Full chapter
Chemical Calculations
Turn formulae, measurements and equations into defensible numerical answers.
O-Level 6092 (2026) / SEC G3 K324 (2027)
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 example
From 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?
- Write N and O with the atom counts as subscripts: N2O4.
- Three molecules contain 3 x 4 = 12 oxygen atoms. Do not simplify N2O4 to NO2 when the actual atoms in one molecule are specified.
N2O4; 12 oxygen atoms in three molecules. A coefficient counts complete molecules; a subscript counts atoms within one.
| Family | Symbols or formulae |
|---|---|
| Common non-metals | H, C, N, O, F, P, S, Cl, Br, I; elemental gases H2, N2, O2, Cl2 |
| Metals and other elements | Li, Na, K, Mg, Ca, Al, Zn, Fe, Cu, Ag, Pb; Si, He, Ne, Ar |
| Acids, alkalis and gases | HCl, HNO3, H2SO4, NaOH, Ca(OH)2, NH3, CO, CO2, SO2, NO, NO2, O3 |
| Common salts and reagents | NaCl, MgCl2, CaCO3, CuSO4, AgNO3, Ba(NO3)2, KI, KMnO4 |
| Oxides and organic examples | CaO, 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.
| Positive ions | Negative 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 understandingThe compound is XCl3 and each chloride ion is Cl-. What is the charge on X?Think it through, then reveal the answer
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 example
Balance combustion
Write the complete combustion of propane.
- Start C3H8(g) + O2(g) -> CO2(g) + H2O(l).
- Balance carbon: 3CO2. Balance hydrogen: 4H2O.
- Products contain ten O atoms, requiring 5O2.
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 example
Find the reacting ions
Aqueous silver nitrate reacts with aqueous sodium chloride.
- AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq).
- Dissolved reactants provide Ag+, NO3-, Na+ and Cl-.
- Na+ and NO3- remain dissolved: cancel them.
Ag+(aq) + Cl-(aq) -> AgCl(s). Total charge is zero on both sides.
Check your understandingCan Mg + O2 -> MgO2 be used to balance formation of magnesium oxide?Think it through, then reveal the answer
The mole is a counting unit
Connect microscopic particles to measurable mass.
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 example
Account 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.
- Mr = 40 + 2(16 + 1) = 74.
- Molar mass = 74 g mol-1.
- n = 3.70/74 = 0.0500 mol.
0.0500 mol of Ca(OH)2 formula units; if fully dissolved, this produces 0.100 mol of OH- ions.
Check your understandingHow many moles of oxygen atoms are present in 0.25 mol of O2 molecules?Think it through, then reveal the answer
Use the equation ratio
Convert to moles before comparing reactants.
- Write and balance the reaction
The coefficients give a mole ratio, not a mass ratio.
- Convert the known quantity to moles
Mass: n = m/M. Gas at room temperature and pressure: n = V/24 with V in dm3.
- Apply the mole ratio
For aA -> bB, n(B) = n(A) x b/a.
- Convert to the requested quantity
Use m = nM or V = 24n at room temperature and pressure.
Worked example
Mass to gas volume
Excess hydrochloric acid reacts with 2.40 g Mg. Calculate hydrogen volume at room temperature and pressure; Mg = 24.
- Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g).
- n(Mg) = 2.40/24 = 0.100 mol; the Mg:H2 ratio is 1:1.
- V(H2) = 0.100 x 24 = 2.40 dm3.
2.40 dm3, or 2400 cm3. The 24 dm3 mol-1 value applies at room temperature and pressure.
Worked example
Mass 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.
- 2Mg(s) + O2(g) -> 2MgO(s). The Mg:MgO ratio is 2:2, or 1:1.
- n(Mg) = 3.60/24 = 0.150 mol, so n(MgO) = 0.150 mol.
- M(MgO) = 24 + 16 = 40 g mol-1; mass = 0.150 x 40 = 6.00 g.
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 example
Identify the limiting reactant
Mix 0.080 mol Mg with 0.100 mol HCl. How much H2 forms?
- The equation needs two moles of HCl for each mole of Mg.
- 0.080 mol Mg would require 0.160 mol HCl; only 0.100 mol is available. HCl is limiting.
- n(H2) = 0.100/2 = 0.050 mol. Mg used = 0.050 mol; Mg remaining = 0.030 mol.
0.050 mol hydrogen, equivalent to 1.20 dm3 at room temperature and pressure.
Check your understandingCan 0.10 mol of Mg react completely with 0.10 mol of HCl?Think it through, then reveal the answer
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 example
A 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.
- H2SO4(aq) + 2NaOH(aq) -> Na2SO4(aq) + 2H2O(l).
- n(NaOH) = 0.150 x 0.0200 = 0.00300 mol.
- n(acid) = 0.00300/2 = 0.00150 mol.
- c(acid) = 0.00150/0.0250 = 0.0600 mol dm-3.
0.0600 mol dm-3. A direct cV = cV shortcut would miss the factor of two.
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 understandingA solution contains 8.0 g NaOH per dm3. Find its molar concentration; M = 40 g mol-1.Think it through, then reveal the answer
Composition, formulae, yield and purity
Distinguish the composition of a compound from the success of a preparation.
Percentage by mass of an element = (its total Ar contribution in the formula / Mr) x 100%. In CaCO3, Mr = 40 + 12 + 48 = 100, so oxygen contributes 48%. An empirical formula is the simplest whole-number atom ratio; a molecular formula gives actual atom counts in a molecule.
Worked example
From percentage composition to molecular formula
A compound contains 40.0% C, 6.7% H and 53.3% O. Its Mr is 180.
- Assume 100 g: C = 40.0/12 = 3.33 mol; H = 6.7/1 = 6.7 mol; O = 53.3/16 = 3.33 mol.
- Divide by the smallest: approximately 1:2:1, giving CH2O.
- Empirical formula mass = 30; multiplier = 180/30 = 6.
Molecular formula C6H12O6. If ratios are near halves, multiply every ratio by two instead of rounding one ratio alone.
Percentage yield = actual product/theoretical product x 100%. The theoretical yield comes from the limiting reactant. Percentage purity = mass of pure substance/mass of sample x 100%. A sample can be very pure even when the preparation gives a poor yield.
Worked example
Keep the denominator meaningful
A reaction should produce 5.00 g of salt but gives 4.10 g. Separately, a 2.50 g limestone sample contains 2.00 g CaCO3.
- Yield = 4.10/5.00 x 100 = 82.0%.
- Purity = 2.00/2.50 x 100 = 80.0%.
82.0% yield; 80.0% purity. Wet crystals can give an apparently excessive yield; impurities can change the measured sample mass.
Worked example
Find purity from the amount that reacts
A 2.80 g sample containing CaCO3 and inert impurities produces 600 cm3 CO2 with excess dilute acid at room temperature and pressure. Find its percentage purity. Use M(CaCO3) = 100 g mol-1; impurities do not produce gas.
- CaCO3(s) + 2HCl(aq) -> CaCl2(aq) + CO2(g) + H2O(l). The carbonate:CO2 mole ratio is 1:1.
- 600 cm3 = 0.600 dm3; n(CO2) = 0.600/24 = 0.0250 mol.
- The sample contains 0.0250 x 100 = 2.50 g pure CaCO3.
- Percentage purity = 2.50/2.80 x 100 = 89.3%.
89.3% CaCO3 by mass. The denominator is the whole impure sample; use the gas to find the mass of the reactive pure component.
Check your understandingWhy cannot percentage composition alone distinguish CH2O from C6H12O6?Think it through, then reveal the answer
Quick revision
Revisit the essentials, then return to an explanation when you need it.
- Formula and equation
Balance before calculating.
- Known quantity to moles
n = m/M; n = cV; n = V/24 at room conditions.
- Mole ratio
Use coefficients, including the limiting reactant.
- Requested quantity
Convert moles and include units.
For ionic equations conserve charge as well as atoms. For concentration use dm3. Percent composition, empirical/molecular formulae, yield and purity are Pure extensions.
Scope and references
Learning outcomes and sources
4. Chemical Calculations (6092 / K324). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
4.1(a) Use chemical symbols and formulae
- Element symbols
- Formulae of named syllabus compounds
4.1(b) Translate atom counts and formulae
- Subscripts, brackets and coefficients
- Reverse interpretation
4.1(c) Balance ion charges to obtain formulae
- Simple and polyatomic ions
- Deduce ion charge from formula
4.1(d) Interpret equations and states
- s, l, g and aq
- Coefficients as mole ratios
4.1(e) Construct balanced equations
- Full equations with states
- Ionic equations and spectators
- Conserve atoms and charge
4.2(a) Define relative atomic mass
- Carbon-12 comparison
- Average atomic mass
4.2(b) Define and calculate relative molecular/formula mass
- Sum relative atomic masses
- Molecular versus ionic terminology
4.2(c) Define the mole using the Avogadro constant
- Specified entities
- Mass-mole-particle relationships
4.2(d) Calculate elemental mass percentages
- Element contribution divided by formula mass
4.2(e) Deduce empirical and molecular formulae
- Mass-to-mole ratios
- Molecular mass multiplier
4.2(f) 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
4.2(g) Use solution concentration
- mol/dm3 and g/dm3
- Titration data
- Unfamiliar reactions with supplied information
4.2(h) Calculate yield and purity
- Actual versus theoretical yield
- Pure substance versus total sample mass
- 2026 Pure Chemistry 6092
Official topic 4, pages 14-15. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.
- 2027 Pure Chemistry K324
Official topic 4, pages 14-15. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.
- 2026 Combined Chemistry 5086 / 5088
Official topic 4, pages 30. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.
- 2027 Combined Chemistry K326 / K328
Official topic 4, pages 30. Original explanations mapped to the stated outcomes; 2026 and 2027 topic content agrees.
- Grail: 6092 Chemistry Complete Notes, Version 1
Background consultation: Chapters 6-7, formulae and mole calculations, pp. 31-37. Teaching additions and examples are original; syllabus scope and chemistry independently checked.