reactants.

For example, 12 g carbon combines with 32

g of oxygen to form 44 g of carbon dioxide:
C  O2  CO2
The idea of tiniest unit of matter, viz. "anu" and 12 g 32 g 44 g
"parmanu", was put forth by Maharishi Kanad in Vedic
period in our country. Another Indian philosopher,
Pakudha Katyayama, elaborated this doctrine and said
that these particles normally exist in combined form,
which gives us various forms of matter.
Democritus, a Greek philosopher, also proposed that
matter is made up of extremely small particles, "the To verify the law of conservation of mass i.e. mass of
atoms". The name comes from Greek word 'atomos' the reactants is same as the mass of the products in
means "indivisible". a chemical reaction.
However, all these theories proposed by the Indian • Prepare 5% aqueous solutions of barium chloride
and Greek philosophers were based on abstract and sodium sulphate in separate test tubes.
thinking and not on experiments. For about 2000 • Take a small amount of sodium sulphate solution in
years, the atomic theory remained a mere speculation a conical flask and some solution of barium chloride in
till John Dalton, an English school teacher, an ignition tube (a small tube made of glass).
propounded his theory on atoms. Dalton's theory was • Hang the ignition tube in the flask carefully; see that
based on the chemical laws of combination known at the solutions do not get mixed. Put a cock on the
that time. These laws of chemical combination were flask.
given by A.L. Lavoisier and Joseph L. Proust and were • Weigh the flask with its content carefully.
based upon extensive study of chemical reactions. • Now, tilt the flask so that the solutions mix with
each other.
Laws of Chemical Combination • Weigh again.
• Observations: As soon as the two solutions mix, a
By studying the quantitative measurements of many chemical reaction takes place to form new products.
reactions it was observed that the chemical reactions Barium chloride + Sodium sulphate  Barium
taking place between various substances are sulphate + Sodium chloride
governed by certain laws. These laws are called the BaCl2  Na2 SO4   BaSO4  2 NaCl
'laws of chemical combination'. These laws formed ( white ppt .)
the basis of Dalton's atomic theory. There are two We put a cork on the mouth of the flask to avoid any
main laws of chemical combination which are as exchange of matter between the flask and its
follows: surrounding.
(1) Law of Conservation of Mass It is observed that the mass of the flask and its
(2) Law of Constant Proportion (also known as the law contents remain the same even after the chemical
of definite composition) reaction has taken place.
• Conclusion: The activity shows that in chemical
Law of Conservation of Mass reaction, mass of the reactants is the same as the
This law deals with the relation between the mass of mass of the products. This proves the law of
the reactants and the products during the chemical conservation of mass.
changes. It was stated by a French chemist, A.
Lavoisier in 1774. The law of conservation of mass
states that during any chemical change, the total mass
of the products is equal to the total mass of the
reactants. In other words, the law of conservation of 1. 10.0 g of CaCO3 on heating gave 4.4 g of CO2
mass means that mass can neither be created nor and 5.6 g CaO. shows that these observations
destroyed during any chemical combination. This law are in agreement with the law of conservation
is also known as law of indestructibility of matter. of mass.
From a large number of experiments, it was Sol.: The above reaction takes place according to
concluded that although substances may undergo the following equation:
chemical changes, the total mass of the products of 
CaO3( s )  CaO( s )  CO2( g ) 
the reaction is exactly equal to the total mass of the
2
 Mass of reactant (CaCO3 )  10 g • Observation: It is observed that 90 g of water on
Mass of products decomposition gives 10 g of hydrogen gas and80 g of
(CaO  CO2 )  (5.6  4.4) g  10 g oxygen gas,
Since, mass of the reactants is equal to the Mass of hydrogen gas obtained 10 g 1
Thus,  
mass of the products, therefore, these Mass of oxygen gas formed 80 g 8
observations are in agreement with the law of From the above two experiments we observe that
conservation of mass. water formed in the laboratory had hydrogen and
oxygen gases combined in the ratio of 1 : 8 by mass.
Law of Constant Proportion
At the same time, water obtained fromany source
This law states that in a pure chemical substance the also had hydrogen and oxygen gases combined in the
elements are always present in definite proportions ratio of 1 : 8 by mass.
by mass. The above results are in accordance with the law of
This law is also known as the law of definite constant composition.
proportions. This means that, whatever, maybe the
source from which a compound is obtained, it is
always made up of the same elements in the same
proportion by mass.

2. Weight of copper oxide obtained, by heating

4.32 g of metallic copper with nitric acid and
subsequent ignition, was 5.40 g. In another
experiment 2.30 g of copper oxide oil
Experiment I (Production of water) reduction yielded 1.84 g of copper. Show
that these findings are in. accordance with the
• Collect the water formed by burning 10 g hydrogen law of constant proportion.
gas completely in 80 g of oxygen. Sol.: Experiment I
• Record the mass of the water so obtained. Weight of copper = 4.32 g
• Observation: The mass of water so obtained is Weight of copper oxide = 5.40 g
found out to be 90 g.  Weight of oxygen = Weight of copper oxide
 Ratio in which hydrogen and oxygen combine Weight of copper
10 1  5.40  4.32  1.08 g
 
80 8 Ratio of copper arid oxygen
Mass of copper 4.32 4
 
Mass of oxygen 1.08 1
 ratio is 4:1

Experiment II
Weight of copper = 1.84 g
Weight of copper oxide = 2.30 g
 Weight of oxygen = Weight of copper oxide
Thus, we see that hydrogen and oxygen combine in Weight of copper
the ratio of 1:8.  2.30  1.84  0.46g
Hydrogen and oxygen combine in the ratio of 1:8 by Ratio of copper and oxygen
mass to form water. Mass of copper 1.84 4
 
Experiment II (Decomposition of water) Mass of oxygen 0.46 1
• Take water from any source (say river, pond, lake,  ratio is 4:1
tap water, well, etc.) and distill it to remove unwanted As the ratio of copper and oxygen (by mass) is same in
salts. the two experiments, law of constant proportion is
• Now take 90 g of pure water and pass electricity verified.
through it to cause its decomposition. Collect the
gases so obtained.

3
 - Atoms of the same element are identical in all
respects, i.e., size, shape, mass and properties.
- Atoms of different elements have different sizes and
3. Copper oxide was prepared by two different masses and also possess different properties.
methods. In one case, 1.75 g of the metal - Atoms of the same or different elements combine
gave 2.19 g of oxide. In the second case, 1.14 together to form molecules or compounds.
g of the metal gage 1.43 g of the oxide. Show - When atoms of different elements combine together
that the given data illustrate the law of to form compounds, they do so in a simple whole
constant proportions. number ratio such as 1 : 1, 2 : 1, 2 : 3 etc.
Sol.: In case I, mass of copper = 1.75 g - Atoms of two different elements may combine in
Mass of copper oxide = 2.19 g different ratios to form more than one compound. For
% of copper in the oxide example, carbon and oxygen may combine to form
Mass of copper carbon monoxide (CO) and carbon dioxide (CO2) in
  100 which the ratios of the combining atoms (C and O) are
Mass of copper oxide
1 : 1and 1 : 2 respectively.
1.75 - The number and lend of atoms in a given compound
 100  79.9%
2.19 is always fixed.
 % of oxygen  100  79.9  20.1% - Atom is the smallest particle that takes place in a
In case II, mass of copper = 1.14 g chemical reaction. In other words, whole atoms rather
Mass of copper oxide = 1.43 g than fractions of atoms take part in a chemical
1.14 reaction.
% of copper in the oxide 100  79.7%
1.43 - An atom can neither be created nor destroyed, i.e.,
 % oxygen  100  79.7  20.3% atom is indestructible.
Thus, copper oxide prepared by any of the given
methods contains copper and oxygen in the same Drawbacks of Dalton's Atomic Theory
proportion by mass (within the experimental error.) (1) Atom is no longer considered as the smallest
Hence, it proves the law of constant proportions. indivisible particles. This is because recent
studies have shown that atom is made up of
4. Calculate the mass of carbon present in 2 g of still smaller particles called electrons, protons
carbon dioxide. and neutrons.
Sol.: Carbon dioxide (CO2 ) contains C and O in the (2) Atoms of the same element may have
fixed proportion by mass, which is 12 : 32, i.e., different masses. For example, there are two
3 : 8. This means that 3 g of carbon combine with 8 g types of atoms of chlorine with masses 35 and
of oxygen to form 11 g of CO2 . In other words, 37 (on the atomic scale, called 'mass
numbers'). Such atoms of the same element
11 g of CO2 contains C  3 g
with different mass numbers are called
3
 2 g of CO2 will contain C   2  0.545 g. isotopes.
11 (3) Atoms of different elements may have same
masses. For example, atoms of potassium and
Dalton's Atomic Theory calcium are known with same mass number
John Dalton provided the basic theory about the (40). Such atoms of different elements with
nature of matter. He took the name atoms for the same mass numbers are known as isobars.
smallest particle of matter as given by the Greek. (4) Substances made up of the same kind of
Based on the laws of chemical combination, he atoms may have different properties. For
proposed a model of atom known as Dalton atomic example, charcoal, graphite and diamond are
theory. all made up of carbon atoms but have
The main postulates of the Dalton's atomic theory different physical properties.
are:
- All matter, whether an element, a compound or a
mixture is made up of extremely small particles called
5. Dalton postulated that “Atoms of same
atoms (i.e., same name was used for the smallest
element are same in all respects.” How was
indivisible particles as used by Greek philosophers).
this statement proved wrong?
4
Sol.: The discovery of isotopes i.e., atoms of same the basis of some important property of the element.
element having same atomic number but However, as more and more elements were
different mass number, proved the above discovered, an International Committee was set up,
statement wrong. called International Union of Pure and Applied
Chemistry (IUPAC),which approved the names of the
What is an Atom? different elements.
The symbol of an element is the "first letter" or "first
The smallest unit of an element, which may or may
letter and another letter" of the English name or
not exist independently, but always takes part in a
Latin name of the element. However, in all cases, the
chemical reaction, is called an atom.
first letter is always capital and another letter (if
Atoms are the building blocks of matter. They are
added) is always a small letter.
smaller than anything you can imagine or compare
with. More than a million atoms when stacked upon
Symbols of elements derived from English names
one another would make a layer as thick as the sheet
of paper of your science book. English name of the Symbol
element
The size of an atom is indicated by its radius which is
called 'atomic radius'. It is measured in 'nanometers'. 1. Argon Ar
One nanometer is one billionth part of a metre. 2. Arsenic As
3. Aluminium Al
1 nanometre (nm)  109 metre (m) or 1 metre (m)
4. Boron B
 109 nanometre (nm)
5. Barium Ba
The table below gives the relative size of the radius of
6. Bromine Br
an atom of hydrogen with respect toother objects.
7. Carbon C
Species Atomic radius 8. Calcium Ca
Hydrogen atom 1010 m 9. Chlorine Cl
Molecule of water 109 m 10. Chromium Cr
Molecule of haemoglobin 108 m 11. Cobalt Co
12. Fluorine F
Grain of sand 104 m
13. Hydrogen H
An ant 102 m
14. Helium He
Watermelon 101 m 15. Iodine I
Symbols Used to Represent Atoms of Different 16. Lithium Li
Elements 17. Magnesium Mg
18. Manganese Mn
In case of elements, symbol means a short method of
19. Nitrogen N
representing the full name of an element. Dalton was
the first scientist to suggest specific symbols in terms 20. Neon Ne
of figures for different elements known at that time 21. Nickel Ni
(which were limited in number). In fact, the symbol 22. Oxygen O
used by him also represented the quantity of the 23. Phosphorus P
element, i.e., one atom of the element. A few of these 24. Platinum Ft
symbols, as proposed by Dalton are given below: 25. Sulphur S
26. Uranium U
27. Zinc Zn

Symbols derived from Latin names

English name of Latin name of the Symbol
the element element
1. Antimony Stibium Sb
2.Copper Cuprum Cu
3. Gold Aurum Au
Different elements have been named either after the 4. Iron Ferrum Fe
name of the place where they were first discovered or 5.Lead Plumbum Pb
after the name of the scientist who discovered it or on
5
 6. Mercury Hydragyrum Hg 23. Platinum Pt 195
7. Potassium Kalium K 24. Gold Au 197
8. Silver Argentum Ag 25. Lead Pb 207
9.Sodium Natrium Na 26. Uranium U 238
10. Tin Stannum Sn
Gram Atomic Mass
Atomic Mass It is the quantity in grams which is numerically equal
Atoms are extremely small particles. Hence, the actual to the atomic mass of an element on a.m.u. scale.
masses of the atoms are so small that, it is difficult to For example, 1 gram atom of hydrogen = 1.008 g,
determine the actual masses of individual atoms. 1 gram atom of sodium = 23 g
In 1961, IUPAC recommended the use of an isotope of The relationship between gram atoms, mass and
carbon with mass number 12 as the standard atomic mass of a substance is,
Mass in grams
reference for measuring atomic masses. It is called Number or gram atoms 
carbon-twelve (C-12) and is represented as 12 C. The Gram atomic mass
atomic mass (or earlier called as "atomic weight") is
now defined as follows:
The atomic mass of an element is the relative mass
of its atoms as compared with the mass of an atom 6. Calculate the number of gram atoms in
of carbon-12 isotope taken as 12 units. (i) 640 g of sulphur
1 (ii) 360 g of magnesium
Atomic mass unit (amu) may be defined as th of Sol.: (i) Atomic mass of sulphur =32 amu
12
the mass of an atom of carbon-12 isotopeon the  One gram atom of sulphur =32g
1 Now, 32 g of stilphiir = 1 grain atom of sulphur
atomic scale, i.e., 1 amu  th of mass of C-12 1
12  640 g of sulphur   640  20 g sulphur
isotope. 32
(ii) Atomic mass of magnesium = 24 amu
Atomic masses of some common elements (in amu or u).  One gram atom of magnesium = 24 g
Element Symbol Atomic Mass Now 24 g of magnesium = 1 g atom
1. Hydrogen H 1 1
 360 g   360  15 g
2. Helium He 4 24
3. Lithium Li 7
4. Boron B 11
5. Carbon C 12 7. An atom of copper has a mass of
6. Nitrogen N 14 1.0625  1022. How, many atoms of copper are
7. Oxygen O 16 there in 16 g?
8. Fluorine F 19 Soln.: In 1.0625 1022 g of copper, number of atoms
9. Neon Ne 20 =1
10. Sodium Na 23  in 16 g of copper, number of atoms
11. Magnesium Mg 24 1
12. Aluminium Al 27  16  1.505 1023
1.0625 1022
13. Phosphorus P 31
14. Sulphur S 32
15. Chlorine Cl 35.5 How do Atoms Exist?
16. Argon Ar 40 The atoms of only a few elements called noble gases
17. Potassium K 39 (such as helium, neon, argon, etc) are chemically un
18. Calcium Ca 40 reactive and exist in free state or single atoms. Atoms
19. Iron Fe 56 of most of the elements are chemically reactive and
20. Copper Cu 63.5 do not exist in Free State or single atom. Atoms
21. Zinc Zn 65 usually exist in two ways:
22. Silver Ag 108 (i) in the form of molecules.
(ii) in the form of ions.
6
 Molecules means that two atoms of oxygen combine together to
A molecule is a group of two or more atoms which are form a molecule. The formula of oxygen molecule is
held together strongly by some kind of attractive
O2 .
forces. Such an attractive force holding the atoms
together is called a chemical bond.
We may also define a molecule as follows: Atomicity
A molecule is the smallest particle of an element or a
compound which can exist freely under ordinary The number of atoms present in one molecule of an
conditions and shows all the properties of that element is called its atomicity.
substance. The atomicity of an element is indicated by writing
the number as a subscript on the right hand side
Molecules of an Element bottom of the symbol.
For example, H 2 shows that the atomicity of
A molecule of an element consists of the same type of hydrogen is 2. P4 shows that the atomicity of
atoms bonded together. For example, a molecule of phosphorus is 4, He shows that the atomicity of
helium is 1.
oxygen is formed when 2 atoms of oxygen combine
On the basis of their atomicities, the elements may be
together. Oxygen atom alone cannot exist classified as monoatomic, diatomic, triatomic, tetra
atomic, etc.
independently. It exists as a diatomic molecule. This
S. No Atomicity Name of the Class Examples
1. 1 Monoatomic (i) Noble gas : Helium (He), Argon (Ar),
Neon (Ne), Krypton (Kr).
(ii)Metals (Exists as large clusters)
Examples : Sodium (Na), Magnesium
(Mg), Aluminium (Al).
(iii) Carbon
2 2 Diatomic (i) Hydrogen (H2) (ii) Oxygen (O2),
(iii) Chlorine (Cl2), (iv) Fluorine (F2),
(v) Nitrogen(N2)
3. 3 Triatomic Ozone (O3)
4. 4 Tetratomic Phosphorus (P4)
5. More than 4 Polyatomic Sulphur (S8), Fullerenes (C60)

Molecules of Compounds proportion. The molecules of compounds may also be

diatomic, triatomic, tetra-atomic and polyatomic in
In the molecules of compounds, the atoms of
nature depending upon the number of the atoms
different elements are combined or bonded together
linked or combined by chemical bonds. For example,
by chemical bonds. These are present in definite
proportion by mass according to law of constant

Compound Combining elements Nature Ratio by mass

Hydrogen chloride (HCl) Hydrogen, Chlorine Diatomic 1 : 35.5
Water (H2O) Hydrogen, oxygen Triatomic 1:8
Ammonia (NH3) Hydrogen, nitrogen Tetra-atomic 14:3
Carbon dioxide (CO2) Carbon, oxygen Triatomic 3:8

The molecules of elements are homoatomic in nature atoms are present in them. For example, a molecule
which means that the atoms present in them are the of nitrogen element is homoatomic in nature (N2). A
same. The molecules of the compounds are molecule of methane (CH4) which is a compound is
heteroatomic in nature in the sense that different heteroatomic in nature.

7
 Ions Information Conveyed by a Chemical Formula

An ion may be defined as an atom or group of atoms By looking at the chemical formula of a substance, we
having positive or negative charge. can gather the following information’s:
• The ion which has one or more positive charges is 1. Name of the substance.
called Cation. At the same time, the ion carrying one 2. Name of various elements present in that
or more negative charges is known as Anions. substance.
• A few positively charged ions or cations are: 3. Chemical formula of a substance represents one
Na+(sodium ion), 1C (potassium ion), Ca+ (calcium ion), molecule of that substance.
Al 3 (aluminium ion). 4. Relative number of atoms of various elements
• A few negatively charged ions or anions are: present in one molecule of that element or
Cl  (chloride ion), S 2 (sulphide ion), (OH  ) compound.
5. Relative masses of various elements in the
(hydroxide ion), ( SO42 ) (sulphate ion). compound.
6. It represents one mole of that substance.
Molecular Formula of a Compound 7. We can calculate the gram molecular mass of that
substance.
The symbolic representation of a molecule of a For example, the chemical formula of CO2 conveys
compound, is called its molecular formula. the following informations:
The molecule of a compound contains two or more 1. The substance is carbon dioxide.
than two types of elements. If a molecule of a 2. Carbon dioxide is composed of the elements;
compound contains two different kinds of elements carbon and oxygen.
only, it is called binary compound. 3. In a molecule of carbon dioxide, carbon and
For example, water ( H 2 O ), sodium chloride ( NaCl ), oxygen atoms combine together in the ratio1 : 2
iron sulphide (FeS), etc. are binary compounds. by number.
4. The ratio of mass of carbon and oxygen is 3 : 8.
To calculate the ratio of number of atoms in a 5. It represents one mole of carbon dioxide
molecule of a compound. molecules.
The ratio of number of atoms in a molecule of a 6. The molecular mass of carbon dioxide is 44 amu.
compound can be calculated, if we know the ratio of
their masses and the atomic masses of the elements, Writing Chemical Formula of Compounds
constituting the molecule of the compound.
Chemical formula of a molecular compound
represents the actual number of atoms of different
elements present in one molecule of the compound,
e.g., chemical formula of water is H2O, that of
8. The ratio of masses of carbon and oxygen is ammonia is NH3, while that of carbon dioxide is CO2
3:8 and their atomic masses are 12 and 16 and so on.
respectively, in molecule of carbon dioxide. Chemical formula of an ionic compound simply
Calculate the ration of atoms of carbon and represents the ratio of the cations and anions present
oxygen in the molecule of carbon dioxide. in the structure of the compound, e.g., the formula
Sol.: Na  Cl  simply represents that sodium chloride
Element Ratio Atomic Mass ratio  Simple contains Na  and Cl  ions in the ratio of 1 : 1
by Mass (u) Atomic Mass ratio (though the actual crystal of sodium chloride consists
Mass of very large but equal number of Na  ions and Cl 
C 3 12 3  12  0.25 1
ions). Similarly, the formula Mg 2 Cl2 simply tells that
O 8 16 8  16  0.50 2
 Ratio of atoms of carbon to oxygen in carbon it contains Mg 2 and Cl  ions in the ratio of 1 : 2 and
dioxide = 1 : 2 so on.

8
 Valency of an Element Sodium Na  Magnesium Iron Fe3
Mg 2
Valency of an element is defined as the combining
Silver Ag  Zinc Zn 2  Gold Au 3
capacity of the element. It is equal to the number of
hydrogen atoms or number of chlorine atoms or Copper Cu  Cobalt Co 2 
double the number of oxygen atoms with which one Gold Au  Copper Cu 2
atom of the element combines. Ammonium Iron Fe2
In addition to the atoms, the ions which are charged ( NH 4 )
species, also have some valencies. Positive ions or
cations have positive valencies. Negative ions or
Some elements have shown more than one valencies.
anions have negative valencies. The valencies of the
In such cases, Roman Numerals are used to denote
poly-valent ions are expressed by enclosing them in
the valencies. These are put in bracket. For example,
bracket and putting the positive or negative signs
copper (I) and copper (II);similarly, iron (II) and iron
outside it. Let us write the valencies of some
(III).
commonly used positive and negative ions.
Valencies of Negative Ions
Valencies of Positive Ions
Positive ions may be monovalent, bivalent, trivalent,
Valencies of Negative Ions
tetravalent etc. depending upon the charge present
Like positive ions, negative ions may also be
on them. These are listed in the following table.
monovalent, bivalent, trivalent, etc. in nature. These
List of some common positive ions (Cations)
are also listed;
Monovalent Bivalent Trivalent

Hydrogen H Barium Ba Aluminium Al 3
2

Potassium K  Calcium Ca 2  Chromium Cr 3

List of some common negative ions (Anions)

Monovalent Bivalent Trivalent
 2
Chloride Cl Sulphide S Nitride N 3
Bromide Br  Oxide O 2 Phosphide P3
 2
Iodide I Carbonate (CO3 ) Phosphate ( PO4 )3
Hydroxide (OH ) Sulphate ( SO4 )2 Borate ( BO3 )3
Nitrate ( NO3 ) Sulphite (SO3 )2 Arsenate ( AsO3 )3
Nitrite ( NO2 ) Manganate (MnO4 )2
Bicarbonate ( HCO3 ) Oxalate (C2O4 )2
Cyanide (CN ) Chromate (CrO4 )2
Permanganate ( MnO4 ) Dichromate (Cr2O7 )2
Chlorate (ClO3 )

Formulae of Ionic Compounds Step I: Write the symbol of the cation showing the
charge on it. Write the symbol of the anion showing
One of the most important points to remember while the charge on it, on the right hand side of the cation.
writing the formula of a chemical compound is that it Step II: If a compound contains polyatomic ions, the
is always electrically neutral. In other words, the formula of the ion is enclosed within brackets before
positive and negative valencies of the ions present in crisscrossing the valencies.
the chemical compound add up to zero. To write a Step III: Divide the valency number by common
formula, follow the steps given below. This method of factor, if any, to get simple ratio. Now ignore the (+)
writing formula is called the crisscross method. and (-) symbols.

9
Step IV: Now, write the valency of the cation at the Step V: If these subscripts are 1, these are not written
bottom right of the anion and the charge number of in the final stoichiometric formulae.
the anion at the bottom of the cation. Thus, the Examples:
symbol of cation is subscribed with the charge
number of the anion and the anion is subscribed with
the charge number of the cation. This is called the
criss-crossing of valencies.

Compound Symbols with Valency Shifting Valency number Formula

Calcium chloride Ca 2 Cl  CaCl2

Mg 2 SO42 or Mg1SO41
Magnesium sulphate MgSO4
(Dividing by C.F .  2 )

Aluminium sulphate
Al 3 SO4 2 Al2 ( SO4 )3

Aluminium phosphate Al 3 PO43 or Al1PO41

AlPO4
(Dividing by C.F .  3 )

Ammounium phosphate NH 43 PO43 ( NH 4 )3 PO4

Potassium dichromate
K 1 Cr2O7 2 K 2Cr2O7

Calcium bicarbonate
Ca 2 HCO31 Ca ( HCO3 ) 2

Formulae of Molecular Compounds 3. Formula of ammonia

While writing the chemical formulae of the molecular

compounds, we write the constituent elements and
their valencies as shown below. Then, we crossover
Formula of ammonia is NH3.
the valencies of the combining atoms.
4. Formula of carbon tetrachloride
1. Formula of hydrogen chloride

Formula of carbon tetrachloride is CCl4 .

Formula of hydrogen chloride is HCl.
Molecular Formulae of Important Acids and Bases
2. Formula of hydrogen sulphide
Molecular Formulae of Important Acids
Acid Molecular Acid Molecular
Formula Formula
Hydrochloric HCl Acetic acid CH 2COOH
Formula of hydrogen sulphide is H2S.
10
acid Sodium NaHCO3 Aluminium Al2 (CO3 )3
Nitric acid HNO3 Carbonic H 2 CO3 hydrogen carbonate
acid carbonate
Sulphuric H 2 SO4 Sulphurous H 2 SO3 Potassium K 2CO3 Zinc ZnCO3
acid acid carbonate carbonate
Phosphoric H 3 PO4 Nitrous acid HNO2 Potassium KHCO3 Iron (II) FeCO3
acid hydrogen carbonate
carbonate
Calcium CaCO3 Lead (II) PbCO3
Molecular Formulae of Important Bases carbonate carbonate
Base Molecular Base Molecular Calcium Ca ( HCO3 ) 2 Tin (II) SnCO3
Formula Formula hydrogen carbonate
Ammonium NH 4OH Aluminium Al (OH )3 carbonate
hydroxide hydroxide Magnesium MgCO3 Copper (II) CuCO3
Sodium NaOH Zinc Zn(OH ) 2 carbonate carbonate
hydroxide hydroxide
Potassium KOH Iron (II) Fe(OH ) 2
hydroxide hydroxide Molecular Formulae of Sulphides
Calcium Ca (OH ) 2 Iron (III) Fe(OH )3 Sulphide Molecular Sulphide Molecular
hydroxide hydroxide Formula Formula
Magnesium Mg (OH ) 2 Copper (II) Cu (OH ) 2 Ammonium ( NH 4 ) 2 S Iron (III) Fe2 S3
hydroxide hydroxide sulphide sulphide
Sodium Na2 S Lead (II) PbS
sulphide sulphide
Molecular Formulae of Important Salts Potassium K 2 S Tin (II) SnS
sulphide sulphide
Molecular formulae of Oxides Calcium CaS Copper (II) CuS
Oxide Molecular Oxide Molecular sulphide sulphide
Formula Formula Magnesium MgS Mercury HgS
Sodium oxide Na2 O Iron (II) FeO sulphide (II)
oxide sulphide
Potassium oxide K 2O Iron (In) Fe2 O3 Zinc ZnS Silver (I) Ag 2 S
oxide sulphide sulphide
Calcium oxide CaO Lead (II) PbO Iron (II) FeS Barium BaS
oxide sulphide sulphide
Magnesium MgO Lead (IV) PbO2
oxide oxide
Aluminium oxide Al2 O3 Copper CaO Molecular Formulae of Nitrates
(II) oxide Nitrate
Zinc oxide ZnO Mercury HgO
(II) oxide Ammonium NH 4 NO3 Iron (II) Fe( NO3 ) 2
nitrate nitrate
Sodium NaNO3 Lead (II) Pb( NO3 ) 2
Molecular Formulae of Carbonates and Hydrogen nitrate nitrate
Carbonates Potassium KNO3 Tin (II) Sn( NO3 ) 2
Carbonate/ Molecular Carbonate/ Molecular nitrate nitrate
Hydrogen Formula Hydrogen Formula Calcium Ca ( NO3 ) 2 Copper (II) Cu ( NO3 ) 2
Carbonate Carbonate nitrate nitrate
Sodium Na2 CO3 Magnesium Mg ( HCO3 ) 2 Magnesium Mg ( NO3 ) 2 Mercury Hg ( NO3 ) 2
carbonate hydrogen nitrate (II) nitrate
carbonate Aluminium Al ( NO3 )3 Silver (I) AgNO3
11
nitrate nitrate chloride chloride
Zinc nitrate Zn( NO3 ) 2 Magnesium MgCl2 Copper (II) CuCl2
chloride chloride
Molecular Formulae of Chlorides Zinc chloride ZnCl2 Mercury HgCl2
Chloride Molecular Chloride Molecular (II) chloride
Formula Formula Aluminium AlCl3 Silver (I) AgCl
Ammonium NH 4 Cl Iron (II) FeCl2 chloride chloride
chloride chloride
Sodium NaCl Iron (III) FeCl3
chloride chloride
Potassium KCl Lead (II) PbCl2
chloride chloride
Calcium CaCl2 Tin (II) SnCl2

Molecular Formulae of Sulphates

Sulphate Molecular Sulphate Miolecular
Formula Formula
Ammonium sulphate ( NH 4 ) 2 SO4 Iron (11) sulphate FeSO4
Sodium sulphate Na2 SO4 Iron (III) sulphate Fe2 ( SO4 )3
Potassium sulphate K 2 SO4 Lead (II) sulphate PbSO4
Calcium sulphate CaSO4 Tin (II) sulphate SnSO4
Magnesium sulphate MgSO4 Copper (II) sulphate CuSO4
Aluminium sulphate Al2 ( SO4 )3 Mercury (II) sulphate HgSO4
Zinc sulphate ZnSO4 Silver (I) sulphate Ag 2 SO4

Molecular Mass Similarly, molecular mass of CO2  atomic mass of

C  2 (atomic mass of oxygen)
Molecular mass of a substance (element or  12  2(16)  12  32  44 a.m.u.
compound) is the average relative mass of its Other examples
molecules as compared with that of an atom of C-12 Molecular mass of H 2  2  atomic mass of hydrogen
isotope taken as 12. In other wards, molecular mass
 2(1)  2 a.m.u.
of a substance represent the number of times the
Molecular mass of NH 3  atomic mass of nitrogen + 3
molecule of that substance is heavier than 1/12th of
the mass of an atom of C-12 isotope. (atomic mass of hydrogen)
 14  3(1)  17 a.m.u.
Calculation of Molecular Mass Formula Mass

As molecules are made up of two or more atoms of In case of the compounds formed by ions (ionic
the same or different elements, and each atom has a compounds) formula of the compound does not
definite atomic mass, therefore, molecular mass of a represent its molecule, but only represents the ratio
molecule of a substance can be calculated by adding of different ions in the compounds.
atomic masses of all the atoms present in one This is called formula unit of the ionic compound.
molecule of the substance. Formula mass of an ionic compounds is obtained by
adding atomic masses of all the atoms in a formula
For example, unit of the compound.
Molecular mass of H 2O  2 (atomic mass of
hydrogen) + atomic mass of oxygen
 2(1)  16  18 a.m.u.

12
For Example, (b) Na2 O
Formula mass of calcium oxide (CaO)  atomic mass (c) Fe2 O3
of calcium + atomic mass of oxygen Sol.: (a) CaCl2 has been formed by
 40  16  56 u

Gram Atomic Mass and Gram Molecular Mass

Therefore, the valency of calcium (i.e., metal)
Atomic mass expressed in grams is called gram atomic is 2.
mass of that element. (b) Na2 O has been formed by
For example,
Atomic mass of hydrogen = 1.0 u
Gram atomic mass of hydrogen = 1.0 g
Atomic mass of oxygen = 16.0 u So, the valency of sodium (i.e., metal) is 1.
Gram atomic mass of oxygen = 16.0 g (c) Fe2 O3 has been formed by

The amount of an element having mass equal to gram

atomic mass is called one "gram atom" (or g atom) of
the element. Therefore, the valency of iron (i.e., metal) is 3.
For example,
11. Write the formula for the following:
1 g atom of hydrogen = 1.0 g
(a) Ferric chloride
1 g atom of oxygen = 16.0 g
(b) Ferrous sulphate
(c) Cupric iodide
Molecular mass expressed in grams is called gram
(d) Cupric oxide
molecular mass of that substance.
(e) Ammonium hydroxide
For example, Molecular mass of H 2  2.0 u
(f) Calcium phosphate
Gram molecular mass of H 2  2.0 g Sol.: (a) FeCl3 (b) FeSO4
Molecular mass of O2  32.0 u (c) Cul2 (d) CuO
Gram molecular mass of O2  32.0 g (d) NH 4OH (f) Ca3 ( PO4 )2

The amount of the substance having mass equal to its 12. Find the molecular mass of the following:
gram molecular mass is called one "gram molecule" (a) O2 (b) Cl2
(or g molecule) of the substance. Thus, (c) C2 H 6 (d) PCl3
1 g molecule of H 2  2.0 g [Atomic masses: O  16, C  12,
1 g molecule of O2  32.0 g Cl  35.5, H  1, P  31 ]
Sol.: O2  2 (atomic mass of oxygen)
 2(16)  32u
(b) Cl2  2 (atomic mass of chlorine)
 2(35.5)  71u
9. Write the formula of compounds formed by:
(c) C2 H 6  2 (atomic mass of C) +6 (atomic mass of
(a) Ag  and S 2
hydrogen)
(b) Cr 3 and O 2   2(12)  6(1)  24  6  30u
(c) Al 3 and SO42 (d) PCl3  atomic mass of phosphorus + 3 (atomic
Sol.: (a) Ag 2 S mass of chlorine)
(b) Cr2 O3  31  3(35.5)
(c) Al2 ( SO4 )3  31  106.5  137.5u

10. What is the valency of the metal in the 13. Find the formula mass of the following:
following formula? (a) ZnO (b) Na2 O (c) CuSO4 .5H 2O
(a) CaCl2 [Atomic masses: Zn  65, O  16,
13
 Na  23, Cu  63.5, H  1 ]
Sol.: (a) ZnO  atomic mass of Zn + atomic mass of
oxygen
 65  16  81 u
14. Calculate the mass percentage composition of
(b) Na2O  2 (atomic mass of sodium) + atomic mass
copper in copper sulphate.
of oxygen [Cu  64, S  32, O  16]
 2(23)  16  46  16  62 u
Sol.: Unit formula mass of CuSO4
(c) CuSO4 .5H 2O  atomic mass of Cu  atomic mass
 1(Cu)  1( S )  4(O)  1(64)  1(32)  4(16)
of S  4 (atomic mass of O) + 5 (molecular mass of
 64  32  64  160 amu.
H 2O )
 Mass % age composition of copper
 63.5  32  4(16)  5(18)
64 100
 249.5 u   40%
160
15. 4.9 g of sulphuric acid contains 0.1 g of
Mass Percentage Composition of an Element in a hydrogen, 1.6 g of sulphur and rest oxygen.
Compound Calculate the mass percentage composition of
The percentage composition of an element in a all the elements of sulphuric acid.
compound can be determined, if we know the Soln.: Mass of sulphuric acid (W) = 4.9 g
molecular mass or unit formula mass and the atomic Mass of hydrogen (1 )  0.1 g
weight of the element. Mass of sulphur (2 )  1.6 g
If M is the molecular mass (or unit formula mass) of a  Mass percentage of hydrogen
compound and A, the atomic mass of its atoms for w 0.1
some particular element then  1 100  100  2.04%
W 4.9
Mass percentage composition of the element =
Mass percentage of sulphur
Sum of atomic masses of the atoms of element A
 100  100 w 1.6
Molecular mass M  2  100   100  32.65%
However, if molecular mass and atomic mass are not W 4.9
given, instead given the mass of compound and mass  Mass percentage of oxygen
of element in it, then percentage composition can be  100  (2.04  32.65)  65.31%
calculated as follows: 16. Calculate the mass of 0.72 gram molecule of
Let W be the mass of a compound which contain  as carbon dioxide (CO2 ).
the mass of an element. Sol.: Molecular mass of CO2  Atomic mass of
 Mass percentage composition or the element = C  2 Atomic mass of
Mass element w
 100   100 O  (12 u  2 16 u)  44 u Gram molecular
Mass of compound W mass of CO2  44 g 1 gram molecule
CO2  44 g
0.72 gram molecule of CO2
(44)
  (0.72)  31.68 g
(1)
 To convert atomic mass into gram atomic mass
Mole Concept
and to convert molecular mass in gram molecular
mass, we have to simply replace 'u' by 'g'. Quite commonly, we use different units for counting
 The masses of hydrogen and oxygen in water can such as dozen for 12 articles, score for 20 articles and
be determined by electrolyzing water. The gross for 144 articles irrespective of their nature.
volumes of hydrogen and oxygen obtained during In a similar way, chemists use the unit 'mole' for
electrolysis are measured, and their masses counting atoms, molecules ions, etc.. A mole is a
calculated. collection, of 6.022 1023 particles. Thus:
A mole represents 6.022 1023 particles.

14
The number 6.022 1023 is called Avogadro number 44
 1021  0.073 g
and is symbolized as N A or N 0 . In other words, a 6.022 1023
mole is an Avogadro number of particles. For example Mass of CO2 left  (0.2  0.073)  0.127 g
1 mole of hydrogen atoms  6.022  1023 hydrogen 0.127 g
Moles of CO2 left 
atoms. 44 g
1 mole of hydrogen molecules  6.022  1023 3
So, 2.8 10 moles of CO2 are left.
hydrogen molecules.
19. How many electrons are present in 1.6 g of
1 mole of sodium ions  6.022  1023 sodium ions. methane?
1 mole of electrons  6.022  1023 electrons. Sol.: Gram-molecular mass of methane (CH4)
The amount of substance that contains the same  12  4  16 g
number of entities (atoms, molecules, ions or other Number of moles in 1.6 g of methane
particles), as the number of atoms present in 0.012 kg
1.6
(or 12 g) of the carbon-12 isotope.   0.1
16
Mole in Terms of mass Number of molecules of methane in 0.1 mole
 0.1 6.022  1023
A mole of atoms is defined as that amount of the  6.022 1022
substance (element) which has mass equal to gram One molecule of methane has  6  4  10
atomic mass (i.e., atomic mass expressed in grams). In electrons
other words, it is equal to one gram atom of the So, 6.022 1022 molecules of methane have
element.  10  6.022 1022 electrons
A mole of molecules is defined as that amount of the  6.022  1023 electrons
substance (element or compound) which has mass 20. Calculate the number of moles in
equal to gram molecular mass (i.e., molecular mass 3.0115  1023 atoms of calcium,
expressed in grams).In other words, it is equal to one
Sol.: 1 mole of any substance contains 6.022 1023
gram molecule of the substance. For example,
particles.
If the number of chemical entities of a substance is
known, then the number of moles of that substance  6.022 1023 atoms of calcium = 1
can be calculated as follows, mole
1
1 atom of calcium 
Number of chemical entities of a subs tan ce ( N ) N
Number of moles  or n  moles
6.022  1023 ( Avogadro ' s Number , N A ) NA
6.022  1023
3.0115  1023 atoms of calcium
1
  3.0115 1023
6.022 1023
 0.5 mole
21. Calculate the mass of
17. What is the number of molecules in 0.25
(a) 0.2 mol molecules of oxygen
moles of oxygen?
(b) one atom of aluminium
Sol.: We know that:
(c) 3.0 mol of Cl ion
1 mole of oxygen contains
(d) 2.5 mol of NaCl
 6.022  1023 molecules (Atomic masses:
So, 0.25 mole of oxygen contains O  16 u, Cl  35.5 u, Na  23 u. Al  27 u )
 6.022 1023  0.25  1.505 1023 molecules Sol.: (a) Molecular mass of oxygen (O2 )  32 u
Thus, 0.25 mole of oxygen contains
 Molar mass of O2  32 g mol 1
1.505 1023 molecules.
Now, 1 .mole of O2 molecule = 32 g
18. From 200 me, of CO2 , 1021 molecules are
 0.2 mole of O2 molecule  32  0.2  6.4 g
removed. How many moles of CO2 are left?
(b) Atomic mass ( Al )  27 u
Sol.: Gram-molecular mass of CO2  44 g
Mass of 1021 molecules of CO2
 Molar mass  27 g mol 1
Now 6.022  1023 atoms of aluminium weigh
15
  27 g Molar mass of iron ( Fe)  56 g mol 1
 1 atom of aluminium will weigh Molar mass of magnesium (Mg )  24 g mol 1
27
  4.48  1023 g Molar mass of H 2  2.0 g mol 1
6.022 10 23

(c) Formula mass of Cl ion  35.5 u Molar mass of O2  32.0 g mol 1

 Molar mass of Cl ions  35.5 g mol 1 Molar mass of H 2O  18.0 g mol 1
Number of moles = 3 (given) Molar mass of NH3  17.0 g mol 1
m( Mass) Molar mass is usually represented by the symbol ‘M’.
Now, n 
M ( Molar mass) Thus, M ( H 2O)  18 g mol 1 and so on.
 m  n  M  3  35.5  106.5 g The volume occupied by one mole of any gas at STP is
(d) Formula mass of NaCl  23  35.5  58.5 u always same and equal to 22400 mL or 22.4 L or
 Molar mass of NaCl (M )  58.5 g mol 1 22.4 dm3 . This volume is called molar volume or gram
molecular volume (G.M.V.).
Number of moles = 2.5 (given)
m
Now, n  Mole concept for Ionic Compounds
M
 m  n M Ionic compounds (like NaCl , KNO3 , etc.) do not
 2.5  58.5
consist of molecules. They consist of a cluster of ions.
 146.25 g
In case of ionic compounds, a mole is defined as
follows:
Mole in Terms of Volume (1) A mole of an ionic compound is that amount
of the substance which has mass equal to
In case of gaseous substances, it is found that gram formula unit mass, i.e., formula unit
Avogadro's number of molecules (i.e., one mole of mass of the ionic compound expressed in
molecules) of any gas under standard conditions of grams.
temperature and pressure, i.e., STP conditions (0°C (2) A mole is defined as that amount of the
and one atmospheric pressure) occupy the same substance which contains Avogadro's number
volume which is equal to 22400 mL or 22.4 litres. of formula units.
Hence, e.g., 1 mole of sodium chloride ( NaCl ) 
A mole of a gaseous substance is defined as that
Gram formula unit mass of NaCl
amount of the substance which has volume equal to
 23  35.5 g  58.5 g
22400 mL at STP conditions.
Thus, 1 mole of CO2 gas = 22400 mL of CO2 at STP = Avogadro's number (6.022 1023 ) of formula units
1 mole of NH3 gas = 22400 mL of NH3 at STF of NaCl
1 mole of SO2 gas = 22400 mL of SO2 at STP  6.022 1023 Na  ions  6.022 1023 Cl ions
Similarly, 1 mole of potassium nitrate
Molar mass and Molar volume ( KNO3 )  39  14  3  16 g
 101 g
The mass of 1 mole of the substance (i.e., Avogadro's
 6.022  1023 formula units of KNO,
number of particles) is called molar mass of that
substance. If the substance is atomic, its molar mass is  6.022 1023 K ions  6.022 1023 NO3ions
equal to gram atomic mass. If the substance is  6.022 1023 K atoms  6.022 1023 N atoms
molecular, its molar mass is equal to gram molecular 3  6.022 1023 O atoms
mass. As it is mass of one mole of the substance, its Combining all the definitions of mole, we arrive at the
units are gram per mole ( g mol 1 ) or kilogram per following results:
mole (kg mol 1 ). 1 mole of atoms = Gram atomic mass  6.022  1023
atoms
Thus,
1 mole of molecules = Gram molecular mass
Molar mass of C-12 isotope
1 1  6.022  1023 molecules
 12 g mol or 0.012 kg mol
= 22.4 L at STP if the substance is a gas.
16
1 mole of an ionic compound = Gram formula unit
mass
 6.022  1023 formula units.

Relationship between mole, Avogadro number and mass.

1 mole of carbon
atoms

atoms of C 12 g of carbon

1 mole of hydrogen
atoms

atoms of H 1 g of hydrogen

1 mole of any particle

(atoms, molecules ion)
number of Relative mass of those
that particle particles in grams

1 mole of
molecules
number of Molecular mass in
that particle grams

Volume of gas
at STP
Recapitulation
Multiply by 22.4 L Divide by 22.4 L

Multiply by
Avogadro’s number
Number of Number of
molecules (N) moles (n)
Divide by
Avogadro’s number

Multiple by molar Divide by molar mass (M)

mass (M)

Mass in
grams

17
 CONCEPT MAP

Atoms Molecules, Ions

Atoms:
Smallest unit of an
element

Atomic mass Atomicity: The number of Mole concept: A mole is

atoms that a molecule of the amount of the
an element contains. substance which
contains the same
number of particles as Molecules: A Ions: An atom
there are atoms group of two having +ve or
more atoms –ve charge
in 12 g of
isotope

Molar volume: The

Mass
volume occupied by 1
Relative Atomic Gram Atomic mole of any gas at STP
Mass: The mass Mass: = 22400 mL
of one atom of
The atomic mass of
the element an element
relative to 1/12th expressed in grams.
the mass of 1 Relative Gram Formula Mass: Molecular
atom of the Molecular Mass: Molecular The sum of mass
element C-12 The mass of a Mass:
atomic
molecule of the
Molecular mass masses of all
substance as
compared to expressed in atoms in a
1/12th the
grams formula unit
Hydrogen Scale: C-12 Scale: mass of an of the
The relative The relative atomic atom of C-12 compound
atomic mass of an mass of an element
element is the is the ratio of the
number of times mass of an atom of
an atom of the the element to
element of the 1/12th the mass of
element is an atom C-12
heavier than an
atom of hydrogen

18
 Mole Concept for ionic Compounds Law of Reciprocal Proportions

• Number of molecules This law was given by Richter in 1794. The law states
No. of particles ( N ) that when definite mass of an element A combines
( n) 
Avogadro ' s number ( N0 ) with two other elements B and C to form two
N
compounds and if B and C also combine to form a
 n …(1) compound, their combining masses are in same
N0
proportion or bear a simple ratio to the masses of B
Also Number of Moles and C which combine with a constant mass of A.
Mass of the subs tan ce (m)
( n) 
Molar mass( M )
m
 n …(2)
M
N m
From (1) and (2) n   - For example, hydrogen combines with sodium and
N0 M
chlorine to form compounds NaH and HCl,
m
N  N0 respectively.
M In NaH , Sodium 23 parts Hydrogen one part
 Number of molecules In HCl, Chlorine 35.5 parts Hydrogen one part
Mass of the subs tan ce
  N0
Molar mass - Sodium and chlorine also combine to form NaCl in
which 23 parts of sodium and 35.5 parts of chlorine
Law of Multiple Proportion are present. These are the same parts which combine
with one part of hydrogen in NaH and HCl
This law was put forward by Dalton in 1808. According respectively.
to this law if two elements combine to form more
than one compound, then the different masses of one
element which combine with a fixed mass of the other
element, bear a simple ratio to one another.
Hydrogen and oxygen combine to form two
compounds H2 (water) and H2O2 (hydrogen peroxide) - Hydrogen combines with sulphur and oxygen to
form compounds H 2 S and H 2 O respectively.
In water, Hydrogen 2 parts Oxygen 16 parts In H 2 S , Hydrogen 2 parts Sulphur 32 parts
In hydrogen peroxide, Hydrogen 2 parts Oxygen 32 parts
In 2
H O , Hydrogen 2 parts Oxygen 16 parts
- The masses of oxygen which combine with same
mass of hydrogen in these two compounds bear a Thus, according to this law, sulphur should combine
simple ratio 1:2. with oxygen in the ratio of 32 : 16 or a simple multiple
Nitrogen forms five stable oxides. of it.
N 2 O Nitrogen 28 parts Oxygen 16 parts Actually, both combine to form SO2 in the ratio of 32
N 2 O2 Nitrogen 28 parts Oxygen 32 parts : 32 or 1 : 1.
- The law of reciprocal proportions is a special case of
N 2O3 Nitrogen 28 parts Oxygen 48 parts
a more general law, the law of equivalent masses,
N 2 O4 Nitrogen 28 parts Oxygen 64 parts
which can be stated as under:
N 2O5 Nitrogen 28 parts Oxygen 80 parts "In all chemical reactions, substances always react in
the ratio of their equivalent masses."
- The masses of oxygen which combine with same
mass of nitrogen in the five compounds bear a ratio Law of Gaseous Volumes
16 : 32 : 48 : 64 : 80 or 1 : 2 : 3 : 4 : 5.
This law was enunciated by Gay-Lussac in 1808.
According to this law, gases react with each other in
the simple ratio of their volumes and if the product is
19
also in gaseous state, the volume of the product also Structural formula
bears a simple ratio with the volumes of gaseous
reactants when all volumes are measured under It represents the way in which atoms of various
similar conditions of temperature and pressure. elements present in the molecule are linked with each
H 2  Cl2  2 HCl other. For example, ammonia is represented as
1 vol 1 vol 2 vol ratio 1 : 1 : 2
2 H 2  O2  2 H 2O
2 vol 1 vol 2 vol ratio 2 : 1 : 2
2CO  O2  2CO2
The formula indicates that three hydrogen atoms are
2 vol 1 vol 2 vol ratio 2 : 1 : 2 linked to one nitrogen atom by three single covalent
N 2  3H 2  2 NH 3 bonds.
1 vol 3 vol 2 vol ratio 1 : 3 : 2
Determination of Empirical and Molecular Formulae

The following steps are followed to determine the

empirical formula of the compound:
Empirical formula - The percentage composition of the compound is
determined by quantitative analysis.
It represents the simplest relative whole number ratio - The percentage of each element is divided by its
of atoms of each element present in the molecule of atomic mass. It gives atomic ratio of the elements
the substance. For example, CH is the empirical present in the compound.
formula of benzene in which ratio of the atoms of - The atomic ratio of each element is divided by the
carbon and hydrogen is 1 : 1. It also indicates that the minimum value of atomic ratio so as to get the
ratio of carbon and hydrogen is 12 : 1 by mass. simplest ratio of the atoms of elements present in the
compound.
Molecular formula - If the simplest ratio is fractional, then values of
simplest ratio of each element is multiplied by a
Molecular formula of a compound is one which smallest integer to get a simplest whole number for
expressed as the actual number of atoms of each each of the element.
element present in one molecule. C6 H 6 is the - To get the empirical formula, symbols of various
molecular formula of benzene indicating that six elements present are written side by side with their
carbon atoms and six hydrogen atoms are present in respective whole number ratio as a subscript to the
the molecule of benzene. lower right hand corner of the symbol.
Thus, Molecular formula  n  Empirical formula - The molecular formula of a substance may be
Molecular formula mass determined from the empirical formula if the
where, n  molecular mass of the substance is known. The
Empirical formula mass
molecular formula is always a simple multiple of
Molecular formula gives the following information’s:
empirical formula and the value of simple multiple is
- Various elements present in the molecule.
obtained by dividing molecular mass with empirical
- Number of atoms of various elements in the
formula mass.
molecule.
Example: A compound of carbon, hydrogen and
- Mass ratio of the elements present in the molecule.
nitrogen contains these elements in the ratio 9:1:3.5.
e.g. The mass ratio of carbon and oxygen in CO2
Calculate the empirical formula. If its molecular mass
molecule is 12 : 32 or 3 : 8.
is 108, what is the molecular formula?
- Molecular mass of the substance.
- The number written before the formula indicates the
number of molecules, e.g., 2CO2 means 2 molecules
of carbon dioxide.

20
Soln.:
Element Element Atomic Relative Simplest
ratio mass number of ratio
atoms
Carbon 9 12 9 0.75
 0.75 3
12 0.25
Hydrogen 1 1 1 1
1 4
1 0.25
Nitrogen 3.5 14 3.5 0.25
 0.25 1
14 0.25
Empirical formula  C3 H 4 N
Empirical formula mass  (3 12)  (4 1)  14  54
Mol. mass 108
n  2
Emp. mass 54
Thus, molecular formula of the compound
 2  empirical formula
 2  C3 H 4 N  C6 H 8 N 2

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