IGCSE OCR Chemistry Unit Test: Elements, Compounds and Mixtures | IGCSE OCR 化学单元测试卷：元素、化合物与混合物

📚 IGCSE OCR Chemistry Unit Test: Elements, Compounds and Mixtures | IGCSE OCR 化学单元测试卷：元素、化合物与混合物

This unit test is designed to assess your understanding of elements, compounds and mixtures — a core topic in IGCSE OCR Chemistry. It includes ten carefully chosen questions covering definitions, separation techniques, atomic structure, chemical formulae and purity. Work through each question, then review the detailed answers and explanations to consolidate your learning. Use this as a self-assessment tool or as a guided revision exercise.

本单元测试卷旨在评估你对元素、化合物与混合物的理解——这是IGCSE OCR化学的核心主题。试卷包含十道精选题目，涵盖定义、分离技术、原子结构、化学式和纯度等考点。请依次作答，然后对照详细答案与解析，以巩固所学知识。你可以将其用作自我测评工具或指引性复习练习。

1. Identifying a Compound | 识别化合物

Question: Which of the following is a compound?

问题： 下列哪项是化合物？

A. Oxygen gas
B. Air
C. Water
D. Brass

A. 氧气
B. 空气
C. 水
D. 黄铜

Answer: C. Water

答案：C. 水

Explanation: A compound is a pure substance made of two or more different elements chemically bonded together in a fixed proportion. Water (H₂O) consists of hydrogen and oxygen atoms covalently bonded. Oxygen gas (O₂) is an element, air is a mixture of gases (mainly N₂ and O₂), and brass is an alloy — a homogeneous mixture of copper and zinc without chemical bonds between the metals.

解析： 化合物是由两种或多种不同元素以固定比例通过化学键结合而成的纯净物。水(H₂O)由氢和氧原子以共价键结合。氧气(O₂)是单质，空气是气体混合物(主要为N₂和O₂)，黄铜是合金——铜与锌的均匀混合物，金属间无化学键。

2. Best Method to Obtain Pure Water from Salt Solution | 从盐溶液中获得纯水的最佳方法

Question: Which separation technique is most suitable to obtain pure liquid water from a sodium chloride solution?

问题： 从氯化钠溶液中获得纯净液态水，最合适的分离技术是？

A. Filtration
B. Evaporation
C. Distillation
D. Chromatography

A. 过滤
B. 蒸发
C. 蒸馏
D. 色谱法

Answer: C. Distillation

答案：C. 蒸馏

Explanation: Distillation separates a solvent from a solution by boiling the liquid and then condensing the vapour. The water evaporates, leaves the salt behind, and pure water is collected after condensation. Simple evaporation would not collect the water; filtration cannot separate dissolved salt. Chromatography separates coloured mixtures, not a salt solution.

解析： 蒸馏通过加热使液体沸腾，再将蒸气冷凝，从而从溶液中分离出溶剂。水蒸发逸出，盐留在烧瓶中，纯水经冷凝后被收集。简单蒸发无法回收液态水；过滤不能分离溶解的盐；色谱法用于分离有色混合物，不适用于盐溶液。

3. Determining the Mass Number | 计算质量数

Question: An atom contains 17 protons and 18 neutrons. What is its mass number?

问题： 某原子含有17个质子和18个中子。它的质量数是多少？

A. 17
B. 18
C. 35
D. 1

A. 17
B. 18
C. 35
D. 1

Answer: C. 35

答案：C. 35

Explanation: Mass number = number of protons + number of neutrons. Here, 17 + 18 = 35. This atom is an isotope of chlorine, specifically chlorine-35. Remember that the number of electrons does not affect the mass number because electrons have negligible mass.

解析： 质量数 = 质子数 + 中子数。本题中，17 + 18 = 35。该原子是氯的一种同位素，即氯-35。注意电子数不影响质量数，因为电子的质量极小，可忽略不计。

4. Describing an Element | 描述元素

Question: Define the term ‘element’ and give one named example.

问题： 请定义术语“元素”，并举出一个命名的例子。

Answer: An element is a pure substance that consists of only one type of atom. All atoms of an element have the same number of protons. Example: carbon (C).

答案： 元素是仅由一种原子组成的纯净物。同一元素的所有原子具有相同的质子数。例子：碳(C)。

Explanation: Elements are the basic building blocks of all matter. There are about 118 known elements. Carbon is found in Group 14 of the periodic table and can exist as diamond, graphite or graphene. Each carbon atom has 6 protons. Oxygen, gold and helium are other common examples.

解析： 元素是构成所有物质的基本单元。已知元素约有118种。碳位于元素周期表第14族，可以以金刚石、石墨或石墨烯等形式存在。每个碳原子有6个质子。其他常见例子包括氧、金和氦。

5. Comparing Mixtures and Compounds | 比较混合物与化合物

Question: Describe two key differences between a mixture and a compound.

问题： 请描述混合物与化合物之间的两个主要区别。

Answer: 1. In a compound, elements are chemically bonded together, whereas in a mixture they are only physically mixed. 2. A compound has a fixed composition by mass (definite ratio of elements), while a mixture can have variable composition.

答案： 1. 化合物中元素通过化学键结合，而混合物中各组分仅为物理混合。2. 化合物具有固定的质量组成(元素比例恒定)，混合物的组成则可以变化。

Explanation: Examples help illustrate this. Water (H₂O) is a compound: always 2 g of hydrogen for every 16 g of oxygen. Air is a mixture: its composition varies with altitude and pollution. Compounds can only be separated into elements by chemical reactions, whereas mixtures can be separated by physical methods such as filtration or distillation.

解析： 实例有助于说明。水(H₂O)是化合物：每16 g氧总是对应2 g氢。空气是混合物：其组成随海拔和污染程度变化。化合物只能通过化学反应分解为元素，而混合物可以通过过滤、蒸馏等物理方法分离。

6. Calculating an Empirical Formula | 计算实验式

Question: A compound contains 40.0 % carbon, 6.7 % hydrogen and 53.3 % oxygen by mass. Determine its empirical formula. (Aᵣ: C = 12, H = 1, O = 16)

问题： 某化合物的质量组成为40.0%碳、6.7%氢和53.3%氧。请确定其实验式。(Aᵣ: C=12, H=1, O=16)

Answer: CH₂O

答案：CH₂O

Explanation: Step-by-step calculation:

解析： 分步计算如下：

moles of C = 40.0 ÷ 12 = 3.33 mol

碳的摩尔数 = 40.0 ÷ 12 = 3.33 mol

moles of H = 6.7 ÷ 1 = 6.7 mol

氢的摩尔数 = 6.7 ÷ 1 = 6.7 mol

moles of O = 53.3 ÷ 16 = 3.33 mol

氧的摩尔数 = 53.3 ÷ 16 = 3.33 mol

Divide each by the smallest number of moles (3.33):

各除以最小的摩尔数(3.33)：

C : H : O = (3.33/3.33) : (6.7/3.33) : (3.33/3.33) = 1 : 2 : 1

C : H : O = 1 : 2 : 1

The simplest whole-number ratio gives the empirical formula CH₂O. This corresponds to compounds like formaldehyde or glucose (which has the empirical formula CH₂O but molecular formula C₆H₁₂O₆).

最简整数比给出实验式CH₂O。对应的化合物可以是甲醛或葡萄糖(葡萄糖的实验式也是CH₂O，但其分子式为C₆H₁₂O₆)。

7. Paper Chromatography of Inks | 墨水的纸色谱分离

Question: Describe how paper chromatography can be used to separate a mixture of coloured inks. Include the terms stationary phase, mobile phase and R_f value.

问题： 描述如何使用纸色谱法分离混合彩色墨水。请使用固定相、流动相和R_f值等术语。

Answer: A spot of ink mixture is placed near the bottom of chromatography paper (stationary phase). The paper is dipped into a suitable solvent (mobile phase) in a beaker, ensuring the spot is above the solvent level. As the solvent rises by capillary action, it carries the ink components at different rates because of their differing solubilities and attractions to the paper. This produces separate coloured spots. The R_f value for each component is calculated by distance moved by spot ÷ distance moved by solvent front.

答案： 将一滴混合墨水点在色谱纸(固定相)的下端附近。把纸放入装有合适溶剂(流动相)的烧杯中，确保墨点位于溶剂液面之上。溶剂因毛细作用上升，同时带动墨水组分以不同速率移动，这是因为它们的溶解度及与纸的吸附力不同，从而形成分离的色斑。每个组分的R_f值 = 斑点移动距离 ÷ 溶剂前沿移动距离。

Explanation: Chromatography relies on the balance between solubility in the mobile phase and retention by the stationary phase. Components that are more soluble and less attracted to the paper travel further, giving higher R_f values. R_f is always less than 1 and can be used to identify substances under identical conditions.

解析： 色谱法基于组分在流动相中的溶解度与固定相的吸附能力之间的平衡。溶解度大且与纸吸附力弱的组分迁移更远，R_f值更大。R_f值始终小于1，并可在相同条件下用于物质鉴定。

8. Recognising Isotopes | 识别同位素

Question: Which statement about isotopes of the same element is correct?

问题： 关于同种元素的同位素，哪项陈述是正确的？

A. They have the same number of protons but different numbers of neutrons.
B. They have different numbers of protons and the same number of neutrons.
C. They have the same mass number.
D. They have different chemical properties.

A. 质子数相同，中子数不同。
B. 质子数不同，中子数相同。
C. 质量数相同。
D. 化学性质不同。

Answer: A

答案：A

Explanation: Isotopes are atoms of the same element, meaning they have the same atomic number (same number of protons and electrons). They differ in the number of neutrons, which changes the mass number. For example, carbon-12 and carbon-14 both have 6 protons, but 6 and 8 neutrons respectively. Chemical properties depend on electrons, so isotopes behave nearly identically in reactions.

解析： 同位素是同一种元素的原子，因此它们具有相同的原子序数(即相同的质子数和电子数)。它们的中子数不同，因而质量数不同。例如，碳-12和碳-14都具有6个质子，但中子数分别为6和8。化学性质取决于电子排布，因此同位素在化学反应中性质几乎相同。

9. Writing a Chemical Formula | 书写化学式

Question: Write the chemical formula for magnesium nitrate.

问题： 写出硝酸镁的化学式。

Answer: Mg(NO₃)₂

答案：Mg(NO₃)₂

Explanation: Magnesium forms Mg²⁺ ions. The nitrate ion is NO₃⁻. To balance the charges, two nitrate ions are needed for each magnesium ion. Hence the formula is Mg(NO₃)₂, indicating one magnesium ion and two nitrate groups. Remember to use brackets when more than one polyatomic ion is present.

解析： 镁形成Mg²⁺离子。硝酸根离子为NO₃⁻。为了使电荷平衡，每个镁离子需要两个硝酸根离子。因此化学式为Mg(NO₃)₂，表示一个镁离子和两个硝酸根基团。当含有两个或更多多原子离子时，应使用括号。

10. Testing Purity Using Boiling Point | 利用沸点检测纯度

Question: A student heats an unknown liquid and records its temperature over time. The liquid boils steadily at 78 °C without any change in temperature during boiling. Suggest whether the liquid is pure, and explain your reasoning.

问题： 某学生加热一种未知液体并记录温度随时间的变化。该液体在78 °C稳定沸腾，沸腾期间温度没有变化。请判断该液体是否纯净，并解释你的理由。

Answer: The liquid is likely a pure substance. A sharp, constant boiling point is characteristic of a pure substance. If the liquid were a mixture, it would boil over a range of temperatures. 78 °C matches the boiling point of pure ethanol, so the liquid could be ethanol; however, any pure substance with a boiling point of 78 °C would also fit this observation.

答案： 该液体很可能是一种纯净物。固定且急剧的沸点是纯净物的特征。如果液体是混合物，它会在一个温度区间内沸腾。78 °C与纯乙醇的沸点吻合，因此该液体可能是乙醇；然而，任何沸点为78 °C的纯净物也会符合这一观察结果。

Explanation: Purity can be assessed by measuring melting or boiling points. Impurities lower the melting point and raise the boiling point, as well as causing the temperature to change during phase transitions. In this case, the steady boiling at one temperature strongly suggests the absence of dissolved impurities. Chromatography or further tests could confirm identity.

解析： 纯度可通过测定熔点或沸点来评估。杂质会降低熔点、升高沸点，并导致相变过程中温度持续变化。本题中，在单一温度下稳定沸腾有力地表明不存在溶解杂质。色谱法或进一步检测可确认具体物质。

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IGCSE OCR Chemistry Unit Test: Elements, Compounds and Mixtures | IGCSE OCR 化学单元测试卷：元素、化合物与混合物