OCR A-Level Chemistry Unified chemistry (03), June 2019: Question 1

9 marks · Medium difficulty · Structured Questions

Answer short questions on dipole moments in CF4, water molecules with Mr of 20, partial pressure of O2, mass of CO2 from combustion of propane, rate dependence on pH, and number of oxygen atoms in P2O5.

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Question

Question 1 consisting of six short parts: (a) explain why CF4 has polar bonds but no overall dipole (2 marks); (b) explain why a small proportion of water molecules have a relative molecular mass of 20 (1 mark); (c) calculate the partial pressure of O2 in a gas mixture with 21% O2 by volume at 1.0 x 10^5 Pa (1 mark); (d) calculate the mass in grams of CO2 formed by complete combustion of 42.0 m^3 of propane at RTP (2 marks); (e) determine the initial rate at pH 3 given the reaction is first order in H+ and rate is 2.4 x 10^-3 mol dm^-3 s^-1 at pH 1 (1 mark); (f) calculate the number of oxygen atoms in 4.26 g of P2O5 (2 marks).
Question text

1 These short questions are from different areas of chemistry.

(a) Explain why a CF4 molecule has polar bonds but does not have an overall dipole.

… [2]

(b) Explain why a small proportion of molecules in water have a relative molecular mass of 20.

… [1]

(c) What is the partial pressure of O2 (in Pa) in a gas mixture containing 21% O2 by volume and

with a total pressure of 1.0 × 105 Pa?

partial pressure of O2 = … Pa [1]

(d) What mass of carbon dioxide (in g) is formed by the complete combustion of 42.0 m3

(measured at RTP) of propane?

mass = … g [2]

(e) A reaction is first order with respect to H+. At a pH of 1, the initial rate is 2.4 × 10−3 mol dm−3 s−1.

What is the initial rate at a pH of 3?

(f) What is the number of oxygen atoms in 4.26 g of P2O5?

initial rate = … mol dm−3 s−1 [1]

number of oxygen atoms = … [2]

Mark scheme

Show the mark scheme Mark scheme for Question 1: (a) F is more electronegative than C, symmetrical/tetrahedral shape so dipoles cancel (2 marks); (b) molecules contain isotopes such as 2H or 18O (1 mark); (c) 21,000 Pa or 2.1 x 10^4 Pa (1 mark); (d) n(C3H8) = 1750 mol, mass of CO2 = 3 x 1750 x 44 = 231,000 g or 2.31 x 10^5 g (2 marks); (e) 2.4 x 10^-5 mol dm^-3 s^-1 (1 mark); (f) n(P2O5) = 0.0300 mol, number of O atoms = 5 x 0.0300 x 6.02 x 10^23 = 9.03 x 10^22 (2 marks). Total 9 marks.

AO

Question Answer Marks Guidance

element

1 (a) Polar bonds 2 AO1.1 Mark independently

F (atom) is more electronegative (than C atom) ×2 ALLOW

OR F is very/the most electronegative C and F have different electronegativities

OR the atoms have different electronegativities

………BUT

DO NOT ALLOW C is more electronegative

ALLOW C–F shown with correct dipole,

i.e. C δ+– Fδ–.

No overall dipole

(CF4 is) symmetrical OR tetrahedral IGNORE square planar

OR dipoles cancel

OR dipoles act in opposite directions IGNORE polar bonds cancel

BUT ALLOW polarities cancel

IGNORE charges cancel

(b) (Molecules) contain 1 AO1.2 ALLOW Molecules contain 18O

• 2H OR deuterium/D

• 3H OR tritium/T

OR O/H atoms have more neutrons (than 1H) Idea of isotopes is critical

… BUT

OR (different) O/H isotopes are present DO NOT ALLOW isotopes of elements different

OR (Molecules are) D2O from H and O (e.g. C)

(c) p(O ) = 0.21 × 1.00 × 105 1 AO2.2

= 21,000 / 2.1 × 104 (Pa)

AO

element

4 ALLOW use of ideal gas equation with a sensible

(d) FIRST, CHECK ANSWER 2

IF answer = 231 000, award 2 marks temperature (20–25ºC) and pressure (100/101 kPa)

-------------------------------------------------------------------- At 20ºC and 100 kPa,

n(C3H8) 100 × 10 × 42.0

n(C3H8) = = 1724… (mol)

42.0 × 103 42.0 × 106 8.314 × 293

n(C3H8) = OR OR 1750 (mol) AO2.2 → ~ 227586 (g) (dependent on roundings)

24.0 24 000

At 25ºC and 100 kPa,

100 × 10 × 42.0

n(C3H8) = = 1695… (mol)

8.314 × 298

Mass of CO2

→ ~ 223767 (g) (dependent on roundings)

mass CO2 = 3 × 1750 × 44 ALLOW use of 8.31 for R

5 ALLOW ECF from n(C3H8)

= 231 000 / 2.31 × 10 (g) ------------------------------------------------------

AO2.6 3

Common errors from 24.0 dm

ALLOW 2 SF, e.g. 230 000 231 → 1 mark No conversion of m3 to dm3

0.231 → 1 mark Confusion of cm3 and dm3

77 000 → 1 mark No 3 × for CO2

(e) Initial rate = 10–2 × 2.4 × 10–3 s–1 1 AO2.2

= 2.4 × 10–5 (mol dm–3 s–1)

(f) FIRST, CHECK ANSWER 2 AO2.2 Alternative approach

IF answer = 9.03 × 1022, award 2 marks 4.26

n(O atoms) = 142.0 × 5 = 0.15

--------------------------------------------------------------------

4.26 O atoms = 0.15 × 6.02 × 1023 = 9.03 × 1022

n(P2O5) = OR 0.03(00) (mol)

142.0

ALLOW ECF from incorrect n(P2O5)

O atoms = 5 × 0.0300 × 6.02 × 1023 ALLOW use of 6.022 × 1023

-------------------------------------------------------

= 9.03 × 1022 Common error

Minimum 3 SF required 1.806 × 1022 OR 1.81 × 1022 → 1 mark No × 5

Total 9

How to answer it

Synoptic Chemistry Drill: Structure, Kinetics & Quantitative Calculations

What this question tests

A rapid-fire set of fundamental principles across physical and inorganic chemistry modules:

  • Electronegativity & Molecular Polarity: Explaining molecular dipoles in symmetrical geometries (CF₄).
  • Isotopes & Mass Spectrometry: Identifying heavy isotopic variants in water molecules.
  • Gas Mixtures & Dalton's Law: Calculating partial pressure from mole/volume fractions.
  • Gas Volumes & Stoichiometry: Multi-step combustion calculation involving unit conversions (m³ to dm³).
  • Logarithmic pH Scale & Rate Orders: Linking rate equations to changes in [H⁺] across pH units.
  • Avogadro Constant & Particles: Calculating total atoms in a given compound mass using mole ratios.
Part (a) • 2 Marks

Bond Polarity vs Overall Molecular Dipole in CF₄

Topic: Shape & Polarity

✅ Correct Marking Points

  • Mark 1: Fluorine is more electronegative than carbon (or F and C have different electronegativities / Cδ+–Fδ− dipole present).
  • Mark 2: The molecule is symmetrical (tetrahedral geometry), so individual bond dipoles cancel (act in opposing directions).

💡 Key Knowledge

A molecule can have polar bonds but be completely non-polar overall if the symmetry causes the vector sum of individual dipoles to equal zero. CF₄ has 4 bonding pairs around central carbon, adopting a symmetrical tetrahedral shape (bond angle 109.5°).

❌ Common Errors & Lost Marks

  • Writing "polar bonds cancel" rather than "dipoles cancel" (examiners ignore "bonds cancel").
  • Writing "charges cancel" (dipoles are partial charges/polarities, not free ionic charges).
  • Stating "carbon is more electronegative than fluorine" (Fluorine is the most electronegative element, Pauling scale 4.0).

🧠 Exam Technique

Always structure two-mark dipole questions into two explicit sentences: (1) state the electronegativity difference that causes individual bond dipoles, and (2) name the geometry/symmetry and explicitly say "the dipoles cancel".

Part (b) • 1 Mark

Isotopes and Heavy Water Molecules

Topic: Mass & Isotopes

✅ Correct Answer

The molecules contain ²H (deuterium / D), ³H (tritium / T), or ¹⁸O isotopes (i.e. heavy isotopes with more neutrons, such as D₂O or H₂¹⁸O).

💡 Key Knowledge

Standard water is ¹H₂¹⁶O with Mr = (2 × 1) + 16 = 18.

To reach Mr = 20, the water molecule must contain naturally occurring heavier isotopes:

  • ¹H₂¹⁸O → (2 × 1) + 18 = 20
  • ²H₂¹⁶O (D₂O) → (2 × 2) + 16 = 20

❌ Examiner Trap

The concept of isotopes is essential. Do NOT refer to isotopes of any element other than hydrogen or oxygen (e.g., carbon contamination). You must explicitly identify heavier hydrogen (²H / deuterium) or heavier oxygen (¹⁸O).

Part (c) • 1 Mark

Partial Pressure of Oxygen

Topic: Gas Mixtures

📐 Step-by-Step Calculation

  1. Identify volume fraction: 21% by volume = mole fraction ( x ) of 0.21.
  2. Apply Dalton's Law:
    p(O₂) = mole fraction × total pressure
  3. Calculate:
    p(O₂) = 0.21 × (1.0 × 10⁵ Pa) = 21 000 Pa (or 2.1 × 10⁴ Pa)

✅ Final Answer

21 000 Pa or 2.1 × 10⁴ Pa

Award 1 mark for the correct numerical value in Pa.
Part (d) • 2 Marks

Mass of CO₂ from Propane Combustion

Topic: Stoichiometry & Molar Gas Volume

📐 Step-by-Step Working

  1. Write the balanced combustion equation:
    C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)
    Stoichiometric ratio: 1 mol C₃H₈ : 3 mol CO₂
  2. Convert gas volume to standard unit (dm³):
    Given: volume = 42.0 m³
    Since 1 m³ = 1000 dm³:
    V(C₃H₈) = 42.0 × 10³ dm³ = 42 000 dm³
  3. Calculate moles of propane, n(C₃H₈):
    n = 42 000 / 24.0 = 1750 mol
    Mark 1: Finding 1750 mol of propane.
  4. Apply mole ratio and calculate mass of CO₂:
    n(CO₂) = 1750 × 3 = 5250 mol
    M(CO₂) = 12.0 + (2 × 16.0) = 44.0 g mol⁻¹
    Mass of CO₂ = 5250 × 44.0 = 231 000 g (or 2.31 × 10⁵ g )
    Mark 2: Correct final mass in grams (allow 2 SF: 230 000 g).

❌ Common Calculation Traps (Lost Marks)

  • Unit Conversion Failure: Using 42.0 dm³ directly gives 231 g (scores 1 mark only). Remember: 1 m³ = 1000 dm³ = 1 000 000 cm³.
  • Cm³/Dm³ Confusion: Dividing by 24 000 without converting to cm³ gives 0.231 g (1 mark).
  • Forgetting the Stoichiometric Ratio: Multiplying by 1 instead of 3 gives 77 000 g (scores 1 mark).
Part (e) • 1 Mark

Reaction Rate at Different pH Values

Topic: Orders of Reaction & pH

💡 Key Knowledge

Recall the definition of pH: pH = −log₁₀[H⁺] , so [H⁺] = 10⁻ᵖᴴ .

  • At pH = 1: [H⁺] = 10⁻¹ = 0.10 mol dm⁻³
  • At pH = 3: [H⁺] = 10⁻³ = 0.0010 mol dm⁻³

Increasing the pH by 2 units decreases [H⁺] by a factor of 10² = 100.

📐 Rate Calculation

Since the reaction is first order with respect to H⁺:

Rate ∝ [H⁺]¹

When [H⁺] decreases by 100 times, the rate also decreases by 100 times:

New Rate = (2.4 × 10⁻³) / 100 = 2.4 × 10⁻⁵ mol dm⁻³ s⁻¹

Award 1 mark for 2.4 × 10⁻⁵ (mol dm⁻³ s⁻¹).
Part (f) • 2 Marks

Number of Atoms in a Compound Mass

Topic: Avogadro Constant & Mole Concept

📐 Step-by-Step Working

  1. Calculate the molar mass of P₂O₅:
    M(P₂O₅) = (2 × 31.0) + (5 × 16.0) = 62.0 + 80.0 = 142.0 g mol⁻¹
  2. Calculate moles of P₂O₅ molecules:
    n(P₂O₅) = 4.26 / 142.0 = 0.0300 mol
    Mark 1: Calculating 0.0300 mol of P₂O₅ (or calculating 0.150 mol of O atoms).
  3. Determine moles and total number of Oxygen atoms:
    Every 1 mole of P₂O₅ contains 5 moles of O atoms:
    n(O atoms) = 0.0300 × 5 = 0.150 mol
    Total O atoms = 0.150 × (6.02 × 10²³) = 9.03 × 10²²
    Mark 2: Correct answer to a minimum of 3 significant figures (9.03 × 10²²).

❌ The Classic Avogadro Mistake

Many students compute the number of molecules of P₂O₅ by multiplying 0.0300 × 6.02 × 10²³ = 1.81 × 10²² , but forget to multiply by 5 for the oxygen atoms. This error forfeits the second mark!

🧠 Exam Technique: Significant Figures

The prompt uses 4.26 g (3 SF). The mark scheme strictly specifies minimum 3 SF for Mark 2. Stating 9.0 × 10²² loses the mark—always give at least 3 SF unless directed otherwise.

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.