OFFICIAL UAT-UK SPEC: C4MODULE: CHEMISTRY2 Tested Topics

C4: Quantitative Chemistry, The Mole and Gas Volumes

Official Cambridge & Imperial ESAT specification requirements for Quantitative Chemistry, The Mole and Gas Volumes on the Chemistry paper. Understand the syllabus boundaries, spot distractor traps engineered by examiners, and master the 2-step Sub-90 shortcuts that compress solve time well below the 88.8-second cutoff.

Official Syllabus Subtopics

Requirements specified in the UAT-UK framework

  • 1The mole concept: n = mass / Mᵣ and Avogadro’s constant (6.02 × 10²³)
  • 2Molar concentration: c = n / V (mol/dm³) and mass concentration (g/dm³)
  • 3Molar gas volume at room temperature and pressure: 24 dm³/mol (24,000 cm³/mol)
  • 4Avogadro’s Law for reacting gas volumes
  • 5Limiting reactants and excess reactant determination
  • 6Percentage yield = (actual yield / theoretical yield) × 100%
  • 7Atom economy = (Mᵣ of desired product / Mᵣ of all reactants) × 100%

Sub-90 Exam Speed Tips

How top candidates solve this spec under the 88.8s clock

Speed Invariant #1

Under identical temperature and pressure, gas volumes react in the EXACT mole ratio of the balanced equation (Avogadro’s Law).

Speed Invariant #2

Atom economy includes stoichiometric balancing coefficients; percentage yield does not.

Examiner Trap Alert

Examiners engineer distractor options using common algebraic expansions and sign slips. Always test structural ratios or invariants before calculating arithmetic values.

Tested Topics in Specification C4

Master the individual question patterns and Sub-90 shortcuts for this specification.

2 Topics Catalogued
ARCH-C-038 Past Qs (3.56%)

Hydrocarbon Combustion Gas Volumes (Avogadro's Law), Excess O2 & Limiting Reactants

By Avogadro's Law, gas volume ratios equal mole ratios directly: for 1 volume of CₓHᵧ burned, x = V(CO2)/V(alkane) immediately, and O2 consumed is (x + y/4) volumes. At rtp, H2O(l) volume = 0.

Classroom: 150s→Sub-90: 40s
+110s Saved per Question
ARCH-C-0418 Past Qs (8%)

Molar Gas Volume (24 dm³ at rtp), Ideal Gas Scaling, Percentage Yield & Atom Economy

Memorise: 1 mol gas at rtp = 24 dm³ = 24,000 cm³, so 1 mmol gas = 24 cm³. Atom economy depends purely on the balanced equation (desired Mᵣ / total reactant Mᵣ), independent of experimental yield.

Classroom: 150s→Sub-90: 40s
+110s Saved per Question
UAT-UK ESAT SPEED REVOLUTION

Ready to Master Specification C4 Under Exam Conditions?

Practice timed questions under the authentic Pearson VUE interface with live split stopwatch pacing, dry-wipe scratchpad, and distractor forensics.