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K326 / K328 / 2027
Chemical Calculations

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Chemical Calculations

Turn formulae, measurements and equations into defensible numerical answers.

O-Level 5086 / 5088 (2026) / SEC G3 K326 / K328 (2027)

01

Read and construct formulae

Subscripts count atoms; ion charges determine ratios.

An element symbol begins with a capital letter; a second letter is lower-case. Co is cobalt, while CO is carbon monoxide. In H2SO4, one formula unit contains 2 H, 1 S and 4 O atoms. Brackets multiply everything inside: Ca(OH)2 contains 1 Ca, 2 O and 2 H. A coefficient multiplies the entire formula: 3H2O represents six H atoms and three O atoms.

Worked exampleFrom atom counts to a formula

A model molecule contains two nitrogen atoms and four oxygen atoms. What is its formula, and how many oxygen atoms are in three molecules?

Try it first. Then show the worked stepsHide the worked steps
  1. Write N and O with the atom counts as subscripts: N2O4.
  2. Three molecules contain 3 x 4 = 12 oxygen atoms. Do not simplify N2O4 to NO2 when the actual atoms in one molecule are specified.
Answer

N2O4; 12 oxygen atoms in three molecules. A coefficient counts complete molecules; a subscript counts atoms within one.

A working symbol and formula bank
FamilySymbols or formulae
Common non-metalsH, C, N, O, F, P, S, Cl, Br, I; elemental gases H2, N2, O2, Cl2
Metals and other elementsLi, Na, K, Mg, Ca, Al, Zn, Fe, Cu, Ag, Pb; Si, He, Ne, Ar
Acids, alkalis and gasesHCl, HNO3, H2SO4, NaOH, Ca(OH)2, NH3, CO, CO2, SO2, NO, NO2, O3
Common salts and reagentsNaCl, MgCl2, CaCO3, CuSO4, AgNO3, Ba(NO3)2, KI, KMnO4
Oxides and organic examplesCaO, MgO, CuO, SiO2; CH4, C2H4, C2H5OH, CH3COOH

An ionic formula must have zero total charge. Al3+ and O2- balance at +6 and -6, giving Al2O3. Mg2+ and NO3- give Mg(NO3)2. Keep a polyatomic ion intact; use brackets only when more than one is needed.

Frequently used ion charges
Positive ionsNegative ions
Na+, K+, Ag+, NH4+Cl-, Br-, I-, OH-, NO3-
Mg2+, Ca2+, Zn2+, Cu2+, Fe2+, Pb2+O2-, CO32-, SO42-
Al3+, Fe3+A Roman numeral gives a metal ion oxidation state; distinguish iron(II) from iron(III).

Check your understanding

A compound has the formula XCl3, and each chloride ion is Cl-. What is the charge on X?

  1. 3+, because three Cl- ions give -3 and the compound is neutral
  2. 3-, because the subscript 3 is the charge on the X ion
  3. 1+, because a metal ion in a chloride is like Na+ in NaCl
Show the answer

3+, because three Cl- ions give -3 and the compound is neutral

Charges in an ionic compound add up to zero. Three Cl- ions contribute -3, so the one X ion must be X3+.

02

Balance atoms, then account for ions

A reaction changes arrangement, not the total number of atoms.

Write correct formulae first, then change coefficients until each element has the same atom count on both sides. Never alter a subscript to balance an equation. State symbols distinguish solid (s), liquid (l), gas (g) and aqueous solution (aq). Aqueous means dissolved in water, not simply liquid.

Worked exampleBalance combustion

Write the complete combustion of propane.

Try it first. Then show the worked stepsHide the worked steps
  1. Start C3H8(g) + O2(g) -> CO2(g) + H2O(l).
  2. Balance carbon: 3CO2. Balance hydrogen: 4H2O.
  3. Products contain ten O atoms, requiring 5O2.
Answer

C3H8(g) + 5O2(g) -> 3CO2(g) + 4H2O(l). Water is shown after cooling; in hot exhaust it is vapour.

An ionic equation removes spectator ions that remain unchanged in solution. Split dissolved ionic compounds into ions; keep solids, gases, water and undissociated molecular substances intact. Cancel identical ions on both sides and check both atoms and total charge.

Worked exampleFind the reacting ions

Aqueous silver nitrate reacts with aqueous sodium chloride.

Try it first. Then show the worked stepsHide the worked steps
  1. AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq).
  2. Dissolved reactants provide Ag+, NO3-, Na+ and Cl-.
  3. Na+ and NO3- remain dissolved: cancel them.
Answer

Ag+(aq) + Cl-(aq) -> AgCl(s). Total charge is zero on both sides.

Check your understanding

To balance the formation of magnesium oxide, a student writes Mg + O2 -> MgO2. Is this acceptable?

  1. Yes. Both sides now have one magnesium atom and two oxygen atoms.
  2. No. It changes the formula of the product; write 2Mg + O2 -> 2MgO.
  3. No. Oxygen should be written as O, so the equation is Mg + O -> MgO.
Show the answer

No. It changes the formula of the product; write 2Mg + O2 -> 2MgO.

Balance by changing the numbers in front of formulae, never the subscripts. Magnesium oxide is MgO (Mg2+ and O2-), so the equation is 2Mg(s) + O2(g) -> 2MgO(s).

Check your understanding

Aqueous barium chloride is added to aqueous sodium sulfate, and a white precipitate forms. Which is the ionic equation?

  1. Na+(aq) + Cl-(aq) -> NaCl(s)
  2. Ba2+(aq) + SO42-(aq) -> BaSO4(s)
  3. BaCl2(aq) + Na2SO4(aq) -> BaSO4(s)
Show the answer

Ba2+(aq) + SO42-(aq) -> BaSO4(s)

Write only the ions that form the precipitate. Na+ and Cl- stay in solution unchanged, so they are spectator ions. The charges balance (2+ and 2- give 0).

03

The mole is a counting unit

Connect microscopic particles to measurable mass.

Make a guess. You are not marked.

You weigh out 12 g of carbon and 12 g of magnesium. Use C = 12 and Mg = 24. Which sample contains more atoms?

  1. Both contain the same number of atoms
  2. The 12 g of magnesium, as its atoms are bigger
  3. The 12 g of carbon, as its atoms are lighter
Show the answer

The 12 g of carbon, as its atoms are lighter

Carbon atoms are half as heavy, so 12 g holds 1 mol of carbon atoms but only 0.5 mol of magnesium atoms.

Relative atomic mass, Ar, compares the average mass of an atom with one twelfth of the mass of a carbon-12 atom. Relative molecular mass, Mr, makes the same comparison for a molecule. Add the Ar values in its formula. For an ionic compound, call this relative formula mass instead. These relative masses have no units.

One mole contains the Avogadro constant number of specified entities: approximately 6.02 x 1023. Specify atoms, molecules, ions or formula units. The molar mass M has units g mol-1 and is numerically equal to the relative molecular or formula mass. Use n = m/M and particle count = n x NA.

Worked exampleAccount for brackets in a formula

Find the relative formula mass of Ca(OH)2 and the amount in 3.70 g. Use Ca = 40, O = 16, H = 1.

Try it first. Then show the worked stepsHide the worked steps
  1. Mr = 40 + 2(16 + 1) = 74.
  2. Molar mass = 74 g mol-1.
  3. n = 3.70/74 = 0.0500 mol.
Answer

0.0500 mol of Ca(OH)2 formula units; if fully dissolved, this produces 0.100 mol of OH- ions.

Try it step by step

Work it out in steps: Molar mass, then Amount in moles.

  1. Problem 1. Find the amount of sodium carbonate, Na2CO3, in 5.30 g. Use Na = 23, C = 12, O = 16. Give the amount in mol to 3 significant figures.
  2. Problem 2. Find the amount of magnesium sulfate, MgSO4, in 6.00 g. Use Mg = 24, S = 32, O = 16. Give the amount in mol to 3 significant figures.
  3. Problem 3. Find the amount of calcium nitrate, Ca(NO3)2, in 4.10 g. Use Ca = 40, N = 14, O = 16. Give the molar mass in g mol-1 and the amount in mol to 3 significant figures.
Show the answer

Problem 1.

Molar mass: M(Na2CO3) = 2(23) + 12 + 3(16) = 106 g mol-1. Count every atom in the formula: two Na, one C and three O.

Amount in moles: n = m / M = 5.30 / 106 = 0.0500 mol of Na2CO3 formula units.

Problem 2.

Molar mass: M(MgSO4) = 24 + 32 + 4(16) = 120 g mol-1.

Amount in moles: n = m / M = 6.00 / 120 = 0.0500 mol. Write 0.0500, not 0.05, to show 3 significant figures.

Problem 3.

Molar mass: M(Ca(NO3)2) = 40 + 2(14 + 3 x 16) = 40 + 2(62) = 164 g mol-1. The 2 outside the bracket doubles everything inside it.

Amount in moles: n = m / M = 4.10 / 164 = 0.0250 mol of Ca(NO3)2.

Check your understanding

How many moles of oxygen atoms are in 0.25 mol of O2 molecules?

  1. 0.50 mol of oxygen atoms, since each molecule has two
  2. 0.125 mol of oxygen atoms, half the molecules
  3. 0.25 mol of oxygen atoms, the same as the molecules
Show the answer

0.50 mol of oxygen atoms, since each molecule has two

Each O2 molecule holds two O atoms, so 0.25 x 2 = 0.50 mol of atoms. The amount of molecules is still 0.25 mol; always say which particle you are counting.

04

Use the equation ratio

Convert to moles before comparing reactants.

Make a guess. You are not marked.

3.60 g of magnesium burns completely in oxygen to make magnesium oxide. How does the mass of magnesium oxide compare with 3.60 g?

  1. It is less than 3.60 g, as some burns away
  2. It is exactly 3.60 g, as mass is conserved
  3. It is more than 3.60 g, as oxygen is added
Show the answer

It is more than 3.60 g, as oxygen is added

The oxide holds all the magnesium plus the oxygen it took from the air, so the product is heavier. The worked example below finds 6.00 g.

A reliable calculation path
  1. Write and balance the reaction

    The coefficients give a mole ratio, not a mass ratio.

  2. Convert the known quantity to moles

    Mass: n = m/M. Gas at room temperature and pressure: n = V/24 with V in dm3.

  3. Apply the mole ratio

    For aA -> bB, n(B) = n(A) x b/a.

  4. Convert to the requested quantity

    Use m = nM or V = 24n at room temperature and pressure.

Worked exampleMass to gas volume

Excess hydrochloric acid reacts with 2.40 g Mg. Calculate hydrogen volume at room temperature and pressure; Mg = 24.

Try it first. Then show the worked stepsHide the worked steps
  1. Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g).
  2. n(Mg) = 2.40/24 = 0.100 mol; the Mg:H2 ratio is 1:1.
  3. V(H2) = 0.100 x 24 = 2.40 dm3.
Answer

2.40 dm3, or 2400 cm3. The 24 dm3 mol-1 value applies at room temperature and pressure.

Try it step by step

Work it out in steps: Moles of the solid, then Moles of gas from the ratio, then Volume of gas.

  1. Problem 1. Excess dilute sulfuric acid reacts with 1.30 g of zinc. Zn(s) + H2SO4(aq) -> ZnSO4(aq) + H2(g). Find the volume of hydrogen at room temperature and pressure, in dm3 to 3 significant figures. Use Zn = 65.
  2. Problem 2. Excess hydrochloric acid reacts with 5.00 g of calcium carbonate. CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + H2O(l) + CO2(g). Find the volume of carbon dioxide at room temperature and pressure, in dm3 to 3 significant figures. Use M(CaCO3) = 100 g mol-1.
  3. Problem 3. Excess hydrochloric acid reacts with 0.540 g of aluminium. 2Al(s) + 6HCl(aq) -> 2AlCl3(aq) + 3H2(g). Find the volume of hydrogen at room temperature and pressure. Give each answer to 3 significant figures. Use Al = 27.
Show the answer

Problem 1.

Moles of the solid: n(Zn) = 1.30 / 65 = 0.0200 mol.

Moles of gas from the ratio: The equation has 1 Zn to 1 H2, so n(H2) = 0.0200 mol.

Volume of gas: V = n x 24 = 0.0200 x 24 = 0.480 dm3 at room temperature and pressure.

Problem 2.

Moles of the solid: n(CaCO3) = 5.00 / 100 = 0.0500 mol.

Moles of gas from the ratio: The equation has 1 CaCO3 to 1 CO2, so n(CO2) = 0.0500 mol. The 2 in front of HCl does not matter here, because the acid is in excess.

Volume of gas: V = 0.0500 x 24 = 1.20 dm3.

Problem 3.

Moles of the solid: n(Al) = 0.540 / 27 = 0.0200 mol.

Moles of gas from the ratio: The equation has 2 Al to 3 H2, so n(H2) = 0.0200 x 3/2 = 0.0300 mol.

Volume of gas: V = 0.0300 x 24 = 0.720 dm3.

Worked exampleMass to mass through the mole ratio

What mass of MgO forms when 3.60 g Mg burns completely in excess oxygen? Use Mg = 24 and O = 16.

Try it first. Then show the worked stepsHide the worked steps
  1. 2Mg(s) + O2(g) -> 2MgO(s). The Mg:MgO ratio is 2:2, or 1:1.
  2. n(Mg) = 3.60/24 = 0.150 mol, so n(MgO) = 0.150 mol.
  3. M(MgO) = 24 + 16 = 40 g mol-1; mass = 0.150 x 40 = 6.00 g.
Answer

6.00 g MgO. The extra 2.40 g is oxygen taken from the air; the 1:1 mole ratio does not mean equal masses.

Worked exampleIdentify the limiting reactant

Mix 0.080 mol Mg with 0.100 mol HCl. How much H2 forms?

Try it first. Then show the worked stepsHide the worked steps
  1. The equation needs two moles of HCl for each mole of Mg.
  2. 0.080 mol Mg would require 0.160 mol HCl; only 0.100 mol is available. HCl is limiting.
  3. n(H2) = 0.100/2 = 0.050 mol. Mg used = 0.050 mol; Mg remaining = 0.030 mol.
Answer

0.050 mol hydrogen, equivalent to 1.20 dm3 at room temperature and pressure.

Check your understanding

Can 0.10 mol of Mg react completely with 0.10 mol of HCl?

  1. No. Mg + 2HCl needs 0.20 mol HCl, so HCl runs out and 0.050 mol Mg is left.
  2. No. HCl is in excess, because one mole of acid gives two moles of H+.
  3. Yes. The amounts are equal, so each reactant is used up at the same time.
Show the answer

No. Mg + 2HCl needs 0.20 mol HCl, so HCl runs out and 0.050 mol Mg is left.

The equation needs 2 mol of HCl for each mol of Mg. 0.10 mol HCl reacts with 0.050 mol Mg, so HCl is limiting and 0.050 mol Mg remains.

05

Concentration and titration

Volume tells you moles only when concentration is known.

Molar concentration c = n/V, with V in dm3. Hence n = cV. Mass concentration in g dm-3 = mass/volume; dividing this by molar mass converts to mol dm-3. Always convert cm3 to dm3 by dividing by 1000 before using n = cV.

Worked exampleA titration with a 1:2 ratio

25.0 cm3 of sulfuric acid requires 20.0 cm3 of 0.150 mol dm-3 NaOH. Find the acid concentration.

Try it first. Then show the worked stepsHide the worked steps
  1. H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l).
  2. n(NaOH) = 0.150 x 0.0200 = 0.00300 mol.
  3. n(acid) = 0.00300/2 = 0.00150 mol.
  4. c(acid) = 0.00150/0.0250 = 0.0600 mol dm-3.
Answer

0.0600 mol dm-3. A direct cV = cV shortcut would miss the factor of two.

Try it step by step

Work it out in steps: Moles of the known solution, then Moles from the ratio, then Unknown concentration.

  1. Problem 1. 20.0 cm3 of hydrochloric acid is neutralised by 22.0 cm3 of 0.100 mol dm-3 sodium hydroxide. NaOH(aq) + HCl(aq) -> NaCl(aq) + H2O(l). Find the concentration of the acid, in mol dm-3 to 3 significant figures.
  2. Problem 2. 25.0 cm3 of sodium hydroxide is neutralised by 12.5 cm3 of 0.0800 mol dm-3 sulfuric acid. H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l). Find the concentration of the sodium hydroxide, in mol dm-3 to 3 significant figures.
  3. Problem 3. 20.0 cm3 of sodium carbonate solution needs 16.0 cm3 of 0.150 mol dm-3 hydrochloric acid. Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). Find the concentration of the sodium carbonate. Give each answer to 3 significant figures.
Show the answer

Problem 1.

Moles of the known solution: n(NaOH) = c x V = 0.100 x 22.0/1000 = 0.00220 mol. Divide cm3 by 1000 to get dm3 first.

Moles from the ratio: The ratio is 1 NaOH to 1 HCl, so n(HCl) = 0.00220 mol.

Unknown concentration: c(HCl) = n / V = 0.00220 / (20.0/1000) = 0.110 mol dm-3.

Problem 2.

Moles of the known solution: n(H2SO4) = 0.0800 x 12.5/1000 = 0.00100 mol.

Moles from the ratio: The ratio is 1 H2SO4 to 2 NaOH, so n(NaOH) = 2 x 0.00100 = 0.00200 mol.

Unknown concentration: c(NaOH) = 0.00200 / (25.0/1000) = 0.0800 mol dm-3.

Problem 3.

Moles of the known solution: n(HCl) = 0.150 x 16.0/1000 = 0.00240 mol.

Moles from the ratio: The ratio is 1 Na2CO3 to 2 HCl, so n(Na2CO3) = 0.00240 / 2 = 0.00120 mol.

Unknown concentration: c(Na2CO3) = 0.00120 / (20.0/1000) = 0.0600 mol dm-3.

Worked exampleA titration with a 250/25 dilution step

2.65 g of anhydrous sodium carbonate, Na2CO3, is dissolved in water and made up to 250 cm3 in a volumetric flask. 25.0 cm3 of this solution needs 22.50 cm3 of hydrochloric acid for neutralisation. Calculate the concentration of the acid. (Mr of Na2CO3 = 106)

Try it first. Then show the worked stepsHide the worked steps
  1. Equation: Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g).
  2. Moles of Na2CO3 in the flask = 2.65 ÷ 106 = 0.0250 mol, in 250 cm3.
  3. Only 25.0 cm3 of the 250 cm3 was titrated, one tenth: 0.0250 × 25.0/250 = 0.00250 mol.
  4. Moles of HCl = 2 × 0.00250 = 0.00500 mol.
  5. Concentration of HCl = 0.00500 ÷ (22.50/1000) = 0.222 mol/dm3 (3 s.f.).
Answer

0.222 mol/dm3. The step students miss is the 25.0/250 factor: the titration uses only one tenth of the solution in the flask.

Try it step by step

Work it out in steps: Moles in the flask, then Moles in the portion titrated, then Moles of acid, then Acid concentration.

  1. Problem 1. 1.06 g of anhydrous sodium carbonate is dissolved and made up to 250 cm3. 25.0 cm3 of this solution needs 20.00 cm3 of hydrochloric acid. Na2CO3(aq) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g). Find the acid concentration, in mol dm-3 to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
  2. Problem 2. 3.18 g of anhydrous sodium carbonate is made up to 250 cm3. 25.0 cm3 of this solution needs 24.00 cm3 of hydrochloric acid. Use the same equation. Find the acid concentration, in mol dm-3 to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
  3. Problem 3. 2.12 g of anhydrous sodium carbonate is made up to 250 cm3. 25.0 cm3 of this solution needs 20.00 cm3 of hydrochloric acid. Use the same equation. Give each answer to 3 significant figures. Use M(Na2CO3) = 106 g mol-1.
Show the answer

Problem 1.

Moles in the flask: n(Na2CO3) in the flask = 1.06 / 106 = 0.0100 mol, in 250 cm3.

Moles in the portion titrated: Only 25.0 cm3 of the 250 cm3 was titrated: 0.0100 x 25.0/250 = 0.00100 mol.

Moles of acid: The ratio is 1 Na2CO3 to 2 HCl, so n(HCl) = 2 x 0.00100 = 0.00200 mol.

Acid concentration: c(HCl) = 0.00200 / (20.00/1000) = 0.100 mol dm-3.

Problem 2.

Moles in the flask: n(Na2CO3) in the flask = 3.18 / 106 = 0.0300 mol.

Moles in the portion titrated: In the 25.0 cm3 portion: 0.0300 x 25.0/250 = 0.00300 mol.

Moles of acid: n(HCl) = 2 x 0.00300 = 0.00600 mol.

Acid concentration: c(HCl) = 0.00600 / (24.00/1000) = 0.250 mol dm-3.

Problem 3.

Moles in the flask: n(Na2CO3) in the flask = 2.12 / 106 = 0.0200 mol.

Moles in the portion titrated: In the 25.0 cm3 portion: 0.0200 x 25.0/250 = 0.00200 mol.

Moles of acid: n(HCl) = 2 x 0.00200 = 0.00400 mol.

Acid concentration: c(HCl) = 0.00400 / (20.00/1000) = 0.200 mol dm-3.

In a real titration, calculate titre from final minus initial burette reading. Repeat until sufficiently close titres are obtained and average concordant results, excluding the rough run. For example, 19.85 and 19.95 cm3 average to 19.90 cm3. Use the supplied equation for an unfamiliar reaction; the calculation path stays the same.

Check your understanding

A solution contains 8.0 g of NaOH per dm3. What is its concentration in mol dm-3? (M = 40 g mol-1)

  1. 320 mol dm-3, since 8.0 x 40 = 320
  2. 0.20 mol dm-3, since 8.0 / 40 = 0.20
  3. 5.0 mol dm-3, since 40 / 8.0 = 5.0
Show the answer

0.20 mol dm-3, since 8.0 / 40 = 0.20

Divide the mass concentration by the molar mass: (8.0 g dm-3) / (40 g mol-1) = 0.20 mol dm-3. The grams cancel and leave mol dm-3.

Exam-style question

10.0 cm3 of a concentrated sulfuric acid solution is diluted with water to 250 cm3. 25.0 cm3 of the diluted acid is neutralised by 20.00 cm3 of 0.100 mol/dm3 sodium hydroxide.
H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l)
Calculate the concentration of the original concentrated acid, in mol/dm3. [4]

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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 average mass of one atom of an element (taking its isotopes into account), compared with 1/12 of the mass of one atom of carbon-12.
Relative molecular mass, MrThe average mass of one molecule of a substance, compared with 1/12 of the mass of one atom of carbon-12. It is the sum of the relative atomic masses of the atoms in the formula.
Pure and Combined: the mole and concentration
TermDefinition to write
MoleThe amount of substance that contains 6.02 × 1023 particles (the Avogadro constant).
Molar volume of a gasOne mole of any gas occupies 24 dm3 at room temperature and pressure.
ConcentrationThe amount of solute in 1 dm3 of solution, in mol/dm3 or g/dm3.
Calculation route
  1. Formula and equation

    Balance before calculating.

  2. Known quantity to moles

    n = m/M; n = cV; n = V/24 at room conditions.

  3. Mole ratio

    Use coefficients, including the limiting reactant.

  4. Requested quantity

    Convert moles and include units.

Scope

Learning outcomes

4. Chemical Calculations (5086 / 5088 / K326 / K328). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 4.1(a) Use chemical symbols and formulae

    • Element symbols
    • Formulae of named syllabus compounds

    Read and construct formulae

  2. 4.1(b) Translate atom counts and formulae

    • Subscripts, brackets and coefficients
    • Reverse interpretation

    Read and construct formulae

  3. 4.1(c) Balance ion charges to obtain formulae

    • Simple and polyatomic ions
    • Deduce ion charge from formula

    Read and construct formulae

  4. 4.1(d) Interpret equations and states

    • s, l, g and aq
    • Coefficients as mole ratios

    Balance atoms, then account for ionsUse the equation ratio

  5. 4.1(e) Construct balanced equations

    • Full equations with states
    • Ionic equations and spectators
    • Conserve atoms and charge

    Balance atoms, then account for ions

  6. 4.2(a) Define relative atomic mass

    • Carbon-12 comparison
    • Average atomic mass

    The mole is a counting unit

  7. 4.2(b) Define and calculate relative molecular/formula mass

    • Sum relative atomic masses
    • Molecular versus ionic terminology

    The mole is a counting unit

  8. 4.2(c) Define the mole using the Avogadro constant

    • Specified entities
    • Mass-mole-particle relationships

    The mole is a counting unit

  9. 4.2(d) Calculate reacting masses and gas volumes

    • Balanced mole ratios
    • 24 dm3 per mol at room temperature and pressure
    • Limiting reactants
    • No variable-temperature gas laws

    Use the equation ratio

  10. 4.2(e) Use solution concentration

    • mol/dm3 and g/dm3
    • Titration data
    • Unfamiliar reactions with supplied information

    Concentration and titration