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The Mole Concept and Stoichiometry

Topic 5 of 5

Solutions, titrations and inferred ratios

Track dilution, aliquots and the reagent left over.

A-Level 9476 (2026-2027)

Concentration connects a measured volume with moles

Use compatible units, then account for every dilution and aliquot.

c = n/V, with V in dm3 when c is in mol dm-3. Divide cm3 by 1000 before using n = cV. Dilution changes concentration and volume but preserves solute amount, provided no solute is lost or reacts: c1V1 = c2V2.

Worked example

Separate titration stoichiometry from dilution

25.00 cm3 of a diluted H2SO4 solution needs 23.60 cm3 of 0.1000 mol dm-3 NaOH for complete neutralisation. The diluted solution was prepared by making 25.00 cm3 stock acid up to 250.0 cm3. Find both acid concentrations.

  1. H2SO4 + 2NaOH → Na2SO4 + 2H2O; n(NaOH) = 0.1000(23.60/1000) = 0.002360 mol.
  2. The acid amount in the titrated aliquot is half this: 0.001180 mol.
  3. Diluted concentration = 0.001180/0.02500 = 0.04720 mol dm-3.
  4. The tenfold dilution makes the stock concentration ten times larger.
Answer

Diluted acid: 0.04720 mol dm-3; stock acid: 0.4720 mol dm-3. The two factors, 2 for reaction stoichiometry and 10 for dilution, have different origins.

An aliquot is a measured fraction of a well-mixed solution. If 25.00 cm3 is taken from 250.0 cm3, the whole flask contains ten times the aliquot amount. Do not multiply the aliquot concentration by ten to obtain the flask concentration: they are the same solution.

Report a final result to the significant figures justified by the measurements or requested by the question. Stoichiometric coefficients and exact unit conversions do not limit significant figures. A burette titre is the difference of two readings; its uncertainty depends on both readings.

Use the measured difference to reveal an unknown ratio

Back titration and loss of water each determine an amount indirectly.

Worked example

Back titration measures acid used by the sample

A 0.5000 g sample containing CaCO3 and an inert impurity reacts with 50.00 cm3 of 0.2000 mol dm-3 HCl. After all CO2 is removed, titrating the entire remaining mixture uses 20.00 cm3 of 0.1000 mol dm-3 NaOH. Find the percentage CaCO3, using M = 100.1 g mol-1.

  1. Initial HCl = 0.01000 mol. Remaining HCl = n(NaOH) = 0.002000 mol.
  2. HCl consumed by the carbonate = 0.008000 mol.
  3. CaCO3 + 2HCl → CaCl2 + CO2 + H2O, so n(CaCO3) = 0.004000 mol.
  4. CaCO3 mass = 0.4004 g; percentage = 0.4004/0.5000 × 100.
Answer

80.08% CaCO3. This assumes the impurity does not consume acid and the remaining acid is all included in the titration. A titrated aliquot would require a separate scaling factor.

Worked example

Infer water of crystallisation

Heating 2.50 g of CuSO4·xH2O to constant mass leaves 1.60 g anhydrous CuSO4. Use M(CuSO4) = 159.6 and M(H2O) = 18.0 g mol-1.

  1. Water lost = 0.90 g, giving 0.90/18.0 = 0.050 mol water.
  2. Anhydrous salt amount = 1.60/159.6 = 0.0100 mol.
  3. Water/salt ratio is approximately 4.99, consistent with x = 5.
Answer

CuSO4·5H2O, assuming the mass loss is only water and the salt itself does not decompose. Incomplete drying makes x too low; loss of solid or decomposition can make it too high.

For an unknown reaction, calculate the amounts actually consumed, divide by their smallest common amount, then infer coefficients. A balanced atom-and-charge check is still necessary. An observed amount ratio can be distorted by excess unreacted reagent, side reactions or incomplete collection; it is not automatically the stoichiometric ratio.