⚗️ C1 · Language of chemistry and structure of matter
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Eduqas A-level Chemistry (A410QS) · C1 — The language of chemistry and structure of matter
Mini-Lesson
C1 · Language of chemistry and structure of matter
This mini-lesson covers the whole of Eduqas C1: writing formulae and balanced equations, chemical calculations (moles, gas volumes, titrations, yield and atom economy), atomic structure (mass spectra, orbitals, ionisation energies) and bonding and structure (VSEPR shapes, electronegativity, intermolecular forces).
Work through each screen, answer every question as you go — the multiple-choice checks, the calculations and the sorting games — and collect ⭐ stars. Press Start when you are ready.
C1 · formulae and equations
Formulae, equations and state symbols
A balanced equation must conserve atoms and charge. State symbols — (s), (l), (g), (aq) — are part of the answer at A level.
Ca(OH)₂(aq) + 2HCl(aq) → CaCl₂(aq) + 2H₂O(l)Ca 1 = 1 · O 2 = 2 · H 4 = 4 · Cl 2 = 2 — every atom is accounted for
An ionic equation strips out the spectator ions. Neutralisation of any strong acid by any strong alkali is really just:
Exam habit: after balancing, check charge as well as atoms. In half-equations such as MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O the left-hand charge is (−1) + (+8) + (−5) = +2, which matches Mn²⁺.
Quick check
Balance it
?Which coefficients balance the combustion of propane: __C₃H₈ + __O₂ → __CO₂ + __H₂O ?
C1 · chemical calculations
The mole, molar mass and concentration
The mole is the amount of substance containing 6.02 × 10²³ particles (the Avogadro constant, L).
n = m ÷ Mn = moles · m = mass in g · M = molar mass in g mol⁻¹
c = n ÷ Vc in mol dm⁻³ · V in dm³ (divide a volume in cm³ by 1000)
Worked example
Moles in 8.00 g of NaOH (M = 23.0 + 16.0 + 1.0 = 40.0 g mol⁻¹):
n = 8.00 ÷ 40.0 = 0.200 mol
Watch the units: 25.0 cm³ is 0.0250 dm³. Forgetting the ÷ 1000 is the single most common titration error.
Calculate
Your turn — moles from mass
1Calculate the amount, in mol, in 12.5 g of calcium carbonate, CaCO₃. (M = 100.1 g mol⁻¹)
mol
Hint: n = m ÷ M = 12.5 ÷ 100.1.
C1 · titrations
Titration calculations
A titration finds an unknown concentration by reacting it with a solution of known concentration. The route is always the same:
1 — moles of the known solution: n = c × V
2 — use the balanced equation to get the mole ratio
3 — moles of the unknown → c = n ÷ V
Accuracy: use concordant titres (within 0.10 cm³ of each other) and take their mean — never average in a rough titre.
Calculate
Your turn — titration
225.0 cm³ of sodium hydroxide solution needed 22.4 cm³ of 0.100 mol dm⁻³ hydrochloric acid for neutralisation. NaOH + HCl → NaCl + H₂O. Calculate the concentration of the NaOH, to 3 significant figures.
mol dm⁻³
Hint: n(HCl) = 0.0224 × 0.100 = 2.24 × 10⁻³ mol. Ratio is 1 : 1, so n(NaOH) = 2.24 × 10⁻³ mol. c = 2.24 × 10⁻³ ÷ 0.0250.
C1 · gases
The ideal gas equation
pV = nRTp in Pa · V in m³ · n in mol · R = 8.314 J K⁻¹ mol⁻¹ · T in K
The units are unforgiving. Convert before you substitute:
kPa → Pa: × 1000. 100 kPa = 1.00 × 10⁵ Pa.
cm³ → m³: ÷ 10⁶. dm³ → m³: ÷ 1000.
°C → K: + 273.
Ideal gas assumptions: the molecules have negligible volume and there are no intermolecular forces. Real gases deviate most at high pressure and low temperature, where both assumptions collapse.
Calculate
Your turn — pV = nRT
3Calculate the volume, in dm³, occupied by 0.250 mol of an ideal gas at 100 kPa and 300 K. (R = 8.314 J K⁻¹ mol⁻¹)
dm³
Hint: V = nRT ÷ p = (0.250 × 8.314 × 300) ÷ (1.00 × 10⁵) = 6.24 × 10⁻³ m³. Then × 1000 to get dm³.
C1 · yield and atom economy
Percentage yield and atom economy
% yield = (actual ÷ theoretical) × 100both as mass, or both as moles — never mix
atom economy = (M of desired product ÷ Σ M of all products) × 100equivalently, ÷ total M of reactants used
They measure different things. Yield is about how well the reaction actually went (losses, side reactions, reversibility). Atom economy is fixed by the equation itself — an addition reaction has 100% atom economy because there is only one product.
Green chemistry: a high atom economy means less waste, cheaper separation and fewer disposal problems — a reaction can have a 95% yield yet a terrible atom economy.
Calculate
Your turn — atom economy
4Limestone is decomposed: CaCO₃ → CaO + CO₂. Calculate the atom economy for making the desired product CaO. (M: CaCO₃ = 100.1, CaO = 56.1, CO₂ = 44.0 g mol⁻¹)
5Heating 12.5 g of CaCO₃ (0.125 mol) could in theory give 7.01 g of CaO. A student actually collects 5.85 g. Calculate the percentage yield.
%
Hint: % yield = (5.85 ÷ 7.01) × 100.
C1 · atomic structure
Atoms, isotopes and the mass spectrometer
Isotopes are atoms of the same element (same atomic number) with different numbers of neutrons. They are chemically identical because chemistry is governed by the electrons.
A time-of-flight mass spectrometer ionises the sample, accelerates the ions, and separates them by mass/charge (m/z). The relative abundances give the relative atomic mass:
Ar = Σ(isotopic mass × % abundance) ÷ 100a weighted mean — which is why Ar values are rarely whole numbers
Calculate
Your turn — relative atomic mass
6Chlorine has two isotopes: ³⁵Cl (75.0%) and ³⁷Cl (25.0%). Calculate the relative atomic mass of chlorine.
Electrons occupy orbitals — regions holding a maximum of 2 electrons of opposite spin. An s orbital is spherical; the three p orbitals are dumb-bell shaped and mutually perpendicular; there are five d orbitals.
Capacity: s = 2, p = 6, d = 10, f = 14.
Aufbau: fill lowest energy first. 4s fills before 3d — but 4s is also lost first on ionisation.
Hund's rule: singly occupy degenerate orbitals before pairing up.
Worked example — iron
Fe (Z = 26): 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Fe³⁺: remove the two 4s electrons first, then one 3d → 1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵, a half-filled d subshell.
The two anomalies: Cr is [Ar] 3d⁵ 4s¹ and Cu is [Ar] 3d¹⁰ 4s¹ — a half-filled or full d subshell is the more stable arrangement.
Quick check
Configuration of an ion
?What is the electron configuration of the Fe³⁺ ion?
C1 · ionisation energies
Ionisation energies
The first ionisation energy is the energy to remove one electron from each atom in one mole of gaseous atoms:
Na(g) → Na⁺(g) + e⁻always endothermic — you are pulling an electron away from a positive nucleus
Three factors control its size: nuclear charge, atomic radius (distance) and shielding by inner shells.
Across a period the first IE generally rises — more protons, similar shielding, smaller radius.
Down a group it falls — more shells, so more distance and more shielding outweigh the extra protons.
Two dips in Period 3: Al < Mg (the outer electron is in a higher-energy 3p orbital) and S < P (in sulfur two electrons are paired in one 3p orbital and repel).
Successive ionisation energies always increase, and a big jump marks the start of a new, closer shell — the number of electrons removed before the jump gives the group number.
Quick check
Explain the dip
?The first ionisation energy of aluminium is lower than that of magnesium, even though aluminium has one more proton. Why?
Calculate
Your turn — successive ionisation energies
7The first five ionisation energies of an element (kJ mol⁻¹) are: 590, 1150, 4940, 6480, 8120. Which group of the Periodic Table is the element in?
(group number)
Hint: look for the big jump — it comes between the 2nd (1150) and 3rd (4940). Two electrons come off easily, so two electrons are in the outer shell.
Covalent — a shared pair of electrons. In a dative (coordinate) bond both electrons come from the same atom, e.g. the fourth N–H bond in NH₄⁺.
Metallic — positive ions in a sea of delocalised electrons.
Electronegativity is the power of an atom to attract the electrons in a covalent bond. It increases across a period and decreases down a group; F is the most electronegative element.
A difference in electronegativity makes a bond polar (δ+ / δ−). But a molecule with polar bonds is only polar overall if the dipoles do not cancel: CO₂ is linear and non-polar; H₂O is bent and polar.
C1 · VSEPR
Shapes of molecules — VSEPR
Valence Shell Electron Pair Repulsion: electron pairs around the central atom repel to get as far apart as possible. Lone pairs repel more strongly than bonding pairs, so each lone pair squeezes the bond angle by roughly 2.5°.
Method: count the electron pairs around the central atom, then subtract for lone pairs. Repulsion order: lone–lone > lone–bond > bond–bond.
Sort it
Sort by shape
Tap a molecule, then tap the shape it adopts. Think: how many electron pairs, and how many are lone pairs?
🔷 Tetrahedral 109.5°
🔺 Trigonal planar 120°
📐 Bent / non-linear
Quick check
The ammonia angle
?Ammonia, NH₃, has four electron pairs around nitrogen but a bond angle of only 107°, not 109.5°. Why?
C1 · intermolecular forces
Intermolecular forces and hydrogen bonding
Three forces act between molecules, all far weaker than a covalent bond:
van der Waals (London / temporary dipole–induced dipole) — present in everything. Stronger with more electrons and a larger contact surface, which is why boiling points rise down the alkanes and why branching lowers them.
Permanent dipole–dipole — between polar molecules.
Hydrogen bonding — the strongest. Requires H bonded directly to N, O or F, plus a lone pair on the N, O or F of the neighbour.
Ice is less dense than water because hydrogen bonds hold the molecules in an open tetrahedral lattice. Hydrogen bonding also explains why HF, H₂O and NH₃ have anomalously high boiling points against the trend of their groups.
C1 · structure
Four kinds of solid structure
Physical properties follow directly from structure and bonding:
Giant ionic (NaCl) — high m.p.; conducts only when molten or aqueous, because the ions must be free to move.
Giant covalent (diamond, SiO₂) — very high m.p., hard, non-conducting. Graphite is the exception: it conducts because one electron per carbon is delocalised, and its layers slide.
Simple molecular (I₂, CO₂) — low m.p. because only weak intermolecular forces are broken on melting; never conducts.
Metallic (Mg) — malleable and conducting because of delocalised electrons.
The classic misconception: melting iodine does not break the I–I covalent bond. It only overcomes the van der Waals forces between I₂ molecules — that is why the melting point is so low.
Match it
Match structure to substance
Tap a description on the left, then the substance on the right.
Structure and bonding
Substance
Recap
The big ideas to know
Equations: balance atoms AND charge; ionic equations remove spectators
Moles: n = m ÷ M · c = n ÷ V · pV = nRT (Pa, m³, K)
Yield vs atom economy: yield = actual ÷ theoretical; atom economy is fixed by the equation
Atomic structure: A(r) is a weighted mean; 4s fills first but is lost first
Ionisation energy: nuclear charge, distance, shielding; dips at Al and S; big jump gives the group