Chapter 3.5 - Chemical formula from Valency

In the previous section, we completed the discussion on valency. In this section, we will see an application of Valency.

From Valency to Chemical formula

Given below are the chemical formulae of some compounds:
• Sodium chloride NaCl • Magnesium chloride MgCl₂ • Aluminium chloride AlCl₃ •  Carbontetrachloride CCl₄  
There are four compounds. Chlorine is present in all four of them. But the number of chlorine atoms present is different in all the four compounds. Why is this so?

We will be able to understand the reason if we analyse the reaction taking place in each of them.
Consider NaCl: 
• Sodium has an electronic configuration 2,8,1. So it needs to lose one electron to attain octet.
• Chlorine has an electronic configuration 2,8,7. So it needs to gain one electron to attain octet.
• So the electron which is lost by one sodium atom is readily accepted by one chlorine atom. 
• The reaction is complete. We have seen it's details here. It is shown again below:

• So one chlorine atom is sufficient to accept the one electron given away by sodium
■ We can bring in 'valency' into the above discussion:
• Sodium has one electron in the outer most shell. So it's valency is 1. 
    ♦ We find that Valency of sodium is equal to the number of chlorine atoms in NaCl 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of sodium atoms in NaCl 
• There seems to be an interchange of valency numbers:
    ♦ Valency of Sodium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of sodium atoms

Now consider MgCl₂  
• Magnesium has an electronic configuration 2,8,2. So it needs to lose two electrons to attain octet.
• A single chlorine atom will not be able to accommodate the two electons given away by magnesium. So one more chlorine atom comes in
• The two chlorine atoms can together complete the reaction, by each of them accepting 'one of the two electrons' from magnesium. We have seen it's details here. It is shown again below: 

• So we see that there will be two chlorine atoms in magnesium chloride
■ We can bring in 'valency' into the above discussion:
• Magnesium has two electrons in the outer most shell. So it's valency is 2. 
    ♦ We find that Valency of Magnesium is equal to the number of chlorine atoms in MgCl₂ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Magnesium atoms in MgCl₂
• There seems to be an interchange of valency numbers:
    ♦ Valency of Magnesium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Magnesium atoms

Now consider AlCl₃
• Aluminium has an electronic configuration 2,8,3. So it needs to lose three electrons to attain octet.
• A single chlorine atom will not be able to accommodate the three electons given away by Aluminium. So two more chlorine atoms comes in
• The three chlorine atoms can together complete the reaction, by each of them accepting 'one of the three electrons' from Aluminium
• So we see that there will be three chlorine atoms in Aluminium chloride
■ We can bring in 'valency' into the above discussion:
• Aluminium has three electrons in the outer most shell. So it's valency is 3. 
    ♦ We find that Valency of Aluminium is equal to the number of chlorine atoms in AlCl₃ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Aluminium atoms in AlCl₃
• There seems to be an interchange of valency numbers:
   ♦ Valency of Aluminium is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Aluminium atoms

Now consider CCl₄
• Carbon has an electronic configuration 2,4. So it needs four more electrons to attain octet.
• Chlorine is also in need for electron. So they form a covalent bond. We have seen it's details here. It is shown again below:

• A single carbon atom needs to form covalent bonds with 4 chlorine atoms. Then only there will be octet. 
• So we find that there will be four chlorine atoms in CCl₄ 
■ We can bring in 'valency' into the above discussion:
• Carbon has four electrons in the outer most shell. So it's valency is 4. 
    ♦ We find that Valency of Carbon is equal to the number of chlorine atoms in CCl₄ 
• Again, Chlorine has 7 electrons in the outermost shell. So it's valency is 8-7 = 1.
    ♦ We find that Valency of chlorine is equal to the number of Carbon atoms in CCl₄
• There seems to be an interchange of valency numbers:
    ♦ Valency of Carbon is equal to the number of chlorine atoms, and in return, valency of chlorine is equal to the number of Carbon atoms

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In all the four cases that we saw above, there is an inter change of valency number and the number of atoms present. Let us write a summary: 
■ In the chemical formula NaCl, 
• The number of atoms of Na present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 1. This '1' is the valency of Na
■ In the chemical formula MgCl₂, 
• The number of atoms of Mg present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 2. This '2' is the valency of Mg
■ In the chemical formula AlCl₃, 
• The number of atoms of Al present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 3. This '3' is the valency of Al
■ In the chemical formula CCl₄, 
• The number of atoms of C present is 1. This '1' is the valency of Cl 
• The number of atoms of Cl present is 4. This '4' is the valency of C

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If there is indeed such an interchange, we get an easy method to write the chemical formulae of any compound. Let us see two more examples:

Consider MgO  
• Magnesium has an electronic configuration 2,8,2. So it needs to lose two electrons to attain octet.
• Oxygen has an electronic configuration 2,6. So it needs to gain two electrons to attain octet.
• So the electrons which are lost by one magnesium atom is readily accepted by one oxygen atom.
• So we see that there will be one magnesium atom, and one oxygen atom in magnesium oxide
We can bring in 'valency' into the above discussion:
• Magnesium has two electrons in the outer most shell. So it's valency is 2. 
• Oxygen has 6 electrons in the outermost shell. So it's valency is 8-6 = 2
• Let us interchange the valency numbers and write the chemical formula. (Here, interchanging does not make any difference because, both valencies are '2'. But we will follow the procedure, and assume that they are interchanged)
• We get the chemical formula as: Mg₂O₂
    ♦ In the above chemical equation, the suffix '2' of magnesium is the valency of Oxygen
    ♦ The suffix '2' of oxygen is the valency of magnesium
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Mg₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is MgO

Consider CO₂
• Carbon has an electronic configuration 2,4. So it needs four more electrons to attain octet.
• Oxygen has an electronic configuration 2,6. So it needs to gain two more electrons to attain octet.
• Both are in need of electrons. So they form a covalent bond as shown in fig.3.20 below:

Fig.3.20

• A single carbon atom needs to combine with 2 oxygen atoms. Then only there will be octet.
• In each bond, 2 pairs (ie., 4 electrons) are shared
• Carbon has a valency of 4, and oxygen has a valency of 2
• Interchange these numbers. So the chemical formula becomes: C₂O₄
• Divide the 'interchanged valencies' by the common factor. '2' and '4' have the common factor '2'. So we get 2 /2 = 1, and 4/2 = 2
• The chemical formula becomes: C₁O₂. If the suffix is '1', it is not usually written
• So the final chemical formula is CO₂ 

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So we find that, just by knowing the valencies of the combining elements, we can write the chemical formula of the compound. The procedure is as follows:

1. Write the symbol of the element with the lower electronegativity first

2. Interchange the valency of each element, and write as suffix

3. Divide each suffix with the common factor. (If there is a common factor)

4. If the suffix of any element is 1, it need not be written

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We will now see some solved examples
Solved example 3.6
Some elements and their valency are given below. 
• Chlorine(Cl), Valency = 1;  • Lithium(Li)  Valency = 1;   • Oxygen(O)  Valency =2;   • Zinc(Zn)  Valency = 2; • Calcium(Ca)  Valency = 2
Write the chemical formulae of the compounds that are formed when they react with each other.
Solution:
In the given list, some elements are metals, and the rest are non-metals. 
• Metals usually do not combine with other metals. 
• Non-metallic elements, some times combine with other non-metallic elements to form compounds
• The most probable cases are those in which metals combine with non-metals
• In the given list, Li, Zn, and Ca are metals. Cl and O are the non-metals

So the possible combinations are:
(i) Li with Cl, (ii) Li with O, (iii) Zn with Cl, (iv) Zn with O, (v) Ca with Cl, (vi) Ca with O, (vii) Cl with O
Let us consider each:
(i) Li with Cl:
• Valency of Li = 1, Electronegativity of Li = 0.98 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Li has the lower Electronegativity. So we write it first: LiCl
• Both valencies are the same. So interchanging gives the same result.We get Li₁Cl₁
• Suffix '1' is not written. So we get the chemical formula as: LiCl 

(ii) Li with O :
• Valency of Li = 1, Electronegativity of Li = 0.98 
• Valency of O = 2, Electronegativity of O = 3.44
• Li has the lower Electronegativity. So we write it first: LiO
• Now we rewrite the above with the interchanged valencies: We get Li₂O₁
• Suffix '1' is not written. So we get the chemical formula as: Li₂O

(iii) Zn with Cl:
• Valency of Zn = 2, Electronegativity of Zn = 1.65 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Zn has the lower Electronegativity. So we write it first: ZnCl
• Now we rewrite the above with the interchanged valencies: We get Zn₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: ZnCl2

(iv) Zn with O:
• Valency of Zn = 2, Electronegativity of Zn = 1.65 
• Valency of O = 2, Electronegativity of O = 3.44
• Zn has the lower Electronegativity. So we write it first: ZnO
• Both valencies are the same. So interchanging gives the same result.We get: Zn₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Zn₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is ZnO

(v) Ca with Cl:
• Valency of Ca = 2, Electronegativity of Ca = 1.00 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Ca has the lower Electronegativity. So we write it first: CaCl
• Now we rewrite the above with the interchanged valencies: We get Ca₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: CaCl2

(vi) Ca with O:
• Valency of Ca = 2, Electronegativity of Ca = 1.00 
• Valency of O = 2, Electronegativity of O = 3.44
• Ca has the lower Electronegativity. So we write it first: CaO
• Both valencies are the same. So interchanging gives the same result.We get: Ca₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Ca₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is CaO

(vii) O with Cl:
• Valency of O = 2, Electronegativity of O = 3.44 
• Valency of Cl = 1, Electronegativity of Cl = 3.16
• Cl has the lower Electronegativity. So we write it first: ClO
• Now we rewrite the above with the interchanged valencies: We get Cl₂O1
• Suffix '1' is not written. So we get the chemical formula as: Cl₂O

Solved example 3.7
Some elements and their valency are given below. 
• Barium (B), Valency = 2; Chlorine(Cl), Valency = 1; • Zinc(Zn)  Valency = 2   • Oxygen(O)  Valency =2; (i) Write the chemical formula of Barium chloride
(ii) Write the chemical formula of Zinc oxide
(iii) Chemical formula of calcium oxide is CaO. What is the valency of calcium?

Solution:
(i) Ba with Cl:
• Valency of Ba = 2
• Valency of Cl = 1
• Basic form of barium chloride is: BaCl
• Now we rewrite the above with the interchanged valencies: We get Ba₁Cl2
• Suffix '1' is not written. So we get the chemical formula as: BaCl2
(ii) Zn with O:
• Valency of Zn = 2
• Valency of O = 2
• Basic form of zinc oxide is: ZnO
• Both valencies are the same. So interchanging gives the same result.We get: Zn₂O₂
• Now divide the 'interchanged valencies' by the common factor. '2' and '2' have the common factor '2'. So we get 2 /2 = 1
• The chemical formula becomes: Za₁O₁. If the suffix is '1', it is not usually written
• So the final chemical formula is ZnO
(iii) The final chemical equation is given as CaO. Here we have to work in a sort of reverse order
• Let valency of Ca = x
• Valency of O = 2
• Basic form can be written as: CaO
• Now we rewrite the above with the interchanged valencies: We get Ca₂Ox
• The final form is given to us as CaO. This means:
    ♦ 2 when divided by the 'common factor of 2 and x', gives 1
    ♦ x when divided by the 'common factor of 2 and x', gives 1
This is possible only if x is equal to 2
• So we can write: valency of Ca = x = 2

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We saw how to determine the chemical formula from the valencies. In the next section, we will see Oxidation and Reduction.

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Chapter 1.2 - Chemical Equations of Reactions

In the previous section, we saw the details about Chemical formula. In this section we will see Chemical Equations.
Consider the reaction between zinc and hydrochloric acid. As a result of the reaction, we will get two substances: zinc chloride and hydrogen. Substances taking part in a reaction are called Reactants. Substances formed as a result of the reaction are called products. In the above reaction:
• Reactants are zinc and hydrochloric acid
• Products are zinc chloride and hydrogen

We have seen that the molecules can be represented by chemical formulae. Using them, we can write 'equations to represent a reaction'. For the reaction between zinc and hydrochloric acid, the equation will be:
Zn + HCl → ZnCl₂ + H₂
Let us analyse the above equation:
• On the left side we have two items. Zn and HCl  
   ♦ Zn represents Zinc, and HCl represents Hydrochloric acid. These are the reactants
• On the right side we have two items. ZnCl₂ and H₂ 
   ♦ ZnCl₂ represents zinc chloride, and H₂ represents hydrogen. These are the products
• So a 'chemical equation' gives us an idea about 'what happens during a reaction'. But some details are missing:
   ♦ On the left side, there is only one chlorine atom. But on the right side, there are two chlorine atoms
   ♦ On the left side, there is only one hydrogen atom. But on the right side, there are two hydrogen atoms
• So the equation is not balanced. It is our job to balance a given equation

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• In the above equation, zinc chloride is present. It is a compound containing Zn (zinc) and Cl (chlorine). Why is zinc chloride represented as ZnCl₂ and not ZnCl ? 
• The reason is that, a molecule of zinc chloride will always consist of:
    ♦ One atom of zinc and
    ♦ Two atoms of chlorine
• So why is it that 'one molecule of  zinc chloride will always consist of One atom of zinc and Two atoms of chlorine'?

• Similarly, in the above equation, hydrochloric acid is present. It is a compound containing H₂ (hydrogen) and Cl (chlorine). Why is hydrochloric acid represented as HCl ? Why not HCl₂ or H₂Cl ?  
• The reason is that, a molecule of hydrochloric acid will always consist of:
    ♦ One atom of hydrogen and
    ♦ One atom of chlorine
• So why is it that 'one molecule of  hydrochloric acid will always consist of One atom of hydrogen and One atom of chlorine'?

Such a question can arise in the case of many compounds. We will learn the answers in later chapters. At present, all we need to learn is, how to balance a 'given equation'. Later, we will learn to write equations on our own, and then balance them.

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To get 2 hydrogen atoms and 2 chlorine atoms on the right side, can we write H₂Cl₂ (instead of HCl) on the left side ? No. because hydrochloric acid is HCl.  H₂Cl₂ represents something else. 
While balancing chemical equations, we must never alter the subscripts. Because altering them will alter the compound. We must alter only the coefficients.

We can follow a definite procedure to balance any chemical equation. We will discuss the procedure with the help of a few examples. First we will take the above equation:

• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Hydrogen has 1 on the reactants side, and 2 on the products side
    ♦ Chlorine has 1 on the reactants side and 2 on the products side
Such differences should be under lined. We do not have to pay attention to the numbers which are not underlined. Among those which are under lined, it is always better to consider elements other than hydrogen and oxygen first. 
• So we consider chlorine. It's number on the left side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of HCl on the left side
• Now write the number of each element again in step 2. We find that hydrogen also becomes 2 on the left side
• There is nothing to underline. The equation is balanced

Another example:

• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Oxygen has 2 on the reactants side, and 1 on the products side
This difference is under lined. We do not have to pay attention to the numbers which are not underlined.  
• So we consider oxygen. It's number on the right side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of H₂O on the right side
• Now write the number of each element again in step 2. We find that oxygen is balanced, but hydrogen is 2 on the left side, and 4 on the right side
• This 2 on the left side must become 4. So put 2 in front of H₂ on the left side
• Now write the numbers for each element in step 3. There is nothing to underline. The equation is balanced

Another example:

• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Oxygen has 2 on the reactants side, and 1 on the products side
This difference is under lined. We do not have to pay attention to the numbers which are not underlined.  
• So we consider oxygen. It's number on the right side must become 2. 
• Remember that we can change only the coefficients. So, in the given equation, put 2 in front of MgO on the right side
• Now write the number of each element again in step 2. We find that oxygen is balanced, but magnesium is 1 on the left side, and 2 on the right side
• This 1 on the left side must become 2. So put 2 in front of Mg on the left side
• Now write the numbers for each element in step 3. There is nothing to underline. The equation is balanced

Another example:

• Write the 'number of atoms' of each element on the reactants side and products side. In the above table, in step 1, we have written the number of atoms according to the 'given equation'. These numbers must be same on both sides. Other wise the equation is not a balanced one. In our case, 
    ♦ Carbon has 3 on the reactants side, and 1 on the products side
    ♦ Hydogen has 8 on the reactants side and 2 on the products side
    ♦ Oxygen has 2 on the reactants side, and 3 on the products side
Such differences should be under lined. We do not have to pay attention to the numbers which are not underlined. Among those which are under lined, it is always better to consider elements other than hydrogen and oxygen first. 
• So we consider carbon. It's number on the right side must become 3 
• Remember that we can change only the coefficients. So, in the given equation, put 3 in front of CO₂ on the right side
• Now write the number of each element again in line with step 2. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen remains in the same unbalanced state: 8,2
    ♦ Oxygen's numbers have changed. But it is still unbalanced: 2,7
• We kept hydrogen and oxygen for the last. So we took carbon first. Now carbon is balanced. Out of hydrogen and oxygen, we must take hydrogen first.
• For hydrogen, the present numbers are: 8,2. The right side also must become 4
• For that, we put 4 in front of H₂O on the right side
• Now write the number of each element again in step 3. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen is balanced: 8,8
    ♦ Oxygen's numbers have changed. But it is still unbalanced: 2,10
• For oxygrn, the 2 must become 10 on the left side
• For that, we put 5 in front of O₂ on the left side
• Now write the number of each element again in line with step 4. We find that
    ♦ Carbon is balanced: 3,3
    ♦ Hydrogen is balanced: 8,8
    ♦ Oxygen is balanced: 10,10
• There is nothing to underline. The equation is balanced.

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Some more solved examples can be seen here. In the next section we will see the 'Details of Atoms'.

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Chapter 1.1 - Chemical Formula of a Compound

In the previous section, we saw the molecules of Elements. In this section we will see the molecules of Compounds. We know that a compound contains more than one element. So there will be more than one type of atoms. For example, in water, there will be atoms of hydrogen and oxygen. 

The different types of atoms in a compound will not exist independently. The atoms will always be in groups. Groups cannot be formed in any manner. There are strict rules for the formation of groups:
• Each group will have atoms of different elements
• The number of atoms of each element will be the same in any given group 
• In the case of compounds also, the name of such groups is Molecule. 
• For example, in water, there will be atoms of two elements: hydrogen and oxygen
• In any given molecule of water, there will be two atoms of hydrogen and one atom of oxygen. This is shown in the fig.1.3 below:

Fig.1.3 Molecule of Water

Representing Atoms and Molecules

The method of representing atoms and molecules can be demonstrated with the help of some examples:
We have seen that the symbol for the element sulphur is S. One atom of sulphur is also represented by the letter S. So if we see the letter S, it indicates one atom of sulphur. But no one atom of sulphur can exist independently. It is always found as groups of 8 sulphur atoms. We have seen that such groups are called molecules. So a group of 8 sulphur atoms is one molecule of sulphur. (sulphur is a polyatomic molecule). One atom of sulphur is represented as S. How will we represent one molecule of sulphur? For that we write the number 8 as a subscript. So one molecule of sulphur is represented as S₈.

If there are more than one sulphur molecule, that number is written on the left side. So, if a sample of sulphur contains 3 molecules of sulphur, it is written as 3S₈. Such a notation will also help us to calculate the total number of sulphur atoms in that sample. The calculation is as follows:
• S₈ indicates that, one Sulphur molecule has 8 sulphur atoms
• 3S₈ indicates 3 molecules of sulphur. 
• So the total number of sulphur atoms in 3 molecules = 3 × 8 = 24
• If the sample contains 2 sulphur molecules, then the total number of sulphur atoms will be 2 × 8 = 16 

Another example: The symbol for the element helium is He. When we write He, it represents one atom of helium. Each of the helium atoms are able to exist independently. So a molecule of helium will consist of only one helium atom. (Helium is a monoatomic molecule). So when we write He, it will represent one molecule of helium also. 

If there are more than one helium molecule, that number is written on the left side. if a sample of helium contains 5 helium molecules, it is written as 5He. As we saw above in the case of sulphur, this type of representation will give us the number of atoms also:
• He indicates that, one helium molecule has 1 helium atom.
• 5He indicates 5 molecules of helium 
• So the total number of helium atoms in 5 molecules = 5 × 1 = 5
• If we see 7He, it will indicate  7 molecules of helium, which is equal to 7 atoms of helium

One more example: We have seen that the symbol for the element hydrogen is H. One atom of hydrogen is also represented by the letter H. So if we see the letter H, it indicates one atom of hydrogen. But no one atom of hydrogen can exist independently. It is always found as groups of 2 hydrogen atoms. We have seen that such groups are called molecules. So a group of 2 hydrogen atoms is one molecule of hydrogen. (hydrogen is a diatomic molecule). One atom of hydrogen is represented as H. How will we represent one molecule of hydrogen? For that we write the number 2 as a subscript. So one molecule of hydrogen is represented as H₂.

If there are more than one hydrogen molecule, that number is written on the left side. So, if a sample of hydrogen contains 3 molecules of hydrogen, it is written as 3H₂. This will also give us the total number of hydrogen atoms in that sample. It can be calculated as follows:
• H₂ indicates that, one hydrogen molecule has 2 hydrogen atoms
• 3H₂ indicates 3 molecules of hydrogen 
• So the total number of hydrogen atoms in 3 molecules = 3 × 2 = 6
• If the sample contains 5 hydrogen molecules, then the total number of hydrogen atoms will be 5 × 2 = 10

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We will now see some solved examples:
Solved example 1.1
Write the number of atoms and molecules in each of the following:
(i) 3N₂ (ii) O₃ (iii) 6Cl₂
Solution: (i) 3N₂
• The subscript 2 indicates that there are two atoms in one molecule
• 3 on the left side indicates that there are a total of 3 molecules
• So the total number of atoms = 3 × 2 = 6
(ii) O₃
• The subscript 3 indicates that there are 3 atoms in one molecule
• There is no number on the left side. It indicates 1 molecule
• So the total number of atoms = 1 × 3 = 3
(iii) 6Cl₂
• The subscript 2 indicates that there are two atoms in one molecule
• 6 on the left side indicates that there are a total of 6 molecules
• So the total number of atoms = 6 × 2 = 12

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So we have seen how the molecules of elements are represented. Now we will see the representation of the molecules of compounds. Consider the example of carbon dioxide. It is a compound, formed when carbon burns in oxygen. It can also form during the decomposition of calcium carbonate. But the ‘method of formation’ of carbon dioxide is not important for our present discussion. We want to know the elements present inside the carbon dioxide, and also quantity of each of those elements.

In one molecule of carbon dioxide:
• Number of carbon atoms = 1
• Number of oxygen atoms = 2

The above numbers will be true for every molecule of carbon dioxide. That is., what ever be the method by which we obtain carbon dioxide, or from which ever source, we obtain the carbon dioxide, every molecule of it will contain one atom of carbon and two atoms of oxygen. We can write in in the form of ratio:
• In every carbon dioxide molecule, the ratio of carbon to oxygen is 1:2. 
If this ratio is not maintained, we will not get carbon dioxide. For example, if 2 atoms of carbon combine with 3 atoms of oxygen, the ratio of carbon to oxygen is 2:3. It will not be carbon dioxide.

As this ratio is always maintained, we can represent carbondioxide as CO₂. It indicates one atom of carbon and 2 atoms of oxygen. This type of representation is called the Chemical formulae of the compound. 

Another example: Any one water molecule will have two atoms of hydrogen and one atom of oxygen. So the chemical formula of water is H₂O

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• A chemical formula can have natural numbers (1, 2, 3 etc.,) in two forms: Regular form and subscript form
• Subscripts are found on the right of each element. It indicates the number of atoms of that element in the molecule. If there is only one atom of an element, it’s subscript ‘1’ is not written
• Numbers in the Regular form is always seen on the extreme left side of the chemical formula. it indicates the total number of molecules. If the total number of molecules is 1, it is not written. Such numbers in the regular form will never be seen in side a chemical formula. it will be only seen on the extreme left side

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With this knowledge, we can analyse any given chemical formula, and write the number of atoms of each element present in it. The following solved examples will demonstrate the procedure
Solved example 1.2:
Determine the number of atoms of each element from the following chemical formulae:
(i) H₂SO₄ (ii) 5H₂O (iii) ZnCl₂ (iv) 7NH₃
Solution: (i) H₂SO₄
■ There is 'no number' on the extreme left. This indicates '1'. So the given chemical formula indicates one molecule. 
• The first element is hydrogen (indicated by H). 
   ♦ It's subscript is 2. So there are 2 atoms of hydrogen
   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 1 × 2 = 2
• The second element is sulphur (indicated by S)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of sulphur atoms = Total number of molecules × number of atoms per molecule = 1 × 1 = 1
• The final element is oxygen (indicated by O). 
   ♦ It's subscript is 4. So there are 4 atoms of oxygen
   ♦ Total number of oxygen atoms = Total number of molecules × number of atoms per molecule = 1 × 4 = 4
(ii) 5H₂O
■ There is '5' on the extreme left. So the given chemical formula indicates 5 molecules. 
• The first element is hydrogen (indicated by H). 
   ♦ It's subscript is 2. So there are 2 atoms of hydrogen
   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 5 × 2 = 10
• The second element is oxygen (indicated by O)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of Oxygen atoms = Total number of molecules × number of atoms per molecule = 5 × 1 = 5
(iii) ZnCl₂
■ There is 'no number' on the extreme left. This indicates '1'. So the given chemical formula indicates one molecule.
• The first element is zinc (indicated by Zn)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of Zinc atoms = Total number of molecules × number of atoms per molecule = 1 × 1 = 1
• The second element is chlorine (indicated by Cl). 
   ♦ It's subscript is 2. So there are 2 atoms of chlorine
   ♦ Total number of chlorine atoms = Total number of molecules × number of atoms per molecule = 1 × 2 = 2
(iv) 7NH₃
■ There is '7' on the extreme left. So the given chemical formula indicates 7 molecules
• The first element is nitrogen (indicated by N)
   ♦ There is 'no subscript'. This indicates '1' atom
   ♦ Total number of nitrogen atoms = Total number of molecules × number of atoms per molecule = 7 × 1 = 7
• The second element is hydrogen (indicated by H). 
   ♦ It's subscript is 3. So there are 3 atoms of hydrogen

   ♦ Total number of hydrogen atoms = Total number of molecules × number of atoms per molecule = 7 × 3 = 21

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In the next section we will see Chemical Equations.

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