The Mole Concept
and Stoichiometry

H2 Chemistry (9476), JC1

By the end of this topic, you can answer

1

An atom is far too small to put on a balance. So how do chemists know a carbon-12 atom has a mass of exactly 12?

2

How many water molecules are there in a single drop of water?

3

Glucose and ethanoic acid have the same percentage composition by mass. How do you tell them apart from data?

4

A reaction should make 10 g of aspirin, but you collect only 7 g. What is the percentage yield, and which reactant ran out first?

5

Burn 10 cm3 of an unknown hydrocarbon gas and measure the gas volumes before and after. How does that give its formula?

6

How can a titration tell you how much ethanoic acid is in a bottle of vinegar?

7

Old roadside breathalysers used orange dichromate that turned green. What is oxidised, what is reduced, and how do you balance the equation?

Recall: what you know from O levels

1Find the Mr of CO₂ (Ar: C 12.0, O 16.0).

12.0 + 2(16.0) = 44.0[1]

2How many moles are in 11.0 g of CO₂?

n = mass ÷ M = 11.0 ÷ 44.0 = 0.250 mol[1]

3Balance:  CH₄ + O₂ → CO₂ + H₂O

CH₄ + 2O₂ → CO₂ + 2H₂O[1]
1

Relative Masses

RecallFrom Atomic Structure: masses are compared with carbon-12, one unit being 1/12 of the mass of a ¹²C atom. Relative masses have no units.

TermDefinitionrelative to 1/12 the mass of one mole of ¹²C atoms
Relative atomic mass, Arthe average mass of one mole of atoms of an element
Relative molecular mass, Mrthe average mass of one mole of molecules
Relative formula mass, Mrthe average mass of one mole of formula units of a compound (ionic compounds)
1

Relative Masses

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%, so 65Cu is (100 − x)%

63.54 = 63x + 65(100 − x)100 → 6354 = 6500 − 2x

x = 73.0, so 63Cu = 73.0% and 65Cu = 27.0%

Your turnBromine has two isotopes, 79Br and 81Br. Ar(Br) = 79.90. Find the percentage of 79Br.

Let the abundance of 79Br be x%.
79.90 = 79x + 81(100 − x)100[1]
7990 = 8100 − 2x
x = 55.0, so 55.0% 79Br (and 45.0% 81Br)[1]
2

The Mole

AnalogyA dozen is a number: 12 eggs, 12 pencils. A mole is also just a number, a very large one, because particles are tiny.

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. A dozen eggs and a dozen melons weigh different amounts, and so do a mole of C and a mole of H₂O.

Note"1 mole of oxygen" is ambiguous: say 1 mol of O atoms or 1 mol of O₂ molecules.
2

The Mole

Everything goes through moles

  • Mass: n = mass ÷ M
  • Particles: N = n × L
  • Gas volume: V = n × Vm, Vm = 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)
  • Solution: n = c × V (V in dm³, 1 dm³ = 1000 cm³)
  • Mole ratio: 1 mol of AmBn contains m mol of A and n mol of B, e.g. 1 mol CaCl₂ has 2 mol Cl⁻
2

The Mole

2.1Particles and gas volume

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

Gas volume → moles:  nNH₃ = 1622.7 = 0.7048 mol

Moles → particles:  N = 0.7048 × 6.02 × 10²³ = 4.24 × 10²³

Your turnHow many oxygen atoms are there in 4.40 g of CO₂ (M = 44.0)?

nCO₂ = 4.40 ÷ 44.0 = 0.1000 mol
Each CO₂ has 2 O atoms: nO = 0.2000 mol[1]
N(O) = 0.2000 × 6.02 × 10²³ = 1.20 × 10²³[1]
2

The Mole

2.2Preparing a solution

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

Your turnWhat volume of 0.500 mol dm⁻³ Na₂CO₃ solution contains 0.0250 mol of Na⁺ ions?

Each Na₂CO₃ gives 2Na⁺: nNa₂CO₃ = 0.01250 mol[1]
V = n ÷ c = 0.01250 ÷ 0.500 = 0.0250 dm³ = 25.0 cm³[1]
2

The Mole

Quick checkWhich sample contains the greatest number of atoms? (Ar: H 1.0, He 4.0, N 14.0, O 16.0)

A2.0 g of H₂
B4.0 g of He
C16.0 g of O₂
D14.0 g of N₂
NoteA: 1.0 mol of H₂ molecules = 2.0 mol of atoms. He: 1.0 mol of atoms. O₂ and N₂: 0.50 mol of molecules = 1.0 mol of atoms. Always check which particle is being counted.
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

e.g. butane: empirical C₂H₅, molecular C₄H₁₀, glucose: CH₂O, C₆H₁₂O₆.

Exam tipWrite C₆H₁₄N₂, never (C₃H₇N)₂.
3

Empirical and Molecular Formulae

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 in 100 g / g63.212.224.6
Amount / mol (÷ Ar)5.26712.201.757
Ratio (÷ smallest, 1.757)371

Step 1Take 100 g, so each % becomes a mass in grams.

Step 2Empirical C₃H₇N, M = 57.0

Step 3114 ÷ 57.0 = 2 ∴ C₆H₁₄N₂

3

Empirical and Molecular Formulae

3.2From combustion data: where do the atoms go?

PredictA compound of C, H and O burns in oxygen. Can we find its mass of O from the CO₂ and H₂O?

  • All the C in the CO₂ came from the sample.
  • All the H in the H₂O came from the sample.
  • The O in the products came from the sample and from the O₂ burnt, so O is found by subtraction: massO = masssample − massC − massH.
3

Empirical and Molecular Formulae

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

  1. 12 g of C in every 44 g of CO₂: massC = 1244 × 3.46 = 0.9436 g
  2. 2 g of H in every 18 g of H₂O: massH = 218 × 1.42 = 0.1578 g
  3. massO = 2.36 − 0.9436 − 0.1578 = 1.259 g
CHO
n / mol0.078630.15780.07869
Ratio121

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

3

Empirical and Molecular Formulae

Your turnA hydrocarbon is 85.7% C by mass and has Mr = 56.0. Find its molecular formula.

In 100 g: C 85.7 g, H 14.3 g
nC = 85.7 ÷ 12.0 = 7.142, nH = 14.3 ÷ 1.0 = 14.30, ratio 1 : 2
Empirical formula CH₂ (Mr = 14.0)[1]
n = 56.0 ÷ 14.0 = 4, so the molecular formula is C₄H₈[1]

Challenge0.600 g of a compound of C, H and O burns to give 0.880 g of CO₂ and 0.360 g of H₂O. Find its empirical formula.

massC = 12.044.0 × 0.880 = 0.2400 g, massH = 2.018.0 × 0.360 = 0.04000 g[1]
massO = 0.600 − 0.2400 − 0.04000 = 0.3200 g[1]
nC : nH : nO = 0.02000 : 0.04000 : 0.02000 = 1 : 2 : 1, so the empirical formula is CH₂O[1]
4

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.

AnalogyA sandwich needs 2 slices of bread and 1 slice of cheese. With 10 bread and 4 cheese, how many sandwiches?

The limiting reagent is used up first, it fixes the maximum amount of product, the theoretical yield. The other reagent is in excess.

4

Stoichiometry, Limiting Reagent and Yield

percentage yield = actual mass (or amount) of producttheoretical mass (or amount) of product × 100%

The actual yield is lower because of incomplete reaction, side reactions and losses during purification.

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.
4

Stoichiometry, Limiting Reagent and Yield

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) = 0.0600 × 0.100 = 6.000 × 10⁻³ mol, nNaCl = 0.0500 × 0.200 = 1.000 × 10⁻² mol
  3. 6.000 × 10⁻³ mol Pb(NO₃)₂ needs 1.200 × 10⁻² mol NaCl, but only 1.000 × 10⁻² mol is present → NaCl is limiting
  1. nPbCl₂ = ½ × 1.000 × 10⁻² = 5.000 × 10⁻³ mol
  2. maximum mass = 5.000 × 10⁻³ × 278.1 = 1.391 g
  3. % yield = 1.15581.391 × 100 = 83.1%
4

Stoichiometry, Limiting Reagent and Yield

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

Your turn5.00 g of CaCO₃ (M = 100.1) is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl.  CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂. Which reagent is limiting? What volume of CO₂ forms at r.t.p.?

nCaCO₃ = 5.00 ÷ 100.1 = 0.04995 mol, nHCl = 0.05000 mol[1]
0.04995 mol of CaCO₃ needs 0.09990 mol of HCl, but only 0.05000 mol is present, so HCl is limiting.[1]
nCO₂ = ½ × 0.05000 = 0.02500 mol
VCO₂ = 0.02500 × 24.0 = 0.600 dm³[1]
4

Stoichiometry, Limiting Reagent and Yield

Quick check2.0 mol of N₂ reacts with 3.0 mol of H₂:  N₂ + 3H₂ → 2NH₃. What is the maximum amount of NH₃?

A4.0 mol
B2.0 mol
C6.0 mol
D1.0 mol
NoteB: 2.0 mol N₂ would need 6.0 mol H₂, so H₂ is limiting. 3.0 mol H₂ gives ⅔ × 3.0 = 2.0 mol NH₃. Option A uses the N₂, which is in excess.
5

Gas Volumes and Combustion

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

CxHy(g) + (x + y4)O₂(g) → xCO₂(g) + y2H₂O(l)

5

Gas Volumes and Combustion

5.2Combustion of hydrocarbons

Worked example10 cm³ of a gaseous hydrocarbon CxHy is exploded 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.

  1. CO₂ 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
5

Gas Volumes and Combustion

Your turn20 cm³ of a gaseous hydrocarbon is exploded with 150 cm³ of O₂ (excess). After cooling to r.t.p. the volume is 110 cm³, and 50 cm³ remains after excess NaOH(aq). Find the formula.

CxHy + (x + y4)O₂ → xCO₂ + y2H₂O
VCO₂ = 110 − 50 = 60 cm³ = 20x, so x = 3[1]
VO₂ used = 150 − 50 = 100 cm³ = 20(x + y4), so y = 8[1]
The hydrocarbon is C₃H₈ (propane).[1]
NoteThe gas left after NaOH(aq) is the unreacted O₂. Water is a liquid at r.t.p., so it takes no part in the gas volume.
6

Volumetric Analysis

TermDefinition
Standard solutiona solution whose concentration is accurately known
Titrant / analytethe solution added from the burette / the 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
6

Volumetric Analysis

6.1Choosing an acid–base indicator

TitrationIndicator
strong acid–strong basemethyl orange or phenolphthalein
strong acid–weak basemethyl orange (pH 3–5)
weak acid–strong basephenolphthalein (pH 8–10)
weak acid–weak baseno suitable indicator

Methyl orange: red below pH 3, yellow above pH 5, orange at its end-point

Phenolphthalein: colourless below pH 8, pink above pH 10

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

6.2Dilution

c₁V₁ = c₂V₂

Adding water does not change the amount of solute, so n before = n after.

Example25.0 cm³ of 2.00 mol dm⁻³ HCl made up to 250 cm³: c₂ = 2.00 × 25.0 ÷ 250 = 0.200 mol dm⁻³
6

Volumetric Analysis

6.3Back-titration

Used when A cannot be titrated directly, e.g. an insoluble solid like CaCO₃ or Mg(OH)₂.

  1. Add a known excess of B: all of A reacts.
  2. Titrate the unreacted B.
  3. Subtract to find the B that reacted with A.

nB reacted with A = nB added − nB unreacted

6

Volumetric Analysis

6.3Back-titration

Worked exampleAn antacid tablet contains impure Mg(OH)₂, which is insoluble. 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.

  1. nHNO₃ added = 0.400 × 0.100 = 0.04000 mol
  2. nHNO₃ left over = nKOH = 0.750 × 0.02050 = 0.01538 mol
  3. nHNO₃ reacted = 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

Your turn2.50 g of impure CaCO₃ is added to 50.0 cm³ of 1.00 mol dm⁻³ HCl (excess). The excess HCl needs 20.0 cm³ of 0.500 mol dm⁻³ NaOH. Find the percentage of CaCO₃ (M = 100.1).

nHCl added = 0.05000 mol, nHCl left over = nNaOH = 0.01000 mol[1]
nHCl reacted = 0.04000 mol, so nCaCO₃ = ½ × 0.04000 = 0.02000 mol[1]
massCaCO₃ = 0.02000 × 100.1 = 2.002 g
% purity = 2.0022.50 × 100% = 80.1%[1]
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
  • DefinitionOxidising agent: oxidises another species and is itself reduced
  • DefinitionReducing agent: reduces another species and is itself oxidised
7

Redox Reactions

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".

Your turnFind the O.N. of: (a) Cr in Cr₂O₇²⁻, (b) S in S₂O₃²⁻, (c) N in NH₄⁺

(a)
2x + 7(−2) = −2, so x = +6[1]
(b)
2x + 3(−2) = −2, so x = +2[1]
(c)
x + 4(+1) = +1, so x = −3[1]
7

Redox Reactions

  • DefinitionDisproportionation: a redox reaction in which one species is simultaneously oxidised and reduced.
Cl₂0+2OH⁻→ Cl+1O⁻+ Cl⁻−1+H₂O
Cl oxidised: 0 → +1Cl reduced: 0 → −1
7

Redox Reactions

7.2Balancing a redox equation by half-equations: MnO₄⁻ + I⁻ in acid

Step 1Split into reduction and oxidation, balance atoms other than O and HMnO₄⁻ → Mn²⁺    2I⁻ → I₂
Step 2Balance O by adding H₂O, then H by adding H⁺MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Step 3Balance charge by adding electronsMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O    2I⁻ → I₂ + 2e⁻
Step 4Multiply so the electrons are equal× 2: 10 e⁻ gained    × 5: 10I⁻ → 5I₂ + 10e⁻
Step 5Add and cancel common species2MnO₄⁻ + 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.
7

Redox Reactions

Your turnBalance the reaction of Cr₂O₇²⁻ with Fe²⁺ in acid. (Cr₂O₇²⁻ → Cr³⁺, Fe²⁺ → Fe³⁺)

Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Oxidation (× 6): Fe²⁺ → Fe³⁺ + e⁻[1]
Overall: Cr₂O₇²⁻ + 14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H₂O + 6Fe³⁺[1]
Exam tipCheck that atoms and total charge both balance: left (−2) + 14 + 12 = +24, right 6 + 18 = +24.
7

Redox Reactions

Common 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⁻.

8

Redox Titrations

TitrationIndicatorEnd-point colour
KMnO₄ (burette) with Fe²⁺ or H₂O₂none (self-indicating)colourless → pale pink
K₂Cr₂O₇ with Fe²⁺N-phenylanthranilic acidgreen → red-violet
S₂O₃²⁻ (burette) with I₂starch, near the enddark blue → colourless

Acidify KMnO₄ titrations with H₂SO₄, not HCl (Cl⁻ is oxidised to Cl₂) or HNO₃ (itself an oxidising agent).

8

Redox Titrations

Iodometric titration: titrate the I₂ that the oxidising agent releases

  • The oxidising agent is added to excess KI, so it reacts completely and the I₂ stays dissolved as I₃⁻.
  • The I₂ released is titrated with S₂O₃²⁻:  I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻
  • Add starch only near the end-point (pale yellow): at high [I₂] the starch–iodine complex precipitates.
8

Redox Titrations

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⁻ = 6 × 4.089 × 10⁻⁴ = 2.453 × 10⁻³ mol
  4. [S₂O₃²⁻] = 2.453 × 10⁻³ ÷ 0.02270 = 0.108 mol dm⁻³
8

Redox Titrations

8.2Finding a new oxidation number

electrons lost = electrons gained

Worked example30.0 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

Your turn25.0 cm³ of 0.100 mol dm⁻³ SO₃²⁻ (S: +4) reacts exactly with 20.0 cm³ of 0.0500 mol dm⁻³ of an oxidising agent containing one atom of metal M at +5. S ends at +6. Find the final O.N. of M.

nSO₃²⁻ = 2.500 × 10⁻³ mol. Each SO₃²⁻ loses 2 electrons as the oxidation number of S increases from +4 to +6, so 5.000 × 10⁻³ mol of electrons are lost.[1]
noxidant = 1.000 × 10⁻³ mol, so each M gains 5 electrons. Final oxidation number of M = +5 − 5 = 0[1]
8

Redox Titrations

Quick checkWhy is KMnO₄ titration acidified with dilute H₂SO₄ rather than HCl?

AHCl is a weaker acid than H₂SO₄.
BMnO₄⁻ would oxidise Cl⁻ to Cl₂, using up extra KMnO₄.
CHCl reacts with Fe²⁺.
DCl⁻ colours the solution, hiding the end-point.
NoteB: MnO₄⁻ is a strong enough oxidising agent to oxidise Cl⁻, so more KMnO₄ would be needed and the titre would be too high.

Exit ticket

1How many Cl⁻ ions are in 0.100 mol of CaCl₂?

0.200 mol × 6.02 × 10²³ = 1.20 × 10²³[1]

2Why can't the mass of O in a C, H, O compound be found from the CO₂ and H₂O directly?

Some O in the products came from the O₂ burnt, so find O by subtraction.[1]

3When is a back-titration used?

When the substance cannot be titrated directly, e.g. an insoluble solid: add excess reagent, titrate the excess.[1]

4O.N. of Mn in MnO₂?

x + 2(−2) = 0 → +4[1]