Introduction

Chemistry is the study of matter and its transformations. Understanding the fundamental laws that govern chemical combinations is crucial for comprehending the behaviour of substances and their reactions. In this lesson, we will explore the fundamental laws of chemical combinations, including the Law of Conservation of Mass, the Law of Definite Proportions, and the Law of Multiple Proportions.

Law of Conservation of Mass

The Law of Conservation of Mass states that in any chemical reaction, the total mass of the reactants is equal to the total mass of the products.

This law emphasizes that atoms are neither created nor destroyed during a chemical reaction.

It is represented by the equation: Mass of Reactants = Mass of Products

Here are a few examples that illustrate this principle:

Combustion: When a piece of wood is burned, it undergoes a chemical reaction with oxygen to produce carbon dioxide and water vapor. Despite the physical and chemical changes that occur during the combustion process, the total mass of the system (including the wood, oxygen, carbon dioxide, and water vapor) remains constant.

Precipitation: When a solution of silver nitrate is mixed with a solution of sodium chloride, a precipitation reaction occurs, leading to the formation of a white solid known as silver chloride. Even though new substances are formed in this chemical reaction, the total mass of the system before and after the reaction remains the same.

Photosynthesis: During photosynthesis, plants convert carbon dioxide and water into glucose (a sugar) and oxygen, using energy from sunlight. Despite the complex biochemical reactions that take place within the plant, the total mass of the system (including the reactants and products) remains constant.

Nuclear reactions: In nuclear reactions, such as radioactive decay or nuclear fission, the nuclei of atoms undergo changes, releasing energy and sometimes transforming into different elements. However, the total mass of the reactants and products involved in the nuclear reaction remains constant.

These examples demonstrate that the law of conservation of mass holds true in various physical and chemical processes, affirming that mass is neither created nor destroyed but simply changes form.

Law of Definite Proportions

The Law of Definite Proportions , also known as the Law of Constant Composition was discovered by Joseph Proust . It states that a pure compound always contains the same elements in definite proportions by mass.

Regardless of the source or method of preparation, the ratio of the masses of elements in a compound remains constant.

This law is exemplified by the formula of water, H2O, which always consists of two hydrogen atoms and one oxygen atom in a fixed ratio by mass.

Here are a few examples that illustrate this law with chemical equations:

Water (H2O):

2H2 + O2 → 2H2O

This equation shows that when hydrogen gas (H2) reacts with oxygen gas (O2) in a 2:1 ratio by volume, it forms water (H2O). Regardless of the source of hydrogen and oxygen, the mass ratio of hydrogen to oxygen in water will always be 1:8.

Carbon Dioxide (CO2):

C + O2 → CO2

When carbon (C) combines with oxygen (O2) in a 1:2 ratio by mass, carbon dioxide (CO2) is formed. The law of definite proportions ensures that regardless of the source of carbon and oxygen, the mass ratio of carbon to oxygen in carbon dioxide will always be 1:2.

Ammonia (NH3):

N2 + 3H2 → 2NH3

When nitrogen gas (N2) reacts with hydrogen gas (H2) in a 1:3 ratio by volume, it produces ammonia (NH3). According to the law of definite proportions, regardless of the source of nitrogen and hydrogen, the mass ratio of nitrogen to hydrogen in ammonia will always be 1:3.

Sodium Chloride (NaCl):

Na + Cl2 → 2NaCl

In this equation, sodium (Na) reacts with chlorine (Cl2) to form sodium chloride (NaCl). The law of definite proportions ensures that the mass ratio of sodium to chlorine in sodium chloride will always be 1:1, regardless of the source of sodium and chlorine.

These examples demonstrate that chemical compounds have fixed mass ratios between their constituent elements, as dictated by the law of definite proportions. Regardless of the scale or source of the reactants, the proportions of elements in a compound remain constant.

Example Sum

Let’s consider the following chemical reaction:

2H2 + O2 -> 2H2O

In this reaction, hydrogen gas (H2) and oxygen gas (O2) combine to form water (H2O).

To prove the law of conservation of mass, we need to show that the total mass of the reactants is equal to the total mass of the products.

Step 1: Calculate the molar masses of each compound:

Molar mass of H2 = 2 grams/mole

Molar mass of O2 = 32 grams/mole

Molar mass of H2O = 18 grams/mole (2 hydrogen atoms with a molar mass of 1 each and 1 oxygen atom with a molar mass of 16)

Step 2: Calculate the total mass of the reactants:

Mass of 2 moles of H2 = 2 moles * 2 grams/mole = 4 grams

Mass of 1 mole of O2 = 1 mole * 32 grams/mole = 32 grams

Total mass of the reactants = 4 grams + 32 grams = 36 grams

Step 3: Calculate the total mass of the products:

Mass of 2 moles of H2O = 2 moles * 18 grams/mole = 36 grams

Total mass of the products = 36 grams

Step 4: Compare the total mass of the reactants and products:

The total mass of the reactants (36 grams) is equal to the total mass of the products (36 grams).

Therefore, this calculation demonstrates the law of conservation of mass, which states that mass is neither created nor destroyed in a chemical reaction. The total mass of the reactants is always equal to the total mass of the products.

Law of Multiple Proportions

The Law of Multiple Proportions states that when two elements combine to form different compounds, the ratio of masses of one element that combine with a fixed mass of the other element can be expressed in small whole numbers.

This law was first proposed by John Dalton and is an essential concept in understanding the atomic nature of matter.

Here are a few examples that illustrate this law with chemical equations:

Carbon Monoxide (CO) and Carbon Dioxide (CO2):

C + O2 → CO2

2C + O2 → 2CO

In these equations, carbon (C) reacts with oxygen (O2) to form carbon dioxide (CO2) and carbon monoxide (CO), respectively. The law of multiple proportions is evident here because the ratio of the masses of oxygen in the two compounds is 2:1, indicating that for a fixed mass of carbon, the masses of oxygen combine in small whole number ratios.

Nitric Oxide (NO) and Nitrogen Dioxide (NO2):

N2 + O2 → 2NO

2NO + O2 → 2NO2

In these equations, nitrogen (N2) reacts with oxygen (O2) to form nitric oxide (NO) and nitrogen dioxide (NO2), respectively. The law of multiple proportions is observed because the ratio of the masses of oxygen in the two compounds is 1:2, indicating that for a fixed mass of nitrogen, the masses of oxygen combine in small whole number ratios.

Hydrogen Peroxide (H2O2) and Water (H2O):

2H2 + O2 → 2H2O

2H2O2 → 2H2O + O2

These equations represent the formation of water (H2O) from the reaction of hydrogen gas (H2) and oxygen gas (O2) and the decomposition of hydrogen peroxide (H2O2) into water (H2O) and oxygen gas (O2). The law of multiple proportions is evident because the ratio of the masses of oxygen in hydrogen peroxide and water is 2:1, indicating that for a fixed mass of hydrogen, the masses of oxygen combine in small whole number ratios.

These examples demonstrate the law of multiple proportions, which shows that when two elements combine to form different compounds, their masses combine in simple ratios, reflecting the ratios of small whole numbers.

Example Sum

Let’s consider the combination of carbon (C) and oxygen (O) to form two different compounds: carbon monoxide (CO) and carbon dioxide (CO2).

Step 1: Determine the molar masses of each element:

Molar mass of C = 12 grams/mole

Molar mass of O = 16 grams/mole

Step 2: Calculate the masses of oxygen in each compound:

Carbon monoxide (CO): Since there is one oxygen atom, the mass of oxygen = 1 * 16 grams = 16 grams.

Carbon dioxide (CO2): Since there are two oxygen atoms, the mass of oxygen = 2 * 16 grams = 32 grams.

Step 3: Compare the ratios of the masses of oxygen in each compound:

The ratio of the masses of oxygen in carbon monoxide to carbon dioxide is 16 grams : 32 grams, which can be simplified to 1 : 2.

This ratio can be expressed as a small whole number ratio (1:2). This demonstrates the law of multiple proportions because the masses of oxygen combine in different compounds in a simple, whole number ratio.

In summary, the law of multiple proportions states that when elements combine to form different compounds, the ratio of their masses can be expressed as small whole numbers. In the case of carbon and oxygen forming carbon monoxide and carbon dioxide, the mass ratio of oxygen is 1:2, demonstrating this law.

Law of Reciprocal Proportions

The law of reciprocal proportions, also known as Richter’s law, states that if two elements combine separately with a fixed amount of a third element, the ratio of the masses in which they combine will be either the same or a simple multiple of the ratio of their masses when they combine with each other.

Here’s an example to illustrate the law of reciprocal proportions:

Let’s consider the elements oxygen (O) and hydrogen (H) combining with chlorine (Cl) to form different compounds: hydrogen chloride (HCl) and water (H2O).

Step 1: Determine the molar masses of each element:

Molar mass of O = 16 grams/mole

Molar mass of H = 1 gram/mole

Molar mass of Cl = 35.5 grams/mole

Step 2: Calculate the masses of oxygen and hydrogen when combined with chlorine:

Hydrogen chloride (HCl): Since there is one hydrogen atom and one chlorine atom, the mass of hydrogen = 1 gram and the mass of chlorine = 35.5 grams.

Water (H2O): Since there are two hydrogen atoms and one oxygen atom, the mass of hydrogen = 2 grams and the mass of oxygen = 16 grams.

Step 3: Compare the ratios of the masses of oxygen and hydrogen when combined with chlorine:

The ratio of the masses of oxygen to hydrogen when combined with chlorine in HCl is 16 grams : 1 gram, which can be simplified to 16:1.

The ratio of the masses of oxygen to hydrogen when combined with chlorine in H2O is 16 grams : 2 grams, which can be simplified to 8:1.

Step 4: Examine the ratios of the masses of oxygen and hydrogen when combined with each other:

The ratio of the masses of oxygen to hydrogen when combined to form water (H2O) is 16 grams : 2 grams, which can be simplified to 8:1.

When we compare the ratios of the masses of oxygen and hydrogen in HCl and H2O, we find that the ratio of their masses in H2O (8:1) is a simple multiple (2 times) of the ratio in HCl (16:1).

This confirms the law of reciprocal proportions, which states that if two elements combine separately with a fixed amount of a third element, the ratio of their masses in which they combine will be either the same or a simple multiple of the ratio of their masses when they combine with each other. In this example, the ratio of oxygen to hydrogen in water (8:1) is a multiple of the ratio in hydrogen chloride (16:1).

Avogadro’s Law

Avogadro’s Law, formulated by Amedeo Avogadro, states that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.

This law is crucial in understanding the relationship between the volume of a gas and the number of molecules it contains.

Avogadro’s Law is represented by the equation: V/n = k, where V is the volume, n is the number of moles, and k is a constant.

Here are a few examples that illustrate Avogadro’s law:

Equal volumes of different gases contain an equal number of molecules:

If we have a container with 1 liter of oxygen gas (O2) and another container with 1 liter of nitrogen gas (N2), both at the same temperature and pressure, Avogadro’s law states that they will contain the same number of molecules. This means that the number of molecules in 1 liter of oxygen gas will be the same as the number of molecules in 1 liter of nitrogen gas.

The effect of molar ratios on volume:

Avogadro’s law also applies to the molar ratios of gases in chemical reactions. For example, when hydrogen gas (H2) reacts with oxygen gas (O2) to form water vapor (H2O) according to the balanced equation:

2H2 + O2 → 2H2O

Avogadro’s law states that the volume of hydrogen gas required to react with a given volume of oxygen gas (at the same temperature and pressure) will be in a ratio of 2:1. This is because the ratio of the coefficients in the balanced equation reflects the molar ratio of the gases involved.

The relationship between volume and moles:

Avogadro’s law can also be applied to the relationship between volume and the number of moles of gas. For example, if we have a certain volume of carbon dioxide (CO2) gas and double the number of moles while keeping the temperature and pressure constant, Avogadro’s law states that the volume of the gas will also double. This means that the volume of gas is directly proportional to the number of moles when other factors are held constant.

These examples illustrate Avogadro’s law, which provides a fundamental understanding of the relationship between volume and the number of particles (moles) of gas at constant temperature and pressure.

Gay-Lussac’s Law of Combining Volumes

Gay-Lussac’s Law of Combining Volumes states that the volumes of gases involved in a chemical reaction, measured at the same temperature and pressure, are always in simple ratios.

These ratios correspond to the stoichiometric coefficients in the balanced chemical equation.

The pressure of a given amount of gas is directly proportional to its temperature when the volume is held constant.

Here are a few examples that demonstrate Gay-Lussac’s law:

Balloon experiment:

If you have a balloon filled with a fixed amount of gas at a certain temperature, and then you heat the balloon by placing it near a heat source, the temperature of the gas inside the balloon increases. According to Gay-Lussac’s law, as the temperature of the gas increases, the pressure inside the balloon also increases. This can be observed by the expansion of the balloon or by measuring the pressure using a pressure gauge.

Aerosol can:

Aerosol cans, such as those containing spray paint or deodorant, rely on Gay-Lussac’s law. The cans are pressurized with gas, and a liquid or solid substance is also present inside. When the valve is opened, the propellant gas escapes, causing a decrease in pressure inside the can. This decrease in pressure, combined with the constant temperature, leads to the liquid or solid substance being forced out of the can as a spray.

Scuba diving:

When scuba diving, divers carry compressed air tanks. The air inside the tank is pressurized to a level higher than the surrounding water pressure. According to Gay-Lussac’s law, as the temperature of the compressed air increases due to the surrounding water temperature, the pressure inside the tank also increases. This allows the diver to breathe the air at a pressure matching the water pressure, enabling them to explore underwater.

Here are a few more examples that illustrate Gay-Lussac’s law:

Hydrogen and Oxygen Reacting to Form Water:

2H2 + O2 → 2H2O

According to Gay-Lussac’s law, the volumes of hydrogen gas and oxygen gas consumed in this reaction will be in a ratio of 2:1. For example, if we have 2 liters of hydrogen gas, we would need 1 liter of oxygen gas to react completely to form 2 liters of water vapor.

Nitrogen and Hydrogen Reacting to Form Ammonia:

N2 + 3H2 → 2NH3

In this reaction, Gay-Lussac’s law states that the volumes of nitrogen gas and hydrogen gas consumed will be in a ratio of 1:3. If we have 2 liters of nitrogen gas, we would need 6 liters of hydrogen gas to react completely to form 4 liters of ammonia gas.

Carbon Monoxide and Oxygen Reacting to Form Carbon Dioxide:

2CO + O2 → 2CO2

According to Gay-Lussac’s law, the volumes of carbon monoxide gas and oxygen gas consumed in this reaction will be in a ratio of 2:1. If we have 2 liters of carbon monoxide gas, we would need 1 liter of oxygen gas to react completely to form 2 liters of carbon dioxide gas.

These examples demonstrate Gay-Lussac’s law, which establishes the relationship between the volumes of gases involved in a chemical reaction. The law states that the ratios of the volumes of reacting gases can be expressed as simple whole number ratios that correspond to the coefficients in the balanced chemical equation.

Dalton’s Atomic Theory

Dalton’s Atomic Theory is a fundamental concept that provides a framework for understanding the laws of chemical combinations.

According to Dalton, elements are made up of tiny, indivisible particles called atoms, which combine in fixed ratios to form compounds.

Atoms of different elements have different masses, and chemical reactions involve the rearrangement of atoms to form new substances.

Here are a few examples that illustrate Dalton’s atomic theory with chemical equations:

Formation of Water (H2O):

According to Dalton’s atomic theory, elements are made up of indivisible particles called atoms. In the case of water formation, the reaction can be represented as follows:

2H2 + O2 → 2H2O

This equation indicates that two molecules of hydrogen gas (H2) react with one molecule of oxygen gas (O2) to form two molecules of water (H2O). Dalton’s theory suggests that each hydrogen molecule consists of two hydrogen atoms (H), and each oxygen molecule consists of two oxygen atoms (O). Therefore, the reaction involves the rearrangement and combination of atoms to form new compounds.

Formation of Ammonia (NH3):

Dalton’s atomic theory also emphasizes that atoms combine in fixed ratios to form compounds. In the case of ammonia formation, the reaction can be represented as follows:

N2 + 3H2 → 2NH3

This equation indicates that one molecule of nitrogen gas (N2) reacts with three molecules of hydrogen gas (H2) to form two molecules of ammonia (NH3). Dalton’s theory suggests that each nitrogen molecule consists of two nitrogen atoms (N), and each hydrogen molecule consists of two hydrogen atoms (H). Therefore, the reaction involves the combination of atoms in fixed ratios to form the compound ammonia.

Decomposition of Water (H2O):

Dalton’s atomic theory also explains the process of chemical decomposition. For instance, the decomposition of water can be represented as follows:

2H2O → 2H2 + O2

This equation indicates that two molecules of water (H2O) can decompose into two molecules of hydrogen gas (H2) and one molecule of oxygen gas (O2). According to Dalton’s theory, this decomposition involves the separation of water molecules into their constituent atoms, with each water molecule consisting of two hydrogen atoms (H) and one oxygen atom (O).

These examples demonstrate how Dalton’s atomic theory provides a framework for understanding chemical reactions and the behavior of atoms in those reactions. The theory highlights the indivisibility of atoms and their role in combining and rearranging to form compounds or decompose into their constituent elements.

Conclusion

Understanding the laws of chemical combinations is crucial for grasping the foundations of chemistry. These laws, including the Law of Conservation of Mass, the Law of Definite Proportions, the Law of Multiple Proportions, Avogadro’s Law, and Gay-Lussac’s Law, provide the basis

Take Quiz

1. Which law states that the ratio of masses of elements in a compound is always constant?
(a) Avogadro’s law
(b) Dalton’s law
(c) Law of definite proportions
(d) Law of multiple proportions

ANSWER

c) Law of definite proportions

2. According to the law of multiple proportions, if two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in:
a) equal proportions
b) variable proportions
c) whole number ratios
d) fractional ratios

ANSWER

c) whole number ratios

3. The law of constant composition is also known as:
a) Dalton’s law
b) Boyle’s law
c) Gay-Lussac’s law
d) Proust’s law

ANSWER

d) Proust’s law

4. According to Avogadro’s law, equal volumes of gases at the same temperature and pressure contain:
a) equal numbers of molecules
b) equal masses of molecules
c) equal numbers of atoms
d) equal masses of atoms

ANSWER

a) equal numbers of molecules

5. The law of combining volumes of gases states that the ratio of volumes of gases involved in a chemical reaction can be expressed as small whole numbers. This is known as:
a) Boyle’s law
b) Charles’s law
c) Avogadro’s law
d) Gay-Lussac’s law

ANSWER

d) Gay-Lussac’s law

6. Which law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of the individual gases?
a) Dalton’s law
b) Boyle’s law
c) Charles’s law
d) Avogadro’s law

ANSWER

a) Dalton’s law

7. The law of conservation of energy states that energy:
a) cannot be created or destroyed, only transformed
b) can be created but not destroyed
c) can be destroyed but not created
d) is irrelevant in chemical reactions

ANSWER

a) cannot be created or destroyed, only transformed

8. According to the law of definite proportions, the ratio of the masses of elements in a compound:
a) varies depending on the reaction conditions
b) is always a multiple of the smallest whole number ratio
c) is different for each reaction
d) is determined by the chemist conducting the experiment

ANSWER

b) is always a multiple of the smallest whole number ratio

9. The law of conservation of mass is not obeyed by a
a) Redox Reaction
b) Double Decomposition Reaction
c) Nuclear reaction
d) Neutralization Reaction

ANSWER

c) Nuclear reaction , since a nuclear reaction is not a chemical reaction

10. Two elements A and B combine chemically to form compounds I , II and III . The ratio of different masses of B combining with a fixed mass of A in I , II and III is 1:3:5 . If 32 parts of mass of A combine with 84 parts by mass of B in II , then in III , 16 parts of A will combine with __
a) 42 parts
b) 70 parts
c) 64 parts
d) 96 parts

ANSWER

b) 70 parts
In II, 32 parts by mass of A combine with 84 parts by mass od B . Thus 16 parts by mass of A (in II) will combine with 42 parts by mass of B . According to the data , different masses of B combining with a fixed mass of A in II and III are in the ratio 3 : 5
Mass of B in II / Mass of B in III = 3/5
Hence , mass of B in III = Mass of B in II X 5/3
= 42 X 5/3 = 70