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# Law of Chemical Combination with Examples

## Key Concepts:

1. Law of conservation of mass
2. Law of constant proportion
3. Atomic mass
4. Molecular mass

## Introduction:

By now we know that the change in which a substance changes its physical properties is called a physical change. Whereas the changes in which one or more new substance is formed is known as chemical change. Chemical reactions occur whenever a chemical change occurs. In a chemical reaction, one or more substances combine to form a new substance entirely different in properties from the original substance.

We have also learned that the number of atoms remains the same before and after the chemical reaction during chemical reaction. Only a rearrangement of atoms takes place in a chemical reaction.

If the number of atoms does not change, the chemical reaction can never change mass.

French chemist A. Lavoisier set the foundation for investigating matter by describing how the substances react with one another by following some laws.

These laws are known as the laws of chemical combination.

The five laws of chemical combination are:

• Law of conservation of mass
• Law of definite proportions
• Law of multiple proportions
• Gay Lussac’s law of gaseous volumes
• Avogadro’s law of chemical combination

### Law of Conversation of Mass

Statement: “The mass can neither be created nor it can be destroyed in a chemical reaction”.

As per this law

In a physical or chemical change, the mass of the products is always equal to the mass of the reactants.

Total mass of the reactant = Total mass of the product

In another word, “In a chemical reaction, the matter is neither produced, nor it can be destroyed.”

The law of mass conservation is also termed the “Law of the indestructibility of matter.”

A French chemist, Antoine Lavoisier, discovered the law of mass action.

Determination Of the Mass of An Atom

### Atomic Mass:

A single atom has a set of a number of protons and neutrons, so the mass is definite (will not change).

Therefore the sum of the number of protons and neutrons in the atom will be always the same.

Electrons contribute so very little mass that they are not counted.

The atomic mass is usually expressed according to the international arrangement in terms of amu, i.e., atomic mass unit.

Definition: Atomic mass of the atom of an element is defined as the average mass of the atom compared to 1/12th of the mass of one carbon-12 atom.

The mass of an atom can be characterized by a sum the mass of protons and the mass of neutrons that is almost equal to the atomic mass.

1 amu = 1.66  x 10-24

Figure: Atomic masses of a few elements

### Molecular Mass

Definition: The molecular mass of a substance is defined as the sum of the masses of each atom present in a molecule of the substance.

Therefore, it is the relative Mass of a molecule expressed in atomic mass units (u).

Example 1: To determine the relative molecular Mass of water (H2O):

The atomic mass of hydrogen = 1u

The atomic mass of oxygen = 16 u

So, the molecular mass of water (which contains two atoms of hydrogen and one atom of oxygen is)

= 2 × 1 + 1 × 16

= 18 u

Therefore, the molecular mass of the water molecule is 18 u.

Example 2: To calculate the relative molecular mass of Nitric acid (HNO3):

HNO3 contain= one hydrogen atom

1 Nitrogen atom

Three oxygen atoms

Therefore, the molecular Mass of HNO3 = (the atomic Mass of H + the atomic Mass of N + 3 × the atomic Mass of O)

= 1 + 14 + 48

= 63 u

Therefore, the molecular Mass of Nitric acid molecule is 63 u.

Examples of Law of Mass Action

Example 1: Consider the formation of the H20 molecule. Here two hydrogen molecule reacts with oxygen to produce water.

In the above case, the total mass of the reactants = total mass of the products.

Example 2: Carbon combines with oxygen at low concentration to form carbon monoxide.

In the above case, the total mass of reactants = the total mass of products.

Example 3: Burning is a chemical process. The flames are produced due to the fuel undergoing combustion (burning).

When a piece of wood burns, the mass of the smoke, ashes, and gases equals the original mass of the total charcoal and the oxygen when it first reacts, so this means that the total mass of any product equals the total mass of the reactant derived.

The Law of conversation of Mass can also be applied to Physical Changes.

Example 4: If 100 grams of ice melted, it produced 100 grams of water.

Solved Problems on Law of Conservation of Mass:

Problem 1. When 10 grams of CaCO3 is heated, it produces 4.4 g of CO2 and  5.6 g of calcium oxide(CaO). Prove that these observations are in agreement with the law of conservation of mass

Solution:

Mass of the reactants, CaCO3 : 10 g

Mass of the products = Mass of CO2 + Mass of  CaO

= 4.4 g + 5.6 g

= 10 g.

Because the Mass of the reactants molecules is equal to the Mass of the product molecules,

The above observations are in agreement with the law of conservation of Mass

Problem 2: Potassium hydroxide (KOH) readily reacts with carbon dioxide (CO2) to produce potassium carbonate (K2CO3) and water (H2O). How many grams of potassium carbonate is formed if 224.4 g of KOH reacted with 88.0 g of CO2. The reaction also produced 36.0 g of water.

Solution: Let the grams of potassium carbonate produced = X

224.4 g KOH + 88.0 g CO2 = X g K2CO3 + 36.0 g H2

312.4 g reactants = X  g K2CO3 +  36.0 g H2

312.4 g reactants – 36.0 g H2O =  X  g K2CO3

Therefore X = 276.4 g of potassium carbonate(K2CO3

### Law Of Constant Proportions

Along with other scientists, Lavoisier noted that many compounds were composed of two or more elements. Each such compound had the same elements in the same proportions, irrespective of where the compound came from or who prepared it.

Statement: This law was stated by Proust as “In a chemical substance the elements are always present in definite proportions by mass”.

Or

All chemical compounds are found to have definite composition irrespective of their method of preparation or sources.

• Two or more elements are always present in the fixed ratio by Mass.
• It does not matter from where the compound comes or how they have prepared it.
• This law is also known as the Law of definite proportion.

Example 1: Water (H2O), its chemical compound. It contains two elements: Hydrogen and Oxygen.

Since mass of H2 = 2g, and oxygen is 16 g and therefore H2 : O = 2 : 16 = 1 : 8

These two elements are always constant proportion by mass which is 1:8, i.e., There will always be 1 gram of hydrogen and 8 grams of oxygen.

It does not matter the source of water, whether it is tap water, water from river or ocean water, or produced by any chemical reaction. (melting, cooling of ice)

Example 2:  Carbon dioxide (CO2) can be obtained by different methods such as,

1. By respiration (as a by-product)
1. By Heating limestone:

CaCO3                      CaO + CO2

1. By heating Sodium bicarbonate-

2NaHCO3                     Na2CO3 +H2O + CO2

1. By the action of Hydrochloric acid (dil.) and  Calcium carbonate-

2CaCO3 + 2HCl                    CaCl2 +H2O + CO2

It has been observed that CO2 obtained by every method contains,

C :  O

12 : 32

3 :  8 (by weight)

Carbon combines with oxygen to form CO and CO2

An illustration representing the mass ratio of elements for a few compounds is given below.

The ratio of the number of atoms of each element is provided below the mass ratio.

For example, in a nitrogen dioxide (NO2) molecule, the ratio of the number of nitrogen and oxygen atoms is 1 : 2, but the mass ratio is 14 : 32 (or 7 : 16).

Example: By using the law of constant composition, if we have to demonstrate that, the two samples of cupric oxide accept by this law.

The first sample is  1.375 g cupric oxide, heated with hydrogen to yield 1.098 g of copper.

CuO  + H2                        Cu   +  H2

1.375 g 1.098 g

In the second sample, 1.179 g of copper atom was dissolved in nitric acid solution to form copper nitrate, consequently burned to form 1.476 g of cupric oxide.

Cu  + HNO3                        Cu(NO3)2                       CuO

1.179-gram  1.476 g

To work the problem, you’d need to find the mass percent of each element in each sample. It doesn’t matter whether you choose to find the percentage of copper or the percentage of oxygen. You’d simply subtract one of the values from 100 to get the percent of the other element.

In the first sample:

Weight of CuO = 1.375 g
wt. of copper = 1.098 g
Weight of oxygen = 1.375 -1.098 = 0.277 g

percent of oxygen atom in copper oxide = (0.277)(100%)/1.375 = 20.15%

Second sample:

Weight of copper = 1.179 g
Weight of copper oxide = 1.476 g
Weight of oxygen atom = 1.476 – 1.179

= 0.297 gram

percent of oxygen in CuO = (0.297)(100%)/1.476 = 20.12%

Therefore, both the samples follow the law of constant composition.

Solved Problem on Law of Constant Proportion

Problem 1: Hydrogen and oxygen atoms combine in the ratio of 1 : 8 by mass to produce water. What will be the mass of oxygen gas required to react completely with 3 g of hydrogen?

1g of Hydrogen gas reacts with 8 g of oxygen to produce water.

1 g Hydrogen → 8 g of oxygen

For 3 g of hydrogen, it requires  =  3 × 8 g of Oxygen

=  24 g of Oxygen gas.

Therefore, 24 g of Oxygen gas will be needed to react completely with 3 g of hydrogen gas.

### Summary:

•  According to the Law of Conservation of Mass, in a chemical reaction, the sum of the masses of the reactants and products remains unchanged.
• Lavoisier discovered the law of conversation of mass.
• The atomic mass of an element s defined as the average mass of the atom, compared to1/12%the mass of one carb atom.
• The mass of an atom can be characterized by the sum of protons and neutrons’ mass.
• The molecular mass of a substance is given by the sum of the atomic mass of each atom molecule of the substance.
• According to law of constant proportions, elements are always presenting a definite proportion by mass in a pure chemical compound.
• Proust discovered this law.
• The law of constant proportions has also been termed the law of definite proportion.

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