Mole Calculations

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / mole calculations and titration analysis, content statements covering reacting-mass stoichiometry; the concentration equation $c = n/V$c=n/V in $\text{mol dm}^{-3}$mol dm−3; limiting-reagent identification; acid–alkali titration analysis; back-titration; and the determination of molar mass from titration data (refer to the official OCR H432 specification document for exact wording). This is the quantitative core of Module 2 and is examined extensively across all three H432 papers.

This is the workhorse lesson of A-Level quantitative chemistry. Three equations — $n = m/M$n=m/M, $c = n/V$c=n/V and the stoichiometric ratio from a balanced equation — combine to solve essentially every quantitative problem that appears in the rest of the course: reacting masses, concentration conversions, titrations, limiting-reagent and yield problems. The framework descends directly from Lavoisier's conservation-of-mass paradigm and Dalton's atomic-ratio postulate, but the moles-divided-by-coefficient trick for identifying the limiting reagent is the procedural move that students get wrong most often. Titration is the experimental anchor: a known-concentration titrant delivered from a burette reacts with an unknown-concentration analyte in a conical flask, with an indicator marking the equivalence point. The 1:1, 1:2 and 2:3 mole-ratio variants are all examined, as are back-titrations (excess + retitrate) and Mr-determination titrations (mass measured, $n$n calculated, $M_r$Mr​ inferred). Every step must show explicit unit conversion: $\text{cm}^3 \rightarrow \text{dm}^3$cm3→dm3 before $c = n/V$c=n/V is the single biggest source of lost marks in OCR's stoichiometry sections.

Key Equation: $n = \dfrac{m}{M}, \qquad c = \dfrac{n}{V}, \qquad n_A : n_B \text{ matches coefficients in the balanced equation}$n=Mm​,c=Vn​,nA​:nB​ matches coefficients in the balanced equation with $V$V in dm³ and $c$c in $\text{mol dm}^{-3}$mol dm−3.

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The Three Core Mole Equations

$n = \frac{m}{M} \qquad c = \frac{n}{V} \qquad N = nL$n=Mm​c=Vn​N=nL

Symbol	Quantity	SI unit
$m$m	mass	g
$M$M	molar mass	$\text{g mol}^{-1}$g mol−1
$c$c	concentration	$\text{mol dm}^{-3}$mol dm−3
$V$V	volume	dm³, not cm³
$N$N	number of particles	(dimensionless)
$L$L	Avogadro constant	$\text{mol}^{-1}$mol−1

Volume conversions

• $1\text{ dm}^3 = 1000\text{ cm}^3 = 1\text{ L} = 1000\text{ mL}$1 dm3=1000 cm3=1 L=1000 mL
• Convert cm³ → dm³ by dividing by 1000
• Convert dm³ → cm³ by multiplying by 1000

Tip: Burettes read in cm³. OCR concentrations are quoted in $\text{mol dm}^{-3}$mol dm−3. Always convert burette readings to dm³ before substituting into $c = n/V$c=n/V — the single most common slip.

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Concentration

Concentration is amount of solute per unit volume of solution: $c = n/V$c=n/V.

Worked Example 1 — Concentration from mass

A $250\text{ cm}^3$250 cm3 solution contains $10.0\text{ g}$10.0 g NaOH. Find $c$c in $\text{mol dm}^{-3}$mol dm−3.

• $M(\text{NaOH}) = 40.0\text{ g mol}^{-1}$M(NaOH)=40.0 g mol−1; $n = 10.0/40.0 = 0.250\text{ mol}$n=10.0/40.0=0.250 mol
• $V = 250/1000 = 0.250\text{ dm}^3$V=250/1000=0.250 dm3
• $c = 0.250/0.250 =$c=0.250/0.250= $1.00\text{ mol dm}^{-3}$1.00 mol dm−3

Worked Example 2 — Mass from concentration

What mass of KOH for $500\text{ cm}^3$500 cm3 of $0.200\text{ mol dm}^{-3}$0.200 mol dm−3 solution?

• $V = 0.500\text{ dm}^3$V=0.500 dm3; $n = cV = 0.100\text{ mol}$n=cV=0.100 mol
• $M(\text{KOH}) = 56.1\text{ g mol}^{-1}$M(KOH)=56.1 g mol−1; $m = 0.100 \times 56.1 =$m=0.100×56.1= $5.61\text{ g}$5.61 g

Worked Example 3 — $\text{g dm}^{-3}$g dm−3 → $\text{mol dm}^{-3}$mol dm−3

$21.2\text{ g dm}^{-3}$21.2 g dm−3 of Na₂CO₃. $M = 106.0$M=106.0.

$c = 21.2/106.0 =$c=21.2/106.0= $0.200\text{ mol dm}^{-3}$0.200 mol dm−3

Worked Example 4 — Dilution

$25.0\text{ cm}^3$25.0 cm3 of $2.00\text{ mol dm}^{-3}$2.00 mol dm−3 HCl is diluted to $250\text{ cm}^3$250 cm3. Find new $c$c.

Moles are conserved on dilution: $n = 2.00 \times 0.0250 = 0.0500\text{ mol}$n=2.00×0.0250=0.0500 mol. New $c = 0.0500/0.250 =$c=0.0500/0.250= $0.200\text{ mol dm}^{-3}$0.200 mol dm−3. Alternative: $c_1V_1 = c_2V_2 \Rightarrow 2.00 \times 25.0 = c_2 \times 250 \Rightarrow c_2 = 0.200$c1​V1​=c2​V2​⇒2.00×25.0=c2​×250⇒c2​=0.200.

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Reacting Masses (Stoichiometry)

The stoichiometric ratio from a balanced equation links moles of reactants to moles of products. The general workflow:

1. Write the balanced equation.
2. Convert the known mass (or volume) to moles using $n = m/M$n=m/M (or $n = cV$n=cV).
3. Use the ratio from the equation to find moles of the target substance.
4. Convert moles back to mass (or volume) as required.

flowchart LR
    A[Mass of A g] -- divide by M_A --> B[Moles of A]
    B -- ratio from balanced equation --> C[Moles of B]
    C -- multiply by M_B --> D[Mass of B g]

Worked Example 5 — Decomposition of CaCO₃

$\text{CaCO}_3(\text{s}) \rightarrow \text{CaO(s)} + \text{CO}_2(\text{g})$CaCO3​(s)→CaO(s)+CO2​(g). From $25.0\text{ g}$25.0 g of CaCO₃:

• $n(\text{CaCO}_3) = 25.0/100.1 = 0.2498\text{ mol}$n(CaCO3​)=25.0/100.1=0.2498 mol
• 1:1 ratio → $n(\text{CaO}) = 0.2498\text{ mol}$n(CaO)=0.2498 mol
• $m(\text{CaO}) = 0.2498 \times 56.1 =$m(CaO)=0.2498×56.1= $14.0\text{ g}$14.0 g

Worked Example 6 — Iron from iron(III) oxide

$\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2$Fe2​O3​+3CO→2Fe+3CO2​. From $80.0\text{ g}$80.0 g Fe₂O₃:

• $n(\text{Fe}_2\text{O}_3) = 80.0/159.6 = 0.5013\text{ mol}$n(Fe2​O3​)=80.0/159.6=0.5013 mol
• Ratio $1:2$1:2 → $n(\text{Fe}) = 1.003\text{ mol}$n(Fe)=1.003 mol
• $m(\text{Fe}) = 1.003 \times 55.8 =$m(Fe)=1.003×55.8= $55.9\text{ g}$55.9 g

Worked Example 7 — CaCO₃ to neutralise H₂SO₄

$\text{CaCO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{CaSO}_4 + \text{H}_2\text{O} + \text{CO}_2$CaCO3​+H2​SO4​→CaSO4​+H2​O+CO2​. From $5.00\text{ g}$5.00 g H₂SO₄:

• $n(\text{H}_2\text{SO}_4) = 5.00/98.1 = 0.05097\text{ mol}$n(H2​SO4​)=5.00/98.1=0.05097 mol
• 1:1 → $n(\text{CaCO}_3) = 0.05097\text{ mol}$n(CaCO3​)=0.05097 mol
• $m = 0.05097 \times 100.1 =$m=0.05097×100.1= $5.10\text{ g}$5.10 g

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Limiting Reagents

When two reactants are present in non-stoichiometric amounts, one runs out first and dictates how much product can form — the limiting reagent. The other is in excess.

Method

1. Calculate moles of each reactant.
2. Divide each by its coefficient in the balanced equation.
3. The smallest result identifies the limiting reagent.
4. Use moles of the limiting reagent (with its ratio) to calculate product.

Worked Example 8 — Zn + HCl

$\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2$Zn+2HCl→ZnCl2​+H2​. $6.5\text{ g}$6.5 g Zn with $30.0\text{ cm}^3$30.0 cm3 of $2.00\text{ mol dm}^{-3}$2.00 mol dm−3 HCl.

• $n(\text{Zn}) = 6.5/65.4 = 0.0994\text{ mol}$n(Zn)=6.5/65.4=0.0994 mol; $n(\text{HCl}) = 2.00 \times 0.0300 = 0.0600\text{ mol}$n(HCl)=2.00×0.0300=0.0600 mol
• $\div$÷ coefficients: Zn → $0.0994/1 = 0.0994$0.0994/1=0.0994; HCl → $0.0600/2 = 0.0300$0.0600/2=0.0300
• HCl is smaller, so HCl is limiting.
• $n(\text{H}_2) = 0.0600/2 = 0.0300\text{ mol}$n(H2​)=0.0600/2=0.0300 mol; $m(\text{H}_2) = 0.0300 \times 2.0 =$m(H2​)=0.0300×2.0= $0.0600\text{ g}$0.0600 g

Worked Example 9 — Ca + O₂

$2\text{Ca} + \text{O}_2 \rightarrow 2\text{CaO}$2Ca+O2​→2CaO. $10.0\text{ g}$10.0 g Ca with $16.0\text{ g}$16.0 g O₂.

• $n(\text{Ca}) = 10.0/40.1 = 0.2494\text{ mol}$n(Ca)=10.0/40.1=0.2494 mol; $n(\text{O}_2) = 16.0/32.0 = 0.500\text{ mol}$n(O2​)=16.0/32.0=0.500 mol
• $\div$÷ coefficients: Ca → $0.1247$0.1247; O₂ → $0.500$0.500. Ca limiting.
• $n(\text{CaO}) = 0.2494\text{ mol}$n(CaO)=0.2494 mol; $m(\text{CaO}) = 0.2494 \times 56.1 =$m(CaO)=0.2494×56.1= $14.0\text{ g}$14.0 g

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Titration Calculations

A titration determines an unknown concentration by reaction with a known-concentration solution. OCR's classic question types are acid–alkali neutralisation.

Worked Example 10 — Standard 1:1 titration

$25.0\text{ cm}^3$25.0 cm3 NaOH requires $23.50\text{ cm}^3$23.50 cm3 of $0.100\text{ mol dm}^{-3}$0.100 mol dm−3 HCl. Find $c(\text{NaOH})$c(NaOH).

• $n(\text{HCl}) = 0.100 \times 0.02350 = 2.350 \times 10^{-3}\text{ mol}$n(HCl)=0.100×0.02350=2.350×10−3 mol
• 1:1 → $n(\text{NaOH}) = 2.350 \times 10^{-3}\text{ mol}$n(NaOH)=2.350×10−3 mol
• $c(\text{NaOH}) = 2.350 \times 10^{-3} / 0.0250 =$c(NaOH)=2.350×10−3/0.0250= $0.0940\text{ mol dm}^{-3}$0.0940 mol dm−3

Worked Example 11 — 2:1 titration (Na₂CO₃ vs HCl)

$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$Na2​CO3​+2HCl→2NaCl+H2​O+CO2​. $25.0\text{ cm}^3$25.0 cm3 Na₂CO₃ requires $18.75\text{ cm}^3$18.75 cm3 of $0.200\text{ mol dm}^{-3}$0.200 mol dm−3 HCl.

• $n(\text{HCl}) = 0.200 \times 0.01875 = 3.750 \times 10^{-3}\text{ mol}$n(HCl)=0.200×0.01875=3.750×10−3 mol
• $n(\text{Na}_2\text{CO}_3) = 3.750 \times 10^{-3} / 2 = 1.875 \times 10^{-3}\text{ mol}$n(Na2​CO3​)=3.750×10−3/2=1.875×10−3 mol
• $c(\text{Na}_2\text{CO}_3) = 1.875 \times 10^{-3} / 0.0250 =$c(Na2​CO3​)=1.875×10−3/0.0250= $0.0750\text{ mol dm}^{-3}$0.0750 mol dm−3

Worked Example 12 — Finding $M_r$Mr​ from titration

$1.59\text{ g}$1.59 g of an unknown diprotic acid $\text{H}_2\text{X}$H2​X is dissolved in water and made up to $250.0\text{ cm}^3$250.0 cm3. A $25.0\text{ cm}^3$25.0 cm3 aliquot requires $20.00\text{ cm}^3$20.00 cm3 of $0.100\text{ mol dm}^{-3}$0.100 mol dm−3 NaOH.

$\text{H}_2\text{X} + 2\text{NaOH} \rightarrow \text{Na}_2\text{X} + 2\text{H}_2\text{O}$H2​X+2NaOH→Na2​X+2H2​O

• $n(\text{NaOH}) = 0.100 \times 0.02000 = 2.00 \times 10^{-3}\text{ mol}$n(NaOH)=0.100×0.02000=2.00×10−3 mol
• $n(\text{H}_2\text{X})$n(H2​X) in aliquot $= 1.00 \times 10^{-3}\text{ mol}$=1.00×10−3 mol
• Total $n(\text{H}_2\text{X})$n(H2​X) in 250 cm³ flask $= 10 \times 1.00 \times 10^{-3} = 1.00 \times 10^{-2}\text{ mol}$=10×1.00×10−3=1.00×10−2 mol
• $M_r = m/n = 1.59 / 0.0100 =$Mr​=m/n=1.59/0.0100= $159$159

Worked Example 12b — Reverse stoichiometry (given product, find limiting reagent)

A reaction $2\text{Al} + 3\text{Cl}_2 \rightarrow 2\text{AlCl}_3$2Al+3Cl2​→2AlCl3​ produces $5.34\text{ g}$5.34 g of AlCl₃. Determine the minimum mass of Al and Cl₂ needed.

• $n(\text{AlCl}_3) = 5.34/133.5 = 0.0400\text{ mol}$n(AlCl3​)=5.34/133.5=0.0400 mol
• $n(\text{Al}) = (2/2)(0.0400) = 0.0400\text{ mol}$n(Al)=(2/2)(0.0400)=0.0400 mol; $m(\text{Al}) = 0.0400 \times 27.0 =$m(Al)=0.0400×27.0= $1.08\text{ g}$1.08 g
• $n(\text{Cl}_2) = (3/2)(0.0400) = 0.0600\text{ mol}$n(Cl2​)=(3/2)(0.0400)=0.0600 mol; $m(\text{Cl}_2) = 0.0600 \times 71.0 =$m(Cl2​)=0.0600×71.0= $4.26\text{ g}$4.26 g

Total mass of reactants = $1.08 + 4.26 = 5.34\text{ g}$1.08+4.26=5.34 g, matching the product mass — conservation of mass confirms the calculation.

Worked Example 12c — Concentration from $\text{g L}^{-1}$g L−1 to $\text{mol dm}^{-3}$mol dm−3

A solution contains $5.00\text{ g L}^{-1}$5.00 g L−1 of KOH. Find $c$c in $\text{mol dm}^{-3}$mol dm−3.

• $1\text{ L} = 1\text{ dm}^3$1 L=1 dm3, so the data is already in g per dm³.
• $M(\text{KOH}) = 56.1\text{ g mol}^{-1}$M(KOH)=56.1 g mol−1
• $c = 5.00/56.1 =$c=5.00/56.1= $0.0891\text{ mol dm}^{-3}$0.0891 mol dm−3

Practical context — making a primary standard

When a titration requires high accuracy (better than $\pm 0.1\%$±0.1%), the titrant is prepared from a primary standard — a substance of high purity, known composition, stable storage and (ideally) high $M_r$Mr​. Anhydrous sodium carbonate (Na₂CO₃) and potassium hydrogenphthalate (KHP, $M_r = 204.2$Mr​=204.2) are common primary standards.

Sodium hydroxide is not a primary standard because it absorbs water and CO₂ from the atmosphere. A NaOH solution made by direct weighing has a concentration that drifts within hours and cannot be relied on to 4 s.f. precision. Instead, the NaOH solution is prepared at approximate concentration and then standardised against KHP via titration; the measured concentration after standardisation is the primary-standard-derived value used in subsequent calculations.

The OCR PAG 2 mark scheme awards method marks for "uses a primary standard" or "standardises against a primary standard" — a discriminator that distinguishes a top-band practical write-up from a routine one.

Worked Example 13 — Back-titration

$2.00\text{ g}$2.00 g of impure MgCO₃ is dissolved in $50.0\text{ cm}^3$50.0 cm3 of $1.00\text{ mol dm}^{-3}$1.00 mol dm−3 HCl (excess). The unreacted HCl requires $22.5\text{ cm}^3$22.5 cm3 of $0.500\text{ mol dm}^{-3}$0.500 mol dm−3 NaOH. Find the % purity.

Step 1 — unreacted HCl (via NaOH):

• $n(\text{NaOH}) = 0.500 \times 0.0225 = 0.01125\text{ mol}$n(NaOH)=0.500×0.0225=0.01125 mol
• $n(\text{HCl unreacted}) = 0.01125\text{ mol}$n(HCl unreacted)=0.01125 mol (1:1)

Step 2 — HCl consumed by MgCO₃:

• $n(\text{HCl total}) = 1.00 \times 0.0500 = 0.0500\text{ mol}$n(HCl total)=1.00×0.0500=0.0500 mol
• $n(\text{HCl reacting with MgCO}_3) = 0.0500 - 0.01125 = 0.03875\text{ mol}$n(HCl reacting with MgCO3​)=0.0500−0.01125=0.03875 mol

Step 3 — moles of MgCO₃: $\text{MgCO}_3 + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\text{O} + \text{CO}_2$MgCO3​+2HCl→MgCl2​+H2​O+CO2​

• $n(\text{MgCO}_3) = 0.03875/2 = 0.01938\text{ mol}$n(MgCO3​)=0.03875/2=0.01938 mol
• $m(\text{MgCO}_3) = 0.01938 \times 84.3 = 1.633\text{ g}$m(MgCO3​)=0.01938×84.3=1.633 g

Step 4 — % purity: $(1.633/2.00) \times 100 =$(1.633/2.00)×100= $81.7\%$81.7%

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Worked Example 13b — Iodine-thiosulfate analysis of bleach

Commercial bleach (sodium chlorate(I), NaClO) is analysed by an iodine-displacement step followed by thiosulfate titration:

• Step 1: $\text{ClO}^- + 2\text{I}^- + 2\text{H}^+ \rightarrow \text{Cl}^- + \text{I}_2 + \text{H}_2\text{O}$ClO−+2I−+2H+→Cl−+I2​+H2​O (acid added to drive the reaction)
• Step 2: $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$I2​+2S2​O32−​→2I−+S4​O62−​ (titrate with thiosulfate to starch endpoint)

A 25.0 cm³ aliquot of bleach is treated with excess KI in acidic solution and the liberated iodine requires 23.50 cm³ of 0.100 mol dm⁻³ thiosulfate to titrate. Find the concentration of NaClO.

• $n(\text{S}_2\text{O}_3^{2-}) = 0.100 \times 0.02350 = 2.350 \times 10^{-3}\text{ mol}$n(S2​O32−​)=0.100×0.02350=2.350×10−3 mol
• $n(\text{I}_2) = 2.350 \times 10^{-3} / 2 = 1.175 \times 10^{-3}\text{ mol}$n(I2​)=2.350×10−3/2=1.175×10−3 mol
• $n(\text{ClO}^-) = n(\text{I}_2) = 1.175 \times 10^{-3}\text{ mol}$n(ClO−)=n(I2​)=1.175×10−3 mol (1:1 in step 1)
• $c(\text{ClO}^-) = 1.175 \times 10^{-3} / 0.0250 =$c(ClO−)=1.175×10−3/0.0250= $0.0470\text{ mol dm}^{-3}$0.0470 mol dm−3

This two-step indirect titration is a workhorse method for quantifying oxidising agents and is examined as a Module 5 (PAG 6) procedure.

Worked Example 13c — Multi-aliquot titration (precision)

A student performs a titration five times and obtains titres of 23.40, 22.95, 23.05, 23.10, 23.05 cm³. Calculate the mean concordant titre.

The 23.40 result is the rough titre — far outside the concordance tolerance ($\pm 0.10\text{ cm}^3$±0.10 cm3) of the other four. The remaining four titres (22.95, 23.05, 23.10, 23.05) span a range of 0.15 cm³, slightly exceeding the strict tolerance. To establish concordance, examine the closest cluster: 23.05, 23.10, 23.05 are within $\pm 0.05\text{ cm}^3$±0.05 cm3 and are concordant. The 22.95 titre is borderline; conservative practice discards it as well.

Mean of concordant titres $= (23.05 + 23.10 + 23.05) / 3 =$=(23.05+23.10+23.05)/3= $23.07\text{ cm}^3$23.07 cm3 (to 2 d.p.).

The discipline is: identify rough titre by inspection; group the remaining titres into a concordant set within $\pm 0.10\text{ cm}^3$±0.10 cm3; take the mean of the concordant set only. OCR's PAG 2 CPAC mark scheme rewards this discrimination explicitly.

Titration Practical Skills (Examinable)

OCR's mark schemes routinely test best-practice titration technique:

• Rinse the burette with the titrant before filling — water dilutes the titrant.
• Rinse the pipette with the solution being measured.
• White tile under the conical flask to see the colour change clearly.
• Swirl continuously during addition.
• Dropwise addition near the end point.
• Eye level + meniscus when reading the burette.
• Repeat until concordant titres (within $\pm 0.10\text{ cm}^3$±0.10 cm3) are obtained.
• Average only the concordant titres for the mean — exclude the rough titre.

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