+→Zn2++Cu2e-
Znloses 2e⁻: oxidised, the reducing agent
Cu²⁺gains 2e⁻: reduced, the oxidising agent
DefinitionOxidising agent: oxidises another species and is itself reduced.
DefinitionReducing agent: reduces another species and is itself oxidised.
DefinitionDisproportionation: a redox reaction in which one species is simultaneously oxidised and reduced, e.g. Cl₂ + 2OH⁻ → ClO⁻ + Cl⁻ + H₂O, with the oxidation numbers of Cl under it: Cl₂ 0 → Cl in ClO⁻ +1 (oxidised) and Cl⁻ −1 (reduced).

7.1Oxidation number rules, in order of priority

Apply them from the top: when two rules disagree, the higher one wins.

  1. Element: 0 (Mg, O₂, P₄)
  2. Sum of O.N. = overall charge of the species (Al³⁺: +3)
  3. Group 1: +1, Group 2: +2, Al: +3
  4. F: −1
  5. H: +1 (so −1 in NaH, where Na comes first)
  6. O: −2 (so −1 in H₂O₂ and +2 in OF₂)
  7. Otherwise, the more electronegative atom takes the negative O.N.

Worked exampleFind the oxidation number of Mn in MnO₄⁻.

x + 4(−2) = −1 → x = +7

Charge is written "2+", oxidation number is written "+2".

7.2Balancing redox equations: MnO₄⁻ + I⁻ in acid

AOHC Half-equation method

ABalance atoms other than O and HMnO₄⁻ → Mn²⁺   2I⁻ → I₂ OBalance O with H₂OMnO₄⁻ → Mn²⁺ + 4H₂O HBalance H with H⁺MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O CBalance charge with e⁻MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O   2I⁻ → I₂ + 2e⁻ AddMatch the e⁻ (× 2 and × 5), add, cancel2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O

ACOH Oxidation number method

ABalance atoms other than O and HMnO₄⁻ + 2I⁻ → Mn²⁺ + I₂ CBalance the change in O.N.: Mn −5 each, I +1 each2MnO₄⁻ + 10I⁻ → 2Mn²⁺ + 5I₂ OBalance O with H₂O2MnO₄⁻ + 10I⁻ → 2Mn²⁺ + 5I₂ + 8H₂O HBalance H with H⁺2MnO₄⁻ + 10I⁻ + 16H⁺ → 2Mn²⁺ + 5I₂ + 8H₂O
NoteIn alkaline solution MnO₄⁻ is reduced to MnO₂. Balance with H⁺ first, then add the same number of OH⁻ to both sides to turn H⁺ into H₂O.
12Topic 06 · The Mole Concept and Stoichiometry
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Redox Titrations

TitrationHalf-equation of oxidising agentIndicatorEnd-point colour
KMnO₄ (burette) with Fe²⁺ or H₂O₂MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂Onone (self-indicating)colourless → pale pink
Acidify with H₂SO₄, not HCl (Cl⁻ is oxidised to Cl₂) or HNO₃ (itself an oxidising agent)
K₂Cr₂O₇ with Fe²⁺Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂ON-phenylanthranilic acidgreen → red-violet
S₂O₃²⁻ (burette) with I₂I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻starch, near the enddark blue → colourless
Iodometric titration: the oxidising agent is added to excess KI, so it reacts completely and the I₂ stays dissolved as I₃⁻. The I₂ released is then titrated with S₂O₃²⁻: the titre gives the amount of I₂, and so the original amount of oxidising agent (e.g. IO₃⁻)
Add starch only near the end-point (pale yellow): at high [I₂] the starch–iodine complex precipitates

8.1Iodometric titration

Worked exampleTo standardise a sodium thiosulfate solution, 0.875 g of KIO₃ is made up to 250 cm³. 25.0 cm³ of this is added to excess KI and acid, and the I₂ formed needs 22.70 cm³ of Na₂S₂O₃. Find [S₂O₃²⁻].

  1. IO₃⁻ + 5I⁻ + 6H⁺ → 3I₂ + 3H₂O,  I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
  2. nIO₃⁻ in 25.0 cm³ = 0.875214.0 × 25.0250 = 4.089 × 10⁻⁴ mol
  3. IO₃⁻ ≡ 3I₂ ≡ 6S₂O₃²⁻ → nS₂O₃²⁻ = 2.453 × 10⁻³ mol
  4. [S₂O₃²⁻] = 2.453 × 10⁻³ ÷ 0.02270 = 0.108 mol dm⁻³

8.2Finding a new oxidation number

electrons lost = electrons gained

30.0 cm³ of 0.0500 mol dm⁻³ Fe²⁺ reacts exactly with 15.0 cm³ of 0.0250 mol dm⁻³ MnO₄⁻, and Fe ends as +3. Find the final oxidation number of Mn.

  1. nFe²⁺ = 1.500 × 10⁻³ mol, each losing 1 e⁻ → 1.500 × 10⁻³ mol e⁻
  2. nMnO₄⁻ = 3.750 × 10⁻⁴ mol → each gains 1.500 × 10⁻³3.750 × 10⁻⁴ = 4 e⁻
  3. Mn: +7 − 4 = +3

8.3Common oxidising and reducing agents

Oxidising agentMediumProduct
MnO₄⁻acidicMn²⁺
MnO₄⁻neutral or alkalineMnO₂
Cr₂O₇²⁻acidicCr³⁺
I₂neutral or acidicI⁻
IO₃⁻acidicI₂
H₂O₂acidicH₂O
Reducing agentMediumProduct
Fe²⁺acidic or neutralFe³⁺
I⁻anyI₂
S₂O₃²⁻neutralS₄O₆²⁻ with I₂
C₂O₄²⁻acidicCO₂
H₂O₂acidicO₂

H₂O₂ can act as an oxidising agent or as a reducing agent. With a stronger oxidising agent such as Cl₂, S₂O₃²⁻ gives SO₄²⁻.

Topic 06 · The Mole Concept and Stoichiometry13