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9476 / 2027
The Mole Concept and Stoichiometry

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The Mole Concept and Stoichiometry

Turn masses, gas volumes and solution measurements into amounts, then let a balanced equation connect them.

A-Level 9476 (2026-2027)

01

Relative masses compare with one common reference

The reference is one twelfth of the mass of a carbon-12 atom.

Four relative masses; none has a unit
QuantityMeaning
Relative isotopic massMass of one atom of a specified isotope divided by one twelfth of the mass of a carbon-12 atom.
Relative atomic mass, ArWeighted mean mass of an atom of the element divided by the same reference mass, for the stated isotopic composition.
Relative molecular mass, MrMass of one molecule divided by the same reference; equal to the sum of its constituent relative atomic masses.
Relative formula mass, MrSum of relative atomic masses in one formula unit; use this term for ionic solids, which have no discrete molecules.

The mass number of an isotope is an integer counting protons and neutrons. Its relative isotopic mass is a measured mass ratio and need not equal that integer exactly. For example, use the isotopic masses supplied in a question rather than silently replacing them with mass numbers.

A relative mass is dimensionless. A molar mass, M, is mass per mole and has a unit such as g mol-1. Their numerical values agree when M is expressed in g mol-1: CO2 has Mr = 44.0 and M = 44.0 g mol-1. For NaCl, say relative formula mass rather than relative molecular mass.

Check your understanding

Why is "relative formula mass" used for sodium chloride, and not "relative molecular mass"?

  1. NaCl has a mass in g mol-1, while relative molecular mass has no unit.
  2. NaCl is a giant ionic lattice with no separate molecules, so it has formula units instead.
  3. NaCl contains two elements, and relative molecular mass is used for single elements.
Show the answer

NaCl is a giant ionic lattice with no separate molecules, so it has formula units instead.

Relative molecular mass applies to discrete molecules. For an ionic solid, add the Ar values for one formula unit, NaCl.

02

A mole counts specified entities

Always name whether the entities are atoms, molecules, ions or formula units.

One mole contains exactly 6.02214076 × 1023 specified elementary entities. This fixes the Avogadro constant NA = 6.02214076 × 1023 mol-1. Use N = nNA and n = m/M. The amount n is measured in mol, not in grams.

Worked exampleOne formula can count several kinds of entity

What does 0.200 mol of CaCl2 represent before and after complete dissolution?

Try it first. Then show the worked stepsHide the worked steps
  1. The solid contains 0.200 mol of CaCl2 formula units, rather than CaCl2 molecules.
  2. Dissolution gives 0.200 mol Ca2+ and 0.400 mol Cl-, assuming complete dissolution and ignoring any minor ion association.
  3. The total amount of ions is 0.600 mol; multiply the amount of the requested entity by the Avogadro constant.
Answer

There are 1.20 × 1023 formula units in the solid sample, and about 3.61 × 1023 dissolved ions under the stated model.

For isotopes, Ar = Σ(isotopic mass × fractional abundance). Percent abundances must first be divided by 100; relative peak intensities can instead be divided by their total. A weighted mean must lie between the masses used and closer to the more abundant isotope.

Worked exampleUse relative abundance as the weighting

An element has isotopes of relative isotopic masses 79.0 and 81.0, with abundances 50.7% and 49.3%. Find its relative atomic mass.

Try it first. Then show the worked stepsHide the worked steps
  1. Convert the abundances to fractions 0.507 and 0.493; they sum to 1.
  2. Calculate 79.0(0.507) + 81.0(0.493) = 79.986.
  3. Round only the final answer to a precision consistent with the supplied masses.
Answer

Ar = 80.0 to three significant figures. The unrounded weighted value is slightly closer to 79.0 than to 81.0.

Check your understanding

Chlorine is 75.8% chlorine-35 (isotopic mass 35.0) and 24.2% chlorine-37 (isotopic mass 37.0). What is its Ar?

  1. 36.5, closer to the heavier isotope
  2. 36.0, the simple mean of the two
  3. 35.5, the abundance-weighted mean
Show the answer

35.5, the abundance-weighted mean

Ar = 35.0 x 0.758 + 37.0 x 0.242 = 26.53 + 8.95 = 35.5. It lies closer to 35.0 because that isotope is more abundant.

03

A formula records an atom ratio, not a mass ratio

Divide each mass by its atomic molar mass before comparing the numbers.

An empirical formula gives the simplest whole-number ratio of atoms of each element. A molecular formula gives the actual numbers in a molecule. Benzene has empirical formula CH and molecular formula C6H6. An ionic formula such as MgCl2 describes a formula-unit ratio, not a molecule.

From mass composition to molecular formula
  1. Choose a convenient sample

    For percentages, assume 100 g. Convert each elemental mass to moles using its atomic molar mass.

  2. Find the simplest ratio

    Divide all amounts by the smallest. If a value is near 1.5 or 1.33, multiply the whole ratio by 2 or 3; do not round these straight to integers.

  3. Use the independent molar mass

    Molecular multiplier = molecular molar mass / empirical-formula molar mass. Multiply every subscript by the same integer.

Worked exampleDistinguish the two formulae

A compound contains 40.0% C, 6.67% H and 53.3% O by mass. Its molar mass is 180 g mol-1. Use C = 12.0, H = 1.00 and O = 16.0.

Try it first. Then show the worked stepsHide the worked steps
  1. In 100 g: n(C) = 40.0/12.0 = 3.333; n(H) = 6.67/1.00 = 6.67; n(O) = 53.3/16.0 = 3.331 mol.
  2. Divide by the smallest to obtain approximately 1 : 2 : 1, so the empirical formula is CH2O.
  3. Empirical-formula molar mass = 30.0 g mol-1; 180/30.0 = 6.
Answer

Empirical formula CH2O; molecular formula C6H12O6. Composition alone cannot choose the multiplier.

Check your understanding

An analysis gives a mole ratio of Fe : O = 1 : 1.49. What is the empirical formula?

  1. FeO2, since 1.49 rounds up to give 1 : 2
  2. FeO, since 1.49 rounds down to give 1 : 1
  3. Fe2O3, as 1 : 1.49 is close to 2 : 3
Show the answer

Fe2O3, as 1 : 1.49 is close to 2 : 3

Multiply by two to get 2 : 2.98, close to 2 : 3. Small differences from whole numbers come from rounding or experimental error.

04

Combustion transfers carbon into carbon dioxide and hydrogen into water

Count atoms from products, then use mass difference only when the remaining element is known.

For complete combustion of a compound containing C and H, every mole of CO2 contains one mole of original carbon atoms, while every mole of H2O contains two moles of original hydrogen atoms. Thus n(C) = n(CO2) and n(H) = 2n(H2O). Oxygen in the products comes from both the compound and the oxygen supply, so product oxygen cannot directly give the original oxygen content.

Worked exampleFind oxygen by a justified mass difference

Complete combustion of 0.460 g of a compound containing only C, H and O produces 0.880 g CO2 and 0.540 g H2O. Its molar mass is 46.0 g mol-1.

Try it first. Then show the worked stepsHide the worked steps
  1. n(C) = 0.880/44.0 = 0.0200 mol; carbon mass = 0.240 g.
  2. n(H) = 2(0.540/18.0) = 0.0600 mol; hydrogen mass = 0.0600 g.
  3. Original oxygen mass = 0.460 - 0.240 - 0.0600 = 0.160 g; n(O) = 0.0100 mol.
  4. The ratio C : H : O is 2 : 6 : 1. Its formula mass is 46.0, so the molecular multiplier is 1.
Answer

C2H6O. Its complete-combustion equation is C2H6O + 3O2 → 2CO2 + 3H2O. The formula does not distinguish ethanol from methoxymethane.

This deduction assumes complete combustion and quantitative product collection. A water absorber can also collect moisture from the incoming gas unless the gas is dried. Incomplete combustion, escaping CO2 or a compound containing an unaccounted element makes the simple mass balance unreliable. State the assumptions when evaluating experimental data.

Check your understanding

A hydrocarbon burns completely to give 1.32 g of CO2 and 0.540 g of H2O. What is its empirical formula? (C = 12.0, H = 1.00, O = 16.0)

  1. CH, since n(C) = 0.0300 and n(H) = 0.0300
  2. CH2, since n(C) = 0.0300 and n(H) = 0.0600
  3. C2H, since n(C) = 0.0600 and n(H) = 0.0300
Show the answer

CH2, since n(C) = 0.0300 and n(H) = 0.0600

n(C) = n(CO2) = 1.32 / 44.0 = 0.0300 mol. n(H) = 2 x n(H2O) = 2 x 0.540 / 18.0 = 0.0600 mol. The ratio C : H = 1 : 2.

05

The balanced equation is the conversion between substances

Balance atoms and charge; change coefficients, never formula subscripts.

A balanced equation conserves each element and total charge. For an ionic equation, remove unchanged spectator ions, then check atoms and charge again. For example, CO32-(aq) + 2H+(aq) → CO2(g) + H2O(l) has zero net charge on both sides. Replacing CO3 with CO2 to balance oxygen would change the reacting species and is invalid.

Mass -> moles -> stoichiometric ratio -> required quantity
  1. Convert the measured amount

    Use n = m/M, n = cV, or pV = nRT as appropriate.

  2. Apply the balanced coefficients

    If aA reacts with bB, the mole ratio n(A)/a = n(B)/b holds for the amounts actually consumed.

  3. Convert back only at the end

    Find the requested mass, volume, concentration or particle count. Keep extra digits in intermediate calculations.

Worked exampleIdentify the limiting reagent before calculating yield

0.540 g Al reacts with 1.065 g Cl2. Find the maximum mass of AlCl3. Use Al = 27.0 and Cl = 35.5.

Try it first. Then show the worked stepsHide the worked steps
  1. Balance: 2Al + 3Cl2 → 2AlCl3.
  2. n(Al) = 0.540/27.0 = 0.0200 mol; n(Cl2) = 1.065/71.0 = 0.0150 mol.
  3. Compare amount/coefficient: 0.0200/2 = 0.0100; 0.0150/3 = 0.00500. Chlorine limits the reaction.
  4. n(AlCl3) = (2/3)(0.0150) = 0.0100 mol. M(AlCl3) = 133.5 g mol-1.
Answer

Maximum mass = 1.34 g to three significant figures. Only 0.0100 mol Al reacts, leaving 0.270 g Al. Theoretical yield assumes the specified reaction goes to completion without side reactions.

Try it step by step

Work it out in steps: Moles of the first reactant, then Moles of the second reactant, then Moles of product from the limiting reagent, then Maximum mass of product.

  1. Problem 1. 2.30 g of sodium reacts with 2.13 g of chlorine. 2Na + Cl2 -> 2NaCl. Find the maximum mass of NaCl, in g to 3 significant figures. Use Na = 23.0 and Cl = 35.5.
  2. Problem 2. 2.80 g of nitrogen reacts with 0.900 g of hydrogen. N2 + 3H2 -> 2NH3. Find the maximum mass of NH3, in g to 3 significant figures. Use N = 14.0 and H = 1.0.
  3. Problem 3. 1.20 g of magnesium burns in 0.640 g of oxygen. 2Mg + O2 -> 2MgO. Find the maximum mass of MgO. Give each answer to 3 significant figures. Use Mg = 24.0 and O = 16.0.
Show the answer

Problem 1.

Moles of the first reactant: n(Na) = 2.30 / 23.0 = 0.100 mol.

Moles of the second reactant: n(Cl2) = 2.13 / 71.0 = 0.0300 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: Na 0.100/2 = 0.0500; Cl2 0.0300/1 = 0.0300. Chlorine is limiting, so n(NaCl) = 2 x 0.0300 = 0.0600 mol.

Maximum mass of product: Mass = 0.0600 x 58.5 = 3.51 g. 0.0400 mol of Na is left over.

Problem 2.

Moles of the first reactant: n(N2) = 2.80 / 28.0 = 0.100 mol.

Moles of the second reactant: n(H2) = 0.900 / 2.0 = 0.450 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: N2 0.100/1 = 0.100; H2 0.450/3 = 0.150. Nitrogen is limiting, so n(NH3) = 2 x 0.100 = 0.200 mol.

Maximum mass of product: Mass = 0.200 x 17.0 = 3.40 g.

Problem 3.

Moles of the first reactant: n(Mg) = 1.20 / 24.0 = 0.0500 mol.

Moles of the second reactant: n(O2) = 0.640 / 32.0 = 0.0200 mol.

Moles of product from the limiting reagent: Compare amount / coefficient: Mg 0.0500/2 = 0.0250; O2 0.0200/1 = 0.0200. Oxygen is limiting, so n(MgO) = 2 x 0.0200 = 0.0400 mol.

Maximum mass of product: Mass = 0.0400 x 40.0 = 1.60 g.

If the isolated dry product is less than this maximum, percentage yield = actual yield / theoretical yield × 100%. Impure or wet product can give an apparent yield above 100%; this is evidence about the measurement, not extra creation of matter. Distinguish product yield from the percentage purity of a starting sample.

06

At the same temperature and pressure, gas volumes follow mole ratios

Water may condense; do not count a liquid as part of the final gas volume.

For gases behaving ideally, V is proportional to n at fixed temperature and pressure. The balanced coefficients therefore also give gas-volume ratios under those conditions. If conditions differ, use pV = nRT first. Do not insert a memorised molar gas volume when the stated temperature and pressure require a different value.

Complete combustion of a hydrocarbon obeys CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O. The carbon dioxide amount gives x; oxygen consumption then gives y. For an oxygen-containing fuel CxHyOz, the oxygen coefficient becomes x + y/4 - z/2.

Worked exampleInfer a hydrocarbon from gas-volume changes

At the same temperature and pressure, 10.0 cm3 of a gaseous hydrocarbon burns completely using 35.0 cm3 O2 and producing 20.0 cm3 CO2. Find the formula.

Try it first. Then show the worked stepsHide the worked steps
  1. The CO2/fuel volume ratio is 2, so x = 2.
  2. The O2/fuel ratio is 3.5, so x + y/4 = 3.5.
  3. Substitute x = 2: y/4 = 1.5 and y = 6.
Answer

C2H6; 2C2H6 + 7O2 → 4CO2 + 6H2O.

Try it step by step

Work it out in steps: Carbon atoms, x, then Hydrogen atoms, y, then Formula.

  1. Problem 1. At the same temperature and pressure, 10.0 cm3 of a gaseous hydrocarbon CxHy burns completely in 50.0 cm3 of O2 and gives 30.0 cm3 of CO2. Use CxHy + (x + y/4)O2 -> xCO2 + (y/2)H2O. Find x and y as whole numbers.
  2. Problem 2. At the same temperature and pressure, 20.0 cm3 of a gaseous hydrocarbon burns completely in 60.0 cm3 of O2 and gives 40.0 cm3 of CO2. Find x and y as whole numbers, then the formula.
  3. Problem 3. At the same temperature and pressure, 15.0 cm3 of a gaseous hydrocarbon burns completely in 97.5 cm3 of O2 and gives 60.0 cm3 of CO2. Find the number of carbon and hydrogen atoms as whole numbers, then the formula.
Show the answer

Problem 1.

Carbon atoms, x: CO2 / fuel = 30.0/10.0 = 3, so x = 3.

Hydrogen atoms, y: O2 / fuel = 50.0/10.0 = 5 = x + y/4, so y/4 = 2 and y = 8.

Formula: The hydrocarbon is C3H8, propane.

Problem 2.

Carbon atoms, x: CO2 / fuel = 40.0/20.0 = 2, so x = 2.

Hydrogen atoms, y: O2 / fuel = 60.0/20.0 = 3 = 2 + y/4, so y = 4.

Formula: The hydrocarbon is C2H4, ethene.

Problem 3.

Carbon atoms, x: CO2 / fuel = 60.0/15.0 = 4, so x = 4.

Hydrogen atoms, y: O2 / fuel = 97.5/15.0 = 6.5 = 4 + y/4, so y/4 = 2.5 and y = 10.

Formula: The hydrocarbon is C4H10, butane.

Check your understanding

In the example, 10.0 cm3 of fuel burns with 50.0 cm3 of O2. After cooling so the water condenses, what dry gas volume remains?

  1. 65.0 cm3: the CO2, the excess O2 and the steam made
  2. 20.0 cm3: just the CO2, since the oxygen is used up
  3. 35.0 cm3: 15.0 cm3 of unused O2 plus 20.0 cm3 of CO2
Show the answer

35.0 cm3: 15.0 cm3 of unused O2 plus 20.0 cm3 of CO2

Measured at the original temperature and pressure, the dry gas is the excess O2 and the CO2 made. Condensed water is liquid, so it is not counted. Absorbing the CO2 would leave 15.0 cm3.

07

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 exampleSeparate 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.

Try it first. Then show the worked stepsHide the worked steps
  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.

Try it step by step

Work it out in steps: Moles of titrant, then Moles in the aliquot, then Concentration of the diluted solution, then Concentration of the stock.

  1. Problem 1. 10.00 cm3 of stock hydrochloric acid is made up to 250.0 cm3. A 25.00 cm3 aliquot of the diluted acid needs 21.50 cm3 of 0.1000 mol dm-3 NaOH. HCl + NaOH -> NaCl + H2O. Find both concentrations in mol dm-3 to 4 significant figures.
  2. Problem 2. 20.00 cm3 of stock sulfuric acid is made up to 250.0 cm3. A 25.00 cm3 aliquot needs 18.40 cm3 of 0.1000 mol dm-3 NaOH. H2SO4 + 2NaOH -> Na2SO4 + 2H2O. Find the stock concentration in mol dm-3 to 4 significant figures.
  3. Problem 3. 50.00 cm3 of stock sodium carbonate solution is made up to 250.0 cm3. A 25.00 cm3 aliquot needs 22.80 cm3 of 0.1000 mol dm-3 HCl. Na2CO3 + 2HCl -> 2NaCl + CO2 + H2O. Give each answer to 4 significant figures.
Show the answer

Problem 1.

Moles of titrant: n(NaOH) = 0.1000 x 21.50/1000 = 0.002150 mol.

Moles in the aliquot: The ratio is 1:1, so n(HCl) in the aliquot = 0.002150 mol.

Concentration of the diluted solution: c(diluted) = 0.002150 / 0.02500 = 0.08600 mol dm-3.

Concentration of the stock: The dilution factor is 250.0/10.00 = 25, so c(stock) = 25 x 0.08600 = 2.150 mol dm-3.

Problem 2.

Moles of titrant: n(NaOH) = 0.1000 x 18.40/1000 = 0.001840 mol.

Moles in the aliquot: n(H2SO4) = 0.001840 / 2 = 0.0009200 mol.

Concentration of the diluted solution: c(diluted) = 0.0009200 / 0.02500 = 0.03680 mol dm-3.

Concentration of the stock: Dilution factor = 250.0/20.00 = 12.5, so c(stock) = 12.5 x 0.03680 = 0.4600 mol dm-3.

Problem 3.

Moles of titrant: n(HCl) = 0.1000 x 22.80/1000 = 0.002280 mol.

Moles in the aliquot: n(Na2CO3) = 0.002280 / 2 = 0.001140 mol.

Concentration of the diluted solution: c(diluted) = 0.001140 / 0.02500 = 0.04560 mol dm-3.

Concentration of the stock: Dilution factor = 250.0/50.00 = 5, so c(stock) = 5 x 0.04560 = 0.2280 mol dm-3.

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.

08

Use the measured difference to reveal an unknown ratio

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

Worked exampleBack 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.

Try it first. Then show the worked stepsHide the worked steps
  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 exampleInfer 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.

Try it first. Then show the worked stepsHide the worked steps
  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.

Check your understanding

In a back titration, a carbonate sample reacts with a known excess of HCl, and the leftover acid is titrated with NaOH. What does the NaOH titre measure?

  1. The acid left over, which is subtracted from the acid added
  2. The carbon dioxide given off, which is dissolved and neutralised by the NaOH
  3. The acid used by the carbonate, which gives the moles of carbonate directly
Show the answer

The acid left over, which is subtracted from the acid added

Moles of HCl used by the sample = moles of HCl added - moles left over. The NaOH titre gives the moles left over.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Exam definitions: learn these in mark-scheme words
TermDefinition to write
Relative atomic mass, ArThe weighted average mass of one atom of an element, relative to 1/12 of the mass of one atom of carbon-12.
Relative isotopic massThe mass of one atom of a particular isotope, relative to 1/12 of the mass of one atom of carbon-12.
Relative molecular mass, MrThe weighted average mass of one molecule of a substance, relative to 1/12 of the mass of one atom of carbon-12. Use relative formula mass for an ionic compound.
MoleThe amount of substance that contains 6.02 × 1023 elementary particles (the Avogadro constant, L).
A reliable calculation route
  1. Identify what was actually measured

    Mass, dry gas, a titre, an aliquot or a loss on heating; write units.

  2. Convert to moles and apply the equation

    Check the limiting reagent, coefficients, atom balance and charge balance.

  3. Return to the requested quantity

    Account for dilution, unreacted excess and significant figures; check whether the result is physically plausible.

Formula from mass: divide by atomic molar masses. Combustion: n(C) = n(CO2), n(H) = 2n(H2O). A gas-volume ratio equals a mole ratio only at matching temperature and pressure. An aliquot changes amount, not the concentration of the solution it came from.

Scope

Learning outcomes

6. The Mole Concept and Stoichiometry. Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 6(a) Define the four relative-mass quantities.

    • Relative atomic, isotopic, molecular and formula mass
    • Carbon-12 reference
    • Relative formula mass for ionic compounds

    Relative masses compare with one common reference

  2. 6(b) Define and use the mole through the Avogadro constant.

    • Specified entities
    • Avogadro constant and particle-number conversion

    A mole counts specified entities

  3. 6(c) Calculate weighted relative atomic mass.

    • Relative isotope abundances
    • Percent or relative-intensity weighting

    A mole counts specified entities

  4. 6(d) Distinguish empirical and molecular formulae.

    • Simplest whole-number ratio
    • Actual atom numbers in a molecule

    A formula records an atom ratio, not a mass ratio

  5. 6(e) Calculate formulae from composition or combustion.

    • Composition by mass
    • Combustion data
    • Independent molar mass for molecular multiplier

    A formula records an atom ratio, not a mass ratioCombustion transfers carbon into carbon dioxide and hydrogen into water

  6. 6(f) Construct balanced chemical equations.

    • Conservation of atoms and charge
    • Correct formulae and coefficients
    • Combustion equations

    Combustion transfers carbon into carbon dioxide and hydrogen into waterThe balanced equation is the conversion between substancesAt the same temperature and pressure, gas volumes follow mole ratios

  7. 6(g) Calculate reacting amounts in solids, gases and solutions.

    • (i) Reacting masses from formulae/equations
    • (ii) Gas volumes, including hydrocarbon combustion
    • (iii) Solution volumes and concentrations
    • Appropriate significant figures

    The balanced equation is the conversion between substancesAt the same temperature and pressure, gas volumes follow mole ratiosConcentration connects a measured volume with molesUse the measured difference to reveal an unknown ratio

  8. 6(h) Deduce stoichiometric relationships from measurements.

    • Use calculated reacting amounts to infer ratios
    • Gas-combustion and indirect mass/solution examples
    • Check experimental assumptions

    At the same temperature and pressure, gas volumes follow mole ratiosUse the measured difference to reveal an unknown ratio