Example Responses for Edexcel IAL Chemistry Unit 1 (CH01) Calculation Questions | Edexcel IAL化学第一单元(CH01)计算题型示例精讲

📚 Example Responses for Edexcel IAL Chemistry Unit 1 (CH01) Calculation Questions | Edexcel IAL化学第一单元(CH01)计算题型示例精讲

Calculation questions form a significant part of Edexcel IAL Chemistry Unit 1 (CH01). They test your ability to apply the mole concept, stoichiometry, gas laws, and volumetric analysis. This article provides example responses to typical calculation questions, with detailed step-by-step solutions to help you master the techniques and avoid common mistakes.

在Edexcel IAL化学第一单元(CH01)中，计算题占据了相当的比例。这些题目考查你运用摩尔概念、化学计量学、气体定律和容量分析的能力。本文提供典型计算题的示例解答，附有详细步骤，助你掌握方法并避免常见错误。

---

1. The Mole and Molar Mass | 物质的量与摩尔质量

The mole is the unit for amount of substance. One mole contains exactly 6.02214076 × 10²³ elementary entities (Avogadro’s constant). Molar mass (M) is the mass of one mole of a substance, expressed in g mol⁻¹.

摩尔是物质的量的单位。1摩尔恰好包含6.02214076 × 10²³个基本单元（阿伏伽德罗常数）。摩尔质量(M)是一摩尔物质的质量，以g mol⁻¹表示。

Key equation: n = m / M, where n is amount in mol, m is mass in g, M is molar mass in g mol⁻¹.

关键公式：n = m / M，其中n为物质的量（mol），m为质量（g），M为摩尔质量（g mol⁻¹）。

Example: Calculate the mass of 0.250 mol of NaOH (Mᵣ = 40.0).

示例：计算0.250 mol NaOH的质量（相对分子质量 = 40.0）。

Solution: mass = amount × molar mass = 0.250 mol × 40.0 g mol⁻¹ = 10.0 g.

解答：质量 = 物质的量 × 摩尔质量 = 0.250 mol × 40.0 g mol⁻¹ = 10.0 g。

---

2. Empirical and Molecular Formulae | 经验式与分子式

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms and is a multiple of the empirical formula.

经验式给出化合物中各元素原子的最简整数比。分子式给出原子的实际数目，是经验式的整数倍。

Example: A compound contains 40.0% carbon, 6.70% hydrogen and 53.3% oxygen by mass. (Aᵣ: H = 1.0, C = 12.0, O = 16.0) Determine its empirical formula. If its relative molecular mass is 180, what is the molecular formula?

示例：某化合物含碳40.0%、氢6.70%、氧53.3%（Aᵣ: H = 1.0, C = 12.0, O = 16.0）。求经验式。若其相对分子质量为180，分子式是什么？

Step 1: Assume a 100 g sample, so masses are: C = 40.0 g, H = 6.70 g, O = 53.3 g.

步骤1：假设样品为100 g，则各元素质量为：C = 40.0 g，H = 6.70 g，O = 53.3 g。

Step 2: Convert masses to moles: n(C) = 40.0 / 12.0 = 3.33 mol; n(H) = 6.70 / 1.0 = 6.70 mol; n(O) = 53.3 / 16.0 = 3.33 mol.

步骤2：将质量转换为物质的量：n(C) = 40.0 / 12.0 = 3.33 mol；n(H) = 6.70 / 1.0 = 6.70 mol；n(O) = 53.3 / 16.0 = 3.33 mol。

Step 3: Divide by the smallest amount (3.33 mol) to get the ratio: C 1.00 : H 2.01 : O 1.00, which rounds to 1:2:1.

步骤3：除以最小数值(3.33 mol)得到比例：C 1.00 : H 2.01 : O 1.00，约为1:2:1。

Step 4: Empirical formula is CH₂O. Formula mass of CH₂O = 12.0 + 2(1.0) + 16.0 = 30.0.

步骤4：经验式为CH₂O。经验式质量 = 12.0 + 2(1.0) + 16.0 = 30.0。

Step 5: Molecular formula multiplier n = Mᵣ / empirical formula mass = 180 / 30.0 = 6. Thus molecular formula = 6 × CH₂O = C₆H₁₂O₆.

步骤5：分子式的倍数 n = 相对分子质量 / 经验式质量 = 180 / 30.0 = 6。因此分子式为C₆H₁₂O₆。

---

3. Reacting Masses and Limiting Reactants | 反应质量与限制反应物

Stoichiometry allows calculation of masses of reactants and products. The limiting reactant is the substance that runs out first and determines the amount of product formed.

化学计量学可用来计算反应物和产物的质量。限制反应物是最先耗尽并决定产物量的物质。

Example: 2.43 g of magnesium is burned in excess oxygen to form magnesium oxide: 2Mg + O₂ → 2MgO. (Aᵣ: Mg = 24.3, O = 16.0) Calculate the theoretical mass of MgO produced.

示例：2.43 g镁在过量氧气中燃烧生成氧化镁：2Mg + O₂ → 2MgO。(Aᵣ: Mg = 24.3, O = 16.0) 计算生成的MgO的理论质量。

Step 1: Moles of Mg = mass / molar mass = 2.43 g / 24.3 g mol⁻¹ = 0.100 mol.

步骤1：Mg的物质的量 = 质量 / 摩尔质量 = 2.43 g / 24.3 g mol⁻¹ = 0.100 mol。

Step 2: From the equation, 2 mol Mg produces 2 mol MgO, so ratio is 1:1. Moles of MgO = 0.100 mol.

步骤2：由方程式知2 mol Mg生成2 mol MgO，比例为1:1。MgO的物质的量 = 0.100 mol。

Step 3: Mass of MgO = moles × Mᵣ = 0.100 mol × (24.3 + 16.0) g mol⁻¹ = 0.100 × 40.3 = 4.03 g.

步骤3：MgO的质量 = 物质的量 × 相对分子质量 = 0.100 mol × (24.3+16.0) g mol⁻¹ = 4.03 g。

If only 3.20 g of O₂ were available, which reagent is limiting? n(O₂) = 3.20 / 32.0 = 0.100 mol. The equation requires 1 mol O₂ for 2 mol Mg, so for 0.100 mol Mg we need only 0.0500 mol O₂. O₂ is in excess, Mg is still the limiting reactant and the theoretical yield remains 4.03 g.

如果只有3.20 g O₂，哪种试剂是限制性的？n(O₂) = 3.20 / 32.0 = 0.100 mol。方程式中2 mol Mg需1 mol O₂，所以0.100 mol Mg只需0.0500 mol O₂。O₂过量，Mg仍是限制反应物，理论产量仍为4.03 g。

---

4. Percentage Yield and Atom Economy | 产率百分数与原子经济性

Percentage yield = (actual yield / theoretical yield) × 100%. Atom economy = (molar mass of desired product / sum of molar masses of all reactants) × 100%.

产率百分数 = (实际产量 / 理论产量) × 100%。原子经济性 = (目标产物的摩尔质量 / 所有反应物摩尔质量之和) × 100%。

Example: In the previous MgO preparation, 3.65 g of MgO was actually collected. Calculate the percentage yield. Then calculate the atom economy for the reaction 2Mg + O₂ → 2MgO, assuming MgO is the desired product.

示例：在上述制备MgO的实验中，实际收集到3.

Published by TutorHao | Chemistry Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）