Atoms and Molecules 2
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yugalkishoreprasad440reactants.
In other words, the law of conservation of
mass means that mass can neither be created nor
destroyed during any chemical combination. This law
The idea of tiniest unit of matter, viz. "anu" and is also known as law of indestructibility of matter.
"parmanu", was put forth by Maharishi Kanad in Vedic From a large number of experiments, it was
period in our country. Another Indian philosopher, concluded that although substances may undergo
Pakudha Katyayama, elaborated this doctrine and said chemical changes, the total mass of the products of
that these particles normally exist in combined form, the reaction is exactly equal to the total mass of the
which gives us various forms of matter. reactants. For example, 12 g carbon combines with 32
Democritus, a Greek philosopher, also proposed that g of oxygen to form 44 g of carbon dioxide:
matter is made up of extremely small particles, "the C O2 CO2
atoms". The name comes from Greek word 'atomos' 12 g 32 g 44 g
means "indivisible".
However, all these theories proposed by the Indian
and Greek philosophers were based on abstract
thinking and not on experiments. For about 2000
years, the atomic theory remained a mere speculation
till John Dalton, an English school teacher,
propounded his theory on atoms. Dalton's theory was To verify the law of conservation of mass i.e. mass of
based on the chemical laws of combination known at the reactants is same as the mass of the products in
that time. These laws of chemical combination were a chemical reaction.
given by A.L. Lavoisier and Joseph L. Proust and were • Prepare 5% aqueous solutions of barium chloride
based upon extensive study of chemical reactions. and sodium sulphate in separate test tubes.
• Take a small amount of sodium sulphate solution in
Laws of Chemical Combination a conical flask and some solution of barium chloride in
an ignition tube (a small tube made of glass).
By studying the quantitative measurements of many • Hang the ignition tube in the flask carefully; see that
reactions it was observed that the chemical reactions the solutions do not get mixed. Put a cock on the
taking place between various substances are flask.
governed by certain laws. These laws are called the • Weigh the flask with its content carefully.
'laws of chemical combination'. These laws formed • Now, tilt the flask so that the solutions mix with
the basis of Dalton's atomic theory. There are two each other.
main laws of chemical combination which are as • Weigh again.
follows: • Observations: As soon as the two solutions mix, a
(1) Law of Conservation of Mass chemical reaction takes place to form new products.
(2) Law of Constant Proportion (also known as the law Barium chloride + Sodium sulphate Barium
of definite composition) sulphate + Sodium chloride
BaCl2 Na2 SO4 BaSO4 2 NaCl
Law of Conservation of Mass ( white ppt .)
This law deals with the relation between the mass of We put a cork on the mouth of the flask to avoid any
the reactants and the products during the chemical exchange of matter between the flask and its
changes. It was stated by a French chemist, A. surrounding.
Lavoisier in 1774. The law of conservation of mass It is observed that the mass of the flask and its
states that during any chemical change, the total mass contents remain the same even after the chemical
of the products is equal to the total mass of the reaction has taken place.
1
• Conclusion: The activity shows that in chemical
reaction, mass of the reactants is the same as the
mass of the products. This proves the law of
conservation of mass.
1. 10.0 g of CaCO3 on heating gave 4.4 g of CO2 Thus, we see that hydrogen and oxygen combine in
and 5.6 g CaO. Shows that these observations the ratio of 1:8.
are in agreement with the law of conservation Hydrogen and oxygen combine in the ratio of 1:8 by
of mass. mass to form water.
Sol.: The above reaction takes place according to
the following equation: Experiment II (Decomposition of water)
CaO3( s ) CaO( s ) CO2( g ) • Take water from any source (say river, pond, lake,
Mass of reactant (CaCO3 ) 10 g tap water, well, etc.) and distill it to remove unwanted
Mass of products salts.
(CaO CO2 ) (5.6 4.4) g 10 g • Now take 90 g of pure water and pass electricity
Since, mass of the reactants is equal to the through it to cause its decomposition. Collect the
mass of the products, therefore, these gases so obtained.
observations are in agreement with the law of • Observation: It is observed that 90 g of water on
conservation of mass. decomposition gives 10 g of hydrogen gas and80 g of
oxygen gas,
Law of Constant Proportion Mass of hydrogen gas obtained 10 g 1
Thus,
This law states that in a pure chemical substance the Mass of oxygen gas formed 80 g 8
elements are always present in definite proportions From the above two experiments we observe that
by mass. water formed in the laboratory had hydrogen and
This law is also known as the law of definite oxygen gases combined in the ratio of 1 : 8 by mass.
proportions. This means that, whatever, maybe the At the same time, water obtained fromany source
source from which a compound is obtained, it is also had hydrogen and oxygen gases combined in the
always made up of the same elements in the same ratio of 1 : 8 by mass.
proportion by mass. The above results are in accordance with the law of
constant composition.
Experiment I (Production of water)
2. Weight of copper oxide obtained, by heating
• Collect the water formed by burning 10 g hydrogen 4.32 g of metallic copper with nitric acid and
gas completely in 80 g of oxygen. subsequent ignition, was 5.40 g. In another
• Record the mass of the water so obtained. experiment 2.30 g of copper oxide oil
• Observation: The mass of water so obtained is reduction yielded 1.84 g of copper. Show
found out to be 90 g. that these findings are in. accordance with the
Ratio in which hydrogen and oxygen combine law of constant proportion.
10 1 Sol.: Experiment I
Weight of copper = 4.32 g
80 8
Weight of copper oxide = 5.40 g
Weight of oxygen = Weight of copper oxide
Weight of copper
5.40 4.32 1.08 g
Ratio of copper arid oxygen
2
Mass of copper 4.32 4 3
2 g of CO2 will contain C 2 0.545 g.
Mass of oxygen 1.08 1 11
ratio is 4:1
Dalton's Atomic Theory
Experiment II John Dalton provided the basic theory about the
Weight of copper = 1.84 g nature of matter. He took the name atoms for the
Weight of copper oxide = 2.30 g smallest particle of matter as given by the Greek.
Weight of oxygen = Weight of copper oxide Based on the laws of chemical combination, he
Weight of copper proposed a model of atom known as Dalton atomic
2.30 1.84 0.46g theory.
Ratio of copper and oxygen The main postulates of the Dalton's atomic theory
Mass of copper 1.84 4 are:
Mass of oxygen 0.46 1 - All matter, whether an element, a compound or a
ratio is 4:1 mixture is made up of extremely small particles called
As the ratio of copper and oxygen (by mass) is same in atoms (i.e., same name was used for the smallest
the two experiments, law of constant proportion is indivisible particles as used by Greek philosophers).
verified. - 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
masses and also possess different properties.
- Atoms of the same or different elements combine
together to form molecules or compounds.
3. Copper oxide was prepared by two different
- When atoms of different elements combine together
methods. In one case, 1.75 g of the metal
to form compounds, they do so in a simple whole
gave 2.19 g of oxide. In the second case, 1.14
number ratio such as 1 : 1, 2 : 1, 2 : 3 etc.
g of the metal gage 1.43 g of the oxide. Show
- Atoms of two different elements may combine in
that the given data illustrate the law of
different ratios to form more than one compound. For
constant proportions.
example, carbon and oxygen may combine to form
Sol.: In case I, mass of copper = 1.75 g
carbon monoxide (CO)and carbon dioxide (CO2) in
Mass of copper oxide = 2.19 g
which the ratios of the combining atoms (C and O) are
% of copper in the oxide
1 : 1and 1 : 2 respectively.
Mass of copper
100 - The number and lend of atoms in a given compound
Mass of copper oxide is always fixed.
1.75 - Atom is the smallest particle that takes place in a
100 79.9%
2.19 chemical reaction. In other words, whole atoms rather
% of oxygen 100 79.9 20.1% than fractions of atoms take part in a chemical
In case II, mass of copper = 1.14 g reaction.
Mass of copper oxide = 1.43 g - An atom can neither be created nor destroyed, i.e.,
1.14 atom is indestructible.
% of copper in the oxide 100 79.7%
1.43
% oxygen 100 79.7 20.3% Drawbacks of Dalton's Atomic Theory
Thus, copper oxide prepared by any of the given
methods contains copper and oxygen in the same (1) Atom is no longer considered as the smallest
proportion by mass (within the experimental error.) indivisible particles. This is because recent
Hence, it proves the law of constant proportions. studies have shown that atom is made up of
still smaller particles called electrons, protons
4. Calculate the mass of carbon present in 2 g of and neutrons.
carbon dioxide. (2) Atoms of the same element may have
Sol.: Carbon dioxide (CO2 ) contains C and O in the different masses. For example, there are two
types of atoms of chlorine with masses 35 and
fixed proportion by mass, which is 12 : 32, i.e.,
37 (on the atomic scale, called 'mass
3 : 8. This means that 3 g of carbon combine with 8 g
numbers'). Such atoms of the same element
of oxygen to form 11 g of CO2 . In other words,
with different mass numbers are called
11 g of CO2 contains C 3 g isotopes.
3
(3) Atoms of different elements may have same of figures for different elements known at that time
masses. For example, atoms of potassium and (which were limited in number). In fact, the symbol
calcium are known with same mass number used by him also represented the quantity of the
(40). Such atoms of different elements with element, i.e., one atom of the element. A few of these
same mass numbers are known as isobars. symbols, as proposed by Dalton are given below:
(4) Substances made up of the same kind of
atoms may have different properties. For
example, charcoal, graphite and diamond are
all made up of carbon atoms but have
different physical properties.
Different elements have been named either after the
name of the place where they were first discovered or
5. Dalton postulated that “Atoms of same after the name of the scientist who discovered it or on
element are same in all respects.” How was the basis of some important property of the element.
this statement proved wrong? However, as more and more elements were
Sol.: The discovery of isotopes i.e., atoms of same discovered, an International Committee was set up,
element having same atomic number but called International Union of Pure and Applied
different mass number, proved the above Chemistry (IUPAC),which approved the names of the
statement wrong. different elements.
The symbol of an element is the "first letter" or "first
What is an Atom? letter and another letter" of the English name or
Latin name of the element. However, in all cases, the
The smallest unit of an element, which may or may
first letter is always capital and another letter (if
not exist independently, but always takes part in a
added) is always a small letter.
chemical reaction, is called an atom.
Atoms are the building blocks of matter. They are
Symbols of elements derived from English names
smaller than anything you can imagine or compare
English name of the Symbol
with. More than a million atoms when stacked upon
element
one another would make a layer as thick as the sheet
of paper of your science book. 1. Argon Ar
The size of an atom is indicated by its radius which is 2. Arsenic As
called 'atomic radius'. It is measured in 'nanometers'. 3. Aluminium Al
One nanometer is one billionth part of a metre. 4. Boron B
1 nanometre (nm) 109 metre (m) or 1 metre (m) 5. Barium Ba
6. Bromine Br
109 nanometre (nm)
The table below gives the relative size of the radius of 7. Carbon C
an atom of hydrogen with respect to other objects. 8. Calcium Ca
9. Chlorine Cl
Species Atomic radius 10. Chromium Cr
Hydrogen atom 1010 m 11. Cobalt Co
Molecule of water 109 m 12. Fluorine F
Molecule of haemoglobin 108 m 13. Hydrogen H
Grain of sand 104 m 14. Helium He
15. Iodine I
An ant 102 m
16. Lithium Li
Watermelon 101 m
17. Magnesium Mg
Symbols Used to Represent Atoms of Different 18. Manganese Mn
Elements 19. Nitrogen N
In case of elements, symbol means a short method of 20. Neon Ne
representing the full name of an element. Dalton was 21. Nickel Ni
the first scientist to suggest specific symbols in terms 22. Oxygen O
4
23. Phosphorus P 12. Aluminium Al 27
24. Platinum Ft 13. Phosphorus P 31
25. Sulphur S 14. Sulphur S 32
26. Uranium U 15. Chlorine Cl 35.5
27. Zinc Zn 16. Argon Ar 40
17. Potassium K 39
Symbols derived from Latin names 18. Calcium Ca 40
English name of Latin name of the Symbol 19. Iron Fe 56
the element element 20. Copper Cu 63.5
1. Antimony Stibium Sb 21. Zinc Zn 65
2.Copper Cuprum Cu 22. Silver Ag 108
3. Gold Aurum Au 23. Platinum Pt 195
4. Iron Ferrum Fe 24. Gold Au 197
5.Lead Plumbum Pb 25. Lead Pb 207
6. Mercury Hydragyrum Hg 26. Uranium U 238
7. Potassium Kalium K Gram Atomic Mass
8. Silver Argentum Ag
It is the quantity in grams which is numerically equal
9.Sodium Natrium Na
to the atomic mass of an element on a.m.u. scale.
10. Tin Stannum Sn
For example, 1 gram atom of hydrogen = 1.008 g,
1 gram atom of sodium = 23 g
Atomic Mass The relationship between gram atoms, mass and
Atoms are extremely small particles. Hence, the actual atomic mass of a substance is,
masses of the atoms are so small that, it is difficult to Mass in grams
Number or gram atoms
determine the actual masses of individual atoms. Gram atomic mass
In 1961, IUPAC recommended the use of an isotope of
carbon with mass number 12 as the standard
reference for measuring atomic masses. It is called
carbon-twelve (C-12) and is represented as 12 C. The 6. Calculate the number of gram atoms in
atomic mass (or earlier called as "atomic weight") is (i) 640 g of sulphur
now defined as follows: (ii) 360 g of magnesium
The atomic mass of an element is the relative mass of Sol.: (i) Atomic mass of sulphur =32 amu
its atoms as compared with the mass of an atom of One gram atom of sulphur =32g
carbon-12 isotope taken as 12 units. Now, 32 g of stilphiir = 1 grain atom of sulphur
1 1
Atomic mass unit (amu) may be defined as th of 640 g of sulphur 640 20 g sulphur
12 32
the mass of an atom of carbon-12 isotopeon the (ii) Atomic mass of magnesium = 24 amu
1 One gram atom of magnesium = 24 g
atomic scale, i.e., 1 amu th of mass of C-12
12 Now 24 g of magnesium = 1 g atom
isotope. 1
360 g 360 15 g
24
Atomic masses of some common elements (in amu or u).
Element Symbol Atomic Mass
1. Hydrogen H 1 7. An atom of copper has a mass of
2. Helium He 4
1.0625 1022. How, many atoms of copper are
3. Lithium Li 7
there in 16 g?
4. Boron B 11
Soln.: In 1.0625 1022 g of copper, number of atoms
5. Carbon C 12
6. Nitrogen N 14 =1
in 16 g of copper, number of atoms
7. Oxygen O 16
1
8. Fluorine F 19 16 1.505 1023
9. Neon Ne 20 1.0625 1022
10. Sodium Na 23
11. Magnesium Mg 24
5
How do Atoms Exist? A molecule of an element consists of the same type of
atoms bonded together. For example, a molecule of
The atoms of only a few elements called noble gases
oxygen is formed when 2 atoms of oxygen combine
(such as helium, neon, argon, etc) are chemically in
together. Oxygen atom alone cannot exist
reactive and exist in free state or single atoms. Atoms
independently. It exists as a diatomic molecule. This
of most of the elements are chemically reactive and
means that two atoms of oxygen combine together to
do not exist in free state or single atom. Atoms usually
form a molecule. The formula of oxygen molecule is
exist in two ways:
O2 .
(i) in the form of molecules.
(ii) in the form of ions.
Atomicity
Molecules
The number of atoms present in one molecule of an
A molecule is a group of two or more atoms which are element is called its atomicity.
held together strongly by some kind of attractive The atomicity of an element is indicated by writing
forces. Such an attractive force holding the atoms the number as a subscript on the right hand side
together is called a chemical bond. bottom of the symbol.
We may also define a molecule as follows: For example, H 2 shows that the atomicity of
A molecule is the smallest particle of an element or a hydrogen is 2. P4 shows that the atomicity of
compound which can exist freely under ordinary phosphorus is 4, He shows that the atomicity of
conditions and shows all the properties of that helium is 1.
substance. On the basis of their atomicities, the elements may be
classified as monoatomic, diatomic, triatomic, tetra
Molecules of an Element atomic, etc.
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 molecule of methane (CH4) which is a compound is
which means that the atoms present in them are the heteroatomic in nature.
same. The molecules of the compounds are
heteroatomic in nature in the sense that different
atoms are present in them. For example, a molecule
of nitrogen element is homoatomic in nature (N2). A
6
Ions 1. Name of the substance.
2. Name of various elements present in that
An ion may be defined as an atom or group of atoms substance.
having positive or negative charge. 3. Chemical formula of a substance represents one
• The ion which has one or more positive charges is molecule of that substance.
called Cation. At the same time, the ion carrying one 4. Relative number of atoms of various elements
or more negative charges is known as Anions. present in one molecule of that element or
• A few positively charged ions or cations are: compound.
Na+(sodium ion), 1C (potassium ion), Ca+ (calcium ion), 5. Relative masses of various elements in the
Al 3 (aluminium ion). compound.
• A few negatively charged ions or anions are: 6. It represents one mole of that substance.
Cl (chloride ion), S 2 (sulphide ion), (OH ) 7. We can calculate the gram molecular mass of that
substance.
(hydroxide ion), ( SO42 ) (sulphate ion).
For example, the chemical formula of CO2 conveys
the following informations:
Molecular Formula of a Compound 1. The substance is carbon dioxide.
2. Carbon dioxide is composed of the elements;
The symbolic representation of a molecule of a carbon and oxygen.
compound, is called its molecular formula. 3. In a molecule of carbon dioxide, carbon and
The molecule of a compound contains two or more oxygen atoms combine together in the ratio 1 : 2
than two types of elements. If a molecule of a by number.
compound contains two different kinds of elements 4. The ratio of mass of carbon and oxygen is 3 : 8.
only, it is called binary compound. 5. It represents one mole of carbon dioxide
For example, water ( H 2O), sodium chloride ( NaCl ), molecules.
iron sulphide (FeS), etc. are binary compounds. 6. The molecular mass of carbon dioxide is 44 amu.
To calculate the ratio of number of atoms in a Writing Chemical Formula of Compounds
molecule of a compound.
The ratio of number of atoms in a molecule of a Chemical formula of a molecular compound
compound can be calculated, if we know the ratio of represents the actual number of atoms of different
their masses and the atomic masses of the elements, elements present in one molecule of the compound,
constituting the molecule of the compound. e.g., chemical formula of water is H2O, that of
ammonia is NH3, while that of carbon dioxide is CO2
and so on.
Chemical formula of an ionic compound simply
represents the ratio of the cations and anions present
8. The ratio of masses of carbon and oxygen is in the structure of the compound, e.g., the formula
3:8 and their atomic masses are 12 and 16 Na Cl simply represents that sodium chloride
respectively, in molecule of carbon dioxide. contains Na and Cl ions in the ratio of 1 : 1
Calculate the ration of atoms of carbon and (though the actual crystal of sodium chloride consists
oxygen in the molecule of carbon dioxide. of very large but equal number of Na ions and Cl
Sol.: ions). Similarly, the formula Mg 2 Cl2 simply tells that
Element Ratio Atomic Mass ratio Simple
by Mass (u) Atomic Mass ratio it contains Mg 2 and Cl ions in the ratio of 1 : 2 and
Mass so on.
C 3 12 3 12 0.25 1
O 8 16 8 16 0.50 2 Valency of an Element
Ratio of atoms of carbon to oxygen in carbon
dioxide = 1 : 2 Valency of an element is defined as the combining
capacity of the element. It is equal to the number of
Information Conveyed by a Chemical Formula hydrogen atoms or number of chlorine atoms or
double the number of oxygen atoms with which one
By looking at the chemical formula of a substance, we atom of the element combines.
can gather the following information’s:
7
In addition to the atoms, the ions which are charged Copper Cu Cobalt Co2
species, also have some valencies. Positive ions or Gold Au Copper Cu 2
cations have positive valencies. Negative ions or Ammonium Iron Fe2
anions have negative valencies. The valencies of the
( NH 4 )
poly-valent ions are expressed by enclosing them in
bracket and putting the positive or negative signs
outside it. Let us write the valencies of some Some elements have shown more than one valencies.
commonly used positive and negative ions. In such cases, Roman Numerals are used to denote
the valencies. These are put in bracket. For example,
Valencies of Positive Ions copper (I) and copper (II);similarly, iron (II) and iron
Positive ions may be monovalent, bivalent, trivalent, (III).
tetravalent etc. depending upon the charge present
on them. These are listed in the following table. Valencies of Negative Ions
List of some common positive ions (Cations)
Monovalent Bivalent Trivalent Valencies of Negative Ions
Like positive ions, negative ions may also be
Hydrogen H Barium Ba 2 Aluminium Al 3
monovalent, bivalent, trivalent, etc. in nature. These
Potassium K Calcium Ca 2 Chromium Cr 3 are also listed;
Sodium Na Magnesium Iron Fe3
Mg 2
Silver Ag Zinc Zn2 Gold Au 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 P 3
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 III: Divide the valency number by common
factor, if any, to get simple ratio. Now ignore the (+)
One of the most important points to remember while and (-) symbols.
writing the formula of a chemical compound is that it Step IV: Now, write the valency of the cation at the
is always electrically neutral. In other words, the bottom right of the anion and the charge number of
positive and negative valencies of the ions present in the anion at the bottom of the cation. Thus, the
the chemical compound add up to zero. To write a symbol of cation is subscribed with the charge
formula, follow the steps given below. This method of number of the anion and the anion is subscribed with
writing formula is called the crisscross method. the charge number of the cation. This is called the
Step I: Write the symbol of the cation showing the criss-crossing of valencies.
charge on it. Write the symbol of the anion showing Step V: If these subscripts are 1, these are not written
the charge on it, on the right hand side of the cation. in the final stoichiometric formulae.
Step II: If a compound contains polyatomic ions, the Examples:
formula of the ion is enclosed within brackets before
crisscrossing the valencies.
8
Compound Symbols with Valency Shifting Valency number Formula
Calcium chloride Ca 2Cl CaCl2
Mg 2 SO42 or Mg1SO41
Magnesium sulphate
(Dividing by C.F . 2 ) MgSO4
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
K2Cr2O7
Calcium bicarbonate Ca( HCO3 )2
Ca 2 HCO31
Formulae of Molecular Compounds
While writing the chemical formulae of the molecular
Formula of carbon tetrachloride is CCl4 .
compounds, we write the constituent elements and
their valencies as shown below. Then, we crossover
the valencies of the combining atoms. Molecular Formulae of Important Acids and Bases
1. Formula of hydrogen chloride Molecular Formulae of Important Acids
Acid Molecular Acid Molecular
Formula Formula
Hydrochloric HCl Acetic acid CH 2COOH
acid
Formula of hydrogen chloride is HCl. Nitric acid Carbonic
HNO3 H 2CO3
acid
2. Formula of hydrogen sulphide
Sulphuric acid H 2 SO4 Sulphurous H 2 SO3
acid
Phosphoric H 3 PO4 Nitrous acid HNO2
acid
Formula of hydrogen sulphide is H2S.
Molecular Formulae of Important Bases
3. Formula of ammonia Base Molecular Base Molecular
Formula Formula
Ammonium NH 4OH Aluminium Al (OH )3
hydroxide hydroxide
Formula of ammonia is NH3. Sodium NaOH Zinc Zn(OH )2
4. Formula of carbon tetrachloride hydroxide hydroxide
Potassium KOH Iron (II) Fe(OH )2
hydroxide hydroxide
9
Calcium Ca(OH )2 Iron (III) Fe(OH )3 sulphide sulphide
hydroxide hydroxide Potassium K 2 S Tin (II) SnS
Magnesium Mg (OH )2 Copper (II) Cu(OH )2 sulphide sulphide
hydroxide hydroxide Calcium CaS Copper (II) CuS
sulphide sulphide
Molecular Formulae of Important Salts Magnesium MgS Mercury HgS
sulphide (II)
Molecular formulae of Oxides sulphide
Oxide Molecular Oxide Molecular Zinc ZnS Silver (I) Ag2 S
Formula Formula sulphide sulphide
Sodium oxide Na2O Iron (II) FeO Iron (II) FeS Barium BaS
oxide sulphide sulphide
Potassium oxide K 2O Iron (In) Fe2O3
oxide
Calcium oxide CaO Lead (II) PbO Molecular Formulae of Nitrates
oxide Nitrate Molecular Nitrate Molecular
Magnesium MgO Lead (IV) PbO2 Formula Formula
oxide oxide Ammonium NH 4 NO3 Iron (II) Fe( NO3 )2
Aluminium oxide Al2O3 Copper CaO nitrate nitrate
(II) oxide Sodium NaNO3 Lead (II) Pb( NO3 )2
Zinc oxide ZnO Mercury HgO nitrate nitrate
(II) oxide Potassium KNO3 Tin (II) Sn( NO3 )2
nitrate nitrate
Molecular Formulae of Carbonates and Hydrogen Calcium Ca( NO3 )2 Copper (II) Cu( NO3 )2
Carbonates nitrate nitrate
Carbonate/ Molecular Carbonate/ Molecular Magnesium Mg ( NO3 )2 Mercury (II) Hg ( NO3 )2
Hydrogen Formula Hydrogen Formula nitrate nitrate
Carbonate Carbonate Aluminium Al ( NO3 )3 Silver (I) AgNO3
Sodium Na2CO3 Magnesium Mg ( HCO3 )2 nitrate nitrate
carbonate hydrogen Zinc nitrate Zn( NO3 )2
carbonate
Sodium NaHCO3 Aluminium Al2 (CO3 )3 Molecular Formulae of Chlorides
hydrogen carbonate Chloride Molecular Chloride Molecular
carbonate Formula Formula
Potassium K 2CO3 Zinc ZnCO3 Ammonium NH 4Cl Iron (II) FeCl2
carbonate carbonate chloride chloride
Potassium KHCO3 Iron (II) FeCO3 Sodium NaCl Iron (III) FeCl3
hydrogen carbonate chloride chloride
carbonate Potassium KCl Lead (II) PbCl2
Calcium CaCO3 Lead (II) PbCO3 chloride chloride
carbonate carbonate Calcium CaCl2 Tin (II) SnCl2
Calcium Ca( HCO3 )2 Tin (II) SnCO3 chloride chloride
hydrogen carbonate Magnesium MgCl2 Copper (II) CuCl2
carbonate chloride chloride
Magnesium MgCO3 Copper (II) CuCO3 Zinc chloride ZnCl2 Mercury HgCl2
carbonate carbonate (II)
chloride
Molecular Formulae of Sulphides Aluminium AlCl3 Silver (I) AgCl
Sulphide Molecular Sulphide Molecular chloride chloride
Formula Formula
Ammonium ( NH 4 )2 S Iron (III) Fe2 S3
sulphide sulphide
Sodium Na2 S Lead (II) PbS
10
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 Ag2 SO4
Molecular Mass represent its molecule, but only represents the ratio
of different ions in the compounds.
Molecular mass of a substance (element or This is called formula unit of the ionic compound.
compound) is the average relative mass of its Formula mass of an ionic compounds is obtained by
molecules as compared with that of an atom of C-12 adding atomic masses of all the atoms in a formula
isotope taken as 12. In other wards, molecular mass unit of the compound.
of a substance represent the number of times the
molecule of that substance is heavier than 1/12th of For Example,
the mass of an atom of C-12 isotope. Formula mass of calcium oxide (CaO) atomic mass
of calcium + atomic mass of oxygen
Calculation of Molecular Mass 40 16 56 u
As molecules are made up of two or more atoms of Gram Atomic Mass and Gram Molecular Mass
the same or different elements, and each atom has a
definite atomic mass, therefore, molecular mass of a Atomic mass expressed in grams is called gram atomic
molecule of a substance can be calculated by adding mass of that element.
atomic masses of all the atoms present in one For example,
molecule of the substance. Atomic mass of hydrogen = 1.0 u
Gram atomic mass of hydrogen = 1.0 g
For example, Atomic mass of oxygen = 16.0 u
Molecular mass of H 2O 2 (atomic mass of Gram atomic mass of oxygen = 16.0 g
hydrogen) + atomic mass of oxygen
2(1) 16 18 a.m.u. The amount of an element having mass equal to gram
Similarly, molecular mass of CO2 atomic mass of atomic mass is called one "gram atom" (or g atom) of
C 2 (atomic mass of oxygen) the element.
12 2(16) 12 32 44 a.m.u. For example,
1 g atom of hydrogen = 1.0 g
Other examples
1 g atom of oxygen = 16.0 g
Molecular mass of H 2 2 atomic mass of hydrogen
2(1) 2 a.m.u.
Molecular mass expressed in grams is called gram
Molecular mass of NH 3 atomic mass of nitrogen + 3 molecular mass of that substance.
(atomic mass of hydrogen) For example, Molecular mass of H 2 2.0 u
14 3(1) 17 a.m.u. Gram molecular mass of H 2 2.0 g
Formula Mass Molecular mass of O2 32.0 u
Gram molecular mass of O2 32.0 g
In case of the compounds formed by ions (ionic
compounds) formula of the compound does not
11
The amount of the substance having mass equal to its (c) C2 H 6 (d) PCl3
gram molecular mass is called one "gram molecule" [Atomic masses: O 16, C 12,
(or g molecule) of the substance. Thus, Cl 35.5, H 1, P 31 ]
1 g molecule of H 2 2.0 g
Sol.: O2 2 (atomic mass of oxygen)
1 g molecule of O2 32.0 g
2(16) 32u
(b) Cl2 2 (atomic mass of chlorine)
2(35.5) 71u
(c) C2 H 6 2 (atomic mass of C) +6 (atomic mass of
hydrogen)
9. Write the formula of compounds formed by:
2(12) 6(1) 24 6 30u
(a) Ag and S 2
(d) PCl3 atomic mass of phosphorus + 3 (atomic
(b) Cr 3 and O 2
mass of chlorine)
(c) Al 3 and SO42 31 3(35.5)
Sol.: (a) Ag2 S 31 106.5 137.5u
(b) Cr2O3
(c) Al2 (SO4 )3 13. Find the formula mass of the following:
(a) ZnO (b) Na2O (c) CuSO4 .5H 2O
10. What is the valency of the metal in the [Atomic masses: Zn 65, O 16,
following formula?
Na 23, Cu 63.5, H 1 ]
(a) CaCl2
Sol.: (a) ZnO atomic mass of Zn + atomic mass of
(b) Na2O
oxygen
(c) Fe2O3 65 16 81 u
Sol.: (a) CaCl2 has been formed by (b) Na2O 2 (atomic mass of sodium) + atomic mass
of oxygen
2(23) 16 46 16 62 u
Therefore, the valency of calcium (i.e., metal) (c) CuSO4 .5H 2O atomic mass of Cu atomic mass
is 2. of S 4 (atomic mass of O) + 5 (molecular mass of
(b) Na2O has been formed by H 2O )
63.5 32 4(16) 5(18)
249.5 u
So, the valency of sodium (i.e., metal) is 1.
(c) Fe2O3 has been formed by Mass Percentage Composition of an Element in a
Compound
The percentage composition of an element in a
compound can be determined, if we know the
Therefore, the valency of iron (i.e., metal) is 3. molecular mass or unit formula mass and the atomic
weight of the element.
11. Write the formula for the following: If M is the molecular mass (or unit formula mass) of a
(a) Ferric chloride compound and A, the atomic mass of its atoms for
(b) Ferrous sulphate some particular element then
(c) Cupric iodide Mass percentage composition of the element =
(d) Cupric oxide Sum of atomic masses of the atoms of element A
(e) Ammonium hydroxide 100 100
Molecular mass M
(f) Calcium phosphate However, if molecular mass and atomic mass are not
Sol.: (a) FeCl3 (b) FeSO4 given, instead given the mass of compound and mass
(c) Cul2 (d) CuO of element in it, then percentage composition can be
(d) NH 4OH (f) Ca3 ( PO4 )2 calculated as follows:
Let W be the mass of a compound which contain as
12. Find the molecular mass of the following: the mass of an element.
(a) O2 (b) Cl2 Mass percentage composition or the element =
12
Mass element w C 2 Atomic mass of
100 100
Mass of compound W O (12 u 2 16 u) 44 u Gram molecular
mass of CO2 44 g 1 gram molecule
CO2 44 g
0.72 gram molecule of CO2
(44)
(0.72) 31.68 g
(1)
Mole Concept
To convert atomic mass into gram atomic mass Quite commonly, we use different units for counting
and to convert molecular mass in gram molecular such as dozen for 12 articles, score for 20 articles and
mass, we have to simply replace 'u' by 'g'. gross for 144 articles irrespective of their nature.
The masses of hydrogen and oxygen in water can In a similar way, chemists use the unit 'mole' for
be determined by electrolyzing water. The counting atoms, molecules ions, etc.. A mole is a
volumes of hydrogen and oxygen obtained during collection, of 6.022 1023 particles. Thus:
electrolysis are measured, and their masses A mole represents 6.022 1023 particles.
calculated. The number 6.022 1023 is called Avogadro number
and is symbolized as N A or N 0 . In other words, a
mole is an Avogadro number of particles. For example
1 mole of hydrogen atoms 6.022 1023 hydrogen
atoms.
14. Calculate the mass percentage composition of 1 mole of hydrogen molecules 6.022 1023
copper in copper sulphate. hydrogen molecules.
[Cu 64, S 32, O 16] 1 mole of sodium ions 6.022 1023 sodium ions.
Sol.: Unit formula mass of CuSO4 1 mole of electrons 6.022 1023 electrons.
The amount of substance that contains the same
1(Cu) 1(S ) 4(O) 1(64) 1(32) 4(16)
number of entities (atoms, molecules, ions or other
64 32 64 160 amu.
particles), as the number of atoms present in 0.012 kg
Mass % age composition of copper
(or 12 g) of the carbon-12 isotope.
64 100
40%
160 Mole in Terms of mass
15. 4.9 g of sulphuric acid contains 0.1 g of
hydrogen, 1.6 g of sulphur and rest oxygen. A mole of atoms is defined as that amount of the
Calculate the mass percentage composition of substance (element) which has mass equal to gram
all the elements of sulphuric acid. atomic mass (i.e., atomic mass expressed in grams). In
Soln.: Mass of sulphuric acid (W) = 4.9 g other words, it is equal to one gram atom of the
Mass of hydrogen (1 ) 0.1 g element.
Mass of sulphur (2 ) 1.6 g A mole of molecules is defined as that amount of the
Mass percentage of hydrogen substance (element or compound) which has mass
w 0.1 equal to gram molecular mass (i.e., molecular mass
1 100 100 2.04% expressed in grams).In other words, it is equal to one
W 4.9
Mass percentage of sulphur gram molecule of the substance. For example,
If the number of chemical entities of a substance is
w 1.6
2 100 100 32.65% known, then the number of moles of that substance
W 4.9 can be calculated as follows,
Mass percentage of oxygen Number of chemical entities of a subs tan ce ( N ) N
Number of moles or n
100 (2.04 32.65) 65.31% 6.022 1023 ( Avogadro ' s Number , N A ) NA
16. Calculate the mass of 0.72 gram molecule of
carbon dioxide (CO2 ).
Sol.: Molecular mass of CO2 Atomic mass of
13
21. Calculate the mass of
(a) 0.2 mol molecules of oxygen
(b) one atom of aluminium
(c) 3.0 mol of Cl ion
17. What is the number of molecules in 0.25 (d) 2.5 mol of NaCl
moles of oxygen? (Atomic masses:
Sol.: We know that: O 16 u, Cl 35.5 u, Na 23 u. Al 27 u )
1 mole of oxygen contains Sol.: (a) Molecular mass of oxygen (O2 ) 32 u
6.022 1023 molecules
Molar mass of O2 32 g mol 1
So, 0.25 mole of oxygen contains
Now, 1 .mole of O2 molecule = 32 g
6.022 1023 0.25 1.505 1023 molecules
Thus, 0.25 mole of oxygen contains 0.2 mole of O2 molecule 32 0.2 6.4 g
1.505 1023 molecules. (b) Atomic mass ( Al ) 27 u
18. From 200 me, of CO2 , 1021 molecules are Molar mass 27 g mol 1
removed. How many moles of CO2 are left? Now 6.022 1023 atoms of aluminium weigh
Sol.: Gram-molecular mass of CO2 44 g 27 g
Mass of 1021 molecules of CO2 1 atom of aluminium will weigh
27
44 4.48 1023 g
1021 0.073 g 6.022 1023
6.022 10 23
(c) Formula mass of Cl ion 35.5 u
Mass of CO2 left (0.2 0.073) 0.127 g
Molar mass of Cl ions 35.5 g mol 1
0.127 g
Moles of CO2 left Number of moles = 3 (given)
44 g
m( Mass)
So, 2.8 103 moles of CO2 are left. Now, n
M ( Molar mass )
19. How many electrons are present in 1.6 g of
m n M 3 35.5 106.5 g
methane?
Sol.: Gram-molecular mass of methane (CH4) (d) Formula mass of NaCl 23 35.5 58.5 u
12 4 16 g Molar mass of NaCl (M ) 58.5 g mol 1
Number of moles in 1.6 g of methane Number of moles = 2.5 (given)
1.6 m
0.1 Now, n
16 M
Number of molecules of methane in 0.1 mole m n M
0.1 6.022 1023 2.5 58.5
6.022 1022 146.25 g
One molecule of methane has 6 4 10
electrons Mole in Terms of Volume
So, 6.022 1022 molecules of methane have
In case of gaseous substances, it is found that
10 6.022 1022 electrons
Avogadro's number of molecules (i.e., one mole of
6.022 1023 electrons
molecules) of any gas under standard conditions of
20. Calculate the number of moles in
temperature and pressure, i.e., STP conditions (0°C
3.0115 1023 atoms of calcium, and one atmospheric pressure) occupy the same
Sol.: 1 mole of any substance contains 6.022 1023 volume which is equal to 22400 mL or 22.4 litres.
particles. Hence,
6.022 1023 atoms of calcium = 1 A mole of a gaseous substance is defined as that
mole amount of the substance which has volume equal to
1 22400 mL at STP conditions.
1 atom of calcium moles
6.022 1023 Thus, 1 mole of CO2 gas = 22400 mL of CO2 at STP
3.0115 1023 atoms of calcium 1 mole of NH3 gas = 22400 mL of NH3 at STF
1 1 mole of SO2 gas = 22400 mL of SO2 at STP
3.0115 1023
6.022 1023
0.5 mole
14
Molar mass and Molar volume (1) A mole of an ionic compound is that amount
of the substance which has mass equal to
The mass of 1 mole of the substance (i.e., Avogadro's gram formula unit mass, i.e., formula unit
number of particles) is called molar mass of that mass of the ionic compound expressed in
substance. If the substance is atomic, its molar mass is grams.
equal to gram atomic mass. If the substance is (2) A mole is defined as that amount of the
molecular, its molar mass is equal to gram molecular substance which contains Avogadro's number
mass. As it is mass of one mole of the substance, its of formula units.
units are gram per mole ( g mol 1 ) or kilogram per e.g., 1 mole of sodium chloride ( NaCl )
mole (kg mol 1 ). Gram formula unit mass of NaCl
23 35.5 g 58.5 g
Thus,
Molar mass of C-12 isotope = Avogadro's number (6.022 1023 ) of formula units
12 g mol or 0.012 kg mol 1
1 of NaCl
Molar mass of iron ( Fe) 56 g mol 1 6.022 1023 Naions 6.022 1023 Cl ions
Similarly, 1 mole of potassium nitrate
Molar mass of magnesium (Mg ) 24 g mol 1
( KNO3 ) 39 14 3 16 g
Molar mass of H 2 2.0 g mol 1 101 g
Molar mass of O2 32.0 g mol 1 6.022 1023 formula units of KNO,
Molar mass of H 2O 18.0 g mol 1 6.022 1023 K ions 6.022 1023 NO3ions
Molar mass of NH3 17.0 g mol 1 6.022 1023 K atoms 6.022 1023 N atoms
Molar mass is usually represented by the symbol ‘M’. 3 6.022 1023 O atoms
Thus, M ( H 2O) 18 g mol 1 and so on. Combining all the definitions of mole, we arrive at the
The volume occupied by one mole of any gas at STP is following results:
always same and equal to 22400 mL or 22.4 L or 1 mole of atoms = Gram atomic mass 6.022 1023
22.4 dm3 . This volume is called molar volume or gram atoms
molecular volume (G.M.V.). 1 mole of molecules = Gram molecular mass
6.022 1023 molecules
Mole concept for Ionic Compounds = 22.4 L at STP if the substance is a gas.
1 mole of an ionic compound = Gram formula unit
Ionic compounds (like NaCl , KNO3 , etc.) do not mass
consist of molecules. They consist of a cluster of ions. 6.022 1023 formula units.
In case of ionic compounds, a mole is defined as
follows:
15
Relationship between mole, Avogadro number and mass.
1 mole of carbon
atoms
6.022 1023 atoms of C 12 g of carbon
1 mole of hydrogen
atoms
6.022 1023 atoms of H 1 g of hydrogen
1 mole of any particle
(atoms, molecules ion)
6.022 1023 number of Relative mass of those
that particle particles in grams
1 mole of
molecules
6.022 1023 number of Molecular mass in
that particle grams
Volume of gas
Recapitulation at STP
Multiply by 22.4 L Divide by 22.4 L
Multiply by
Avogadro’s number ( N0 )
Number of Number of
molecules (N) moles (n)
Divide by
Avogadro’s number ( N0 )
Multiple by molar Divide by molar mass (M)
mass (M)
Mass in
grams
16
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
group of two having +ve or
there are C - 12 atoms more atoms –ve charge
in 12 g of C - 12
isotope
N A 6.022 1023
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: 2 V.D
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
17
Mole Concept for ionic Compounds with two other elements B and C to form two
compounds and if B and C also combine to form a
• Number of molecules compound, their combining masses are in same
No. of particles ( N ) proportion or bear a simple ratio to the masses of B
( n) and C which combine with a constant mass of A.
Avogadro ' s number ( N 0 )
N
n …(1)
N0
Also Number of Moles
Mass of the subs tan ce (m)
( n)
Molar mass( M ) - For example, hydrogen combines with sodium and
m chlorine to form compounds NaH and HCl ,
n …(2) respectively.
M
N m In NaH , Sodium 23 parts Hydrogen one part
From (1) and (2) n In HCl , Chlorine 35.5 parts Hydrogen one part
N0 M
m
N N0 - Sodium and chlorine also combine to form NaCl in
M which 23 parts of sodium and 35.5 parts of chlorine
Number of molecules are present. These are the same parts which combine
Mass of the subs tan ce with one part of hydrogen in NaH and HCl
N0
Molar mass respectively.
Law of Multiple Proportion
This law was put forward by Dalton in 1808. According
to this law if two elements combine to form more
than one compound, then the different masses of one - Hydrogen combines with sulphur and oxygen to
element which combine with a fixed mass of the other form compounds H 2 S and H 2O respectively.
element, bear a simple ratio to one another. In H 2 S , Hydrogen 2 parts Sulphur 32 parts
Hydrogen and oxygen combine to form two In H 2O, Hydrogen 2 parts Oxygen 16 parts
compounds H2 (water) and H2O2 (hydrogen peroxide)
Thus, according to this law, sulphur should combine
In water, Hydrogen 2 parts Oxygen 16 parts
with oxygen in the ratio of 32 : 16 or a simple multiple
In hydrogen peroxide, Hydrogen 2 parts Oxygen 32 parts
of it.
- The masses of oxygen which combine with same Actually, both combine to form SO2 in the ratio of 32
mass of hydrogen in these two compounds bear a : 32 or 1 : 1.
simple ratio 1:2. - The law of reciprocal proportions is a special case of
Nitrogen forms five stable oxides. a more general law, the law of equivalent masses,
N 2O Nitrogen 28 parts Oxygen 16 parts which can be stated as under:
"In all chemical reactions, substances always react in
N 2O2 Nitrogen 28 parts Oxygen 32 parts
the ratio of their equivalent masses."
N 2O3 Nitrogen 28 parts Oxygen 48 parts
N 2O4 Nitrogen 28 parts Oxygen 64 parts Law of Gaseous Volumes
N 2O5 Nitrogen 28 parts Oxygen 80 parts
This law was enunciated by Gay-Lussac in 1808.
- The masses of oxygen which combine with same According to this law, gases react with each other in
mass of nitrogen in the five compounds bear a ratio the simple ratio of their volumes and if the product is
16 : 32 : 48 : 64 : 80 or 1 : 2 : 3 : 4 : 5. also in gaseous state, the volume of the product also
bears a simple ratio with the volumes of gaseous
Law of Reciprocal Proportions reactants when all volumes are measured under
similar conditions of temperature and pressure.
This law was given by Richter in 1794. The law states H 2 Cl2 2HCl
that when definite mass of an element A combines 1 vol 1 vol 2 vol ratio 1 : 1 : 2
18
2H 2 O2 2H 2O 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
2CO O2 2CO2 bonds.
Determination of Empirical and Molecular Formulae
2 vol 1 vol 2 vol ratio 2:1:2
N2 3H 2 2 NH3
The following steps are followed to determine the
1 vol 3 vol 2 vol ratio 1:3:2 empirical formula of the compound:
- The percentage composition of the compound is
determined by quantitative analysis.
- The percentage of each element is divided by its
atomic mass. It gives atomic ratio of the elements
Empirical formula
present in the compound.
- The atomic ratio of each element is divided by the
It represents the simplest relative whole number ratio
minimum value of atomic ratio so as to get the
of atoms of each element present in the molecule of
simplest ratio of the atoms of elements present in the
the substance. For example, CH is the empirical
compound.
formula of benzene in which ratio of the atoms of
- If the simplest ratio is fractional, then values of
carbon and hydrogen is 1 : 1. It also indicates that the
simplest ratio of each element is multiplied by a
ratio of carbon and hydrogen is 12 : 1 by mass.
smallest integer to get a simplest whole number for
each of the element.
Molecular formula
- To get the empirical formula, symbols of various
elements present are written side by side with their
Molecular formula of a compound is one which
respective whole number ratio as a subscript to the
expressed as the actual number of atoms of each
lower right hand corner of the symbol.
element present in one molecule. C6 H 6 is the
- The molecular formula of a substance may be
molecular formula of benzene indicating that six determined from the empirical formula if the
carbon atoms and six hydrogen atoms are present in molecular mass of the substance is known. The
the molecule of benzene. molecular formula is always a simple multiple of
Thus, Molecular formula n Empirical formula empirical formula and the value of simple multiple is
Molecular formula mass obtained by dividing molecular mass with empirical
where, n
Empirical formula mass formula mass.
Molecular formula gives the following information’s: Example: A compound of carbon, hydrogen and
- Various elements present in the molecule. nitrogen contains these elements in the ratio 9:1:3.5.
- Number of atoms of various elements in the Calculate the empirical formula. If its molecular mass
molecule. is 108, what is the molecular formula?
- Mass ratio of the elements present in the molecule. Soln.:
e.g. The mass ratio of carbon and oxygen in CO2 Element Element Atomic Relative Simplest
ratio mass number of ratio
molecule is 12 : 32 or 3 : 8.
atoms
- Molecular mass of the substance.
Carbon 9 12 9 0.75
- The number written before the formula indicates the 0.75 3
number of molecules, e.g., 2CO2 means 2 molecules 12 0.25
Hydrogen 1 1 1 1
of carbon dioxide. 1 4
1 0.25
Structural formula Nitrogen 3.5 14 3.5 0.25
0.25 1
14 0.25
It represents the way in which atoms of various Empirical formula C3 H 4 N
elements present in the molecule are linked with each Empirical formula mass (3 12) (4 1) 14 54
other. For example, ammonia is represented as
Mol. mass 108
n 2
Emp. mass 54
Thus, molecular formula of the compound
2 empirical formula
2 C3 H 4 N C6 H8 N2
19
(iii) CuSO4 5H 2O
Given atomic masses: C 12.0 u, H 1.0 u,
O 16.0 u, Al 27.0 u, S 32.0 u,
1. A piece of copper weighs 0.635 g. How many Cu 63.5.
atoms of copper does it contain? Sol.: (i) Molecular mass of C12 H 22O11
Sol.: Gram atomic mass of copper 63.5 g 12 At. mass of C 22 At. mass of
Number of moles in 0.635 g of copper H 11 At. mass of O
0.635 12 12.0 u 22 1.0 u 1116.0 u
0.01 moles
63.5 144 22 176 u 342 u
Number of copper atoms in one mole (ii) Molecular mass of Al2 (SO4 )3
6.022 1023 2 At. mass of Al 3 (At. mass of
Number of copper atoms in 0.01 moles S 4 At. mass of O)
0.01 6.022 1023 2 27.0 u 3 (32.0 u 4 16.0 u)
6.022 1021 54 u 3 (32 64)u 54 3 96 u 54
2. Calculate the mass of a single atom of sulphur 288 u 342 u.
and a single molecule of carbon dioxide? (iii) Molecular mass of CuSO4 5H 2O
Sol.: Gram atomic mass of sulphur 32 g At. mass of Cu At. mass of S 4 At.
Mass of one sulphur atom mass of O
Gram atomic mass 32 5 (2 At. mass of H 1 At. mass of O)
5.31 1023 g
6.022 10 23
6.022 1023 63.5 u 32.0 u 4 16.0 u 5(2 1.0 u 16.0 u)
Formula of carbon dioxide CO2 63.5 32.0 64.0 90.0 u 249.5 u.
Molecular mass of CO2 12 2 16 44 5. An element X shows a variable valency of 3
Gram molecular mass of CO2 44 g and 5. What are the formulae of the oxide
formed by it?
Mass of one molecule of
Sol.: (i) Formula of oxide when X has valency = 3
Gram atomic mass 44
CO2 7.306 1023 g
6.022 10 23
6.022 1023
3. What mass of silver nitrate will react with (ii) Formula of the oxide when X has valency =
5.85 g of sodium chloride to produce 14.35 g 5
of silver chloride and 8.5 g of sodium nitrate if
the law of conservation of mass is true?
Sol.: The reaction is:
Silver nitrate + Sodium chloride Silver 6. Calculate the formula unit mass of
chloride + Sodium nitrate Na2CO3 .10H 2O.
If law of conservation of mass is true, Atomic masses: Na 23.0 u, C 12.0 u,
Total mass of reactants = Total mass of O 16.0 u, H 1.0 u.
products Sol.: Formula mass of Na2CO3 .10H 2O
i.e., Mass of AgNO3 Mass of NaCl Mass
2 At. mass of Na At. mass of C 3 At.
of AgCl Mass of NaNO3 mass of O 10(2 At. mass of
We are given: Mass of NaCl 5.85 g , H At. mass of O)
Mass of AgCl 14.35 g , 2 23.0 u 12.0 u 3 16.0 u 10(2 1.0 16.0)u
Mass of NaNO3 8.5 g 46.0 12.0 48.0 180 u 286 u.
Substituting these values in the above 7. 1022 atoms of an element 'X' are found to have
equation, we get a mass of 930 mg. Calculate the molar mass of
Mass of AgNO3 5.85 g 14.35 g 8.5 g the element 'X'.
Mass of AgNO3 (14.35 8.5) 5.85 g Sol.: Molar mass of an element is the mass of
22.85 5.85 17.0 g. Avogadro's number of atoms.
4. Calculate the molecular masses of the 1022 atoms of the element have mass
following: 930
930 mg g 0.930 g
(i) C12 H 22O11 (ii) Al2 (SO4 )3 1000
20
6.022 1023 atoms will have mass No. of moles of sodium
Mass of sodium in grams
0.930
22 6.022 1023 g 56.0 g
10 Gram atomic mass or sodium
Molar mass of the element 56 g mol 1 11.5 g
0.5 moles
N 23 g
Alternatively, n For calcium
N0
Gram atomic mass of calcium = 40 g
m
Also, n Mass of calcium = 15 g
M No. of moles of calcium
m N Mass of calcium in grams
or M m 0 0.930
n N Gram atomic mass of calcium
6.022 1023 atoms mol 1
56.0 g mol 1.
15 g
0.375 moles
1022 atoms 40 g
As, calcium has less number of moles (0.375
8. Calculate the number of molecules present in
moles) as compared to sodium (0.5 moles).
1 litre of water assuming that density of water
Therefore, sodium has more number of atoms
is 1 g mL1. than calcium.
Sol.: 1 Litre of water 1000 mL 11. Calculate the number of aluminium ions in
Mass of 1000 mL of water = Volume 0.056 g of aluminium oxide
Density ( Al1O3 ).[ Al 27; O 16]
1000 mL 1 g mL1 1000 g Sol.: Molar mass of
No. of moles in 1000 g Al2O3 2( Al ) 3(O) 2(27) 3(16)
H 2O
m 1000
5.55 moles 54 48 102 g / mol.
M 18 Number of moles 0.056 g of
No. of molecules ( N ) n N0 55.55 Mass of Al2O3
Al2O3
6.022 1023 Molar mass of Al2O3
3.345 10 molecules.
25
0.056 g
9. Calculate the mass percentage of composition 5.49 1049 mol
102 g / mol
of sulphur in calcium sulphate [CaSO4 ].
Now, Al2O3 molecule has 2 moles of Al 3
[Ca 40; S 32; O 16]
ions.
Sol.: Molecular mass of
CaSO4 1(Ca) 1(S ) 4(O) or 1 mol of Al2O3 2 moles of Al 3 ions
1 (40) 1 (32) 4 (16) 1 mol of Al2O3 2 6.022 1023 Al 3 ions
40 32 64 136 amu. 5.49 104 mol of Al2O3
Also mass of sulphur in 5.49 104 2 6.022 1023 Al 3 ions
CaSO4 1(32) 32 amu 6.611020 Al 3 ions.
Mass percentage of sulphur in 12. Calculate the number of atoms of each type in
32 3.42 g of sucrose (C12 H 22C11 ).
CaSO4 100 23.53%
136 Sol.: Molecular mass of sucrose
10. Which amongst the following has more 12 12 22 1 1116 342 u.
number of atoms? Molar mass 342 g mol 1
(i) 11.5 g of sodium, or [ Na 23; Ca 40].
m
(ii) 15 g of calcium. Now, Number of molecules ( N ) N0
M
Sol.: We know that equal number of moles of
Number of molecules of sucrose in
different elements contain equal number of
3.42
atoms. Thus, we shall convert masses of 3.42 g 6.022 1023
sodium and calcium to find which has more 342
number of moles. 6.022 1021 molecules
For sodium Now,
gram atomic mass of sodium = 23 g 1 molecule of sucrose (C12 H 22C11 )
Mass of sodium = 11.5 g contains = 12 atoms of C
21
So, 6.022 1021 molecules will contain Molar mass of CO2 12 2 15 44 g
12 6.022 1021 atoms 44 g of CO2 represent 1 mol
72.26 1021 atoms of C 22 g of CO2 represent
1 molecule of sucrose (C12 H 22C11 ) (22 g )
1(mol ) 0.5mol
contains = 22 atoms of H (44 g )
6.022 1021 molecules will contain Now, 1 mole of CO2 contain oxygen atoms
22 6.022 1021 atoms of hydrogen 2 N0 2 6.022 1023
132.5 1021 atoms of hydrogen 0.5 mole of CO2 contain oxygen atoms
1 molecule of sucrose (C12 H 22C11 ) 2 6.022 1023 0.5 6.022 1023 atoms.
contains = 11 atoms of oxygen Step II. Weight of carbon monoxide (CO)
6.022 1021 molecules will contain 6.022 1023 atoms of oxygen are present in
11 6.022 1021 atoms of oxygen 6.022 1023 molecules of carbon monoxide
66.24 1021 atoms of oxygen (CO).
13. Calculate the mass of sodium which contains By definition,
same number of atoms as those present in 4 g 6.022 1023 molecules of CO have mass =
of calcium. (Atomic masses: Ca 40, Na 23 Molar mass of CO 12 16 28 g.
) 16. (a) An element shows variable valencies 4 and
m 6. Write the formulae of its two oxides.
N N0 (b) An element forms an oxide A2O5 .
M
Sol.: Number of atoms in 4 g of (i) What is the valency of the element A?
4 (ii) What will be the formula of the chloride of
Ca 6.022 1023 the element?
40
Sol.: (a) Let the element be represented by the
6.022 1022 atoms symbol E.
Also. 6.022 1023 atoms of sodium = 23 g Formula of oxide in which valency of E is
6.022 1022 atoms of sodium will have mass 4 E2O4 or EO2
23 Formula of oxide in which valency of E is
6.022 1022 2.3 g
6.022 1023 6 E2O6 or EO3
14. Calculate the number of moles in the (b) Formula of oxide of the element A2O5
following: (i) The valency of the element A in the oxide =
(i) 28 g of He (ii) 46 g of Na. (iii) 60 g of Ca. 5
Given gram atomic mass of (i) He = 4 g (ii) Na (ii) The formula of the chloride of the element
= 23 g (iii) Ca = 40 g. A ACl5 .
Sol.: (i) 28 g of He
17. A flask P contains 0.5 mole of oxygen gas.
The no. of moles
Another flask Q contains 0.4 mole of ozone
Mass of He in grams m (28 g )
7mol gas. Which of the two flasks contains greater
Gram atomic mass M (4 g ) number of oxygen atoms?
(ii) 46 g of Na Sol.: 1 molecule of oxygen (O2 ) 2 atoms of
The no. of moles oxygen
Mass of Na in grams m (46 g )
2mol In flask P: 1 molecule of ozone (O3 ) 3
Gram atomic mass M (23g ) atoms of oxygen
(iii) 60 g of Ca 1 mole of oxygen gas 6.022 1023 molecules
The no. of moles 0.5 mole of oxygen gas 6.022 1023 0.5
Mass of Ca in grams m (60 g ) molecules
1.5mol
Gram atomic mass M (40 g ) 6.022 1023 0.5 2 atoms 6.022 1023
atoms
15. Calculate the weight of carbon monoxide In flask Q: 1 mole of ozone gas
having the same number of oxygen atoms as 6.022 1023 molecules
are present in 22 g of carbon dioxide. 0.4 mole of ozone gas 6.022 1023 0.4
Sol.: Step I. No. of the oxygen atoms in 22 g of molecules
carbon dioxide (CO2)
22
6.022 1023 0.4 3 atoms 7.23 1023 The formula NaCl represents only the
atoms simplest formula of sodium chloride.
Flask Q has a greater number of oxygen
atoms as compared to flask F.
18. What weight of calcium contains the same
number of atoms as are present in 3.2 g of
sulphur? 1. In a reaction, 5.3 g of sodium carbonate
Sol.: Step I. No. of atoms in 3.2 g of sulphur reacted with 6 g of ethanoic acid. The
Gram atomic mass of S = 32 g products were 2.2 g of carbon dioxide, 0.9 g
32 g of sulphur contain atoms 6.022 1023 water and 8.2 g of sodium ethanoate. Show
3.2 e of sulphur contain atoms that the observations are in agreement with
6.022 1023 the law of conservation of mass.
3.2 g 6.022 1022 Sodium carbonate + ethanoic acid sodium
32 g
ethanoate + carbon dioxide + water
Step II. Weight of 6.022 1022 atoms of Ans.: Total mass of reactants = mass of sodium
calcium carbonate + mass of ethanoic acid
Gram atomic mass of Ca 40 g = 5.3 g+6g= 11.3 g
6.022 1023 atomic of Ca weigh = 40 g Total mass of products = mass of sodium
6.022 1022 atoms of Ca weigh ethanoate + carbon dioxide + water
40 g = 8.2 g + 2.2 g + 0.9 g = 11.3 g
6.022 1022 4 g Thus, the mass of reactants is equal to the
6.022 1023
19. Why is the statement "1 mole of hydrogen" mass of products, therefore the observations
not correct? What should be the correct are in agreement with the law of conservation
statement? of mass.
Sol.: 1 mole of hydrogen does not tell whether we 2. Hydrogen and oxygen combine in the ratio of
have 1 mole of hydrogen atoms or 1 mole of 1 : 8 by mass to form water. What mass of
hydrogen molecules. The correct statement oxygen gas would be required to react
should be "1 mole of hydrogen atoms" or "1 completely with 3 g of hydrogen gas?
mole of hydrogen molecules". Ans.: 1 g of hydrogen reacts with oxygen 8 g
20. What do you understand by 'variable 3 g of hydrogen reacts with oxygen
valency'? Give two examples of metals and 8 3 24 g
two examples of non-metals showing variable Thus 24 g of oxygen gas would be required to
valency. react completely with 3 g of hydrogen gas.
Sol.: When an element shows more than one 3. Which postulate of Dalton's atomic theory is
valency in different compounds, it is said to the result of the law of conservation of mass?
exhibit variable valency. Ans.: The postulate that "atoms can neither be
Examples of metals showing variable valency: created nor destroyed in a chemical reaction"
Cu( I ), Cu( II ); Fe( II ), Fe( III ) is the result of the law of conservation of
Examples of non-metals showing variable mass.
valency: 4. Which postulate of Dalton's atomic theory can
Sulphur in H 2 S , SO2 and shows valencies of explain the law of definite proportions?
2, 4 and 6 respectively. Ans.: The postulate that "A chemical compound
Phosphorus in PCl3 and PCl5 shows a always consists of the same elements
combined together in the same proportion by
valency of 3 and 5 respectively.
mass" is the law of definite proportions.
21. Does NaCl represent the actual formula of
5. Define the atomic mass unit.
sodium chloride? Why or why not?
Ans.: Atomic mass unit is defined as the mass unit
Sol.: NaCl does not represent the actual formula
equal to exactly one-twelfth (1/12th) of the
of sodium chloride. This is because a crystal of
mass of one atom of carbon-12. It is denoted
sodium chloride consists of a very large but
by u (unified mass).
equal number of Na ions and Cl ions. i.e. 1 u = 1.66 x 10-24 g
Hence, the actual formula would be 6. Why is it not possible to see an atom with
( Na Cl )n when n is a very large number. naked eyes?
23
Ans.: It is not possible to see an atom with naked 11. Calculate the molecular masses of
eye because of its extremely small size H 2 , O2 , Cl2 , CO2 , CH 4 , C2 H 6 , C2 H 4 , NH 3 , CH 2 OH
(atomic radius is of the order of 10-10 m). Ans.: Molecular mass of H 2 2 atomic
7. Write down the formulae of (i) sodium oxide,
mass of H 2 1 u 2u
(ii) aluminium chloride, (iii) sodium sulphide,
(iv) magnesium hydroxide Molecular mass of O2 2 atomic mass of
Ions present Formula O 2 16 u 32 u
Ans.: Molecular mass of Cl2 2 atomic mass of
(i) Sodium oxide Na2O Cl 2 35.5 u 71 u
Molecular mass of CO2 atomic mass of
C 2 atomic mass of O
Ions present Formula
12 (2 16) (12 32) u 44 u
Molecular mass of CH 4 atomic mass of
(ii) Aluminium chloride AlCl3 C 4 atomic mass of H
12 (4 1) u (12 4) u 16 u
Molecular mass of C2 H 6 2 atomic mass of
Ions present Formula C 6 atomic mass of H
(2 12 6 1) u (24 6) u 30 u
(iii) Sodium sulphide Na2 S Molecular mass of C2 H 4 2 atomic mass of
C 4 atomic mass of H
(2 12 4 1) u (24 4) u 28 u
Ions present Formula
Molecular mass of NH 3 atomic mass of
(iv) Magnesium hydroxide Mg (OH )2 N 3 atomic mass of H
(14 3 1) u (14 3) u 17 u
8. Write down the names of compounds Molecular mass of CH3OH atomic mass of
represented by the following formulae: C + 3 x atomic mass of H + atomic mass of O +
(i) Al2 (SO4 )3 atomic mass of H
(ii) CaCl2 (12 3 1 16 1) u (12 3 17) u 32 u
(iii) K 2 SO4 12. Calculate the formula unit masses of
ZnO, Na2O, K2CO3 . Given atomic masses of
(iv) KNO3
Zn = 65 u,
(v) CaCO3
Zn 65 u, Na 23u, K 39 u, C 12 u .
Ans.: (i) Aluminium sulphate (ii) Calcium chloride
(iii) Potassium sulphate (iv) Potassium nitrate Ans.: Formula unit mass of ZnO atomic mass of
(v) Calcium carbonate Zn atomic mass of O
9. What is meant by the term chemical (65 16) u 81u
formulae? Formula unit mass of Na2O 2 atomic mass
Ans.: The chemical formula of a compound is a of Na atomic mass of O
symbolic representation of its composition. (2 23 16) u 62 u
e.g. formula of calcium oxide. Formula unit mass of K2CO3 2 atomic
mass of K atomic mass of C 3 atomic
mass of O
(2 39 12 3 6) u (78 12 48) u 138 u
Thus, formula of calcium oxide CaO . 13. If one mole of carbon atoms weighs 12 gram/
10. How many atoms are present in a what is the mass (in gram) of 1 atom of
(i) H 2 S molecule and (ii) PO43 ion? carbon?
Ans.: (i) 3 atoms because H 2 S molecule has two Ans.: Molecular mass of carbon = 12 g
atoms of hydrogen and one atom of sulphur. 6.022 1023 atoms of carbon have mass
(ii) 5 atoms because PO43 ion has one atom 12 g
of phosphorus and four atoms of oxygen.
24
1 atom of carbon has mass Ans.: A polyatomic ion is a group of atoms carrying
12 positive or negative charge.
g 1.99 1023 g
6.022 1023 e.g.
14. Which has more number of atoms, 100 grams
of sodium or 100 grams of iron (given atomic Polyatomic ions Symbol
mass of Na 23u, Fe 56 u) ? Ammonium NH 4
Ans.: 100 g of sodium: Hydroxide OH
Number of sodium atoms
Nitrate NO3
given mass 100
6.022 1023 6.022 1023 HCO3
molar mass 23 Hydrogen carbonate
= 2.62 x 1024 atoms.
Sulphate SO4 2
100 g of iron :
Number of iron atoms Sulphite SO32
given mass 100 PO43
6.022 1023 6.022 1023 Phosphate
molar mass 56 CO32
Carbonate
= 1.07 x 1024 atoms
18. Write the chemical formulae of the following:
Thus, 100 g of sodium has more atoms.
(a) Magnesium chloride
15. A 0.24 g sample of compound of oxygen and
(b) Calcium oxide
boron was found by analysis to contain 0.096
(c) Copper nitrate
g of boron and 0.144 g of oxygen. Calculate
(d) Aluminium chloride
the percentage composition of the compound
(e) Calcium carbonate
by weight.
Ans. (a) Magnesium chloride (b) Calcium oxide
Ans.: We know that, % of any element in a
Mass of element
compound 100
Mass of compound
0.096
% of boron 100 40%
0.24
0.144
% of oxygen 100 60% (c) Copper nitrate (d) Aluminium chloride
0.24
16. When 3.0 g of carbon is burnt in 8.00 g
oxygen, 11.00 g of carbon dioxide is
produced. What mass of carbon dioxide will
be formed when 3.00 g of carbon is burnt in
50.00 g of oxygen?
Which law of chemical combination will
(e) Calcium carbonate
govern your answer?
Ans.: C O2 CO2
3 g 8 g 11 g
Total mass of reactants = mass of C mass
of O 3 8 11 g
Total mass of reactants = total mass of 19. Give the names of the elements present in the
products following compounds.
Hence, the law of conservation of mass is (a) Quick lime
proved. (b) Hydrogen bromide
Further, it also shows carbon dioxide contains (c) Baking powder
carbon and oxygen in a fixed ratio by mass, (d) Potassium sulphate
which is 3 : 8. Thus it also proves the law of Ans.: (a) Quick lime is CaO . Elements present are
constant proportions. 3 g of carbon must also calcium and oxygen.
combine with 8 g of oxygen only. This mean (b) Hydrogen bromide is HBr . Elements
that (50 - 8) = 42 g of oxygen will remain present are hydrogen and bromine.
unreacted.
17. What are polyatomic ions? Give examples.
25
(c) Baking powder is NaHCO3 . Elements Number of moles in
present are sodium, hydrogen, carbon and 20
20 g H 2O 1.11mole
oxygen 18
(d) Potassium sulphate is K 2 SO4 . Elements (c) Molar mass of
present are potassium, sulphur and oxygen. CO2 12 16 2 12 32 44 g
20. Calculate the molar mass of the following Number of moles in
substances. 22
22 g CO2 0.5 mole
(a) Ethyne, C2 H 2 44
(b) Sulphur molecule, S8 23. What is the mass of
(a) 0.2 mole of oxygen atoms?
(c) Phosphorus molecule, P4 (atomic mass of
(b) 0.5 mole of water molecules?
phosphorus = 31) Ans.: (a) Mass of 1 mole of oxygen atom = 16 u
(d) Hydrochloric acid, HCl Mass of 0.2 mole of oxygen atoms
(e) Nitric acid, HNO3 . (16 0.2) u 3.2 u
Ans.: (a) Molar mass of C2 H 2 2 atomic mass of (b) Mass of 1 mole of H 2O molecule
C 2 atomic mass of H (1 2 16) 18 u
(2 12 2 1) u (24 2) u 26 u Mass of 0.5 mole of H 2O molecule
(b) Molar mass of S8 8 atomic mass of S (18 0.5) u 9 u
(8 32) u 256 u 24. Calculate the number of molecules of sulphur
(c) Molar mass of P4 4 atomic mass of P ( S8 ) present in 16 g of solid sulphur?
(4 31) u 124 u Ans.: Mass of 1 mole of S8 8 32 256 g
(d) Molar mass of HCl atomic mass of H + 256 g of S8 contains 6.022 1023 molecules
atomic mass of Cl (1 35.5) u 36.5 u
16 g of S8 contains
(e) Molar mass of HNO3 atomic mass of H
6.022 1023
+ atomic mass of N 3 atomic mass of O 16 3.76 1022 molecules.
256
(1 14 3 16) u (1 14 48) u 63 u 25. Calculate the number of aluminium ions
21. What is the mass of present in 0.051 g of aluminium oxide.
(a) 1 mole of nitrogen atoms? (Hint: The mass of an ion is the same as that
(b) 4 moles of aluminium atoms (atomic mass of an atom of the same element. Atomic mass
of aluminium = 27)? of Al 27 u )
(c) 10 moles of sodium sulphite ( Na2 SO3 ) ? Ans.: Molar mass of
Ans.: (a) Mass of nitrogen atom = number of moles Al2O3 2 27 3 16 54 48 102 g
atomic mass 114 14 g
102 g of Al2O3 contains
(b) Mass of aluminium atom 4 atomic
2 6.022 1023 Al 3 ions
mass 4 27 108 g
0.051 g of Al2O3 contains
(c) Mass of
Na2 SO3 2 23 32 16 3 46 32 48 126 g 2 6.022 10
23
0.051
1 mole of Na2 SO3 126 g 102
6.022 1020 Al 3 ions.
10 moles of Na2 SO3 (10 126) g 1260 g
22. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Ans.: (a) Molar mass of O2 2 16 32 g
Multiple Choice Questions (MCQs)
Number of moles in
Given mass 12 1. Which of the following correctly represents
12 g O2 0.375 mole
Molar mass 32 360 g of water?
(b) Molar mass of H 2O 2 1 16 18 g (i) 2 moles of H 2O
(ii) 20 moles of water
(iii) 6.022 x 1023 molecules of water
26
(iv) 1.2044 x 1025 molecules of water 3. The chemical symbol for nitrogen gas is
(a) only (i) (b) (i) and (iv) (a) Ni (b) N 2
(c) (ii) and (iii) (d) (ii) and (iv) (c) N + (d) N
Ans. (b) Nitrogen molecule is diatomic molecule,
1mole 6.022 10 atoms/molecules/ions of a
23 therefore, it exists as N; molecules.
substance = gram molecular mass 4. The chemical symbol for sodium is
Ans. (d) (ii) and (iv) points correctly represent 360 (a) So (b) Sd (c) NA (d) Na
g of water. Ans. (d) The chemical symbol for sodium is derived
(ii) From point, from its Latin name 'Natrium'. In a 'two letter'
1 mole of water = molar mass of water =18g symbol, the first letter is the 'capital letter'
20 moles of water = 18 g x 20 = 360g but the second letter is the 'small letter'.
(iv) From point, Therefore, its symbol is 'Na'.
6.022 x 1023 molecules of water = 1 mole = 5. Which of the following would weigh the
18g of water highest?
1.2044 x 1025 molecules of water (a) 0.2 mole of sucrose (C12H22O11 )
18 g 1.2044 1025 (b) 2 moles of CO2
360 g
6.022 1023 (c) 2 moles of CaCO3 (d) 10 moles of H 2O
Therefore, points (ii) and (iv) represent 360 g Thinking Process
of water. 1 mote = gram molecular mass of the
Students can check other points also, as compound.
(i)1 mole of water = molar mass of water = 18 Ans. (c) 2 moles of CaCO3 would weigh the
g of water. highest.
2 mole of water = 18x2 = 36g (a) Mass of 1 mole of sucrose
(iii) 6,022 x1023 molecules of water =18g (C12 H22O11 )(12 12) (1 22) (16 11) 342 g
water.
Mass of 0.2 mole of sucrose
2. Which of the following statements is not true
342 0.2 68.4g
about an atom?
(a) Atoms are not able to exist independently Mass of 0.2 mole of sucrose = 342 x 0.2 =
(b) Atoms are the basic units from which 68.4g
molecules and ions are formed (b) Mass of 1 mole of
(c) Atoms are always neutral in nature CO2 12 (16 2) 44 g
(d) Atoms aggregate in large numbers to form Mass of 2 moles of CO2 44 2 88g
the matter that we can see, feel or touch (c) Mass of 1 mole of
Ans. (d) This statement is not true CaCO3 40 12 (16 3) 100 g
The correct statement is as The molecules
Mass of 2 moles of CaCO3 100 2 200 g
and ions aggregate together in large number
to form the matter. We cannot see the (d) Mass of 1 mole of H 2O 2 16 18g
individual molecules/ions with our eyes. Mass of 10 moles of H 2O 18 10 180 g
only we can see the various substances which Therefore, mass of 2 moles of CaCO3 is the
are a big collection of molecules/ions, So, highest, .e., 200g.
option (d) is incorrect. 6. Which of the following has maximum number
Atoms of most of the elements are chemically of atoms?
very reactive and do not exist in the free (a) 18 g of H 2O (b)18 g of O2
state. The atoms of only noble gases are
(c)18 g of CO2 (d) 18 g of CH 4
chemically uncreative and exist in free state.
Atoms usually exist in the two forms
(i) molecules and (ii) ion's Number of atoms
Therefore, atoms are the basic unit from mass
which molecules and ions are formed. 6.022 1023 P
molecular mass
When atoms form molecules or ions, they
where, P = number of atoms in the molecule
become stable because they acquire the
Ans. (d) Option (d) is the correct answer, i.e., 18 g
stable noble gas electron arrangement.
Therefore, they are neutral in nature of CH 4 has maximum number of atoms. As,
27
(a) Number of atoms in 18 g of 8. Mass of one atom of oxygen is
18 16 32
H 2O 6.022 1023 3 (a) g (b) g
18 6.023 1023 6.023 1023
18.066 1023 1.8066 1024 (c)
1
(d) 8 u
g
(b) Number of atoms in 18 g of 6.023 1023
18
O2 6.022 1023 2
32 1 mole of oxygen = 6.023 x 1023 atoms of
3.387 1023 2 6.774 1023 oxygen = gram atomic mass of oxygen
(c) Number of atoms in 18g of Ans. (a) Mass of 6.023 x 1023 atoms of oxygen =
18 gram atomic mass of oxygen mass of oxygen
CO2 6.022 1023 3 7.390 1023
44 Mass of atoms of oxygen
(d) Number of atoms in 18 g of Mass of 6.023 1023 atoms of oxygen = 16
18 16
CH 4 6.022 1023 5 3.387 1024 Mass of atom of oxygen g
16 6.023 1023
Thus, 18 g of CH 4 , contains the maximum 9. 3.42 g of sucrose are dissolved in 18 g of
number of atoms. water in a beaker. The number of oxygen
7. Which of the following contains maximum atoms in the solution are
number of molecules? (a) 6.68 x1023 (b) 6.09 x1022
(a) 1g CO2 (b) 1g N 2 (c) 6.022 x 10 (d) 6.022 x1021
(c) 1g H 2 (d) 1g CH 4
(i) First we find out the number of O-atom in
3.42 g of sucrose.
In 1 mole number of molecules = 6.022 x 1023
(ii) Then, we find out the number of O-atoms
Ans. (c) Option (c) the correct answer, i.e., 1 g H 2
in 18 g of water.
contains maximum number of molecules. (iii) By adding these two, we will find out the
(a) Number of molecules in total number of0-aio'ms in the solution.
44 g CO2 6.022 1023 (molar mass of Ans. (a)
N2 44 g ) Step 1. Molar mass of sucrose,
Number of molecules in C12 H 22O11 12 12 1 22 16 11 342 g
6.022 10 23 or 342 g = 1 mole of sucrose
1g CO2 1.37 1022 3,42 g = 0.01 mole of sucrose
44
(b) Number of molecules in 1 mole of sucrose (C12 H 22O11 ) contains 0-
28gN2 6.022 1023 (molar mass of N2 28g atoms = 11 x 6.022 x1023 atoms
) 0.01 mole of sucrose will contain 0-atoms
Number of molecules in = 0.01 x 11 x 6.022x 1023 atoms = 6.6242 x
1022
6.022 1023
1g N 2 2.15 1022 Step 2. 18 g of water ( H 2O) = 1 mole of
28
(c) Number of molecules in water 1 mole of water ( H 2O) contains 0-
2 g H 2 6.022 1023 (molar mass of H 2 2 g ) atoms = 6.022 x 1023 atoms
Number of molecules in Step 3. By adding the number of I-atoms
present in 3.42 g of sucrose and 18 g of water
6.022 1023
1gH 2 3.0111023 we get 6.022x 1023 + 6.6242 x 1022 = 1022
2 (60,22 +6.6242) = 66.844 x 1022 = 6.68x 1023
(d) Number of molecules in
atoms
16 g CH 4 6.022 1023 (molar mass of 10. A change in the physical state can be brought
CH 4 16 g ) about
Number of molecules in (a) only when energy is given to the system
6.022 10 23 (b) only when energy is taken out from the
1 g CH 4 3.76 1022 system
16
(c) when energy is either given to, or taken
Thus, 1g H 2 contains the maximum number
out from the system
of molecules.
(d) without any energy change
28
(ii) with SO42 Cu 2 SO42 CuSO4
The heat energy which has to be supplied to (iii) with PO43 Cu 2 PO43 Na3 PO4
change the state of a substance is called its Compounds of Fe3
latent heat. It does not raise the temperature.
(i) with Cl Fe3 Cl1 FeCl3
It is used up in overcoming the forces of
attraction between the particles of a (ii) with SO42 Fe3 SO42 Fe2 (SO4 )3
substance during the change of state. (iii) with PO43 Fe3 PO43 FePO4
Ans. (c) A change in the physical state can be
Compounds of Fe3
brought about when energy is either given to,
or taken out from the system. It is because, (i) with Cl Fe3 Cl1 FeCl3
energy change helps in changing the (ii) with SO42 Fe3 SO42 Fe2 (SO4 )3
magnitude of attraction forces between the (iii) with PO43 Fe3 PO43 FePO4
particles, thus helps in changing the physical
14. Write the cat ions and anions present (if any)
states i.e., solid, liquid, gas) of matter.
in the following compounds.
(a) CH3COONa (b) NaCl
(c) H 2 (d) NH 4 NO3
11. Which of the following represents a correct
chemical formula? Name it.
(a) CaCI (b) BiPO4 Positively charged ions are called cations
(c) NaSO4 (d) NaS while negatively charged ions are called
onions.
Thinking Process
Ans.
(i) Option (a) has correct formula — CaCl2 — S. No. Compounds Cat ion Anion
Calcium chloride (a) CH3COONa Na CH 3COO
(ii) Option (c) has correct formula — Na2 SO4
(b) NaCl Na Cl
— Sodium sulphate
(c) H2 - -
(iii) Option (d) has correct formula — Na2 S —
Sodium supplied
(d) NH 4 NO3 NH 4 NO3
Ans. (b) BiPO4 , is the correct formula, its name is 15. Give the formulae of the compounds formed
bismuth phosphate. from the following sets of elements.
12. Write the molecular formulae for the (a) Calcium and fluorine
following compounds (b) Hydrogen and sulphur
(a) Copper (II) bromide (c) Nitrogen and hydrogen
(b) Aluminium (III) nitrate (d) Carbon and chlorine
(c) Calcium (II) phosphate (e) Sodium and oxygen
(d) Iron (III) sulphide (f) Carbon and oxygen
(e) Mercury (II) chloride Ans.
(f) Magnesium (II) acetate (a) Calcium and fluorine Ca 2 F 1 CaF2
Ans. (a) Cu 2 Br1 CuBr2 (b) Hydrogen and H 1 S 2 H 2 S
(b) Al 3 NO31 Al ( NO3 )3 sulphur
(c) Ca 2 Cl1 HgCl2 (c) Nitrogen and N 3 H 1 NH 3
hydrogen
(d) Fe3 S 2 Fe2 S3 (d) Carbon and chlorine C 4 C1 CCl4
2 1
(e) Hg Cl HgCl2 (e) Sodium and oxygen Na1 O2 Na2O
1 2
(f) CH 3COO Mg (CH3COO)2 Mg (f) Carbon and oxygen 4 2
C O C2 O4 or CO2
13. Write the molecular formulae of all the
compounds that can be formed by the 16. Which of the following symbols of elements
combination of following ions. are incorrect? Give their correct symbols.
(a) Cobalt CO (b) Carbon c
Cu 2 , Na , Fe3 , Cl , SO42 , PO43
(c) Aluminium AL (d) Helium He
Ans. Compounds of Cu 2 (e) Sodium So
(i) with Cl Cu 2 Cl1 CuCl2
29
Ans. (a) Cobalt CO is incorrect symbol. Its correct Ans. In water molecule ( H 2O) , number of
symbol is Co. neutrons = [(number of neutrons in H) x 2 +
(b) Carbon c is incorrect symbol. Its correct (number of neutrons in O)]
symbol is C. = 0 x 2 + 8=8 (as number of neutrons in H = 0)
(c) Aluminium AL is incorrect symbol, Its Mass of 8 neutrons = 8 x 1.00893 = 8.07
correct symbol is Al. ( mass of one neutron = 1.008934)
(e) Sodium So is incorrect symbol. Its correct Molar mass of water = 1.008x2+16.0 =18.016u
symbol is Na. (It is derived from Latin name Fraction of mass due to neutrons
'Natrium'). massof total neutronsin water×100
(d) Helium, He is the correct symbol. =
molar massof water
17. Give the chemical formulae for the following
8.07
compounds and compute the ratio by mass of 100 44.8%
the combining elements in each one of them. 18.016
(a) Ammonia 20. Does the solubility of a substance change with
(b) Carbon monoxide temperature? Explain with the help of an
(c) Hydrogen chloride example.
(d) Aluminium fluoride Ans. Yes, the solubility of a substance changes with
(e) Magnesium sulphide temperature,
Ans. The maximum amount of a solute which can
S. Compounds Chemical Ratio by mass of the be dissolved in 100 g of a solvent at a
No. formula combining elements specified temperature is called its solubility,
(a) Ammonia NH 3 N : H 14: 3 Effect of temperature on solubility
(i) The solubility of solids in liquids usually
(b) Carbon CO C : O 12:16 3: 4 increases on increasing the temperature and
monoxide
decreases on decreasing the temperature.
(c) Aluminium HCl H : Cl 1:35.5 (ii)The solubility of gases in liquids usually
fluoride decreases on increasing the temperature and
(d) Aluminium AlF3 Al : F 27 :57 9:19 increases on decreasing the temperature.
fluoride Let us take an example of copper sulphate.
(e) Magnesium MgS Mg : S 24 : 32 3: 4 The solubilities of copper sulphate in water at
sulphide various temperatures are given below
18. State the number of atoms present in each of
the following chemical species Temperature Solubility of copper
(a) CO32 (b) PO43 sulphate
0o C 14 g
(c) P2O5 (d) CO
10o C 17 g
Ans. (a) Number of atoms in CO32 = number of C-
20o C 21 g
atoms + number of O-atoms = 1 + 3 = 4
30o C 24 g
(b) Number of atoms in PO43 = number of P-
40o C 29 g
atoms + number of O-atoms = 1 + 4 = 5
50o C 34 g
(c) Number of atoms in P2O5 = number of P-
60o C 40 g
atoms + number of O-atoms = 2 + 5 = 7
70o C 47 g
(d) Number of atoms in CO = number of C-
atoms + number of O-atoms = 1 + 1 = 2
19. What is the fraction of the mass of water due From the above data, it is clear that, when the
to neutrons? temperature is increased from 0°C to 70'C,
the solubility of copper sulphate in water
increases from 14 g to 47g. The above data
(i) First we will find out the total number of shows that the solubility of a salt increases on
neutrons in water molecule ( H 2O) . increasing the temperature.
(ii) Then, we will find out the mass of total 21. Classify each of the following on the basis of
number of neutrons. Then the fraction of the their atomicity.
mass of water due to neutrons. (a) F2 (b) N 02
30
(c) N 2 0 (d) C2 H 6
(e) P4 (f) H 2 02
(g) P4 010 (h) 03 24. Verify by calculating that
(i) HCI (j) CH 4 (a) 5 moles of C 02 and 5 moles of H 2 0 do
(k) He (I) Ag not have the same mass.
(b) 240 g of calcium and 240 g magnesium
elements have a mole ratio of 3 : 5.
The number of atoms present in one molecule
of an element is called its atomicity.
On the basis of atomicity, elements can be 1 mole = molar mass of the substance.
classified as monoatomic (having 1-otom Ans. (a) Molar mass of
molecules'), diatomic (having 2-atoms CO2 12 2 16 12 32 44 g mol1
molecules) and polyatomic (having more than 1 mole of CO2 has mass = 44 g
2-atoms molecules). 5 moles of CO2 have mass = 44 x 5 = 220 g
Ans. They are classified as
Similarly, molar mass of
Monoatomic (k) He, (Z) Ag Diatomic (a) F2 , (i)
H 2O 2 1 16 18g mol1
HCl
1 mole of H 2O has mass = 18 g
Polyatomic (b) NO2 , (c) N 2O, (d) C2 H 6 , (e) P4 ,
5 moles of H 2O have mass = 18 x 5= 90 g
(f) H 2O2 , (g) P4O10 , (h) O3 , (j) CH 4
Both the values are not same.
22. You are provided with a fine white coloured
So, it is verified.
powder which is either sugar or salt. How
(b) Molar mass of calcium = 40 g
would you identify it without testing?
40 g of Ca has number of moles = 1 mol
Ans. There are many ways by which we can
240 g of Ca has number of moles
differentiate between sugar and salt without
1
testing Dissolve salt and sugar separately in 240 6 mol
alcohol, sugar will be dissolved in it while salt 40
will not be dissolved. Molar mass of magnesium =24 g
(ii) Heat these salts separately, sugar will melt 24 g of Mg has number of moles = 1 mol
readily while salt will not. 240 g of Mg has number of moles
(iii)Dissolve these two separately in water. 1
240 10 mol
Salt solution will conduct electricity due to the 24
presence of Nation and Cl ion while sugar Number of moles of
6 3
solution will be a non-conductor. So, simply Number of moles of Mg 10 5
test a drop of the solution with an ohmmeter So, it is verified.
and you can immediately tell the difference. 25. Find the ratio by mass of the combining
23. Calculate the number of moles of magnesium elements in the following compounds.
present in a magnesium ribbon weighing 12g. (a) CaC 03 (b) MgCl2
Molar atomic mass of magnesium is 24 g (c) H 2 S 04 (d) C2 H 5 0H
mol1 .
(e) NH 3 (f) Ca(0H )2
Ans. (a) CaCO3 Ca : C : O 40 :12 : 48 10 : 3:12
1 mole = molar mass of the substance
(b) MgCl2 Mg : Cl 24: 2 35.5 24: 71
Ans. Given that molar atomic mass of Mg = 24 g
(c) H 2 SO4 H : S : O 2 1:32: 4 16 2
mol1
24 g of Mg =1 mol :32: 64 1:16:32
112 1 (d) C2 H5OH C : H : O 2 12: 6 1
12 g of Mg 0.5mol
24 2 1:16 24: 6:16 12:3:8
Alternative Method (e) NH3 N : H 14: 3 1 14: 3
Number of moles of (f) Ca(OH )2 Ca : O : H 40: 2 16
Mg
given mass
12 g
0.5mol : 2 1 40:32: 2 20:16:1
molar mass 24 g mol 1 26. Calcium chloride when dissolved in water
dissociates into its ions according to the
following equation.
31
CaCl2 (ag ) Ca 2 (aq) 2Cl (aq) 225 g of HgS will contains
Calculate the number of ions obtained from 200.6 225
Hg 194.05 g
CaCl2 when 222 g of it is dissolved in water. 232.6
29. The mass of one steel screw is 4.11 g. Find the
Thinking Process
mass of one mole of these steel screws.
1mole 6.022 1023 ions Compare this value with the mass of the earth
Ans. Molar mass of CaCl2 40 2 35.5 (5.98 x 1024 kg). Which one of the two is
40 71 111 g mol1 heavier and by how many times?
CaCl2 ionizes in water as Ans. Given that, mass of one steel screw = 4.11 g
Mass of earth 5.98 1024 kg 5.98 1027 g
Ca 2 (aq ) 2Cl (aq )
CaCl2 (aq)
1mol 2mol We know that, 1 mole = 6.022 x 1023
3mol atoms/molecules/ions
Mass of 1 mole of screws = 6.022x 1023 x 4.11
111g of CaCl2 produces ions = 3 mol = 3x
g = 2.48x 1024 g
6.022x 1023 ions Mass of 1 mole screw 2.48x 1024 g
222 g of CaCl2 produces ions Mass of mole screw 2.48 10 g
24
3 6.022 10 23
5.98 10 g
27
222 Mass of earth
111 1 1
36.132 1023 3.6132 1024 ions 2.41 10
3
2410
27. The difference in the mass of 100 moles each Ratio of 1 mole of screw and mass of earth
of sodium atoms and sodium ions is 5.48002 = 1:2410
g. Compute the mass of an electron. Earth is heavier than screw by 2410 times.
Thinking Process 30. A sample of vitamin C is known to contain
in actual, mass of an electron 9.11028 g or 2.58 x 1024 oxygen atoms.
9.11031 kg How many moles of oxygen atoms are present
Ans. Sodium atom loses an electron to form in the sample?
Ans. Given that, number of oxygen atoms in the
sodium ion, Na Na + e-
Sodiumatom Sodiumion Electron given sample =2.58 x 1024 We know that,
For 100 moles of each sodium atom and 6.022x 1023 oxygen atoms = 1 mol
sodium ion there would be difference of 2.58 1024 oxygen atoms
100e 2.58 10
24
4.28 mol
Mass of 100 moles of electron = 5.48002 g 6.022 10
23
Mass of 1 mole of electron 5.48002g 31. Raunak took 5 moles of carbon atoms in a
6.022 x 1023 electrons are present in 1 mole container and Krish also took 5 moles of
of electron sodium atoms in another container of same
[as 1 mole =6.022 x 1023 electrons] i.e., mass weight.
of 6.022 x1023 electrons = 5.48002/100 g (a) Whose container is heavier?
Mass of 1 electron (b) Whose container has more number of
5.48002 26
atoms?
9.1 10 g
100 6.022 10
23 Ans. (a) 1 mole = molar mass of a substance.
28. Cinnabar (HgS) is a prominent ore of mercury. mole of carbon atoms weigh =12g
How many grams of mercury are present in 5 moles of carbon atoms will weigh =12 x 5
225 g of pure HgS? Molar mass of Hg and S = 60 g
are 200.6 g mol-1 and 32g mol-1 respectively. Hence, Raunak's container has weigh = 60 g In
the same way, mole of sodium atoms weigh
Ans. Given that, molar mass of Hg =200.6 g mol-1
=23 g
and that of S = 32 g mol-1
5 moles of sodium atoms will weigh = 23 x 5
Molar mass of HgS = molar mass of the Hg +
= 115 g
molar mass of the S
Hence, Krish's container has weigh =115g
Molar mass of HgS = 200.6 + 32 = 232.6 g
Thus, Krish's container is heavier than
232.6 got HgS contains Hg= 200.6 (as
Raunak's container.
molar mass of Hg= 200.6 g mol-1 ) As, 1 mole = 6.022 x 1023 atoms
32
(b) Here, both containers have 5 moles of 33. The visible universe is estimated to contain
each carbon and sodium, therefore, both 1022 stars. How many moles of stars are
containers have equal number of atoms, i.e., present in the visible universe?
5 x 6.022 x 10 atoms. Ans. 6.022 x 1023 stars = 1 mole of stars.
or 3.011 x 1024 atoms in each. 11022
32. Fill in the missing data in the following table. 1022 stars 1.67 102 mol
6.022 1023
Species H 2O CO2 Na- MgCl2 34. What is SI prefix for each of the following
property atom multiples and submultiples of a unit?
Number of 2 - - 0.5 (a) 103 (b) 101
particles (c) 10 2
(d) 106
Number of - 3.011 - -
(e) 109 (f) 1012
particles 10 22
Ans. (a) 10 kg
3
(b) 101 deci
Mass 36 g - 115 g -
Ans. For H2O (c) 102 centi (c) 106 micro
Given, number of moles = 2 and mass = 36 g (e) 109 nano (f) 1012 pico
Number of particles = number of moles 35. Express each of the following in kilograms
6.022 1023 (a) 5.84 103 mg (b) 58.34 g
= 2 x 6.022 x 1023 = 1.2044 x 1024 (c) 0.584 g (d) 5.873 1021 g
For CO2
Given, number of particles = 3.011 x 1023
Number of moles of 1kg 103 g 106 mg and 103 kg 1g
CO2
number of particles Ans. (a) 106 mg 1kg
6.022 1023 1 5.84 10
3
3.0111023 584 103 mg 9
kg 5.84 10 kg
0.5mol 10
6
6.022 1023
Mass of CO2 moles x molar mass (b) 103 g 1kg
= 0.5 x 44 = 22 g
1 58.34
( molar mass of CO2 = 12 + 2 x 16 = 44) 58.34 g kg 58.34 109 kg
103
For Na-atom
Given, mass = 115 g 5.834 102 kg
mass 115 1 0.584 3 4
Number of moles 5mol (c) 0.584 g 3
kg 0.584 10 kg 5.84 10 kg
molar mass 23 10
5.873
Number of particles = 5 x 6.022 x1023 = (d) 5.873 1021 g 3
kg 0.5.873 10
24
kg
3.011 x 1024 10
36. Compute the difference in masses of 103
For MgCl2
moles each of magnesium atoms and
Given, number of moles = 0.5 magnesium ions.
Number of particles = 0.5 x 6.022 x1023 =
(Mass of an electron 9.11031 kg )
3.011 x 1023
Mass = number of moles x molar mass
(molar mass of MgCl2 = 24 + 2 x 35,5 = 24 + (i) First we find out the number of Mg atoms
71 = 95) = 0.5 x 95 = 47.5 g and number of Mg ions in 103 motes of the
Thus, the completed table is as Mg atom.
Species Number Number of Mass (ii) Then we calculate the difference in their
property of particles masses.
particles Ans. 103 moles of Mg atoms =103 x 6.022 x 1023
H 2O 2 1.2044 1024 36 g 6.022 1026 Mg atoms
CO2 0.5 3.0111023 22 g 103 moles of Mg 2 ions = 103 x 6.022 x 1023
Na-atom 5.0 3.01110 24 115 g 6.022 1026 Mg 2 ions
MgCl2 05 3.01110 23 47.5 g One Mg 2 ion is formed from one Mg atom by
loss of 2 electrons, As Mg Mg 2 2e i.e.,
33
difference in the mass of 1 Mg atom and 1 Atomic mass of gold (Au) = 197 g
Mg 2 ion = mass of 2 electrons 197 g of gold have number of atoms
Difference in the mass of 6.022 x 1026 Mg 0.9 g of gold will have number of atoms
atoms and Mg 2 ions 6.022 1023 0.9
2.75 1021
= mass of 2 x 6.022 x 1026 electrons 197
40. What are ionic and molecular compounds?
= 2 x 6.022 x 1026 x 9.1 x 1031 kg.
Give examples.
(as mass of an electron = 9.1 x 1031 kg)
Ans. Ionic compounds are those compounds which
= 109.6004 x 10-5 kg = 1.096 x 103 kg = 1.096 g are made up of ions. In ionic compounds, the
37. Which has more number of atoms? positively charged ions (cations) and
100 g of N3 or 100 g of NH 3 negatively charged ions (anions) are held
Ans. Molar mass of N3 = 2x 14 = 28 g together by the strong electrostatic forces of
28 g of N 2 has number of molecules = attraction, which are known as ionic bonds or
6.022 x 1023 electrovalent bonds, e.g., sodium chloride
100 g of N 2 has number or molecules (NaCI), calcium oxide (CaO), etc.
Molecular compounds are those compounds
6.022 1023 100
2.11024 in which the atoms of the elements share
28 electrons via covalent bonds . Thus, in these
Atoms in 100 g of N2 2.11024 2 compounds, covalent bonds are present.
4.2 1024 atoms e.g., methane (CH 4 ) , water ( H 2O) etc.
Similarly, molar mass of 41. Compute the difference in masses of one
NH3 14 3 1 17 g mole each of aluminium atoms and one mole
17 g NH 3 has number of molecules = 6.022 of its ions (Mass of an electron is 9.1 1028 g ).
x 1023 Which one is heavier?
100 g of NH 3 has number of molecules Ans. Ionizations of Al atom occurs as
6.022 10 100
23 Al Al 3 3e
3.54 1024
17 Therefore, Al 3 ion is formed from Al atom by
Atoms in 100 g of loss of 3 electrons. Difference in mass of mole
NH3 3.54 1024 1024 4 1.416 1025 atoms and 1 mole of Al 3 ions
Thus, 100 g of NH 3 has more number of = mass of 3 x 6.022 x 1023 electrons
atoms. (3 6.022 1023 ) (9.11028 g )
38. Compute the number of ions present in 5.85 g (as mass of an electron 9.11028 g )
of sodium chloride. 164.4 105 g 1.644 103 g
Ans. Molar mass of sodium chloride (NaCI) = 23 +
mole of Al atoms is heavier than 1 mole of
35.5 = 58.5 g
58.5 g of NaCI has number of ions = 6.02 x Al 3 ions.
1023 x 2 It is because 1 mole of Al atom have
As, NaCl Na + Cl- e 13 6.022 1023 electrons and 1 mole of
Sodium
chloride
Sodium ion Chlorideion Al ion have e- = 10 x 6.022 x 1023 electrons
5.85 g of NaCI has number of ions 42. A silver ornament of mass 'm' gram is polished
with gold equivalent to 1% of the mass of
6.022 1023 2 5.85
silver. Compute the ratio of the number of
58.5 atoms of gold and silver in the ornament.
12.044 1022 1.2044 1023
39. A gold sample contains 90% of gold and the
rest copper. How many atoms of gold are 1 mole =6.022 x1023 atoms or molecules =
present in one gram of this sample of gold? molar/atomic mass of substance
Ans. Let, the mass of the sample is 100 g Ans. Mass of silver (Ag) ornament = m g
Mass of gold = 90 g Mass of gold used for polishing
and mass of copper =(100-90)= 10 g 1
m g 0.01mg
100 g of sample has gold = 90 g 100
90 Atomic mass of Ag = 108 u
1 g of this sample has gold 0.9 g 1 mole of Ag = 108 g = 6.022 x 1023 atoms
100
34
Thus, 108 g of Ag have atoms = 6.022 x 1023 Ans. (a) In a chemical reaction, the sum of the
mg of Ag will have atoms masses of the reactants and products remains
6.022 1023 unchanged. This is called law of conservation
m of mass.
108
Similarly, atomic mass of gold (Au) = 197 u (b) A group of atoms carrying a fixed charge
mole of Au = 197 g = 6.022 x1023 atoms on them is called polyatomic ion.
Thus, 197 g of Au have atoms = 6.022 x 1023 (c) The formula unit mass of Ca3 ( PO4 )2 is 310
0.01 m g of Au will have atoms g.
6.022 1023 0.01m (d) Formula of sodium carbonate is Na2CO3
197 and that of ammonium sulphate is
Ratio of the number of atoms of gold and ( NH 4 )2 SO4 .
6.022 1023 6.022 1023 Complete the following crossword puzzle
silver 0.01m : m
197 108 (Figure) by using the name of the chemical
1 1 elements. Use the data given in the table
: =108:19700
19700 108 following.
43. A sample of ethane (C2 H 6 ) gas has the same
mass as 1.5 x 1020 molecules of methane 45. Complete the following crossword puzzle
(figure) by using the name of the chemical
(CH 4 ) . How many C2 H 6 molecules does the
elements. Use the data given in table
sample of gas contain? following.
(i) First we will find out the mass of 15 1020 Across Down
molecules of methane. 2. The element used by 1. A white lustrous metal
(ii) Then, we will find out the mass of ethane, Rutherford during his - used for making ornaments
which is equal to the mass of methane. scattering experiment. and which tends to get
tarnished black in the
(iii) from this mass, we will then find out the
presence of moist air.
number of C2 H 6 molecules. 3. An element which forms 4. Both brass and bronze are
Ans. 6.022 x 1023 molecules of methane have rust on exposure to moist alloys of the element.
mass = molar mass of methane (CH 4 ) air.
= 12 + 4 x 1 = 16 g 5. A very reactive non- 6. The metal which exists in
metal stored under water. the liquid state at room
1.5 10 molecules of methane have mass
20
temperature.
16 1.5 1020
3.98 103 g 7. Zinc metal when treated 8. An element with symbol
6.022 1023 with dilute hydrochloric Pb.
This is also the mass of ethane (C2 H 6 ) acid produces a gas of this
element which when
Molar mass of ethane (C2 H6 ) 2 12 6 1
tested with burning splinter
24 6 30g As 30 g of ethane has number produces a pop sound.
of molecules = 6.022 x 1023
3.98 103 g of ethane has number of
6.022 1023 3.98 103
molecules g
30 g
7.99 1019 8 1019 molecules
44. Fill in the blanks.
(a) In a chemical reaction, the sum of the
masses of the reactants and products remains
unchanged. This is called ......... ,
(b) A group of atoms carrying a fixed charge
on them is called ......... .
(c) The formula unit mass of Ca3 ( PO4 ) 2 is…….
Ans.
(d) Formula of sodium carbonate is ......... and
2. Gold 1. Silver
that of ammonium sulphate is ......... .
3. Iron 4. Copper
35
5. Phosphorus 6. Mercury
7. Hydrogen 8. Lead
(b) Total number of inert gases in this
crossword puzzle is 6. Their names and
symbols are as follows
S. No. Name of gas Symbol
46. (a) In this crossword puzzle (Figure), names of 1. Helium He
11 elements are hidden. 2. Neon Ne
Symbols of these are given below. Complete 3. Argon Ar
the puzzle.
4. Krypton Kr
1 Cl 2. H 3. Ar 4. 0
5. Xe 6. N 7. He 8. F 5. Xenon Xe
9. Kr 10. Rn 11. Ne 6. Radon Rn
47. Write the formulae for the following and
calculate the molecular mass for each one of
them.
(a) Caustic potash (b) Baking powder
(c) Lime stone (d) Caustic soda
(e) Ethanol
(f) Common salt
Baking soda is sodium hydrogen carbonate
NaHCO3 Baking powder is a mixture,
therefore, does not have a formula, in baking
powder, there is o dry acid also which can be
(b) Identify the total number of inert gases, tartaric acid from grapes or citric acid or
their names and symbols from this cross ward others.
puzzle. Ans.
Ans. (a) the names of these elements are given as S. Compound Formul Molecular mass
follows No. a
(a) Caustic KOH 39 16 1 56u
1. Cl- Chlorine 7. He – Helium
potash
2. H-Hydrogen 8. F- fluorine
(b) Baking NaHCO3 23 1 12 3 16 84u
3. Ar-Argon 9. Kr- Krypton powder
4. O- Oxygen 10. Rn- Radon
(c) Lime stone CaCO3 40 12 3 16 100u
5. Xe- Xenon 11. Ne-Neon
6. N-Nitrogen (d) Caustic NaOH 23 16 1 1 40u
soda
(e) Ethanol C 2 H 5 OH 2 12 5 1 16 1 46u
(f) Common NaCl 23 35.5 58.5u
salt
36
48. In photosynthesis, 6 molecules of carbon (c) 1 atom of He
dioxide combine with an equal number of (d) 1 mole atoms of He
water molecules through a complex series of 4. Which of the following is a triatomic
reactions to give a molecule of glucose having molecule?
a molecular formula C6 H12O6 . How many (a) carbon dioxide (b) ammonia
grams of water would be required to produce (c) helium (d) sugar
18 g of glucose? Compute the volume of 5. Which of the following is the correct symbol
water so consumed assuming the density of of an element?
water to be 1g cm3 . (a) SN (b) CO
(c) Cu (d) PB
Ans. During photosynthesis, following reaction
6. The mass of 1 amu is
takes place 6CO2 6H 2O C6 H12O6 6O2 1 1 1
6(2 1 16) (a) g (b) g
12 6.022 10 23
6.022 1023
6 18 108g 12
(c) g (d) 6.022 1023 g.
Molecular formula of glucose 6.022 10 23
(C6 H12O6 ) 6 12 12 1 6 16 7. All samples of carbon dioxide contain carbon
72 12 96 180g and oxygen in the mass ratio 3 : 8. This is in
180 g of glucose require the amount of agreement with the law of
water = 108 g (a) conservation of mass
18 g of glucose will require the amount of (b) constant proportions
108 g 18 g (c) multiple proportions
water 10.8 g (d) gaseous volumes.
180 g
8. The indivisibility of an atom was proposed by
As the amount of water consumed (m) = 10.8
(a) Rutherford (b) Dalton
g
(c) Bohr (d) Einstein
Given that, density (d) of water 1g cm3 9. Which of the following is a tetraatomic
Volume (V) of water consumed molecule?
m m 10.8 g (a) O2 (b) O3
Ad d 10.8 cm3
d V 1g cm3 (c) NO2 (d) SO3
10. The number of atoms present in a molecule of
a substance is called
(a) molecularity (b) atomicity
(c) valency (d) reactivity
11. The valency of nitrogen in ammonia ( NH 3 ) is
(a) 2 (b) 0 (c) 3 (d) 4
Multiple Choice Questions
12. A sample of pure water, irrespective of its
source contains 11.1% hydrogen and 88.9%
1. Which one of the phrases would be incorrect
oxygen. The data supports
to use?
(a) law of constant proportions
(a) a mole of an element
(b) law of conservation of mass
(b) a mole of a compound
(c) law of reciprocal proportions
(c) an atom of an element
(d) law of multiple proportions
(d) an atom of compound
13. In compound A, 1.00 g nitrogen combines
2. The law of multiple proportions is illustrated
with 0.57 g oxygen. In compound B, 2.00 g
by the pair of compounds
Nitrogen combines with 2.24 g oxygen. In
(a) sodium chloride and sodium bromide
compound C, g nitrogen combines with 5.11 g
(b) water and heavy water
oxygen.
(c) sulphur dioxide and sulphur trioxide
These results obey the following law
(d) magnesium hydroxide and magnesium
(a) law of constant proportion
oxide
(b) law of multiple proportion
3. Which of the following has smallest mass?
(c) law of reciprocal proportion
(a) 4 g of He
(d) law of partial pressure
(b) 6.022 1023 atoms of He 14. The mass of a molecule of water is
37
(a) 3 1026 Kg (b) 3 1025 Kg 23. One mole of a gas occupies a volume of 22.4L.
This is derived from
(c) 1.5 1026 Kg (d) 2.5 1026 Kg
(a) Berzelius' hypothesis
15. The volume occupied by 4.4 g of CO2 at STP is (b) Gay-Lussac's law
(a) 22.4 L (b) 2.24 L (c) Avogadro's law
(c) 0.224 L (d) 0.1L (d) Dalton's law
16. Which of the following molecule has highest 24. Which of the following represents a
number of particles? polyatomic ion?
(a) 8 g of CH 4 (b) 4.4 g of CH 2 (a) Sulphide (b) Chloride
(c) 34.2 g of C12 H12O11 (d) 2 g of H 2 (c) Sulphate (d) Nitride
25. Which one of the following pair of gases
17. A metal M and Cl2 in different proportions to contains the same number of molecules?
form two compounds, A and B. The mass ratio (a) 16 g of O2 and 14 g of N 2
M: Cl is 0.89: 1 in A and 1791: 1 in B. This
(b) 8 g of O2 and 22 g of CO2
example explains the
(a) law of reciprocal proportions (c) 28 g of N 2 and 22 g of CO2
(b) law of definite proportions (d) 32 g of O2 and 32 g of N 2
(c) law of partial pressure 26. 6.022 1020 atoms of silver (at. mass 108 u)
(d) law of multiple proportions Weight is
18. Molecular mass is defined as the (a) 108 103 g (b) 108 g
(a) mass of one atom compared with the mass
(c) 0.108 g (d) 10.8 g
of one molecule
(b) mass of one atom compared with the mass 27. The number of carbon atoms in 1 g of CaCO3
of one atom of hydrogen is
(c) mass of one molecule of any substance (a) 6.022 1023 (b) 6.022 1021
compared with the mass of one atom of (c) 3.0125 10 22
(d) 1.204 1023
C 12 28. H 2 represents
(d) none of the above (a) 1 mole of hydrogen atoms
19. Which of the following elements has a symbol (b) 1 g of hydrogen
with two letters? (c) both (a) and (b)
(a) Tin (b) Uranium (d) none of these.
(c) Carbon (d) Boron 29. The amount of a substance is measured in
20. The molecular formula P;05 means that (a) mole (b) kilogram
(a) a molecule contains 2 atoms of P and 5 (c) seconds (d) gram.
atoms of O
30. Valency of silver in Ag 2 S is
(b) the ratio of the mass of P to the mass of O
(a) 1 (b) 2
in the molecule is 2 : 5
(c) 0 (d) 3
(c) there are twice as many P atoms in me
31. Latin name for gold is
molecule as there are O atoms
(a) argentum (b) aurum
(d) the ratio of the mass of P to the mass of O
(c) kalium (d) none of these
in the molecule is 5 : 2.
32. Atomicity of sulphur is
21. The formula of chloride of a metal M is MCl3 ,
(a) 8 (b) 4 (c) 2 (d) 1
then the formula of the phosphate of metal 33. The atomic mass of calcium (Ca) is 40 g. The
M will be
number of moles in 60 g of calcium are
(a) MPO4 (b) M 2 PO4
(a) 0.5 mol (b) 2.0 mol
(c) M 3 PO4 (d) M 2 ( PO4 )3 (c) 1.5 mol (d) 0.75mol
22. An atom is the 34. A chemical equation is always balanced to
(a) smallest particle of matter known fulfill the condition of
(b) smallest particle of a gas (a) Dalton's atomic theory
(c) smallest particle of an element that can (b) Law of constant composition
Takes part in a chemical change (c) Law of multiple proportions
(d) radioactive emission (d) Law of conservation of mass
38
35. In carbon disulphide (CO2 ) , the mass of (c) a charged particle (d) a form of light.
sulphur in combination with 3.0 g of carbon is 45. Number of moles of water present in 180 g of
(a) 4.0 g (b) 6.0 g water will be
(c) 64.0 g (d) 16.0 g (a) 5 (b) 10
36. What mass of carbon dioxide (CO2 ) will (c) 15 (d) 18
th
contain 3.0111023 molecules? 1
46. The mass of a carbon atom is called
(a) 11.0 g (b) 22.0 g 12
(c) 4.4 g (d) 44.0 g (a) mass number
37. Which is not correct according to Dalton's (b) atomic mass
atomic theory? (c) atomic mass unit
(a) Atoms are indivisible. (d) relative atomic mass.
(b) Atoms combine in simple whole number 47. "All matter is made up of very small particles
ratios. which cannot be further broken down. These
(c) All atoms of an element may not have particles are called atoms". This statement is
same mass. one of the assumptions of
(d)Atoms of different elements have different (a) Rutherford's nuclear theory
masses (b) Bohr's theory
38. Chemical formula of ferric oxide is (c) Dalton's atomic theory
(a) FeO (b) Fe2O3 (d) Kinetic theory of gases.
(c) Fe3O4 (d) None of these. 48. Two gaseous samples were analyzed. One
39. Mass of 1 mole of nitrogen atoms is contained 1.2 g of carbon and 3.2 g of oxygen.
(a) 28 g (b) 10.0 g The other contained 27.3% carbon and 72.7°o
oxygen. The experimental data are in
(c) 28amu (d) 14amu
accordance with
40. The atomic mass of sodium is 23. The number (a) law of conservation of mass
of moles in 46 g of sodium are (b) law of definite proportions
(a) 1 (b) 2 (c) law of reciprocal proportions
(e) 2.3 (d) 4.6 (d) law of multiple proportions
41. Which of the following represents 1amu ? 49. 20.8 g of BaCl2 on reaction with 9.8 g of
(a) Mass of C 12 atom
H 2 SO4 produces 7.3 g of HCl and some
(b) Mass of O 16 atom
(c) 1/12th of mass of C 12 atom amount of BaSO4 . The amount of BaSO4
(d) Mass of hydrogen molecule formed is
42. In which of the following the valence of each (a) 23.3 g (b) 20.8 g
of the constituent elements is equal to the (c) 9.8 g (d) 10.4 g
total number of atoms in one molecule of the 50. How many grams of H 2 SO4 are present in
compound? 0.25 mole of H 2 SO4 ?
(a) HCl (b) H 2 S (a) 2.45 (b) 24.5
(c) CaO (d) MgCl2 (c) 0.245 (d) 0.25
43. Which one of the following statements is 51. An oxide of iodine (1 127) contains 25.4 g of
true? iodine and 8 g of oxygen. Its formula could be
(a) Mass of 0.5 mole of N 2 gas > Mass of 0.5 (a) l2O3 (b) l2O
mole of N atoms (c) l2O5 (d) l2O9
(b) Mass of 0.5 mole of N 2 gas = Mass of 0.5 52. 52u of He contains
mole of N atoms (a) 4 6.022 1023
(c) Mass of 0.5 mole of N 2 gas < Mass of 0.5 (b) 13 atoms
mole of N atoms (c) 13 6.022 1023 atoms
(d) Mass of 0.5 mole of N 2 gas = Mass of 0.5 (d) 4 atoms.
mole of 0; gas. 53. Number of atoms in 4.25 g of NH, is nearly
44. 'Ion' is a three-letter word. It means (a) 11023 (b) 1.5 1023
(a) a mixture of iodine, oxygen and nitrogen (c) 2 1023 (d) 6 1023
(b) an alloy of iron
39
54. How many molecules are present in one gram (a) Ca2 PO4
of hydrogen? (b) Ca2 ( PO4 )3
(a) 6.022 1023 (b) 6.022 1022
(c) Ca3 ( PO4 )2 (d) CaPO4
(c) 3.01 x 1023 (d) 3.0125 1022
65. Element which shows variable valency is
55. Which one of the following statements is
(a) iron (b) tin
incorrect?
(c) copper (d) all of these
(a) One gram atom of carbon -12 contains
66. Number of gram molecules in 63 g HN03 is
avogadro's number of atoms.
(a) 1 (b) 2
(b) One mole of oxygen gas contains
(c) 3 (d) 4
avogadro's number of molecules.
67. The symbol ' u ' means
(c) One mole hydrogen gas contains
(a) atomic mass unit (b) unified mass
avogadro's number of atoms.
(c) atomic mass (d) none of these
(d) One mole of electron stands for 6.02 1023 68. Mass of 3 moles of NaOH in grams is
electrons. (a) 240 g (b) 120 g
56. The volume occupied by 1 mole atom of a (c) 100 g (d) 69 g
diatomic gas at STP is 69. In a hydrocarbon, the mass ratio of hydrogen
(a) 22.4 L (b) 11.2 L to carbon is 1 : 3. The empirical formula of the
(c) 5.6 L (d) 44.8 L hydrocarbon is
57. The empirical formula of the compound
(a) CH 4 (b) C2 H 2
containing hydrogen, carbon, oxygen and
nitrogen in the mass ratio 1 : 3 : 4 : 7 is (c) C2 H 3 (d) CH 3
(a) C3 HO4 N7 (b) CH 2ON 70. Empirical formula of a compound is CH 2O
(c) CH 4OH 2 (d) C2 H8OH 2 and its molecular weight is 120. The molecular
formula of the compound is
58. Dalton's atomic theory contained
(a) Boyle's law (b) Charle's law (a) C2 H 4O2 (b) C3 H 6O3
(c) Mass conservation law (c) C4 H8O4 (d) CH 2O
(d) Both (a) and (b) 71. A sample of phosphorus trichloride ( PCl3 )
59. A particle which maintains its chemical contains 1.4 moles of the substance. How
Identity even after physical and chemical Many atoms are there in the sample?
changes is (a) 4 (b) 8.4 1023
(a) atom (b) molecule
(c) 3.372 10 24
(d) 5.6
(c) compound (d) none of these
72. The correct formula of aluminum sulphate is
60. Symbol of an element consists of one or two
(a) AlSO4 (b) Al2 SO4
letters derived from ....... name of element
(a) Common (b) Latin (c) Al3 (SO4 )2 (d) Al2 (SO4 )3
(c) Greek (d) Any of the three 73. The mass of one C atom is
61. The atomic symbols for mercury and (a) 6.023 1023 g (b) 1.99 1023 g
potassium are respectively (c) 2.00 g (d) 12 g
(a) Mr and P (b) K and Hg 74. The chemical symbol for barium is
(c) Hg and P (d) Hg and K (a) B (b) Ba
62. Molecules of phosphorus and ammonia axe (c) Be (d) Bi
respectively 75. What are the total number of moles of atoms
(a) monoatomic and triatomic in 4.32 g of Sc( NO3 )3
(b) monoatomic and diatomic (a) 0.0132 (b) 0.324
(c) tetra-atomic and triatomic (c) 0.0187 (d) 0.243
(d) tetra-atomic and tetra-atomic 76. How many atoms in total are present in
63. A cation is CoCl3 .6H 2O ?
(a) negatively charged ion
(a) 17 (b) 22
(b) neutral atom
(c) 8 (d) 18
(c) positively charged ion
77. The mass in grams of 5 moles of Fe is
(d) group of atoms
(a) 280 (b) 260
64. Formula of calcium phosphate is
(c) 320 (d) 180
40
78. How many grams of O is present in 50 g of 4. During a chemical reaction, the sum of the.....
CaCO3 ? of the reactants and products remains
(a) 50 g (b) 150 g unchanged.
(c) 48 g (d) 24 g 5. The number of P atoms in one mole of P4
79. What is the mass of 0.5 moles of methane is………
(CH 4 ) ? 6. The number of moles of 100 g of Sulphur ( S8 )
(a) 16 g (b) 8 g is..............
(c) 80 g (d) 0.8 g. 7. The atomicity of ozone is.............
80. How many moles are present in 11.5 g of 8. 1 amu stands for ............ of a carbon atom.
sodium? 9. 4.4 g of CO2 is equal to........... mol.
(a) 0.05 mole (b) 5.0 mole 10. The ratio by mass of S and O in SO2 is........ .
(c) 1.5 mole (d) 0.5 mole
11. Clusters of atoms that act as an ion called......
81. The weight of a molecule of the compound
ion.
C60 H122 is
12. The formula mass of ( NH 4 )2 SO4 is................
(a) 1.4 1021 13. Atoms of two or more elements join together
(b) 1.09 1021 g during chemical combination to form .......... .
(c) 5.025 1023 g 14. An atom is the .......... particle of an element
(d) 16.023 10 g
23 which can take part in .......... .
15. The smallest particle which can exist in a free
82. The percentage of copper and oxygen in
state is called a ........... .
samples of CuO obtained by different
16. One mole of sodium sulphate contains ........
methods were found to be the same. This
atoms of sodium,............. atom of sulphur and
illustrates the law of
.......... atoms of oxygen.
(a) constant proportions
17. Gram atomic mass of an element is expressed
(b) conservation of mass
in............ .
(c) multiple proportions
18. The value of Avogadro constant is........... .
(d) reciprocal proportions
19. The number of atoms present in a molecule is
83. The atomic symbols for mercury, antimony
called......... .
And tungsten are respectively
20. The volume of one mole of a gas at standard
(a) Mr , At , Tn (b) Hg , Sb,W
temperature and pressure is .......... litres.
(c) Hg , At , Tn (d) Hg , Sb,Wo
84. Silica is a.....
(a) monoatomic element
(b) diatomic element
(c) triatomic compound 1. Two elements sometimes can form more than
(d) tetra-atomic compound one compound.
85. How many atoms of phosphorus are present 2. The smallest particle of a compound is
in phosphine? element
(a) 4 (b) 2 3. The symbol of silver is Si.
(c) 1 (d) 3 4. The molecular formula of a compound is
determined by the valency of the elements
present in the compound.
5. Noble gas molecules are diatomic.
6. The elements having valency 2 are called
divalent.
1. In a chemical compound, the constituent 7. Formula mass of Na2O is 62amu .
elements are present in ........ proportions by 8. Those particles which have more or less
weight. electrons than the normal atoms are called
2. The mass of 5 moles of ammonia (CH 3 ) ions.
is....... 9. The relative atomic mass of an element has
3. The valency of Cu in cuprous chloride No unit.
(Cu2Cl2 ) is........
41
10. Both 106 g of sodium carbonate and 12 g of (B) Sulphite (q) SO42
carbon have the same number of carbon
(C) Sulphate (r) PO43
atoms.
11. 1 mol of gold and iron contain same number (D) Carbonate (s) CO32
of atoms. 4. Column I Column II
12. The molecular formula of ferrous sulphate is (A) Chloride ion (p) Divalent, negative
Fe2 (SO4 )3 and ferric sulphate is FeSO4 . (B) Calcium ion (q) Trivalent, positive
13. Hydrogen and oxygen combine in the ratio of (C) Aluminium ion (r) Divalent, positive
1: 8 by mass to form water. (D) Oxide ion (s) Monovalent,
14. Different proportions of oxygen in various negative
oxides of nitrogen prove the law of multiple 5. Column I Column II
proportions. (A) K 2CO3 (p) 62u
15. Carbon dioxide is heavier than oxygen and (B) Na2O (q) 138 u
nitrogen. (C) HNO3 (r) 64 u
16. Molar mass of ethyne (C2 H 2 ) is 28 g/mol. (D) SO2 (s) 63 u
17. The formula for arsenic acid is H 3 AsO4 . The 6. Column I Column II
name of its sodium salt Na3 AsO4 is sodium (p) 6.022 10 molecules
23
(A) 0.25 mole oxygen
arsenate. (q) 1.505 10 molecules
23
(B) 18 g water
18. In ionic compounds, the charge on each ion is (C) 46 g Na atom (r) 6.022 10 atoms
23
used to determine the molecular formula of (D) 1 mole C atom (s) 12.044 10 atoms
23
the compound. 7. Column I Column II
19. 32 g of CO2 consists of two mole. (A) Silver (p) Hg
20. The chemical symbol P stands for polonium. (B) Gold (q) Pb
(C) Mercury (r) Ag
(D) Lead (s) Au
8. Column I Column II
(A) Phosphorus (p) Pd
In this section, each question contains
(B) Potassium (q) Pt
statements given in two columns which have
(C) Platinum (r) P
to be matched. Statements (A, B,C,D) in
(D) Palladium (s) K
Column-1have to be matched with
9. Column I Column II
statements (p, a, r, s) in Column-II the
(A) Water (p) 14: 3
answers to these questions have to be
(B) Ammonia (q) 1: 8
appropriately bubbled as illustrated in the
(C) Carbon dioxide (r) 1:1
following example. If the
(D) Sulphur dioxide (s) 3: 8
correct matches are A-a,
10. Column I Column II
A-r, B-p, B-s, C-r, C-s and
(A) Calcium (p) 14
D-q, then the Correctly
(B) Nitrogen (q) 16
bubbled matrix will look
(C) Oxygen (r) 23
like as shown.
(D) Sodium (s) 40
1. Column I Column II
(A) 52 g of He (p) 2 moles
(B) 8 g of O2 (q) 1 mole Directions: In each of the following questions, a
(C) 2 g of H 2 (r) 0.25 mole statement of Assertion (A) is given followed by a
(D) 56 g of H 2 (s) 13 mole corresponding statement of Reason (R) just below it.
2. Column I Column II Of the statements, mark the correct answer as
(A) Monoatomic (p) Sulphur (a) If both assertion and reason are true and
(B) Diatomic (q) Helium Reason is the correct explanation of assertion
(C) Terra-atomic (r) Oxygen (b) If both assertion and reason are true but
(D) Poly-atomic (s) Phosphorus Reason is not the correct explanation of
3. Column I Column II assertion
(A) Phosphate (p) SO32 (c) If assertion is true but reason is false
42
(d) If assertion is false but reason is true. Reason: It is the average mass of an atom
1. Assertion: Empirical formula of glucose is taking care of relative abundance of all its
HCHO possible isotopes.
Reason: Molecular formula of glucose will 12. Assertion: Atomicity of O3 is 3.
also be HCHO Reason: 1 mole of an element contains
2. Assertion: When 10 g of CaCO3 is 6.023 1023 atoms,
decomposed, 5.6 g of residue is left and 4.4 g 13. Assertion: Both 44 g CO2 and 16 g CH 4 have
of CO2 escapes. same number of carbon atoms.
Reason: Law of conservation of mass is Reason: Both contain 1 g atom of carbon
followed. which contains 6.023 1023 carbon atoms.
3. Assertion: Pure water obtained from different 14. Assertion: S.I unit of atomic mass and
sources such as river, well, spring, sea etc. molecular mass is kg.
always contains hydrogen and oxygen Reason: It is equal to the mass of 6.023 1023
combined in the ratio of 1 : 8 by mass. atoms.
Reason: A chemical compound always contain 15. Assertion: In CO molecule 12 parts by mass of
elements combined in same fixed proportions carbon combine with 16 parts by mass of
by mass.
oxygen and in CO2 12 parts by mass of carbon
4. Assertion: Molecular weight of SO2 is double
combine with 32 parts by mass of oxygen.
to that of O2 . Reason: When two elements combine
Reason: One mole of SO2 contains double separately with a fixed mass of third element,
the number of molecules present in one mole then the ratio of their masses in which they
of O2 . do so is either the same or whole number
5. Assertion: 1 mole of H; and Oz each occupy multiple of the ratio in which they combine
22.4 L at standard temperature and pressure. with each other.
Reason: Molar volume for all gases at the
standard temperature and pressure has the
same value.
6. Assertion: Mass can neither be created nor
destroyed. PASSAGE 1: A mole of an atom is a collection of atoms
Reason: The volume occupied by one mole of whose total mass is the number of grams equal to the
any gas at STP is always same. atomic mass. Since equal number of moles of
7. Assertion: Atomic mass of aluminium is 27. different elements contain an equal number of atoms
Reason: An atom of aluminium is 27 it becomes convenient to express the amounts of the
Times heavier than l/12th of elements in terms of moles. A mole represents a
the mass of carbon-12 atom. definite number of particles viz, atoms, molecules,
8. Assertion: One atomic mass unit (amu) is ions or electrons. This definite number is called
mass of an atom equal to exactly one-twelfth Avogadro number or Avogadro constant which is
of the mass of a carbon- 12 atom. equal to 6.022 1023 Hence a mole represents
Reason: Carbon-12 isotope was selected as 6.022 1023 particles of the substance. One mole of
standard. substance represents one gram-formula of the
9. Assertion: Number of gram-molecules of substance. One mole of a gas at standard
SO2Cl2 in 13.5 g of sulphuryl chloride is 0.2. temperature and pressure occupies 22.4 litres.
Reason: Gram - molecules is equal to those 1. How many grams of sodium must be taken to
molecules which are expressed in gram. get 1 mole of the element?
10. Assertion: Percentage of carbon in Na2CO3 is (a) 23 g (b) 35.5 g
(c) 63.5 g (d) 46 g
11.32%.
Mass of carbonelement
2. What is the mass in grams of a single atom of
% of Carbon 100 chlorine? (Atomic mass of chlorine = 35.5)
Molecular mass of Na2CO3
11. Assertion: Atomic mass has no unit but (a) 6.54 1023 g (b) 5.9 10 23 g
expressed in amu. (c) 0.0025 g (d) 35.5 g
3. How many number of moles are there in 5.75
g of sodium ? (Atomic mass of sodium = 23)
43
(a) 0.25 (b) 0.5 (c) All the atoms of a given element are
(c) 1 (d) 2.5 identical.
4. What is the mass in grams of 2.42 mol of zinc? (d) During chemical combination, atoms of
(Atomic mass of Zn 65.41 ) different elements combine in simple ratios.
(a) 200 g (b) 25 g 2. Which postulate of Dalton's atomic theory
(c) 85 g (d) 158 g explains law of definite proportions?
(a) Atoms of an element do not change during
PASSAGE 2: The percentage composition of elements a chemical reaction.
in a compound is calculated from the molecular (b) The elements consist of atoms having fixed
formula of the compound. The molecular mass of a mass and the number and kind of atoms in a
compound is calculated from the atomic masses of given compound is fixed.
the various elements present in the compound. The (c) Different elements have different kind of
percentage of mass of each element is then calculated atoms.
by the help of the following relation. Percentage (d) Atoms are of various kinds.
mass of the element in the compound Mass of the 3. "If 100 g of calcium carbonate (whether in the
Mass of theelement form of marble of chalk) are decomposed, 56
element 100
Mass of thecompound g of calcium oxide and 44 g of carbon dioxide
are formed." Which law of chemical
1. What is the percentage of calcium in calcium
carbonate? combination is illustrated by this statement?
(a) 32.65 (b) 40 (a) Law of constant proportions
(c) 88.9 (d) 18 (b) Law of conservation of mass
2. Calculate the mass of carbon present in 2 g of (c) Law of multiple proportions
carbon dioxide. (d) Law of conservation of energy.
(a) 12 g (b) 6 g 4. When 5 g calcium is burnt in 2 g oxygen, 7 g of
calcium oxide is produced. When 5 g of
(c) 0.545 g (d) 5.45 g
calcium is burnt in 20 g of oxygen, then also 7
3. What is the molecular mass of calcium
chloride? g of calcium oxide is produced. Which law of
chemical combination is being followed?
(a) 88 u (b) 111 u
(c) 65 u (d) 100 u (a) Law of conservation of mass
4. Percentage of S in aluminum sulphate is (b) Law of multiple proportions
(a) 28 (b) 56 (c) Law of constant proportions
(e) 65 (d) 75 (d) No law is being followed.
PASSAGE 3: According to Dalton's atomic theory, all PASSAGE 4: The molecular mass of a substance is the
relative mass of its molecule as compared with the
matter whether an element, a compound or a mixture
mass of a carbon-12 atom taken as 12 units. The
is composed of small particles called atoms which can
molecular mass of a substance indicates the number
neither be created nor destroyed during a chemical
reaction. Dalton's theory provides a simple of times one molecule of a substance is heavier than
1
explanation for the laws of chemical combination. He of C 12 atom. It is equal to the sum of atomic
used his theory to explain law of conservation of 12
masses, law of constant proportions and law of masses of all the atoms present in a molecule.
multiple proportions, based on various postulates of Depending on the number of atoms of same or
the theory. Dalton was the first scientist to use the different elements present in the molecule, it can be
symbols for the elements in a very specific sense. monoatomic, diatomic, triatomic, tetra- atomic or a
When he used a symbol for an element he so meant a poly-atomic molecule.
definite quantity of that element, that is one atom of 1. Which is an example of a poly-atomic
that element. molecule?
1. Which postulate of Dalton's atomic theory is (a) S8 (b) HNO3
the result of the law of conservation of mass? (c) C2 H 5OH (d) All of these.
(a) Atoms can neither be created nor 2. Ozone is a ...... molecule.
destroyed. (a) monoatomic (b) diatomic
(b) Each element is composed of extremely (c) triatomic (d) tetra-atomic.
small particles called atoms.
44
3. Carbon dioxide, hydrogen sulphide, calcium 3. Calculate the volume occupied by 2.8 g of N 2
chloride and sodium oxide are examples of at STP.
(a) Triatomic molecules 4. The molecular formula of ferric sulphate is
(b) Triatomic and tetra-atomic molecules Fe2 (SO4 )3 . (Atomic mass:
(c) Diatomic and triatomic molecules
Fe 56, S 32, O 16)
(d) Tetraatomic molecules.
(a) Calculate the molar mass of Fe2 (SO4 )3 .
(b) How many moles of each element is there
in 40 g of ferric sulphate?
5. What is the mass of 0.5 mole of water ( H 2O) ?
(Atomic masses: H 1u, O 16 u .
6. Calculate the number of atoms of each
1. Give two examples of trivalent metal ions. element present in 122.5 g of KClO3
2. Give one example of a polyatomic cation. 7. Calculate the number of moles in 2800 mL of
3. Calculate formula mass of sodium carbonate? oxygen gas at STP.
4. Write formula for the following compounds 8. Calculate the number of moles in 12.044 1023
aluminium sulphide, tin (II) fluoride helium atoms.
magnesium sulphate and ammonium nitrate. 9. Why does not the atomic mass of an element
5. How many atoms are present in 1 mole of represents the actual mass of its atom?
hydrogen gas? 10. What is the difference between the actual
6. Write the names of the following compounds: mass of a molecule and gram molecular
H 2 S , CO, N2O4 and PCl5 mass?
7. From the given list identify the diatomic 11. The atomic mass of an element is in fraction.
molecules Ne, Ar , O2 , F2 , Cl2 , H 2O, P4, S8 , O3 . What does it mean?
12. How many moles are present in 11.5 g of
8. Calculate the molar mass of ethyl alcohol
sodium?
(C2 H5OH ) .
13. How many particles are represented by 0.25
9. What happens to an element ' A ' if its atom mole of an element?
gains two electrons? 14. Difference between an atom and an ion.
10. Formula of the carbonate of a metal M is 15. Write the formula and names of compounds
M 2CO3 . Write the formula of its chloride. formed by
11. Calculate molar mass of water. 16. (a) Na and HCO3 (b) K and CO32
12. What is the percentage of sulphur in sulphuric
(c) Cu 2 and SO42 (d) Cu 2 and O 2
acid?
13. Calculate the molecular mass of sugar (e) Na and SO42 (f) NH 4 and CO32
(C12 H 22O11 )
14. What is an atomicity of a molecule of an
element?
15. Calculate the number of moles in
1. What are the postulates of Dalton's atomic
(i) 36 g of water (ii) 10 g of hydrogen
theory?
[Atomic mass: O 16, H 1]
2. A sample of carbon dioxide contains carbon
and oxygen in the mass ratio of 3 : 8. Find the
formula of carbon dioxide. [Relative atomic
masses : C 12, O 16]
1. Carbon and oxygen react with each other in
the ratio 3 : 8 by mass. What weight of carbon 3. 25.4 g of iodine and 14.2 g of chlorine are
should be used to react completely with 40 g made to react completely to yield a mixture of
of oxygen? ICI and ICI . Calculate the ratio of the moles
2. Give the chemical formula, for each of the of ICI and ICI 3 .
following acids: 4. 8.4 g of sodium bicarbonate on reaction with
(a) Nitric acid (b) Hydrogen bromide 20 g of acetic acid liberated 4.4 g of carbon
(c) Carbonic acid (d) Hypochlorous acid dioxide gas into the atmosphere. What is the
(e) lodic acid (f) Sulphurous acid mass of the residue left?
45
5. A sample of ammonia ( NH 3 ) weight 2.00 g.
What mass of sulphur dioxide ( SO2 ) contains
the same number of molecules as are in 2.00
g of ammonia? Multiple Choice Questions
6. What is the mass of one atom of hydrogen?
(atomic mass of hydrogen 1u ) 1. D 2. C 3. C 4. A 5. C 6. B 7. B
7. Calculate the number of the constituent 8. B 9. D 10. B 11. C 12. A 13. B 14. A
atoms in 53 g of Na2CO3 . (Atomic masses: 15. B 16. D 17. D 18. C 19. A 20. A 21. A
Na 23u , C 12 u, O 16 u ) 22. C 23. C 24. C 25. A 26. C 27. B 28. D
29. A 30. A 31. B 32. A 33. C 34. D 35. D
8. Calculate the number of moles of phosphorus
36. B 37. C 38. B 39. D 40. B 41. C 42. C
( P) atoms in 100 g of phosphorus. If
43. A 44. C 45. B 46. C 47. C 48. B 49. A
phosphorus is considered to contain P4 50. B 51. C 52. C 53. B 54. C 55. C 56. B
molecules, then how many moles of P4 57. C 58. C 59. A 60. D 61. D 62. D 63. C
molecules are there? 64. C 65. D 66. A 67. B 68. B 69. A 70. C
71. C 72. D 73. B 74. B 75. C 76. B 77. A
78. D 79. B 80. D 81. A 82. A 83. B 84. C
85. C
This section contains 5 questions. The answer Fill in the Blanks
to each of the questions is a single digit
integer, ranging from o to 9. 1. fixed 2. 85 amu
If the correct answers to question numbers 3. 1 4. masses
X , Y , Z and W (say) are 6, 0, 9 and 2 5. 2.41 x 1024 6. 0.39
respectively, then the correct darkening of 7. 3
8.
th
1
bubbles will look like the following.
12
9. 0.1 10. 1 : 1
11. polyatomic 12. 132 u
13. molecule
14. smallets, chemical reaction
15. molecule 16. two, one, four
17. grams 18. 6.022 x 1023
19. atomicity 20. 22.4
True or False
1. The mass of 0.5 moles of methane is
2. The mass of one molecule of water is 1. True 2. False: Molecules is the smallest
x 1023 g . The value of x is particle of a substance (compound).
3. False: The symbol of silver is Ag. 4. True
3. The ratio of molecules present in 6.6 g of CO2
5. False: Noble gases are 6. True
and 3.2 g of SO2 is monoatomic.
4. A person spends ten thousand rupees per 7. True 8. True 9. True 10. True 11. True
second, x 1012 years will take approximately 12. False: Molecular formula of ferrous sulphate is FeSO4
to spend one mole of rupees. The value of x is and ferric sulphate is Fe(SO4)3
The weight of 11022 molecules of 13. True 14. True 15. True
CuSO4 .5H 2O is 16. False: Molar mass of ethyne is 26 g/mol
17. True 18. True
Mass of CO2 gram
19. False: No. of moles
Gram molecular mass
32
0.7
44
20. False: The chemical symbol P stands for phosphorus
46
Matrix Match Type
1. As Br Cq Dp.
2. Aq Br Cs Dp.
3. Ar Bp Cq Ds.
4. As Br Cq Dp.
5. Aq Bp Cs Dr.
6. Aq Bp Cs Dr.
7. Ar Bs Cp Dq.
8. Ar Bs Cq Dp.
9. Aq Bp Cs Dr.
10. As Bp Cq Dr.
Assertion and Reason Type
1. C 2. A 3. A 4. C 5. A 6. B 7. A
8. A 9. D 10. A 11. A 12. B 13. A 14. D
15. B
Passage Comprehension
Passage : 1
1. A 2. D 3. A 4. D
Passage : 2
1. B 2. C 3. B 4. A
Passage : 3
1. A 2. B 3. B 4. C
Passage : 4
1. D 2. C 3. A
Integer Answer Type
1. 8 2. 3 3. 3 4. 2 5. 4
47
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