AQA A-Level Chemistry AS Paper 1, June 2023: Question 4

5 marks · Medium difficulty · State/Explain/Numerical

Determine the relative formula mass of a Group 2 nitrate and identify metal M using the ideal gas equation and stoichiometry from its thermal decomposition.

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Question

Question 04 states: M is a Group 2 metal that forms the nitrate M(NO3)2. 0.320 g of M(NO3)2 is heated strongly and decomposes completely according to the equation 2 M(NO3)2(s) -> 2 MO(s) + 4 NO2(g) + O2(g). The mixture of gases formed has a volume of 225 cm³ at 450 °C and 101 000 Pa. Determine the Mr of M(NO3)2 and identify M. Gas constant R = 8.31 J K⁻¹ mol⁻¹. Answer spaces are provided for Mr of M(NO3)2 and Identity of M, total 5 marks.
Question text

04 M is a Group 2 metal that forms the nitrate M(NO3)2

0.320 g of M(NO3)2 is heated strongly and decomposes completely.

2M(NO3)2(s) → 2MO(s) + 4NO2(g) + O2(g)

The mixture of gases formed has a volume of 225 cm3 at 450 °C and 101 000 Pa

Determine the Mr of M(NO3)2

Identify M.

The gas constant, R = 8.31 J K–1 mol –1

[5 marks]

Mr of M(NO3)2

Identity of M

Mark scheme

Show the mark scheme Mark scheme for Question 04 showing a 5-mark breakdown: M1 converts units to V = 225 x 10^-6 m³ and T = 723 K. M2 calculates total moles of gas as n = pV / RT = (101 000 x 225 x 10^-6) / (8.31 x 723) = 3.78 x 10^-3 mol. M3 calculates moles of M(NO3)2 using the 2:5 mole ratio: n = M2 x (2/5) = 1.51 x 10^-3 mol. M4 calculates Mr = mass / mol = 0.320 / (1.51 x 10^-3) = 211.9 (accepting 211.5 to 212). M5 calculates Ar(M) = 211.9 - 124.0 = 87.9 and identifies M as strontium (Sr).

Question Marking guidance Additional Comments/Guidelines Mark

M1 V = 225×10–6 m3 ; T = 723 K M1 converts units of V and T

M2 M2 calculates the amount, in moles, of gas

produced

V 101 000 × 225 × 10−6

p –3

n = = = 3.78 × 10 mol

RT 8.31 × 723

M3 calculates the amount, in moles, of M(NO3)2

M3 2

n = (M2 × )

2 –3

n = M2 × = 1.51 × 10 mol

04 (4 x AO2,

M4 1 x AO3)

M4 calculation of Mr

mass 0.320

Mr = = = 211.9 (allow 211.5 to 212)

mol ( M3)

M5 M5 determination of Metal, M using M4

Consequential to Mr value on answer line: must be

Ar (M) = 211.9 – 124.0 = 87.9, so a Group 2 metal

M is Sr

How to answer it

Identifying an Unknown Group 2 Metal via Gas Decomposition

📋 What this question tests

This 5-mark unstructured question evaluates your synoptic problem-solving skills across physical and inorganic chemistry:

  • Ideal Gas Equation: Rearranging pV = nRT and converting non-standard units (cm³ to m³, °C to K).
  • Reaction Stoichiometry: Identifying all gaseous products and using stoichiometric ratios correctly (5 mol gas : 2 mol nitrate).
  • Molar Mass & Periodic Table deduction: Calculating Mr , removing the anion fragment mass (2 × NO₃⁻), and matching relative atomic mass ( Ar ) to Group 2.
Question 04 • 5 Marks Total

Thermal Decomposition & Gas Calculations

Reaction: 2 M(NO₃)₂(s) → 2 MO(s) + 4 NO₂(g) + O₂(g)

📐 Step-by-Step Calculation

Step 1: Mark 1 Convert measurements to standard SI units

  • Volume ( V ): 225 cm³ = 225 ÷ 1,000,000 = 225 × 10⁻⁶ m³ (or 2.25 × 10⁻⁴ m³)
  • Temperature ( T ): 450 °C = 450 + 273 = 723 K
  • Pressure ( p ): already in Pascals = 101 000 Pa

Step 2: Mark 2 Calculate total amount of gaseous products ( n )

  • Rearrange: n = pV / RT
  • n(gas) = (101 000 × 225 × 10⁻⁶) / (8.31 × 723)
  • n(gas) = 22.725 / 6008.13 = 3.78 × 10⁻³ mol (keep unrounded in your calculator)

Step 3: Mark 3 Determine moles of M(NO₃)₂ using stoichiometry

  • Look at the balanced equation: 2 moles of M(NO₃)₂ produce (4 + 1) = 5 moles of gas.
  • Molar ratio of M(NO₃)₂ to gas = 2 : 5 (or × 0.4)
  • n(M(NO₃)₂) = 3.782 × 10⁻³ × (2 / 5) = 1.51 × 10⁻³ mol

Step 4: Mark 4 Calculate relative formula mass ( Mr )

  • Mr = mass / moles
  • Mr = 0.320 g / 1.513 × 10⁻³ mol = 211.9 (allowable range: 211.5 to 212)

Step 5: Mark 5 Identify the Group 2 metal M

  • Formula mass of two nitrate ions, 2 × NO₃: 2 × [14.0 + (3 × 16.0)] = 2 × 62.0 = 124.0
  • Ar(M) = Mr − 124.0 = 211.9 − 124.0 = 87.9
  • Consult Periodic Table Group 2: Mg (24.3), Ca (40.1), Sr (87.6), Ba (137.3).
  • Metal M is Strontium (Sr).

✅ Final Answers Required

  • Mr of M(NO₃)₂: 211.9 (allow 211.5 – 212)
  • Identity of M: Sr (or Strontium)
Full 5/5 marks awarded for full mathematical progression and correct identification.

💡 Key Knowledge Points

  • SI Units for pV = nRT: p in Pa, V in m³, T in K, R = 8.31 J K⁻¹ mol⁻¹ .
  • Gaseous species count: MO(s) is solid, NO₂(g) and O₂(g) are both gases: 4 + 1 = 5 moles of gas formed.
  • Group 2 nitrates: Decompose upon heating to give the metal oxide, brown NO₂ fumes, and O₂ gas.

❌ Common Errors & Traps

  • Ignoring O₂ in gas ratio: Many candidates assumed NO₂ was the only gas, using a 2:4 ratio instead of 2:5, leading to incorrect mole and Mr values.
  • Unit conversion fails: Dividing cm³ by 1,000 (giving dm³) rather than 1,000,000 (giving m³).
  • Celsius trap: Forgetting to add 273 to convert 450 °C into 723 K.
  • Subtracting only one NO₃: Subtracting 62 instead of 124 (since the formula is M(NO₃)₂ with two nitrate groups).
  • Premature rounding: Rounding intermediate values to 2 s.f. too early, which pushes the final Mr outside the allowed 211.5 – 212 range.

🧠 Exam Technique & Examiner Tips

  • Consequential Marking (ECF): Mark 5 is strictly consequential on M4. If you made an arithmetic error in M4 but used your calculated Mr to subtract 124 and correctly picked the nearest Group 2 metal, you can still gain M5!
  • Check your Group: Always verify your deduction belongs to Group 2 as specified in the question stem. Candidates who chose non-Group 2 elements forfeited the final mark.
  • Write every step explicitly: Showing clear substitutions for M1, M2, and M3 safeguards method marks even if a keying error happens on the calculator.

Topics

Physical Chemistry · Inorganic Chemistry · 3.1.2 Amount of Substance · 3.2.2 Group 2, The Alkaline Earth Metals

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.