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Water - Composition of Water

Grade 7ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Water is a chemical compound composed of two elements: Hydrogen and Oxygen, represented by the chemical formula H2OH_2O.

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Composition by Volume: When water is decomposed by electrolysis, the ratio of the volume of Hydrogen gas to Oxygen gas produced is 2:12:1.

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Composition by Mass: The ratio of Hydrogen to Oxygen in water by mass is always 1:81:8. This is derived from the atomic masses: H=1 uH = 1 \text{ u} and O=16 uO = 16 \text{ u}. For H2OH_2O, mass is (2×1):16=2:16=1:8(2 \times 1) : 16 = 2:16 = 1:8.

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Electrolysis of Water: Pure water is a non-conductor of electricity. To make it conduct, a small amount of dilute sulphuric acid (H2SO4H_2SO_4) is added to form 'acidulated water'.

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During electrolysis, Hydrogen gas (H2H_2) is liberated at the negative electrode (Cathode), and Oxygen gas (O2O_2) is liberated at the positive electrode (Anode).

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Chemical Test for Water: Anhydrous Copper Sulphate (CuSO4CuSO_4), which is white, turns blue when it comes into contact with water, forming hydrated copper sulphate (CuSO4⋅5H2OCuSO_4 \cdot 5H_2O).

📐Formulae

2H2+O2→2H2O2H_2 + O_2 \rightarrow 2H_2O

Ratio by Mass of H:O=2×116=1:8\text{Ratio by Mass of } H:O = \frac{2 \times 1}{16} = 1:8

CuSO4+5H2O→CuSO4⋅5H2OCuSO_4 + 5H_2O \rightarrow CuSO_4 \cdot 5H_2O

2H2O→Electrolysis2H2↑+O2↑2H_2O \xrightarrow{\text{Electrolysis}} 2H_2 \uparrow + O_2 \uparrow

💡Examples

Problem 1:

Calculate the amount of Oxygen required to react completely with 5 g5 \text{ g} of Hydrogen to form water.

Solution:

Since the ratio of Hydrogen to Oxygen by mass in water is 1:81:8, 1 g1 \text{ g} of Hydrogen reacts with 8 g8 \text{ g} of Oxygen. Therefore, 5 g5 \text{ g} of Hydrogen will react with 5×8=40 g5 \times 8 = 40 \text{ g} of Oxygen.

Explanation:

According to the Law of Constant Proportions, elements in a compound are always present in definite proportions by mass. For water, this ratio is 1:81:8.

Problem 2:

During the electrolysis of water, if 30 cm330 \text{ cm}^3 of Hydrogen is collected at the cathode, what volume of Oxygen will be collected at the anode?

Solution:

The volume ratio of Hydrogen to Oxygen in water is 2:12:1. If Hydrogen is 30 cm330 \text{ cm}^3, the volume of Oxygen = 302=15 cm3\frac{30}{2} = 15 \text{ cm}^3.

Explanation:

In the decomposition of water (2H2O→2H2+O22H_2O \rightarrow 2H_2 + O_2), two molecules of Hydrogen are produced for every one molecule of Oxygen, resulting in a 2:12:1 volume ratio.