Class IX, CHEMISTRY, "Chemical Combinations and Chemical Equations"

CHEMICAL COMBINATIONS
&
CHEMICAL REACTION

Laws of Chemical Combinations

There are four laws of chemical combinations these laws explained the general feature of chemical change. These laws are:
1. Law of Conservation of Mass
2. Law of Definite Proportions
3. Law of Multiple Proportions
4. Law Reciprocal Proportions

Antoine Lavoiser has rejected the worn out ideas about the changes that take place during a chemical reaction. He made careful quantitative measurements in chemical reactions and established that mass is neither created nor destroyed in a chemical change.

1. Law of Conservation of Mass

Statement

It is presented by Lavoiser. It is defined as:

 “Mass is neither created nor destroyed during a chemical reaction but it only changes from one form to another form.”

In a chemical reaction, reactants are converted to products. But the total mass of the reactants and products remains the same. The following experiment easily proves law of conservation of mass.

Practical Verification (Landolt Experiment)

German chemist H. Landolt, studied about fifteen different chemical reactions with a great skill, to test the validity of the law of conservation of mass. For this, he took H-shaped tube and filled the two limbs A and B, with silver nitrate (AgNO₃) in limb A and Hydrochloric Acid (HCl) in limb B. The tube was sealed so that material could not escape outside. The tube was weighed initially in a vertical position so that the solution should not intermix with each other. The reactant were mixed by inverting and shaking the tube. The tube was weighed after mixing (on the formation of white precipitate of AgCl). He observed that weight remains same.

HCl + AgNO₃ ———> AgCl + NaNO₃

2. Law of Definite Proportions

Statement

It is presented by Proust. It is defined as:

“When different elements combine to give a pure compound, the ratio between the masses of these elements will always remain the same.”

Proust proved experimentally that compound obtained from difference source will always contain same elements combined together in fixed proportions.

Example

Water can be obtained from different sources such as river, ocean, well, canal, tube well, rain or by the chemical combination of hydrogen and oxygen. If different samples of water are analyzed, it will have two elements, hydrogen and oxygen and the ratio between their mass is 1:8.

3. Law of Multiple Proportions

Statement

This law is defined as:

“When two elements combine to give more than one compounds, the different masses of one element, which will combine with the fixed mass of other element, will be in simple whole number ratio.”

Two different elements can combine to form more than one compound. They can do so by combining in different ratios to give different compounds.

Example

Hydrogen and oxygen combine with one another to form water (H₂O) and hydrogen peroxide (H₂O₂). In water and hydrogen oxide 2 g of hydrogen combine with 16g and 32g of oxygen respectively. According to law of multiple proportions, the different masses of oxygen (16g and 32g) which have reacted with fixed mass (2g) of hydrogen will have a simple ratio between each other i.e. 16:32 or 1:2. It means that hydrogen peroxide contains double the number of oxygen atoms than water. This law proves this point of Dalton’s Atomic Theory that atoms do not break in a chemical reaction.

4. Law of Reciprocal Proportions

Statement

This law is defined as:

“When two element A, B combine separately, with the mixed mass of the third element E, the ratio in which these elements combine with E is either the same or simple multiple of the ratio in which A and B combine with each other.”

Example

Hydrogen and Nitrogen separately combine to form ammonia (NH₃) and dinitrogen oxide (N₂O), in these compounds, fixed mass of nitrogen is 14g and combines with 8 g of oxygen and 3 g of hydrogen. The ratio between the mass of oxygen and hydrogen is 8:3. Hydrogen and oxygen also combine with one another to form water (H₂O). The ratio between hydrogen and oxygen in water is 16:2. These ratios are not same. Let us observe whether these ratios are simple multiple to each other or not following mathematical operation is carried out.

8:3 ::16:2
8/3 : 16/2
or
8/3 x 2/16
or
1/3 => 1:3

Atom

It is the smallest particle of an element which can exist with all the properties of its own element but it cannot exist in atmosphere alone.

OR

The smallest particle of an element which cannot exist independently and take part in a chemical reaction is known as Atom.

Examples

Hexogen(H), Carbon (C), Sodium (Na), Gold (Au) etc.

Molecule

When two or more than two atoms are combined with each other a molecule is formed. It can exist freely in nature.

OR

The particle of a substance (Element or Compound) which can exist independently and show all the properties of that substance is called molecule.

Atoms of the same or different elements react with each other and form molecule.
Atoms of some elements can exist independently, since they have property of molecule so they are called mono atomic molecule.

Examples

Examples of Molecules of the elements are Hydrogen (H₂). Nitrogen (N₂), Oxygen (O₂) etc.
Molecules of different elements are called compounds. For example HCl, H₂O, CH₄ etc.

Valency

The combining capacity of all elements with other elements is called valency.

Example
H = 1
C = 4
Al = 3
Mg = 2
Na = 1

Chemical Formula

“A brief name used for full chemical name at a compound is called Chemical Formula.”

A chemical formula is used to represent an element or a compound in terms of symbols. It also represents the number and type of atoms of elements present in the smallest unit of that substance.

Example

The chemical formula of hydrogen sulphide is H₂S. It shows two types of elements (H and S) and number of
atoms of element (2H and 1S). Similarly the formula of NaCl show number and type of different atoms present in its smallest unit.

Formula Weight

It is the sum of the weights of the atoms present in the formula of a substance.

Molecular Weight

It is the sum of the atomic masses of all the atoms present in a molecule.

Empirical Formula

Definition  It is the simplest formula of a chemical compound which represents the element present of the compound and also represent the simplest ratio between the elements of the compound.
Examples

·         The empirical formula of benzene is “CH”. It indicates that the benzene molecule is composed of two elements carbon and hydrogen and the ratio between these two elements is 1:1.

·         The empirical formula of glucose is “CH₂O”. This formula represents that glucose molecule is composed of three elements carbon, hydrogen and oxygen. The ratio between carbon and oxygen is equal but hydrogen is double.

Determination of Empirical Formula

To determine the empirical formula of a compound following steps are required.
1. To detect the elements present in the compound.
2. To determine the masses of each element.
3. To calculate the percentage of each element.
4. Determination of mole composition of each element.
5. Determination of simplest ratio between the element of the compound.

Illustrated Example of Empirical Formula

Consider an unknown compound whose empirical formula is to be determined is given to us. Now we will use the above five steps in order to calculate the empirical formula.

Step I – Determination of the Elements

By performing test it is found that the compound contains magnesium and oxygen elements.

Step II – Determination of the Masses

Masses of the elements are experimentally determined which are given below.

·         Mass of Mg = 2.4 gm

·         Mass of Oxygen = 1.6 gm

Step III – Estimation of the Percentage

The percentage of an element may be determined by using the formula.
% of element = Mass of element / Mass of compound x 100
In the given compound two elements are present which are magnesium and oxygen, therefore mass of compound is equal to the sum of the mass of magnesium and mass of oxygen.
Mass of compound = 2.4 + 1.6 = 4.0 gm
% of Mg = Mass of Mg / Mass of Compound x 100
               = 2.4 / 4.0 x 100
               = 60%
% of O    = Mass of Oxygen / Mass of Compound x 100
                = 1.6 / 4.0 x 100
                = 40%

Step IV – Determination of Mole Composition

Mole composition of the elements is obtained by dividing percentage of each element with its atomic mass.
Mole ratio of Mg = Percentage of Mg / Atomic Mass of Mg
Mole ratio of Mg = 60 / 24
                             = 2.5
Mole ratio of O    = Percentage of Oxygen / Atomic Mass of Oxygen
                             = 40 / 16
                             = 2.5

Step V – Determination of Simplest Ratio

To obtain the simplest ratio of the atoms the quotients obtained in the step IV are divided by the smallest quotients.
Simple ratio of Element = Value of Mole ratio / minimum Value of mole ratio
Simple ratio of  Mg = 2.5 / 2.5
                                = 1
Simple ratio of O    = 2.5 / 2.5
                                = 1
Thus the empirical formula of the compound is MgO
Note
If the number obtained in the simplest ratio is not a whole number then multiply this number with a smallest number such that it becomes a whole number maintain their proportion.

Molecular Formula

Definition
The formula which shows the actual number of atoms of each element present in a molecule is called molecular formula.

OR

It is a formula which represents the element ratio between the elements and actual number of atoms of each type of elements present per molecule of the compound.

Examples

The molecular formula of benzene is “C₆H₆″. It indicates that
1. Benzene molecule is composed of two elements carbon and hydrogen.
2. The ratio between carbon and hydrogen is 1:1.
3. The number of atoms present per molecule of benzene are 6 carbon and 6 hydrogen atoms.
The molecular formula of glucose is “C₆H₁₂O₆″. The formula represents that
1. Glucose molecule is composed of three elements carbon, hydrogen and oxygen.
2. The ratio between the atoms of carbon, hydrogen and oxygen is 1:2:1.
3. The number of atoms present per molecule of glucose are 6 carbon atoms. 12 hydrogen atoms and 6 oxygen atoms.

Determination of Molecular Formula

The molecular formula of a compound is an integral multiple of its empirical formula.
Molecular formula = (Empirical formula) _n
Where n is a digit = 1, 2, 3 etc.
Hence the first step in the determination of molecular formula is to calculate its empirical formula by using the procedure as explained in empirical formula. After that the next step is to calculate the value of n
value of n = Molecular Mass / Empirical Formula Mass

Example

The empirical formula of a compound is CH₂O and its molecular mass is 180.
To calculate the molecular formula of the compound first of all we will calculate its empirical formula mass.
Empirical formula mass of CH₂O = 12 + 1 x 2 + 16
                                                      = 30
Value of n = Molecular Mass / Empirical Formula Mass
                  = 180 / 30
                  = 6
Molecular formula = (Empirical formula) _n
                               = (CH₂O) ₆
                               = C₆H₁₂O₆

Molecular Mass

Definition
The sum of masses of the atoms present in a molecule is called as molecular mass.

OR

It is the comparison that how mach a molecule of a substance is heavier than 1/12th weight or mass of carbon atom.
Example
The molecular mass of CO₂ may be calculated as

Molecular mass of CO₂ = Mass of Carbon + 2 (Mass of Oxygen)
                                       = 12 + 2 x 16
                                       = 44 a.m.u

Molecular mass of H₂O = (Mass of Hydrogen) x 2 + Mass of Oxygen
                                        = 1 x 2 + 16
                                        = 18 a.m.u

Molecular mass of HCl = Mass of Hydrogen + Mass of Chlorine
                                      = 1 + 35.5
                                      = 36.5 a.m.u

Gram Molecular Mass
Definition
 
The molecular mass of a compound expressed in gram is called gram molecular mass or mole.
Examples
1. The molecular mass of H₂O is 18. If we take 18 gm H₂O then it is called 1 gm molecular mass of H₂O or 1 mole of water.
2. The molecular mass of HCl is 36.5. If we take 36.5 gm of HCl then it is called as 1 gm molecular mass of HCl or 1 mole of HCl.

Mole
Definition
It is defined as atomic mass of an element, molecular mass of a compound or formula mass of a substance expressed in grams is called as mole.

OR

The amount of a substance that contains as many number of particles (atoms, molecules or ions) as there are atoms contained in 12 gm of pure carbon.
Examples
1. The atomic mass of hydrogen is one. If we take 1 gm of hydrogen, it is equal to one mole of hydrogen.
2. The atomic mass of Na is 23 if we take 23 gm of Na then it is equal to one mole of Na.
3. The atomic mass of sulphur is 32. When we take 32 gm of sulphur then it is called one mole of sulphur.
From these examples we can say that atomic mass of an element expressed in grams is called mole.
Similarly molecular masses expressed in grams is also known as mole e.g.
The molecular mass of CO₂ is 44a.m.u. If we take 44 gm of CO₂ it is called one mole of CO₂ or the molecular mass of H₂O is 18a.m.u. If we take 18 gm of H₂O it is called one mole of H₂O.
When atomic mass of an element expressed in grams it is called gram atom
While
The molecular mass of a compound expressed in grams is called gram molecule.
According to the definition of mole.
One gram atom contains 6.02 x 10²³atoms
While
One gram molecule contain 6.02 x 10²³ molecules.

Molar Mass
The mass of one mole of a substance is called molar mass.
Example
1 mole of Hydrogen atom (H) = 1.008g
1 mole of Hydrogen molecule (H₂) = 2.016g
Thus mass of substance is related to the particles by mole.

Avogadro’s Number
An Italian scientist, Avogadro’s calculated that the numbers of particles (atoms, molecules) in one mole of a substance are always equal to 6.02 x 10²³. This number is known as Avogadro’s number and represented as N_A.
Example
1 gm mole of Na contain 6.02 x 10²³ atoms of Na.
1 gm mole of Sulphur = 6.02 x 10²³ atoms of Sulphur.
1 gm mole of H₂SO₄ = 6.02 x 10²³ molecules H₂SO₄
1 gm mole of H₂O = 6.02 x 10²³ molecules of H₂O
On the basis of Avogadro’s Number “mole” is also defined as
Mass of 6.02 x 10²³ molecules, atoms or ions in gram is called mole.

Determination Of The Number Of Atoms Or Molecules In The Given Mass Of A Substance
Example 1
Calculate the number of atoms in 9.2 gm of Na.
Given:
Mass of Na = 9.2 gm
Atomic mass of Na = 23 a.m.u
Number of Atoms = N_A = ?
Solutions:
            Using Formula
N_A = mass of substance / molecular Mass x 6.02 x 10²³
      = 9.2 / 23 x 6.02 x 10²³
      =  2.41 x 10²³ Atoms

Determination Of The Mass Of Given Number Of Atoms Or Molecules Of A Substance
Example 2
Calculate the mass in grams of 3.01 x 10²³ molecules of glucose (C₆H₁₂O₆).
Given:
Molecular mass of glucose = 180 a.m.u
Number of Molecules = N_A = 3.01 x 10²³ molecules
Mass of Glucose = ?
Solutions:
            Using Formula
N_A = mass of substance / molecular Mass x 6.02 x 10²³
Mass of Glucose = N_A x Molecular Mass / 6.02 X10²³
                            = 3.01 x 10²³ x 180 / 6.02 X10²³
                            = 90 gm

Chemical Reaction
A chemical change in which reactants are converted to products is called chemical reaction.

Zn + 2HCl ——–> ZnCl₂ + H₂

The fact that a chemical reaction is taking place can be inferred from the following observation.
1. Evolution of a gas
2. Change in colour
3. Change in temperature.
4. Emission of light.

Types of Chemical Reaction
The chemical reaction is classified into following types:

1. Displacement Reaction
The reaction in which an atom or group of atoms is displaced by another atom or group of atoms in a compound is called displacement reaction.

Fe + CuO ———> Cu + FeO

2. Double Displacement Reactions
The reactions in which reacting substances exchange their radicals or ions are double displacement reaction. Insoluble salts are formed by mixing soluble salts.

3. Addition Reactions
When two different compounds or elements react together to give only one confound, the reaction will be called addition reaction.

2Mg + O₂ ——–> 2MgO

4. Decomposition Reaction
The reaction in which some compounds may decompose into elements or simpler compounds on heating is called decomposition reaction.

CaCO₃ ———> CaO + CO₂ (Heat)

Chemical Equation
Symbolic representation of chemical change in terms of symbols and formulae is called Chemical Equation.

Method of Equation Writing
A chemical equation can be written as follows:
1. Write the formulae and symbols of the reactants on the left hand side.
2. Write the formulae and sympols of the products on the right hand side.
3. Separate the reactants and products by an arrow which is directed towards the products.

Characteristics of Chemical Equation
1. Chemical equation must be representative of a chemical reaction.
2. It should represent molar quantities.
3. It should be balanced in terms of atoms/molecules of reactants and products.

Reactants
Those substances, which react together in a chemical reaction, are called reactants.

Zn + 2HCl ——> ZnCl₂ + H₂

In the above reaction Zn and HCl are the reactants.

Products
Those substances, which are formed in a chemical reaction, are called products.

Zn + 2HCl ——> ZnCl₂ + H₂

In the above reaction, ZnCl₂ and H₂ are products.
Information obtained from a Chemical Equation
1. A balanced equation indicates that which reactant undergo chemical change. It indicates that which products are formed.
2. It indicates that how many moles of reactants under go chemical change. It indicates that how many moles of products are formed.

Why are Chemical Equations Balanced
A chemical equation must be balanced in order to satisfy the law of conservation of matter, which states that matter can neither be created nor be destroyed during a chemical reaction.

 Mass – Mass Relationship
In this relationship we can determine the unknown mass of a reactant or product from a given mass of the substance involved in the chemical reaction by using a balanced chemical equation.
Example
Calculate the mass of CO₂ that can be obtained by heating 50 gm of limestone (CaCO₃).
Solution
Step I  (Write a Balanced Equation)

CaCO₃  ——> CaO + CO₂

Step II  (Write Down The Molecular Masses And Moles Of Reactant & Product)

CaCO₃  ——> CaO + CO₂

Method I – MOLE METHOD
Number of moles of 50 gm of CaCO3 = 50 / 100 = 0.5 mole
According to equation
1 mole of CaCO₃ produce 1 mole of CO₂
0.5 mole of CaCO₃ produce 0.5 mole of CO₂
Now,
                    We Know That
Mole = Mass of Substance / Molecular Mass
Therefore,
Mass of CO₂ = Moles x Molecular Mass
                     = 0.5 x 44
                     = 22 gm

Method II – FACTOR METHOD
From equation we may write as
100 gm of CaCO₃ gives = 44 gm of CO₂
1 gm of CaCO₃ will give = (44/100) gm of CO₂
50 gm of CaCO₃ will give = (50 x 44 / 100) gm of CO₂
50 gm of CaCO₃ will give = 22 gm of CO₂