OCR A-Level Chemistry AS Depth in chemistry (02), June 2019: Question 1

✅ Correct Answer

1. (Weighted) mean/average mass of an atom (do not accept 'element').
2. Compared with 1/12th mass of carbon-12 (or compared with mass of carbon-12 which is 12).

❌ Common Errors

Students frequently lose the first mark by stating "mean mass of an element" instead of specifying an atom. Precision in chemical definitions is strictly required by examiners.

📐 Calculation Steps

1. Identify total percentage: 82.9% (Sr-88) + 6.9% (Sr-87) = 89.8%.
2. Calculate abundance of the third isotope: 100% - 89.8% = 10.2%.
3. Set up the Ar equation:
   87.73 = [(88 × 82.9) + (87 × 6.9) + (Mass × 10.2)] / 100
4. Rearrange to solve for Mass:
   Mass = [(100 × 87.73) - (7295.2) - (600.3)] / 10.2
5. Solve: 86.03 → Round/identify as isotope mass 86 (Sr-86).

💡 Key Knowledge

Mass spectrometry data allows identification of isotopic compositions. Always check that your calculated abundances total 100% before solving for missing atomic masses.

🧠 Exam Technique & Diagram Requirements

To score full marks on the explanation of melting point and electrical conductivity, your answer must tie structure directly to forces and mobile charge carriers.

• Diagram: Draw a regular 2D/3D lattice of positive ions (labelled Sr²⁺ or + ions) surrounded by a sea of delocalised electrons (labelled e⁻ or electrons). Minimum two rows of ions required.
• High Melting Point: Strong electrostatic attraction between positive ions and delocalised electrons requires a lot of energy to overcome.
• Electrical Conductivity: Delocalised electrons are mobile and able to move through the structure to carry a current.

❌ Common Errors

Using the term intermolecular forces when describing metals. Metals feature metallic bonding (electrostatic attraction between ions and delocalised electrons), not intermolecular forces which only exist between covalent molecules.

💡 Key Knowledge (Group 2 Trends)

Reactivity increases down Group 2 as atomic radius increases and shielding increases, making outer electrons easier to lose.

✅ Any Two Acceptable Observations (vs. Calcium)

1. More vigorous fizzing / bubbling / effervescence.
2. Dissolves more quickly / faster reaction.
3. Resulting solution has a higher pH / is more alkaline (due to greater solubility of Sr(OH)₂ compared to Ca(OH)₂).
4. Precipitate forms less readily / less soluble precipitate.

📐 Step-by-Step Calculation

1. Mass of anhydrous SrCl₂ (residue):
   17.58g (constant final mass) - 15.96g (crucible) = 1.62 g
2. Mass of water lost:
   18.65g (initial crucible + hydrate) - 17.58g (crucible + residue) = 1.07 g
3. Moles of anhydrous SrCl₂:
   Mr of SrCl₂ = 87.6 + (35.5 × 2) = 158.6 g mol⁻¹
   Moles = 1.62 / 158.6 = 0.01021 mol
4. Moles of H₂O:
   Mr of H₂O = 18.0 g mol⁻¹
   Moles = 1.07 / 18.0 = 0.05944 mol
5. Ratio (x):
   Ratio H₂O : SrCl₂ = 0.05944 / 0.01021 = 5.82...
6. Final Answer (2 sig fig): x = 5.8

❌ Calculation Traps & Significant Figures

Watch out for using non-constant mass readings. Always use the final constant mass (17.58 g) for calculations. Ensure your final value is rounded to exactly 2 significant figures as requested in the prompt.

✅ (dii) Taking Four Readings

To ensure all water of crystallisation has been completely removed (to check for constant mass).

✅ (diii) Reducing Percentage Uncertainty

1. Use a larger mass of hydrated strontium chloride.
2. Use a balance with a higher precision (e.g., to 3 or 4 decimal places instead of 2).