📚 Case Study Practice for Year 12 OCR Chemistry | Year 12 OCR 化学案例分析实战演练
Welcome to this case-study practice article designed for Year 12 OCR Chemistry. We will work through ten typical exam-style scenarios, developing problem-solving skills across quantitative and qualitative topics. Follow each worked solution carefully and note the OCR-specific terminology.
欢迎阅读这篇为 Year 12 OCR 化学设计的案例分析实战练习。我们将演练十个典型考试风格的场景，锻炼定量与定性问题的解决能力。请仔细研读每道题的解题步骤，并注意 OCR 专用术语。
1. Empirical and Molecular Formula from Combustion Analysis | 燃烧分析与分子式确定
A student carried out combustion analysis of an organic compound containing C, H and O. When 0.370 g of the compound was completely burnt, 0.880 g of CO₂ and 0.450 g of H₂O were collected. The molar mass of the compound is 74 g mol⁻¹. Determine the empirical and molecular formulae.
一名学生对一种含碳、氢、氧的有机化合物进行了燃烧分析。将 0.370 g 该化合物完全燃烧后，收集到 0.880 g CO₂ 和 0.450 g H₂O。该化合物摩尔质量为 74 g mol⁻¹。试确定其实验式和分子式。
) = 44 g mol⁻¹; moles of CO₂ = 0.880 ÷ 44 = 0.020 mol. Mass of C = 0.020 × 12.0 = 0.240 g.
步骤1：计算 CO₂ 中碳的质量。CO₂ 摩尔质量为 44 g mol⁻¹；CO₂ 物质的量 = 0.880 ÷ 44 = 0.020 mol。C 的质量 = 0.020 × 12.0 = 0.240 g。
mol. Each H₂O contains 2 H, so moles of H = 0.050 mol. Mass of H = 0.050 × 1.0 = 0.050 g.
步骤2：计算 H₂O 中氢的质量。H₂O 摩尔质量为 18 g mol⁻¹；H₂O 物质的量 = 0.450 ÷ 18 = 0.025 mol。每分子 H₂O 含 2 个 H，因此 H 的物质的量 = 0.050 mol。H 的质量 = 0.050 × 1.0 = 0.050 g。
Step 3: Mass of oxygen = total mass – (mass of C + mass of H) = 0.370 – (0.240 + 0.050) = 0.080 g.
步骤3：氧的质量 = 总质量 – (碳质量 + 氢质量) = 0.370 – (0.240 + 0.050) = 0.080 g。
Step 4: Moles of O = 0.080 ÷ 16.0 = 0.0050 mol.
步骤4：O 的物质的量 = 0.080 ÷ 16.0 = 0.0050 mol。
-number ratio C : H : O = 0.020 : 0.050 : 0.0050 = 4 : 10 : 1. Empirical formula = C₄H₁₀O.
步骤5：求最简单整数比 C : H : O = 0.020 : 0.050 : 0.0050 = 4 : 10 : 1。实验式 = C₄H₁₀O。
0 × 1) + 16 = 74 g mol⁻¹. Since molar mass is also 74 g mol⁻¹, molecular formula = C₄H₁₀O.
步骤6：实验式质量 = (4 × 12) + (10 × 1) + 16 = 74 g mol⁻¹。由于摩尔质量也是 74 g mol⁻¹，分子式即为 C₄H₁₀O。
2. Hess’s Law and Enthalpy Cycles | 赫斯定律与焓变循环
tion to calculate the standard enthalpy of combustion of butane, C₄H₁₀(g). ΔfH⁰ (CO₂(g)) = –394 kJ mol⁻¹, ΔfH⁰ (H₂O(l)) = –286 kJ mol⁻¹, ΔfH⁰ (C₄H₁₀(g)) = –126 kJ mol⁻¹.
利用以下标准生成焓变数据，计算丁烷 (C₄H₁₀(g)) 的标准燃烧焓。ΔfH⁰ (CO₂(g)) = –394 kJ mol⁻¹，ΔfH⁰ (H₂O(l)) = –286 kJ mol⁻¹，ΔfH⁰ (C₄H₁₀(g)) = –126 kJ mol⁻¹。
rite the equation for the combustion of butane: C₄H₁₀(g) + 6½ O₂(g) → 4 CO₂(g) + 5 H₂O(l).
步骤1：写出丁烷燃烧的化学方程式：C₄H₁₀(g) + 6½ O₂(g) → 4 CO₂(g) + 5 H₂O(l)。
Step 2: Apply Hess’s Law. ΔcH⁰ = Σ ΔfH⁰(products) – Σ ΔfH⁰(reactants). The enthalpy of formation of O₂ is zero.
步骤2：应用赫斯定律。ΔcH⁰ = Σ ΔfH⁰(生成物) – Σ ΔfH⁰(反应物)。O₂ 的生成焓为零。
Step 3: Σ ΔfH⁰(products) = [4 × (–394)] + [5 × (–286)] = –1576 – 1430 = –3006 kJ mol⁻¹.
步骤3：生成物的 Σ ΔfH⁰ = [4 × (–394)] + [5 × (–286)] = –1576 – 1430 = –3006 kJ mol⁻¹。
Step 4: Σ ΔfH⁰(reactants) = ΔfH⁰(C₄H₁₀) = –126 kJ mol⁻¹.
步骤4：反应物的 Σ ΔfH⁰ = ΔfH⁰(C₄H₁₀) =
Published by TutorHao | Year 12 Chemistry Revision Series | aleveler.com
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