Stoichiometry — Science Sage

Review: The Mole and Key Formulas

The mole (mol) is the SI unit for amount of substance. One mole contains exactly 6.022 × 10²³ particles — this is Avogadro's number (Nₐ). It links the macroscopic world (grams, litres) to the microscopic world (atoms, molecules).

Formula	Variables	Use when…
n = m / M	n in mol, m in g, M in g·mol⁻¹ (molar mass from periodic table)	Converting between mass and moles of a pure substance
n = c × V	c in mol·L⁻¹, V in L	Working with solutions (titrations, dilutions)
n = V / Vm	Vm = 22.4 L·mol⁻¹ at STP (0°C, 101.3 kPa)	Gas volumes at standard temperature and pressure only

n = m / M

n = c × V

n = V / Vm

Concentration and Standard Solutions

Concentration (c) expresses how much solute is dissolved per litre of solution:

c = n / V (mol·L⁻¹, also written mol·dm⁻³)

Preparing a standard solution: Weigh the exact mass of pure solute (primary standard), dissolve it in a small volume of distilled water in a beaker, then transfer quantitatively into a volumetric flask and make up to the calibration mark. This gives a precisely known concentration.

Dilution: When more solvent is added, the number of moles of solute stays constant but the volume increases, so concentration decreases. Use:

c₁V₁ = c₂V₂

For very low concentrations, perform a serial dilution — dilute a dilution — to reach the desired concentration without large errors.

Unit conversion reminder: 1 L = 1 000 cm³ = 1 000 mL. Titration volumes are usually given in cm³ — divide by 1 000 to get litres before using n = cV. For example, 25.00 cm³ = 0.02500 L.

Volumetric Analysis — Titration

A titration determines the unknown concentration of one solution (the analyte) using a solution of known concentration (the titrant or standard solution).

Equipment and roles:

Burette — holds the titrant; delivers precise, variable volumes; read to ±0.05 cm³
Pipette — transfers an exact fixed volume of analyte into the flask
Conical flask (Erlenmeyer flask) — reaction vessel; easy to swirl without spilling
Indicator — added in drops to detect when the reaction is complete (the endpoint)

Acid-base titration procedure:

1. Rinse the burette with a small volume of titrant; rinse the pipette with analyte
2. Use the pipette to transfer the exact volume of analyte into the conical flask
3. Add 2–3 drops of indicator to the flask
4. Fill the burette with titrant and record the initial reading
5. Add titrant slowly, swirling constantly; slow to a drop at a time near the endpoint
6. Stop when the colour changes permanently; record the final burette reading
7. Titre volume = V_final − V_initial; repeat until concordant results (within 0.10 cm³)

Equivalence point — the exact stoichiometric point where moles of H⁺ = moles of OH⁻ (or more generally, where the reacting species are in exact stoichiometric ratio). Endpoint — when the indicator changes colour. These should coincide as closely as possible.

Indicator selection:

Strong acid + strong base: equivalence pH = 7; phenolphthalein or methyl orange both acceptable
Strong base + weak acid: equivalence pH > 7; use phenolphthalein (changes pH 8.2–10.0)
Strong acid + weak base: equivalence pH < 7; use methyl orange (changes pH 3.1–4.4)

Worked Example — Finding concentration by titration:

25.00 cm³ of HCl(aq) is titrated with 0.100 mol·L⁻¹ NaOH(aq). 22.50 cm³ of NaOH was needed to reach the endpoint.

Step 1: n(NaOH) = c × V = 0.100 × 0.02250 = 2.25 × 10⁻³ mol
Step 2: HCl + NaOH → NaCl + H₂O; mole ratio 1:1 → n(HCl) = 2.25 × 10⁻³ mol
Step 3: c(HCl) = n / V = 2.25 × 10⁻³ / 0.02500 = 0.0900 mol·L⁻¹

Back Titration

Used when the analyte reacts slowly with the titrant, does not react directly, or the endpoint is difficult to detect. The analyte is reacted with a known excess of reagent A, then the unreacted excess of A is titrated with standard solution C.

n(reacted with analyte) = n(total A added) − n(excess A titrated)

Worked Example — Purity of CaCO₃ in limestone:

Step 1: Add excess HCl (measured volume and concentration) to the limestone sample.
CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂

Step 2: Titrate the excess HCl with standardised NaOH; find n(excess HCl).

Step 3: n(HCl reacted with CaCO₃) = n(HCl total) − n(excess HCl)

Step 4: From the equation, n(CaCO₃) = ½ × n(HCl reacted)

Step 5: mass(CaCO₃) = n × 100.09 g·mol⁻¹; %purity = mass(CaCO₃)/mass(sample) × 100

Gravimetric Analysis

Gravimetric analysis determines the amount of an analyte by converting it into an insoluble precipitate of known composition, which is then filtered, dried, and weighed precisely.

Add a reagent that reacts with the analyte to form an insoluble precipitate
Filter using quantitative filter paper or a sintered glass crucible
Dry to constant mass (often heated/ignited in a furnace)
Weigh the precipitate; calculate moles using n = m/M
Use stoichiometry to find moles (and hence mass) of analyte

Example: To determine sulfate (SO₄²⁻) in a solution, add excess BaCl₂:
Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s) ↓
Weigh the dried BaSO₄ precipitate. Since the ratio is 1:1, n(SO₄²⁻) = n(BaSO₄) = mass(BaSO₄) / 233.4 g·mol⁻¹

Gas Stoichiometry

At STP (0°C = 273.15 K, 101.325 kPa), one mole of any ideal gas occupies 22.4 L·mol⁻¹ (the molar volume, V_m). For non-STP conditions, use the Ideal Gas Law:

PV = nRT R = 8.314 J·mol⁻¹·K⁻¹ T(K) = T(°C) + 273

Rearrangements:

n = PV / RT (find moles from pressure, volume, temperature)
P = nRT / V (find pressure)
V = nRT / P (find volume)

IMPORTANT unit requirements for PV = nRT:
P must be in Pascal (Pa): 1 kPa = 1 000 Pa; 1 atm = 101 325 Pa
V must be in cubic metres (m³): 1 L = 0.001 m³ = 1 × 10⁻³ m³
T must be in Kelvin (K): T(K) = T(°C) + 273
n in mol, R = 8.314 J·mol⁻¹·K⁻¹ (= Pa·m³·mol⁻¹·K⁻¹)

For a fixed amount of gas changing conditions, use the Combined Gas Law:

P₁V₁ / T₁ = P₂V₂ / T₂ (n constant)

Worked Example — Gas produced in a reaction:

2.50 g of CaCO₃ is heated: CaCO₃(s) → CaO(s) + CO₂(g)
What volume of CO₂ is produced at 25°C and 100 kPa? (M(CaCO₃) = 100.09 g·mol⁻¹)

Step 1: n(CaCO₃) = 2.50 / 100.09 = 0.02498 mol
Step 2: Mole ratio 1:1 → n(CO₂) = 0.02498 mol
Step 3: Convert: T = 25 + 273 = 298 K; P = 100 000 Pa
Step 4: V = nRT / P = (0.02498 × 8.314 × 298) / 100 000 = 6.19 × 10⁻⁴ m³ = 0.619 L

Percentage Purity

Real samples are rarely 100% pure. Percentage purity expresses what fraction of a sample is the desired substance:

% purity = (mass of pure substance / total mass of sample) × 100

The mass of the pure substance is usually found via titration (back or direct) or gravimetric analysis first. Percentage purity is always ≤ 100% and refers to mass percentage, not mole percentage.

IEB Extension: Potentiometric Titrations & Buffers

Potentiometric titration: A pH electrode measures the pH continuously as titrant is added. The data is plotted as a titration curve (pH vs. volume added).

Buffer region — flat part of curve where pH changes slowly; mixture of weak acid and conjugate base
Half-equivalence point — halfway to equivalence; [A⁻] = [HA]; pH = pKa
Equivalence point — steepest part / inflection point of the curve

Buffer solutions: A mixture of a weak acid (HA) and its conjugate base (A⁻) in similar amounts. They resist large pH changes on addition of small amounts of acid or base. Described by the Henderson-Hasselbalch equation:

pH = pKa + log([A⁻] / [HA])

EDTA complexometric titrations: EDTA (ethylenediaminetetraacetic acid) forms stable 1:1 complexes with metal ions regardless of charge. Used extensively to determine water hardness (Ca²⁺ and Mg²⁺ concentrations) with a metallochromic indicator such as Eriochrome Black T.

Virtual Titration — HCl(aq) vs NaOH(aq)

NaOH Volume Added

Volume (mL) 0.0 mL

HCl: 25.00 mL of 0.0900 mol·L⁻¹ | NaOH: 0.100 mol·L⁻¹

Indicator

Volume NaOH

0.0mL

1.05

Stage

Before EQ

Question 1 — NSC

25.00 cm³ of 0.200 mol·L⁻¹ HCl is titrated with 0.100 mol·L⁻¹ NaOH. What volume of NaOH is needed to reach the equivalence point?

n(HCl) = 0.200 × 0.02500 = 5.00 × 10⁻³ mol. Mole ratio 1:1. n(NaOH) needed = 5.00 × 10⁻³ mol. V = n/c = 5.00 × 10⁻³ / 0.100 = 0.0500 L = 50.00 cm³.

Question 2 — NSC

In a back titration, excess HCl is used to dissolve a CaCO₃ sample (molar mass 100 g·mol⁻¹). The excess HCl requires 18.00 cm³ of 0.500 mol·L⁻¹ NaOH to neutralise. If 40.00 cm³ of 0.500 mol·L⁻¹ HCl was originally added, what is n(CaCO₃)?

n(HCl total) = 0.500 × 0.040 = 0.020 mol. n(excess HCl) = 0.500 × 0.018 = 0.009 mol. n(HCl reacted with CaCO₃) = 0.020 − 0.009 = 0.011 mol. Since CaCO₃ + 2HCl → …, n(CaCO₃) = 0.011/2 = 0.0055 mol.

Question 3 — NSC

A gas occupies 2.00 L at 300 K and 150 kPa. How many moles of gas are present? (R = 8.314 J·mol⁻¹·K⁻¹)

PV = nRT → n = PV/RT = (150 000 × 0.002) / (8.314 × 300) = 300 / 2494.2 = 0.120 mol. Remember: P in Pa (150 kPa = 150 000 Pa), V in m³ (2.00 L = 0.002 m³).

Question 4 — NSC

A sample of impure iron ore contains 3.00 g of Fe₂O₃ in 5.00 g total. What is the percentage purity? (M(Fe₂O₃) = 160 g·mol⁻¹)

%purity = (3.00/5.00) × 100 = 60.0%. No molar mass calculation is needed here — percentage purity by mass is simply mass of pure compound ÷ total mass × 100.

Question 5 — NSC

In a gravimetric analysis, BaSO₄ precipitate (M = 233.4 g·mol⁻¹) weighing 0.4668 g is obtained. How many moles of SO₄²⁻ were in the original solution?

n(BaSO₄) = 0.4668/233.4 = 2.00 × 10⁻³ mol. Stoichiometry: Ba²⁺ + SO₄²⁻ → BaSO₄ (1:1 ratio), so n(SO₄²⁻) = n(BaSO₄) = 2.00 × 10⁻³ mol.

Question 6 — NSC

Which indicator is most suitable for a titration between a strong acid and a weak base (where the equivalence point is at pH ≈ 5)?

The indicator must change colour within the steep part of the titration curve near pH 5. Methyl orange changes too early (pH 3.1–4.4), phenolphthalein too late (pH 8.2–10.0), and litmus is not sharp enough. Bromocresol green (pH 3.8–5.4) spans the equivalence point well.

Question 7 — NSC

In the thermite reaction 2Al(s) + Fe₂O₃(s) → Al₂O₃(s) + 2Fe(l), 54.0 g of Al is reacted with 96.0 g of Fe₂O₃. (M(Al) = 27 g·mol⁻¹, M(Fe₂O₃) = 160 g·mol⁻¹, M(Fe) = 55.8 g·mol⁻¹) What mass of Fe is produced?

n(Al) = 54.0/27 = 2.00 mol. n(Fe₂O₃) = 96.0/160 = 0.600 mol. The equation requires 2 mol Al per 1 mol Fe₂O₃, so 0.600 mol Fe₂O₃ would need only 1.200 mol Al — we have 2.00 mol Al available (excess), so Fe₂O₃ is the limiting reagent. Using the 1:2 ratio of Fe₂O₃:Fe, n(Fe) = 2 × 0.600 = 1.200 mol. Mass = 1.200 × 55.8 = 67.0 g. (Wrongly assuming Al is limiting gives 111.6 g; forgetting the 1:2 ratio gives 33.5 g.)

Question 8 — NSC

In the Haber process: N₂(g) + 3H₂(g) → 2NH₃(g). 28.0 g of N₂ reacts with excess H₂. Only 28.9 g of NH₃ is actually obtained. What is the percentage yield? (M(N₂) = 28 g·mol⁻¹, M(NH₃) = 17 g·mol⁻¹)

n(N₂) = 28.0/28 = 1.00 mol. The mole ratio N₂:NH₃ = 1:2, so the theoretical n(NH₃) = 2.00 mol, giving a theoretical mass = 2.00 × 17 = 34.0 g. Percentage yield = (actual/theoretical) × 100 = (28.9/34.0) × 100 = 85.0%. Percentage yield always compares the actual mass obtained to the theoretical maximum calculated from the limiting reagent — since H₂ is in excess here, N₂ determines the theoretical yield.

IEB Extension Questions

Question 7 — IEB

A buffer solution contains 0.10 mol·L⁻¹ acetic acid (pKa = 4.74) and 0.20 mol·L⁻¹ sodium acetate. What is the pH? (Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]))

pH = 4.74 + log(0.20/0.10) = 4.74 + log(2) = 4.74 + 0.301 = 5.04. When [A⁻] > [HA], pH > pKa — the solution is more basic than the pure acid.

Question 8 — IEB

A potentiometric titration of a weak acid with strong base shows the half-equivalence point at pH = 4.2. This means the pKa of the weak acid is:

At the half-equivalence point, exactly half the acid has been neutralised, so [A⁻] = [HA]. Therefore pH = pKa + log(1) = pKa + 0 = pKa = 4.2.

These questions require full working. Show all steps: write the formula, substitute values with units, calculate the answer, and include the unit in your final answer. Concordant titration results should agree to within 0.10 cm³.

Question 1 [4 marks]

In a titration, 20.00 cm³ of H₂SO₄ of unknown concentration reacts exactly with 32.00 cm³ of 0.150 mol·L⁻¹ NaOH solution. The equation is:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

Calculate the concentration of the H₂SO₄. Show all working.

Question 2 [6 marks]

A limestone sample of mass 2.000 g is treated with 50.0 cm³ of 1.00 mol·L⁻¹ HCl (excess). The excess acid requires 22.0 cm³ of 0.500 mol·L⁻¹ NaOH to neutralise.

Calculate:
(a) n(HCl total)
(b) n(excess HCl)
(c) n(HCl that reacted with limestone)
(d) n(CaCO₃) [CaCO₃ + 2HCl → CaCl₂ + H₂O + CO₂]
(e) Mass of CaCO₃ (M = 100.09 g·mol⁻¹)
(f) Percentage purity of the limestone sample

Question 3 [4 marks]

0.500 mol of nitrogen gas (N₂) is sealed in a container at 27°C and 200 kPa. (R = 8.314 J·mol⁻¹·K⁻¹)

(a) What volume does it occupy? (Express answer in litres.)
(b) If the gas is heated to 127°C at constant volume, what is the new pressure?

Question 4 [3 marks]

An iron ore sample is analysed by gravimetric analysis. Fe²⁺ ions in solution are precipitated as Fe(OH)₂. The dried precipitate has a mass of 0.810 g.
(M(Fe(OH)₂) = 89.86 g·mol⁻¹; M(Fe) = 55.85 g·mol⁻¹)

(a) Calculate n(Fe(OH)₂)
(b) State n(Fe²⁺) with reasoning
(c) Calculate the mass of Fe in the sample

Question 5 [4 marks]

Explain the difference between the equivalence point and the endpoint of a titration. Why should these be as close as possible, and how does the choice of indicator affect this? In your answer, refer to the pH ranges of at least two named indicators.

Question 6 [7 marks]

25.00 cm³ of NaOH solution of unknown concentration is titrated against 0.100 mol·L⁻¹ HCl. A rough run is done first, followed by two accurate runs. The burette readings are recorded below:

Run	Initial reading (cm³)	Final reading (cm³)	Titre (cm³)
1 (rough)	0.00	27.80	?
2	0.50	27.85	?
3	0.30	27.65	?

(a) Complete the table by calculating the titre (final − initial reading) for each run. (2 marks)
(b) Two of the three titres are concordant (agree to within 0.10 cm³). Identify which two runs these are, and explain why the rough run (Run 1) should be excluded from the average. (2 marks)
(c) Calculate the average titre using only the concordant runs. (1 mark)
(d) Using the equation HCl + NaOH → NaCl + H₂O, calculate the concentration of the NaOH solution. (2 marks)

Review: The Mole and Key Formulas

Concentration and Standard Solutions

Volumetric Analysis — Titration

Back Titration

Gravimetric Analysis

Gas Stoichiometry

Percentage Purity

Virtual Titration — HCl(aq) vs NaOH(aq)

Ideal Gas Law Calculator — PV = nRT