Laws of Chemical Combination (With Numerical Illustrations)
The laws of chemical combination are the foundational principles that explain how substances combine in fixed proportions to form compounds. Let’s explore each law in detail with explanations, illustrations, and numerically solved problems.
1. Law of Conservation of Mass
Statement:
The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants equals the total mass of the products.
Illustration:
Consider the decomposition of calcium carbonate (CaCO₃):
CaCO3 → CaO + CO2
Here:
- Mass of CaCO₃ (reactant) = 100 g
- Mass of CaO (product) = 56 g
- Mass of CO₂ (product) = 44 g
Total mass before reaction: 100 g (reactant)
Total mass after reaction: 56 g (CaO) + 44 g (CO₂) = 100 g (products)
Explanation:
In this reaction, the total mass of the reactants (100 g) is exactly equal to the total mass of the products (100 g), demonstrating that mass is conserved.
Numerical Problem:
If 10 g of calcium carbonate is decomposed and produces 4 g of carbon dioxide, calculate the mass of calcium oxide formed.
Solution:
According to the law of conservation of mass:
Mass of CaCO₃ = Mass of CaO + Mass of CO₂
Given:
- Mass of CaCO₃ = 10 g
- Mass of CO₂ = 4 g
So,
Mass of CaO = 10 g – 4 g = 6 g
Hence, 6 g of calcium oxide is formed.
2. Law of Definite Proportions (Law of Constant Proportions)
Statement:
A chemical compound always contains the same elements in a fixed ratio by mass, irrespective of its source or method of preparation.
Illustration:
Water (H₂O) always contains hydrogen and oxygen in a mass ratio of 1:8.
Let’s break it down:
- Atomic mass of hydrogen (H) = 1 u
- Atomic mass of oxygen (O) = 16 u
In a molecule of water (H₂O):
- Mass of hydrogen = 2 × 1 u = 2 u
- Mass of oxygen = 1 × 16 u = 16 u
Mass ratio of hydrogen to oxygen:
2 u / 16 u = 1/8
Explanation:
No matter where the water is obtained (whether from rain, river, or lab preparation), it will always have hydrogen and oxygen in a 1:8 mass ratio.
Numerical Problem:
Calculate the mass of oxygen required to react with 4 g of hydrogen to form water.
Solution:
Given:
- Mass ratio of hydrogen to oxygen in water = 1:8
Let the mass of oxygen be x.
According to the ratio:
4 g / x = 1/8
Cross-multiplying gives:
x = 4 × 8 = 32 g
So, 32 g of oxygen is required to react with 4 g of hydrogen.
3. Law of Multiple Proportions
Statement:
If two elements combine to form more than one compound, the different masses of one element that combine with a fixed mass of the other element are in a ratio of small whole numbers.
Illustration:
Carbon and oxygen form two compounds: carbon monoxide (CO) and carbon dioxide (CO₂).
- In CO, the mass ratio of carbon to oxygen is 12:16.
- In CO₂, the mass ratio of carbon to oxygen is 12:32.
Let’s express the mass of oxygen in both compounds for a fixed mass of 12 g of carbon:
- In CO: Mass of oxygen = 16 g
- In CO₂: Mass of oxygen = 32 g
The ratio of the masses of oxygen in CO and CO₂ is:
16/32 = 1:2
Explanation:
The ratio is a simple whole number (1:2), demonstrating the law of multiple proportions.
Numerical Problem:
Carbon forms two oxides. In the first oxide, 1.5 g of oxygen combines with 3 g of carbon. In the second oxide, 4 g of oxygen combines with 3 g of carbon. Show that these data illustrate the law of multiple proportions.
Solution:
For the first oxide:
- Mass of oxygen = 1.5 g
- Mass of carbon = 3 g
For the second oxide:
- Mass of oxygen = 4 g
- Mass of carbon = 3 g
The ratio of the masses of oxygen in the two oxides is:
1.5 g / 4 g = 3/8 = 1:2.67
This ratio approximates to small whole numbers (1:3), confirming the law of multiple proportions.
4. Gay-Lussac’s Law of Gaseous Volumes
Statement:
When gases react together, they do so in volumes which bear a simple whole number ratio to each other and to the volumes of the products, if all volumes are measured under similar conditions of temperature and pressure.
Illustration:
Consider the reaction between hydrogen and oxygen to form water vapor:
2 H2(g) + 1 O2(g) → 2 H2O(g)
Volume ratio:
- Hydrogen : Oxygen : Water vapor = 2 : 1 : 2
Explanation:
This simple whole number ratio (2:1:2) aligns with Gay-Lussac’s law, showing that gases react in definite volume proportions.
Numerical Problem:
If 10 liters of hydrogen gas react with oxygen, how many liters of oxygen are required and how many liters of water vapor are produced?
Solution:
From the equation:
2 H2(g) + 1 O2(g) → 2 H2O(g)
The volume ratio of hydrogen to oxygen to water vapor is 2:1:2. Therefore:
- Volume of hydrogen = 10 liters
- According to the ratio, volume of oxygen required = 10 liters × (1/2) = 5 liters
- Volume of water vapor produced = 10 liters × (2/2) = 10 liters
Hence, 5 liters of oxygen are required, and 10 liters of water vapor are produced.
5. Avogadro’s Law
Statement:
Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules.
Illustration:
If 1 liter of hydrogen gas and 1 liter of oxygen gas are measured at the same temperature and pressure, both will contain the same number of molecules, despite their different masses.
Explanation:
This principle allows us to determine the molecular count of gases based on volume alone, provided conditions of temperature and pressure are constant.
Numerical Problem:
Calculate the volume of oxygen gas required to react with 22.4 liters of hydrogen gas at standard temperature and pressure (STP) to form water vapor.
Solution:
From the balanced equation:
2 H2(g) + 1 O2(g) → 2 H2O(g)
Volume ratio of hydrogen to oxygen is 2:1. Thus:
- Volume of hydrogen = 22.4 liters
- Volume of oxygen required = 22.4 liters × (1/2) = 11.2 liters
Hence, 11.2 liters of oxygen gas are required to react with 22.4 liters of hydrogen gas.
Conclusion
Understanding these fundamental laws of chemical combination helps in comprehending the behavior of substances during chemical reactions. Each law provides critical insights into the nature of matter and the principles governing chemical reactions.
