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SOME BASIC CONCEPTS OF CHEMISTRY
INTRODUCTION
Matter can be classified as mixtures and pure substances
Mixtures contain two or more substances in any ratio.
These substances are called components of the mixture and
can be separated by physical or mechanical means without
loosing their identity. A mixture may be heterogeneous or
homogeneous.
Heterogeneous mixtures do not have uniform composition
through out for example charcoal in water, sugar in salt etc.
Homogeneous mixtures have uniform composition for
example air, salt in water, alloys etc.
Pure Substances have same composition and properties.
Their components cannot be separated by simple physical
methods for e.g. iron, gold, water, glucose etc. These are
subdivided into elements and compounds.
Elements are pure substances that cannot be decomposed
into simpler substances by chemical changes. It consists of
only one type of atoms. e.g. sodium (Na), silver (Ag), oxygen
(O2 ), Chlorine (Cl 2 ) etc. Presently 118 different elements are
known, out of which 92 are naturally occurring.
Compounds are formed when two or more atoms of
different elements combine together in a fixed ratio e.g.
carbon dioxide (CO 2 ) water (H 2 O) ammonia (NH 3 ) etc.
The properties of the constituent atoms are completely
different from that of the formed compound. For example
water is formed by the reaction of hydrogen and oxygen at
very high temperature. Hydrogen is combustible whereas
oxygen helps in combustion but the product water is neither
combustible nor helps in combustion. It is used for
extinguishing fire.
MATTER
(has mass, occupies space)
MIXTURES
• Variable composition
• Components retain their characteristic properties.
• May be separated into pure substances by physical
methods
• Mixtures of different compositions may have
widely different properties.
HOMOGENEOUS
• Have same composition
throughout i.e. one phase
• Components
indistinguishable
are
HETEROGENEOUS
• Do not have same
composition throughout
i.e. two or more phase
• Components
are
distinguishable
Properties of Matter: Properties are the characteristic
quantities of different kinds of matters by which they can
be easily recognized. These properties can be classified as
physical properties and chemical properties.
Physical properties of a substance can be measured or
observed without changing their identity or composition.
e.g. colour, odour, boiling point, melting points, density etc.
Chemical properties of substances are those in which they
undergo change in composition to form new substances.
e.g. reaction of different substances.
2H 2 + O 2 → 2H 2 O
Chemistry -1(Some Basic Concepts of Chemistry)
PURE SUBSTANCES
• Fixed composition
• Cannot be separated into simpler substances by
physical methods
• Can only be changed in identity and properties by
chemical methods.
• Properties do not vary.
COMPOUNDS
• Can be decomposed
into simpler substances
by chemical changes,
always in a definite ratio
ELEMENTS
• Cannot be decomposed
into simpler substances
by chemical changes
FUNDAMENTAL AND DERIVED UNITS
The quantitative scientific observation generally requires
the measurement of one or more physical quantities such as
mass, length, density, volume, pressure, temperature, etc.
A physical quantity is expressed in terms of a number and a
unit. Without mentioning the unit, the number has no
meaning.
Characteristic of a good unit
1.
It should be invariable with place and time.
2.
It should be easily reproducible.
3.
It should be universally acceptable.
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Common system of units based upon three basic units of
length, mass and time are tabulated below.
Physical
quantity
Length
Mass
Time
CGS
centimeter
gram
second
Units
FPS
foot
pound
second
MKS
Meter
kilogram
second
The three basic units, i.e., units of mass, length and time are
independent units and cannot be derived from any other
units, hence they are called fundamental units. The three
fundamental units cannot describe all the physical
quantities such as temperature, intensity of luminosity,
electric current and the amount of the substance. Thus,
seven units of measurement, namely mass, length, time,
temperature, electric current, luminous intensity and
amount of substance are taken as basic units.
All other units can be derived from them and are, therefore,
called derived units. The units of area, volume, force,
work, density, velocity, energy, etc., are all derived units
SI Units of Measurement: The SI system has seven basic
units. The various fundamental quantities that are expressed
by these units along with their symbols are tabulated below.
Basic physical quantity
Length
Mass
Time
Temperature
Electric current
Luminous intensity
Amount of substance
Unit
Metre
Kilogram
Second
Kelvin
Ampere
Candela
Mole
Symbol
m
kg
s
K
A
cd
mol
Sometimes sub multiples and multiples are used to reduce
or enlarge the size of the different units.
Prefix
deci
centi
milli
micro
nano
pico
femto
atto
Symbol
d
c
m
µ
n
p
f
a
Value
10−1
10−2
10−3
10−6
10−9
10−12
10−15
10−18
Multiples
Prefix Symbol
deca
da
hecto
h
kilo
k
mega
M
giga
G
tera
T
peta
P
exa
E
Chemistry -1(Some Basic Concepts of Chemistry)
(b)
Volume = length × breadth × height
= m × m × m = m3 [cubic metre]
(c)
Density =
(d)
Speed =
(e)
Acceleration =
(f)
Force = mass × acceleration
= kg × ms−2 = kg ms−2 [newton (N)]
Pressure = force per unit area
kg ms −2
=
= kg m−1s−2 or Nm−2 (pascal−Pa)
2
m
Energy = force × distance travelled
= kg ms−2 × m = kg m2 s−2 (joule − J)
(g)
(h)
mass
kg
= 3 = kg m−3
volume
m
distance covered
metre
=
= ms−1
time
time
ms −1
change in velocity
=
= ms−2
time taken
s
(i)
Electric charge = Current × Time
= A × s = Coulomb (C)
(j)
Electric potential =
(k)
Concentration =
(l)
Electrochemical equivalent (Z) =
Energy
2
–2 –1
= kg × m × s A
Charge
= JA–1s–2 = volt (V) or JC–1
Mole
–3
= mol m
3
m
E
= kg ⋅ C–1
F
LAWS OF CHEMICAL COMBINATION
The use of SI system is slowly growing; however, older
systems are still in use.
Sub multiples
SI Units for Some Common Derived Quantities
2
(a) Area = length × breadth = m × m = m [square metre]
Value
10
102
3
10
106
109
1012
1015
1018
Experimental results point out that chemical combination is
governed by following laws.
LAW OF CONSERVATION OF MASS (LAVOISIER)
(LAW OF INDESTRUCTIBILITY OF MATTER)
Statement: This law states that during a chemical change
no gain or loss of mass can be measured.
OR
Mass can neither be created nor be destroyed in a chemical
reaction i.e. total mass of the system remains constant.
Thus when reactants are completely converted into
products, then
Total mass of reactants = Total mass of products
In case reactants are not consumed completely the above
relationship will be
Total mass of reactants = Total mass of products +
Masses of unreacted reactants
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SOLVED EXAMPLE
1.
LAW OF MULTIPLE PROPORTIONS (JOHN DALTON)
What weight of sodium chloride would be
decomposed by 4.9 g of sulphuric acid, if 6 g of
sodium bisulphite and 1.825 g hydrogen chloride
were produced in the reaction?
Solution:
NaCl + H 2 SO 4 → NaHSO 4 + HCl
According to law of conservation of mass
W (NaCl) + W ( H 2SO 4 ) = W ( NaHSO4 ) + W (HCl)
W (NaCl) + 4.9 = 6 + 1.825
LAW OF RECIPROCAL PROPORTIONS OR
EQUIVALENT PROPORTIONS (RICHTER)
W (NaCl) = 6 + 1.825 – 4.9 = 2.925 ≈ 2.9 g
LAW OF CONSTANT PROPORTIONS OR
DEFINITE COMPOSITION (JOSEPH PROUST)
Statement: A given chemical compound always contains
the same elements combined together in definite proportion
by mass, i.e., it has a fixed composition and does not
depend on the method of its preparation or the source from
which it has been obtained. This law is also known as ‘Law
of definite proportions’.
E.g. pure water obtained from different sources such as
river, well, spring, sea or in a chemical reaction etc. always
contains hydrogen and oxygen combined together in the
ratio 1 : 8 by weight.
SOLVED EXAMPLE
2.
In an experiment, 2.4g of iron oxide on reduction
with hydrogen yield 1.68g of iron. In another
experiment 2.9g of iron oxide give 2.03g of iron on
reduction with hydrogen. Show that the above
data illustrate the law of constant proportions.
Solution :
In the first experiment
W (FeO) = 2.4 g,
W (Fe) = 1.68 g
W ( O 2 ) = W (FeO) − W (Fe) = (2.4 − 1.68) = 0.72 g
Ratio of oxygen and iron = 0.72 : 1.68 = 1 : 2.33
In the second experiment
W (FeO) = 2.9 g,
W (Fe) after reduction = 2.03 g
W ( O 2 ) = (2.9 − 2.03) = 0.87 g
Ratio of oxygen and iron = 0.87 : 2.03 = 1 : 2.33
Thus, the data illustrate the law of constant proportions,
as in both the experiments the ratio of oxygen and iron is
the same.
Chemistry -1(Some Basic Concepts of Chemistry)
Statement: When two elements combine to form two or
more than two compounds, the weight of one of the
elements which combines with a fixed weight of the other,
bears a simple whole number ratio.
Explanation : Hydrogen and oxygen combine to form two
compounds water (H 2 O) and hydrogen peroxide (H 2 O 2 ).
H2O
H2O2
2 : 16
2 : 32
Ratio of the masses of O which combine with fixed mass of
H is
16 : 32
or
1:2
Statement: The weight ratio of two elements A and B
which combine with the fixed weight of C separately is
either the same or some simple whole number multiple of
the weight ratio in which A and B combines together.
Explanation: Consider three
2 H2
elements sulphur, oxygen and
hydrogen. Both sulphur and
H2S
H2O
oxygen separately combine to
form H 2 S and H 2 O respectively.
32
16
They also combine with each
O
S
32 SO2 32
other to form SO 2 as shown below
In H 2 S, 2 parts by weight of hydrogen combine with 32
parts by weight of sulphur.
In H 2 O, 2 parts by weight of hydrogen combine with 16
parts by weight of oxygen.
∴ The ratio of the weights of sulphur & oxygen which, combine
separately with the fixed weight i.e. 2 parts of hydrogen is,
S :O
32 : 16
2 : 1
……… (i)
Now the ratio of sulphur and oxygen in SO 2
S:O
32 : 32
1 : 1
………(ii)
The two ratios (i) and (ii) are related to each other as
2 1
: or 2 : 1 (Simple multiple of each other)
1 1
Sulphur and oxygen also react to form SO 3 which is also in
accordance with the law.
Now the ratio of sulphur and oxygen in SO 3
S:O
32 : 48
2 : 3
………(iii)
The two ratios (i) and (iii) are also related to each other as
2 2
: or 3 : 1 (Simple multiple of each other)
1 3
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SOLVED EXAMPLE
3.
Copper sulphide contains 66.5% Cu, copper oxide
contais 79.9% Cu and sulphur trioxide contains
40% S. show that the date illustrates the law of
reciprocal proportions.
Solution :
In CuS, % of Cu = 66.5
% of S = 100 – 66.5 = 33.5
66.5 g of Cu combine with 33.5 g of sulphur
In CuO, % of Cu = 79.9
% of O = 100 – 79.9 = 20.1
79.9 g of Cu combine with 20.1 g of oxygen
Let us fix the mass of copper as 1 g
In CuS, 66.5 g of Cu combine with S = 33.5 g
33.5
= 0.503 g
1 g Cu will combine with S =
66.5
In CuO, 79.9 g of Cu combine with O = 20.1 g
20.1
= 0.251 g
1 g of Cu will combine with O =
79.9
The ratio of weights of S and O which combine with the
fixed mas sof copper (1 g)
S : O
0.503 : 0.251
2 : 1
… (i)
In SO 3 , % of S = 40,
% of O = 60
Thus, S : O
40 : 60
2 : 3
… (ii)
The ratio (i) and (ii) are
2 2
: or 3 : 1
1 3
This is a simple ratio. Hence law of reciprocal proportions
is illustrated.
Formation of Ammonia: N 2(g) + 3H 2(g) → 2NH 3(g)
1 volume 3 volumes
The ratio of volumes is N 2 : H 2 : NH 3 :: 1 : 3 : 2.
When the volumes of gases are measured under similar
conditions of temperature and pressure it is found in
the above reaction that 1 volume of nitrogen gas combines
with three volumes of H 2 gas to form two volumes of
ammonia gas.
SOLVED EXAMPLE
4.
1 volume
1 volume
2 volumes
2C 2 H 2(g) + 5O 2(g)
→
4CO 2(g)
2 vol
5 vol
4 vol
40 mL
5
×40 mL
2
4
×40 mL
2
40 mL
100 mL
80 mL
+ 2H 2 O ()
So, for complete combustion of 40 mL of acetylene,
100 mL of oxygen are required and 80 mL of carbon
dioxide is formed.
PROBLEMS
1.
0.44 g of a hydrocarbon on complete combustion
with oxygen gave 1.8 g water and 0.88 g carbon
dioxide. Show that these results are in accordance
with the law of conservation of mass.
2.
10 mL of hydrogen combine with 5 mL of oxygen to
yield water. When 200 mL of hydrogen at NTP are
passed over heated CuO, the CuO loses 0.144 g of its
mass. Do these results correspond to the law of
constant proportions?
3
Carbon and oxygen are known to form two
compounds. The carbon content in one of these
compounds is 42.9% while in the other, it is 27.3%.
Show that this data is in the agreement with the law
of multiple proportions.
4.
Water contains 88.90% oxygen and 11.10% of
hydrogen, ammonia contains 82.35% of nitrogen and
17.65% of hydrogen and dinitrogen trioxide contains
63.15% oxygen and 36.85% of nitrogen. Show that
these data illustrate law of reciprocal proportions.
For the above reaction when volumes are measured under
similar conditions of temperature and pressure, one volume
of hydrogen combines with one volume of chlorine to form
two volumes of hydrogen chloride gas. Thus the proportion
of H 2(g) and Cl 2(g) and HCl (g) by volume is 1: 1 : 2
Chemistry -1(Some Basic Concepts of Chemistry)
How much volume of oxygen will be required for
complete combustion of 40 mL of acetylene
(C 2 H 2 ) and how much volume of carbon dioxide
will be formed? All volumes are measured at
NTP.
Solution:
GAY LUSSAC’S LAW OF COMBINING VOLUMES
OF GASES
Statement: Under same conditions of temperature and
pressure, the volumes of the gaseous reactants and products
are in simple ratio of small whole numbers, when the gases
participate in chemical reactions.
Explanation:
Formation of HCl gas: H 2(g) + Cl 2(g) → 2 HCl (g)
2 volumes
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Chemistry -1(Some Basic Concepts of Chemistry)
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DALTON’S ATOMIC THEORY
AVOGADRO’S HYPOTHESIS
Dalton’s atomic theory convincingly explained the laws of
chemical combination. Postulates of this theory are
i.
Matter is made up of minute, indivisible particles
called atoms and has independent existence.
ii.
Atoms of same element have identical properties such
as same mass, size, shape and chemical properties.
iii.
Atoms of different elements differ in their properties
and have different masses and size.
iv.
Compounds are formed when atoms of different
elements combine with each other in simple and
fixed numerical ratio such as 1 : 1, 1 : 2, 2 : 3 etc.
v.
Atom cannot be created, destroyed or transformed
into atom of other element in the chemical process.
A chemical reaction simply change the way in which
atoms are grouped together.
Statement: Equal volumes of all gases under similar
conditions of temperature and pressure contain equal
number of gaseous molecules, regardless their chemical
nature and physical properties.
For example, if we enclose equal volumes of three gases
hydrogen (H 2 ), oxygen (O 2 ) and chlorine (Cl 2 ) in different
flasks of the same capacity under similar conditions of
temperature and pressure, we find that all the flasks have
the same number of molecules. However, these molecules
may differ in size and mass.
Gay-Lussac and Avogadro’s laws can be illustrated as
follows:
Gay-Lussac’s law
Avogadro’s law
+
2H 2(g)
2H 2 O (g)
2 vol
2 mol
→
O 2(g)
1 vol
1 mol
2 vol
2 mol
Limitations and Drawbacks: Dalton’s atom still has its
significance as the unit participating in chemical reaction,
but later researchers proved that Dalton’s theory was not
wholly correct. His theory has following drawbacks.
Avogadro’s law : It may be stated as, “At constant
pressure and temperature, volume of a gas is directly
proportional to the number of molecules”.
i.
V∝N
ii.
iii.
iv.
With the discovery of sub-atomic particles i.e.
electrons, protons, neutrons the atom can no longer
be indivisible.
Discovery of isotopes indicated that all atoms of the
same elements are not perfectly identical. They may
differ in their masses.
Atoms of different elements may posses the same
mass (isobar) but they always have different atomic
numbers and differ in chemical properties.
It could not explain why certain substances all
containing atoms of the same element are differ in
their properties. For example charcoal, graphite and
diamond all are made up of only C-atom, but still
their properties are quite different.
Mathematically, we can say
or
where N is the number of molecules.
V
=k
N
V = volume,
N is proportional to number of moles n
∴
V
=k
n
n=
Mass of gas
Molar mass of gas
At STP we can calculate the volume per mol i.e.
V
n
According to general gas equation
PV = nRT
Vm =
V
RT
=
n
P
v.
Atoms of one element can be transmuted into atoms
of other element (nuclear reactions).
P = 101.325 kPa,
vi.
In certain organic compounds like proteins, starch
cellulose etc. the ratio in which the atoms combine
cannot be regarded as simple.
Vm =
Chemistry -1(Some Basic Concepts of Chemistry)
k = proportionality constant
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–1 –1
R = 8.314 J mol K ,
T = 273 K
8.314 × 273
= 22.41 dm3 mol–1 = 22.41 L
101.325
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3
Thus, the volume 22.4 Litres or 22.4 dm at NTP
corresponds to the molecular weight of the gas, and is
referred as gram molecular volume.
6.
ATOMICITY: “The number of atoms present in one
molecule of a substance is called atomicity”.
Relationship between molecular weight and vapour
density :
Example: The atomicity of oxygen can be calculated by
considering the reaction between hydrogen and oxygen to
form water vapour. It has been experimentally found that 2
volumes of hydrogen react with 1 volume of oxygen to
form 2 volumes of water vapour.
Vapour density is defined as the ratio of the weight of a
certain volume of a gas to the weight of same volume of
hydrogen gas, under similar conditions of temperature and
pressure, is called vapour density (V.D).
Hydrogen
2 volumes
+ Oxygen → Water vapour
1 volume
2 volumes
n molecules
2n molecules
2 molecules
1 molecule
2 molecules
1 molecule
1/2 molecule
Weight of certain volume of gas
Weight of same volume of hydrogen
If the given gas contains ‘x’ molecules in a certain volume,
then according to Avogadro’s Law:
V.D =
Weight of x molecules of the gas
Weight of x molecules of hydrogen gas
V.D =
Weight of 1 molecule of the gas
Weight of 1 molecule of hydrogen gas
1 molecule
Thus, one molecule of water contains one molecule of
hydrogen (2 atoms) and ½ molecule of oxygen. The
molecular mass of water has been found to be 18 amu.
Since one molecule of hydrogen contains two atoms of
hydrogen, therefore, weight of oxygen in one molecule of
water is 18 – 2 = 16 amu. This corresponds to the weight of
one atom of oxygen. Thus, one atom is present in half
molecule of oxygen. Therefore,
1/2 molecule of O 2 = 1 atom or 1 molecule of O 2 = 2 atoms
Thus, atomicity of oxygen is 2.
Importance of Avogadro’s Law

In this law, the word ‘molecule’ was used for the first time.
2.
The law suggested the matter consists of two types of
particles namely atoms and molecules.
Avogadro’s law made a clear demarcation between
atoms and molecules
Hydrogen is diatomic,
V.D =
Weight of 1 molecule of the gas
Weight of 2 atoms of Hydrogen
∴ V.D. =
1
Weight of 1 molecule of the gas
×
2
Weight of 1 atom of Hydrogen
∴ V.D. =
Molecular wt. of gas
2

1.
3.
Thus,
V.D =
Applying Avogadro’s hypothesis
2n molecules
It helped in the determination of actual number of
molecules (Avogadro’s number) in gram molecular
volume of a gas.
Weight of 1 molecule of the gas
= molecular wt.
Weight of 1 atom of Hydrogen
∴ Molecular weight of a gas = 2 × Vapour density.
This formula can be used for the determination of
molecular weight of volatile substances.
CONCEPT OF ATOMS & MOLECULES
4.
It explained Gay−Lussac’s law of combining
volumes which led to the following conclusions:
a.
Molecular weight of a gas = 2 × Vapour density.
b.
Gram molecular volume of all gases is equal to
−3 3
22.4 × 10 m or 22.4 dm3 at S.T.P or N.T.P.
Hydrogen atom is the smallest with mass 1.667 × 10
c.
Atomicity of elements, determination of molecular
weights, atomic weights, equivalent weight,
molecular formula etc can be obtained.
5.
It helped in modification of Dalton’s atomic theory.
Molecule: It is the smallest particle of a substance (element
or compound), which has free or independent existence and
possesses all characteristic properties of the substance.
A molecule of an element is composed of like atoms while
Chemistry -1(Some Basic Concepts of Chemistry)
Atom: The smallest particle of an element that can take
part in chemical change and may or may not exist freely, as
such. Every atom of an element has a definite mass of the
order of 10–26 kg and radius of the order of 10–15 m.
–26
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a molecule of a compound contains fixed number of atoms
of two or more different elements.
A molecule may be broken down into its constituent atoms
but the atom is indivisible during a chemical change.
ATOMIC AND MOLECULAR MASSES
Atomic mass (mass of one mole of atom): The present
system of atomic masses is based on carbon-12 as the standard.
Thus, atomic mass of an element is defined as, “the
number which indicates how many times the mass of one
atom of the element is heavier than that of 1/12th part of the
mass of one atom of carbon-12”.
Atomic mass of element
Mass of one atom of the element
=
th
(1 / 12) part of the mass of one atom of C - 12
Mass of one atom of the element
× 12
=
Mass of one atom of carbon - 12
x
y
×a+
× b +………
100
100
xa + yb + .......
=
100
x, y = percentage abundance of the isotopes.
SOLVED EXAMPLES
Average isotopic mass =
5.
Relative abundances and masses of three isotopes
of carbon are listed below.
Isotope
Relative abundance
(%)
Atomic mass
(amu)
98.892
1.108
2 × 10–10
12
13.00335
14.00317
12
C
C
14
C
13
Calculate average atomic mass of carbon.
Solution:
Atomic masses of some elements on the basis of C-12
Avg. at. mass of C
H – 1.008 u
Mg – 24.305 u
Na – 22.989 u
C × % 13 C + 14 C × % 14 C
100
98
.
892
12
1
.
108
13.00335 + 2 × 10 −10 × 14.00317
×
+
×
=
100
O – 16.00 u
Cu – 63.546 u
Zn – 65.38 u
Cl – 35.453 u
Fe – 55.847 u
Ag – 107.868 u
In this system 12C is assigned a mass of exactly 12 atomic
mass unit (amu). One atomic mass unit is defined as
“a mass exactly equal to one-twelfth the mass of one
C-12 atom”.
12
=
12
C+
13
= 12.001 u
6.
Boron has two isotopes boron-10 and boron-11,
whose percentage abundance are 19.6 and 80.4
respectively. What is the average atomic mass of
boron?
Actual mass of one atom of C = 1.9924 × 10–23g
1.9924 ×10 −23
–24
= 1.66056 × 10 g
12
Today amu has been replaced by ‘u’ (Unified Mass)
C×%
1 amu =
Solution:
Thus,
Actual mass of an atom of an element
= At. Mass in amu × 1.66 × 10–24 g
Contribution of boron-10 = 10.0 × 0.196 = 1.96 amu
Contribution of boron-11 = 11.0 × 0.804 = 8.844 amu
Adding both = 1.96 + 8.844 = 10.804 amu
Thus, the average atomic mass of boron is 10.804 amu
Hence, actual mass of H-atom = 1.008 ×1.66 × 10–24
= 1.6736× 10–24 g
7.
Boron occurs in nature in the form of two isotopes
having atomic mass 10 and 11. What are the
percentage abundances of two isotopes in a sample
of boron having average atomic mass 10.8?
Using Dulong and Petit’s law approximate atomic weight
can be calculated.
Atomic weight × specific heat ≈ 6.4
Solution:
Average Atomic Mass: In calculations, we actually use
average atomic masses of elements instead of atomic mass.
Many naturally occurring elements exist as a constant
mixture of isotopes (atoms of the same element having
same atomic number but different mass number).
Let the % abundance of B10 isotope = x
The average relative atomic mass depends upon the isotopic
composition of that particular element.
Thus,
Chemistry -1(Some Basic Concepts of Chemistry)
11
Then % abundance of B isotope = 100 – x
x × 10 + (100 − x ) × 11
100
But, average atomic mass = 10.8
The average atomic mass =
∴
x × 10 + (100 − x ) × 11
= 10.8
100
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Equivalent weight : The number of parts by weight of
a substance which combines with or displaces one part
by weight of hydrogen or 8 parts by weight of oxygen or
35.5 parts by weight of chlorine.
Or 10x + 1100 – 11x = 10.8 × 100
– x = – 1100 + 1080
Or
x = 20
10
11
Thus, percentage abundance; B = 20, B = 80
Molecular Mass: It may be defined as,” the mass of a
molecule of a substance relative to the mass of one atom of
carbon taken as 12 or it is a number which indicates how
many times one molecule of a substance is heavier in
comparison to 1/12th of the mass of one atom of carbon 12”.
Molecular mass (M) =
Mass of one molecule of the substance
th
1
mass of one atom of carbon - 12
12
Gram equivalent weight: The equivalent weight of a
substance expressed in grams is called gram equivalent
weight of that substance.
METHODS OF DETERMINATION OF EQUIVALENT WEIGHT
a.
Eq.wt. of metal =
b.
The mass of a molecule is equal to the sum of the masses of
the atoms present in a molecule. One molecule of water
consists of 2 atoms of hydrogen and one atom of oxygen.
Thus, molecular mass of water
= (2 × 1.008) + 16.00 = 18.016 amu.
Wt. of metal
× 1.008
Wt. of H 2 displaced
Equivalent weight of metal by chloride formation
method:
Eq.wt. of metal =
Wt. of metal
× 35.5
Wt. of chlorine combined
c.
Equivalent weight of metal by oxide formation
method:
Weight of metal
×8
Eq. wt. of metal =
Weight of oxygen combined
d.
Equivalent weight =
Gram-molecular mass or Gram molecule:
Molecular mass of a substance expressed in grams is called
gram-molecular mass or gram molecule or 1 mole.
For example, the molecular mass of chlorine is 71 and,
therefore, its gram-molecular mass or gram molecule is 71 g.
Hydrogen displacement method:
Mol.wt. / At. wt. / Formula wt.
Valence
Approximate At. wt.
Eq. wt.
Similarly, molecular mass of oxygen (O 2 ) is 32, i.e.,
2 × 16 = 32 amu.
Valency =
Gram molecular mass of oxygen = 32 g
Correct At. wt. = Eq. weight × corrected valency.
Gram molecular mass should not be confused with the mass
of one molecule of the substance in grams. The mass of one
molecule of a substance is known as its actual mass. For
example, the actual mass of one molecule of oxygen is
equal to 32 × 1.66 × 10–24 g = 5.32 × 10–23 g.
SOLVED EXAMPLE
8.
SOLVED EXAMPLES
9.
Solution:
Calculate the mass of 1.5 gram molecule of
sulphuric acid.
Mass of the chlorine in the metal chloride = 49.5%
Mass of metal = 100 – 49.5 = 50.5
Solution:
Molecular mass of H 2 SO 4 = 2 × 1 + 32 + 4 × 16 = 98.0
amu
Equivalent mass of the metal =
Gram molecular mass of H 2 SO 4 = 98 g
=
Mass of 1.5 gram molecule of H 2 SO 4 = 98.0 × 1.5 = 147.0
g
EQUIVALENT WEIGHT
Chemistry -1(Some Basic Concepts of Chemistry)
A chloride of an element contains 49.5% chlorine.
The specific heat of the element is 0.056. Calculate
the equivalent mass, valency and atomic mass of
the element.
Mass of metal
× 35.5
Mass of chlorine
50.5
× 35.5 = 36.21
49.5
According to Dulong and Petit’s law
Approximate atomic mass of the metal =
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=
6.4
0.056
= 114.285 ≈ 114.3
Valency =
Approximate atomic mass 114.3
=
= 3.1 = 3
Equivalent mass
36.21
Hence, exact atomic mass = 36.21 × 3 = 108.63
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10.
One gram of the chloride was found to contain
0.835 g of chlorine. Its vapour density is 85.
Calculate its molecular formula.
Solution:
Mass of metal chloride = 1 g
The mole is the amount of a substance which contains
Avogadro’s number of particles namely 6.023 × 1023.
Mass of chloride = 0.835 g
−3
Avogadro’s number: The number present in 12 × 10 kg
23
(12g) of 12
6 C isotope which is found to be 6.023 × 10
particles is called Avogadro’s number. These units can be
atoms, molecules, ions, electrons etc.
Mass of metal = 1 – 0.835 = 0.165 g
Equivalent mass of metal =
Valency of the metal =
=
Mole: The quantity of matter which contains as many
−3
particles as there are in exactly 12g (12× 10 kg) of
12
6 C isotope.
Or
0.165 × 35.5
= 7.01
0.835
2 × V.D.
E + 35.5
SOLVED EXAMPLE
11.
A piece of copper weighs 0.635 g. How many
atoms of copper does it contain?
Solution:
Gram atomic mass of copper = 63.5 g
2 × 85
170
=
= 3.99 ≈ 4
7.01 + 35.5
42.51
Formula of the chloride = MCl 4
PROBLEMS
5.
6.
7.
Calculate the molecular mass of glucose (C 6 H 12 O 6 )
molecule. Given At. masses of H = 1.008 amu,
C = 12.011 amu, O = 16.0 amu.
(180.162 amu)
Thallium has two isotopes 203Tl and 205Tl. Knowing
that the atomic weight of thallium is 204.4, which
isotope is more abundant of the two?
205
( Tl = 70%)
The oxide of an element contains 32.33 percent of the
element and the vapour density of its chloride is 79.
Calculate the atomic mass of the element.
(15.28)
No. of moles in 0.635 g of copper =
No. of copper atoms in one mole = 6.023 × 1023
No. of copper atoms in 0.01 moles = 0.01 × 6.023 × 1023
= 6.023 × 1021
8.
23
No. of moles =
iii.
No. of moles =
(0.1)
What is the mass of 3.01 × 1022 molecule of
(0.85 g)
ammonia? (At. wt. N = 14, H = 1)
10.
Calculate the mass of
(2.32 × 10–23 g)
i.
1 atom of C14
ii.
1 molecule of N 2
iii.
1 molecule of water
iv.
100 molecules of sucrose (C 12 H 22 O 11 ).
(5.68 × 10–20 g)
(4.65 × 10–23
g)
1 mole = 6.023 × 10 particles = Molecular mass
ii.
PROBLEMS
Calculate the gram atoms in 2.3 g of sodium.
[Atomic weight of sodium = 23 g]
9.
MOLE CONCEPT
i.
0.635
= 0.01
63.5
Mass in grams
Mass of one mole in grams
(2.99 × 10–23 g)
No. of particles
6.023 × 10 23
PERCENTAGE COMPOSITION OF COMPOUNDS
Percentage composition of the compound is the relative
mass of the each of the constituent element in 100 parts of
it.
Gram atom of an element
iv.
Mass of one atom =
v.
Mass of one molecule =
vi.
No. of molecules =
6.023 × 10
23
Gram molecular mass
6.023 × 10 23
% of element =
Volume of gas (in L)
× 6.023 × 1023
22.4 L
Chemistry -1(Some Basic Concepts of Chemistry)
X
× 100
M
Where, X = Mass of an element in 1 mole of the compound
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M = Molecular mass of the compound
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SOLVED EXAMPLE
12.
It is found that 16.5 g of metal combines with
oxygen to form 35.60 g of metal oxide. Calculate
the percentage of metal and oxygen in the
compound.
Solution:
Mass of oxygen in oxide = (35.60 – 16.50) = 19.10 g
16.50
× 100 = 46.3
35.60
19.10
% of oxygen =
× 100 = 53.7
35.60
% of metal =
11.
PROBLEMS
Calculate the percentage composition of carbon in
the following alkanes. (At. no. of C = 12, H = 1)
12.
(i) C 2 H 6
(ii) CH 4
(iii) C 3 H 8
(iv) C 4 H 10
(i. 80%, ii. 75%, iii. 85.7%, iv. 83.7%)
Determine the percentage of water of crystallisation,
iron, sulphur, oxygen in pure ferrous sulphate
(FeSO 4 ⋅ 7H 2 O).
(At. no. of Fe = 56, S = 32, O = 16, H = 1)
(45.32%, 20.14%, 11.51%, 23.02%)
EMPIRICAL AND MOLECULAR FORMULAE
Calculation of empirical formula :Empirical formula is
calculated from the percentage composition. The steps
involved in the calculation are :
Step 1 : The percentage of each element is divided by its
atomic mass. This gives the relative number of different
atoms present in the molecule.
Step 2 : The relative numbers of different atoms obtained
in Step 1 are divided by the lowest one amongst them as to
get simple ratio of atoms present in the molecule.
Step 3 : The values obtained in Step 2 may or may not be
whole numbers. In case one or more values are fractional,
these are multiplied by a suitable integer to get simplest
ratio in whole numbers. Minor fractions are neglected.
Step 4 : The symbols of each element present are written
side by side in a line with the number of atoms as
determined in Step 2 or Step 3 as subscripts to the lower
corner of each. This gives the empirical or simplest formula.
Calculation of molecular formula: Knowing the empirical
formula, the molecular formula can be ascertained if the
molecular mass of the substance is known. It may be the
same as the empirical formula of the substance or an exact
multiple of it.
Molecular formula = n × (Empirical formula)
The value of ‘n’ can be determined if molecular mass of the
substance is known.
Empirical Formula (E.F) : The empirical formula of a
compound is the simplest formula which expresses the
simple whole number ratio of the atoms of constituent
elements present in the molecule.
The simplest formula of substance capable of expressing its
percentage composition can be called its empirical formula.
For example, CH 2 O is the empirical formula of acetic acid.
It expresses that the simplest whole number ratio between
carbon, hydrogen and oxygen atoms present in one
molecular of acetic acid is 1 : 2 : 1.
n=
SOLVED EXAMPLES
13.
Element
Water
Carbon
Methane
Hydrogen Peroxide
Acetylene
Glucose
Diborane
Tetraphosphorus decoxide
M.F
H2O
C
CH 4
H2O2
C2H2
C 6 H 12 O 6
B2 H6
P 4 O 10
Chemistry -1(Some Basic Concepts of Chemistry)
%
At.
mass
Relative
no.
of
atoms
Simplest
ratio
26.6
0.68
K
26.6 39.1
39.1
0.68
= 0.68
=1
35.4
0.68
Cr
35.4 52.0
52
0.68
= 0.68
=1
38.1
2.38
O
38.1 16.0
0.68
16
= 2.38
= 3.5
Therefore, empirical formula is K 2 Cr 2 O 7 .
Empirical and Molecular formula of some compounds
E.F.
H2O
C
CH 4
HO
CH
CH 2 O
BH 3
P2O5
Calculate the empirical formula for a compound
that contains 26.6% potassium, 35.4% chromium
and 38.1% oxygen.
[Given At. no. of K = 39.1; Cr = 52; O = 16]
Solution :
Molecular Formula (M.F) : The formula which gives the
actual number of atoms of various elements present in the
molecule of the substance is termed as molecular formula.
Compound
Molecular mass
Empirical formula mass
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number
ratio
1×2=2
1×2=2
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14.
An organic compound contains 40% C, 6.66% H
and rest oxygen. Its molar mass is 60. Calculate its
empirical and molecular formulas.
Solution :
Element %
C
40.0
At.
mass
12
H
6.66
1
O
53.34 (by 16
difference)
Relative no. Simplest
of atoms
ratio
3.33
40
= 3.33
=1
12
3.33
6.66
6.66
= 6.66
=2
1
3.33
53.34
3.33
= 3.33
=1
16
3.33
Empirical formula = CH 2 O
Empirical formula mass = (12 + 2 + 16) = 30
Mol. mass
60
n=
=
=2
30
Emp. mass
14.
1 vol. N 2 + 3 vol. H 2 → 2 vol. NH 3
Thus, calculations based on chemical equations are of three
types:
i.
Calculations based on mass−mass relationship.
ii.
Calculations based on mass−volume relationship.
iii.
Calculations based on volume−volume relationship.
Calculations using Chemical Equations:
A balanced chemical equation gives quantitative
information in terms of molecules, masses, moles and
volumes between the different reactants and products
involved in the reaction. This is called stoichiometry.
The quantitative information conveyed by a chemical
equation helps in a number of calculations.
Complete the following blanks for the reaction indicated
Molecular formula = 2 × (Empirical formula)
= 2 × (CH 2 O) = C 2 H 4 O 2
13.
Volume interpretation:
CaH 2(s) + 2H 2 O (g) → Ca(OH) 2(s) + 2H 2(g)
PROBLEM
A compound of carbon, hydrogen and nitrogen
contains the elements in the ratio of 18 : 2 : 7.
Calculate its empirical formula. If the molecular mass
is 108, what is its molecular formula?
(C 6 H 8 N 2 )
Calculate the empirical formula of a mineral which
has the following percentage composition
CuO = 44.82%, SiO 2 = 34.83% and water = 20.35 %
(at. wt. of Cu = 63.5, Si = 28).
(CuO. SiO 2 . 2
H 2 O)
STOICHIOMETRY
QUANTITATIVE RELATIONS IN CHEMICAL
REACTIONS:
Stoichiometry is the calculation of the quantities of
reactants and products involved in a chemical reaction. It is
based on the chemical equation and on the relationship
between masses and moles.
A chemical equation can be interpreted as follows :
N 2(g) + 3H 2(g) → 2NH 3(g)
i.
Moles: 2 Moles + ……. → ……. + …….
ii.
Grams: 42 g + ……. → ……. + …….
iii.
Hydrogen atoms: 6.02 × 1023 + … → … + …
i.
SOLVED EXAMPLE
15.
Step I: Write the formula of reactants with plus sign on the
left hand side. Draw arrow from left to right and write the
formula of products right side of the arrow.
MgO + HCl → MgCl 2 + H 2 O
Step II: Balance the masses of each atom in the chemical
formula of reactants and products by proper whole number
coefficients to get the balance chemical equation.
MgO + 2HCl → MgCl 2 + H 2 O
1 mol
1 molecule N 2 + 3 molecules H 2 → 2 molecules NH 3
Molar interpretation:
Mass interpretation:
28 g N 2 + 6 g H 2 → 34 g NH 3
Chemistry -1(Some Basic Concepts of Chemistry)
Calculate the number of grams of magnesium
chloride that could be obtained from 17.0 g of HCl
when HCl is reacted with excess of magnesium oxide.
Solution:
Molecular interpretation:
1 mol N 2 + 3 mol H 2 → 2 mol NH 3
Mass-mass relationship: Here the weight of the
reactants and products are considered.
2 mol
2 × 36.5 g
73 g
1 mol
(24 + 71) g
95 g
1 mole
73 g of HCl produce MgCl 2 = 95 g
1 g of HCl produce MgCl 2 =
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17 g of HCl will produce MgCl 2 =
95
× 17 = 22.12 g
73
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ii.
Mass-Volume relationship: Here such problems are
considered which involve both weights and volumes.
Weight of the solid substances may be compared
with the volume of gases. Gram-molecular volume
law, according to which a gram molecule (mol. wt in
gram) of every gas at NTP occupies 22.4 litres is
used in finding solutions to such problems.
SOLVED EXAMPLE
16.
Calculate the volume of chlorine that can be
obtained at S.T.P. by reaction of 1.58 g of KMnO 4
and excess of hydrochloric acid.
Solution:
LIMITING REAGENT:
When a reaction is carried out in such a way, that the
reactants are simultaneously consumed completely, we can
say that the reactants are in stoichiometric proportions.
The reactant or reagent that is completely consumed
determines the quantity of products formed and is called the
limiting reactant.
The reactant which is taken in excess than the limiting
reactant is the excess reactant.
SOLVED EXAMPLE
18.
Balanced equation
2KMnO 4 + 16HCl
2KCl + 2MnCl 2 + 8H 2 O + 5Cl 2
2 mol
(316 g)
5 mol
(5 × 22.4 L at S.T.P.)
Thus, volume of Cl 2 produced at S.T.P.
5 × 22.4 × 1.58
=
= 0.560 L or 560 mL.
316
Carbon monoxide reacts with oxygen to form
carbon dioxide according to the equation,
2CO + O 2 → 2CO 2 . In an experiment 400 ml of
carbon monoxide and 180 ml of oxygen were
allowed to react to produce carbon dioxide. All the
volumes were measured under the same conditions of
temperature and pressure. Find out the composition
of the final mixture. Identify the limiting reagent.
Solution:
iii.
Volume-Volume relationship : Gay-Lussac law of
gases supplemented by Avogadro hypothesis is made
use of in this type of calculations in which, reactions
among gases are studied. Molecules of reacting gases
& the products formed thereof are compared to their
volumes i.e. the ratio amongst the molecules is taken
as the ratio amongst their volumes, according to the
reverse of Avogadro's law.
SOLVED EXAMPLE
17.
Calculate the volume of O 2 at STP required to
completely burn 200 ml of acetylene.
Solution:
The stoichiometric reaction is represented as
+
Moles of CO
=
O 2(g)
2CO 2(g)
Moles of O 2
400
= 0.179
22400
=
180
= 8.04 × 10–3
22400
According to above equation 2 moles of CO require 1 mole
of O 2 for the reaction. Hence for 0.179 mol of CO, the
moles of O 2 required will be
0.179 ×
1
= 0.0895 moles
2
But we have only 8.04 × 10–3 mole of O 2 . Hence, O 2 is the
limiting reagent in this case.
CO 2 would be formed only from that amount of available
O 2 i.e. 8.04 × 10–3. Since 1 mole of O 2 gives 2 mole of CO 2 .
2CH ≡ CH + 5O 2 → 4CO 2 + 2H 2 O
2 mol
5 mole
2 × 22.4 litre 5 × 22.4 L
200 ml
x ml
∴x=
2CO (g)
8.04 × 10 mol O 2 × 2 = 0.0161 mole CO 2
–3
1 mol CO 2 = 22.4 L
0.0161 mole CO 2 = 22.4 × 0.0161
5 × 22.4lit
× 200 ml = 500 ml
2 × 22.4lit
= 0.361 L = 361 ml
Volume of O 2 = 500 ml
So the composition of the final mixture is
(Note: Experimental conditions for the volume of C 2 H 5 are
not mentioned. So, it is assumed to be at STP)
CO = 400 – 361 = 39 ml
Chemistry -1(Some Basic Concepts of Chemistry)
CO 2 = 361 ml
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19.
In a reaction A + B 2 → AB 2 , identify the
limiting reagent, if any, in the following reaction
mixtures:
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B
19.
(Mol. wt. of PCl 3 = 137.5, Cl 2 = 71, P 4 O 7 = 236,
POCl 3 = 153.5)
(PCl 3 )
1 litre of oxygen and 3 litres of carbon monoxide
measured at STP are reached to get carbon dioxide.
Calculate the volume of the gases after the reaction
and also the weight of carbon dioxide formed in the
reaction.
(0 litre, 1 litre, 2 litre, 44 g, 3.928 g)
Solution:
(i)
The given reaction is A + B 2 → AB 2
Here 300 atoms of a requires 300 molecules of B,
since there are only 200 molecules of B provided
∴ B is the limiting reagent.
(ii)
3 mol B requires 3 mol A. Since only 2 mol of A are
provided, therefore A is the limiting reagent.
(iii) 100 atom of A + 100 molecules of B constitute a
Stoichiometric mixture. Neither A nor B is the
limiting reagent.
(iv)
B is the limiting reagent as 5 mol A requires 5 mol B
but only 2.5 mol B are given.
(v)
A is the limiting reagent as 5 mol B requires 5 mol A,
but only 2.5 mol A are provided.
PROBLEMS
15.
Calculate the volume of carbon dioxide at NTP
evolved by strong heating of 20g calcium carbonate.
(4.48 litres)
16.
The hydrated salt Na 2 SO 4 ⋅ nH 2 O, undergoes 55.9 %
loss in weight on heating and becomes anhydrous.
(10)
Calculate the value of n.
17.
The reaction, 2C + O 2 → 2CO is carried out by
taking 24 g of carbon and 96 g of oxygen. Find out
i.
ii.
iii.
18.
Which reactant is left in excess and how much?
How many moles of CO are formed?
How many gram of other reactant should be
taken so that nothing is left at the end of the
reaction?
(i. 64 g O 2 , ii. 56 g, iii. 72
g)
Phosophoryl chloride, POCl 3 , is used in the
manufacture of fire retardants, gasoline additives and
hydraulic fluids. One reaction by which it can be
prepared is
6 PCl 3() + 6 Cl 2(g) + P 4 O 10(s) → 10 POCl 3()
If 1.00 kg each of PCl 3 , Cl 2 and P 4 O 10 are allowed
to react, which is the limiting reactant?
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KEY IDEAS
1.
2.
3.
4.
5.
6.
Matter is anything that occupies space has mass.
Law of conservation of mass (Lavoisier in 1774).
During any physical or chemical change, the total
mass of the products formed is equal to the total
mass of reactants consumed.
Law of constant composition (Proust in 1798). A
chemical compound always contains same elements
combined together in same proportion by mass.
The smallest particle of an element that takes part
in chemical reactions is atom.
8.
The smallest particle of a substance that has
independent existence is molecule.
9.
One twelfth (1/12) of the mass of an atom of C-12
is atomic mass unit. It is equal to 1.66 × 10–27 kg.
10. The average relative mass of an atom of element as
compared with mass of a carbon atom (C-12) taken
as 12 a.m.u. is atomic mass.
Law of multiple proportions (John Dalton in
1803). When two elements combine together to
form two or more than two compounds then the
masses of one of elements that combine with fixed
mass of the other bear a simple whole number ratio
to one another.
11. The average relative mass of a molecule of the
substance as compared with mass of a carbon atom
(C-12) taken as 12 a.m.u. is molecular mass.
12. One mole is 6.022 × 1023 specified particles.
Gay Lussac’s law (1808). When gases react with
each other they do so in volumes which bear a
simple whole number ratio to one another and to
the volume of the products, if they are also gases,
provided all volumes are measured under similar
condition of temperature and pressure.
Avogadro’s law: Equal volume of all gases under
similar conditions of temperature and pressure
contain equal number of molecules.
6.022 × 1023
particles
7.
13. Mass percentage composition of a compound gives
the mass of each element expressed as percentage
of the total mass.
14. The formula which gives the simplest whole number
ratio of atoms of different elements present in the
molecule of a compound is empirical formula.
15. The reagent that is completely consumed in the
reaction is called limiting reagent.
in terms of
particles
1 mole of O-atoms = 6.022 × 1023 atoms
1 mole of O2 molecules = 6.022 × 1023 molecules
in terms of
volume
1 MOLE
in terms
of mass
1 gram atom
of element
1 gram molecule
of substance
12 g
C
18 g
H2O
Mass of
substance
23 g
Na
44 g
CO2
Multiplied by
Multiplied by
MOLE
Divided by
22.4 L of a
gas at N.T.P.
Divided by
1 gram formula
weight of substance
58.5 g
NaCl
100 g
CaCO3
Number of
particles
AVOGADRO NUMBER
MOLAR MASS
Multiplied by
22.4 L
Divided by
22.4 L
Vol. at N.T.P. in litres
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SOLVED EXAMPLES ON MULTIPLE CHOICE QUESTIONS
Example 1:
The number of gram molecules of oxygen in 6.022 × 1024
molecules of CO is
(A)
(C)
10 g-moles
5 g-moles
(B)
(D)
1 g-moles
0.5 g-moles
Example 5:
In a compound C, H & N are present in the ratio
9 : 1 : 3.5 (by weight). Molecular mass of the compound is
108. Formula of the compound is
(A)
(C)
C2H6N2
C3H4N
Elament
6.022 × 10
= 10
6.022 × 10 23
24
Mole of O in CO =
∴
C
10
= 5 g-mole
2
H
N
The correct alternative is (C).
Example 2:
Tap water is
(A)
(C)
(B)
(D)
a mixture
none of these
N=
NaOH & KOH
H2O & D2O
1
= (C 3 H 4 N) 2
(B)
(D)
KCl & KI
SO 2 & SO 3
∴
The correct alternative is (B).
Example 6:
The reaction of calcium with water is represented by the
equation:
Ca + 2H 2 O → Ca(OH) 2 + H 2
The volume of hydrogen measured at S.T.P. that would be
liberated when 8 g calcium completely reacts with water is
(A)
(C)
4480 cm3
0.2 cm3
Solution:
4.6 × 1022 atoms of certain element weights 13.8 g, the
atomic mass of the element is
Ca + 2H 2 O →
290.5
34.4
(B)
(D)
40 g
180.6
10.4
(B)
(D)
22.42 = 22400 cm3
(1 mL = 1 cm3)
40 g Ca produces = 22400 cm3 of H 2
6.02 × 1023 = 1 mole
4.6 × 1022 atom = 13.8 g
8 g Ca will produce =
13.8 × 6.02 × 10 23
= 180.6
4.6 × 10 22
2240 cm3
0.4 cm3
Ca(OH) 2 + H 2
Solution:
∴
4
= C6H8N2
The correct alternative is (D).
6.02 × 1073 atom =
3
Mol.wt. 108
=
=2
E.F.W.
54
Example 4:
(A)
(C)
Simple ratio
M..F. = (E.F.) n
The correct alternative is (B).
Solution:
In SO 2 and SO 3 two same elements (S and O) combine to
form two different compounds (SO 2 and SO 3 ).
∴
At. ratio
66.6
= 5.5
12
7.4
= 7.4
1
26.0
= 1.86
14
E.F.W. = 36 + 4 + 14 = 54
a compound
an element
Example 3:
Which of the following pairs of compounds illustrates the
law of multiple properties?
(A)
(C)
% Elament
9
= 66.6
13
1
= 7.4
13
3.5
= 26.0
13
Empirical formula = C 3 H 4 N
Solution:
Tap water contains soluble impurities hence it is a
homogeneous mixture.
∴
C6H8N2
C 9 H 12 N 3
Solution:
Solution:
Mole of CO =
(B)
(D)
∴
22400
× 8 = 4480 cm3 of H 2
40
The correct alternative is (A).
The correct alternative is (B).
Chemistry -1(Some Basic Concepts of Chemistry)
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XI + XII + Entrance - 2015
C-21
Vidyasagar Classes
THEORY QUESTIONS
1.
Briefly discuss the importance of studying chemistry.
OR
17.
What is gram molar volume?
18.
Define the terms element, symbol, compound and
formula.
19.
Distinguish the following
What is the need of studying chemistry?
(i) Atom from molecule
(ii) Atom from gram–atom
(iii) Mole from molecule.
2.
What are fundamental units? List their SI units of
measurement.
3.
What are derived units?
4.
What is meant by SI units? Name the seven basis SI
units.
(i)
Why can atomic masses be referred to as
relative numbers?
5.
List the SI units for the following derived quantities.
(ii)
Why the atomic masses of most of the elements
are not whole numbers?
20.
(i) Force
(ii) Pressure
(iii) Energy
Give reasons
21.
Explain the need of term average atomic mass.
22.
What do you understand by the empirical and
molecular formulae of a substance? How two are
related to each other?
23.
What is atomic mass unit?
State and explain Gay Lussac’s Law of gaseous
volumes and mention its important application.
24.
Write the postulates of Dalton’s theory.
25.
Give the limitations of Dalton’s theory.
7.
State and explain Avogadro’s hypothesis with a
suitable example. What is the importance of this law?
26.
Write a note on chemical stoichiometry.
8.
Show that the volume of 1 mole of any gas at STP is
22.4 L.
27.
What are limiting reagents?
9.
Briefly explain the mole concept.
6.
State and explain the Gay Lussac’s law of combining
volumes of gases with a suitable example.
OR
OR
Define a mole what is its importance in chemistry?
10.
Define :
(i)
(ii)
(iii)
(iv)
11.
Mole
Empirical formula
Molecular formula
Avogadro’s number
Describe the following laws
(i) Conservation of mass
(ii) Conservation of energy
(iii) Conservation of mass and energy
12.
State the law of constant proportions. Illustrate with
an example.
13.
State the law of multiple proportions. Illustrate with
an example.
14.
Define atomic mass and molecular mass.
15.
Define atom and molecule.
16.
What is Avogadro’s number?
Chemistry -1(Some Basic Concepts of Chemistry)
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C-22
Vidyasagar Classes
8.
EXERCISE
1.
The % composition of four hydrocarbons is as
follows:
(i)
(ii)
(iii)
(iv)
%C
75
80
85.7
91.3
%H
25
20
14.3
8.7
2.
Boron has two isotopes B and B whose relative
abundances are 20% and 80% respectively. Atomic
weight of boron is
22.4 lit
16.8 lit
(B)
(D)
11.2 lit
5.6 lit
13.
M
3M
(B)
(D)
2M
4M
2
8
(B)
(D)
15.
16g of O 3
48g of SO 3
3 gram molecules of CO 2
18 × 1023 CO 2 molecules
6 × 1023 oxygen atoms
6 × 1023 carbon atoms
Chemistry -1(Some Basic Concepts of Chemistry)
2.88 × 10–3
4.54 × 10–3
(B)
(D)
1.66 × 10–3
1.66 × 10–2
C3H3
C2H4
(B)
(D)
C2H2
C6H6
70
100
(B)
(D)
140
65
4
2
(B)
(D)
6
3
Four oxides of nitrogen are given below.
(ii)
(iv)
NO 2
N2O5
Oxides having 30.5% nitrogen are
(A)
(C)
One mole of CO 2 contains
(A)
(B)
(C)
(D)
One mole of oxygen
One molecule of sulphur trioxide
100 amu of uranium
44 g of carbon dioxide
(i) NO
(iii) N 2 O 4
(B) 16g of SO 3
(D) 1g of hydrogen
1.084 × 1020
4.84 × 1017
Caffeine has a molecular weight of 194. If it contains
28.9 % by mass of nitrogen, number of atoms of
nitrogen in one molecule of caffeine is
(A)
(C)
4
16
(B)
(D)
An organic compound made of C, H and N contains
20 % nitrogen. What will be its molecular mass if it
contains only one nitrogen atom in it?
(A)
(C)
14.
6.023 × 1019
6.023 × 1023
A compound contains 92.3% of carbon, and the rest
hydrogen. The molecule of the compound is 39 times
heavier than hydrogen molecule. The molecular
formula of the compound is
(A)
(C)
The gas having same number of molecules as 16g of
oxygen is (at. wt S = 32)
(A)
(C)
7.
12.
11
10.8
If molecular mass and atomic mass of sulphur are
256 and 32 respectively then its atomicity is
(A)
(C)
6.
(B)
(D)
M gram of a substance when vaporised occupy a
volume of 5.6 L at NTP. The molecular masses of the
substance will be
(A)
(C)
5.
10.2
10.5
3.125 × 10–2
2.5 × 10–2
If 1021 molecules are removed from 200 mg of CO 2
then the number of moles of CO 2 left will be
(A)
(C)
The total volume of mixture of 2 g of helium and 7 g
of nitrogen under S.T.P. conditions is
(A)
(C)
4.
11.
11
(B)
(D)
Which among the following is the heaviest?
(A)
(B)
(C)
(D)
constant proportion
conservation of mass
multiple Proportions
reciprocal Proportions.
0.02
1.25× 10–2
How many water molecules are there in one drop of
water (volume 0.0018 ml) at room temperature?
(A)
(C)
10.
10
(A)
(C)
3.
(A)
(C)
9.
The data illustrates the law of
(A)
(B)
(C)
(D)
How many moles of magnesium phosphate,
Mg 3 (PO 4 ) 2 , will contain 0.25 mole of oxygen atoms?
16.
(B)
(D)
(ii) and (iii)
(i) and (iv)
20 g of sulphur on burning in air produced 11.2 L of
SO 2 at STP. The percentage of unreacted sulphur is
(A)
(C)
We teach success
(i) and (ii)
(iii) and (iv)
80
60
(B)
(D)
20
40
XI + XII + Entrance - 2015
C-23
Vidyasagar Classes
17.
30 g of Mg (At. wt. = 24) combines with 30 g of
oxygen in a sealed tube. The composition of residual
mixture will be
(A)
(B)
(C)
(D)
18.
19.
CH
CH 4
The limiting reagent for this reaction is NaOH
0.08 moles of Fe(OH) 3 is formed
0.12 mole of FeCl 3 is left unreacted
all of the above
X2Y3
XY 2
(B)
(D)
(B)
(D)
155.8
55.8
In the reaction 4A + 2B + 3C → A 4 B 2 C 3 ,
what will be the number of moles of product formed,
starting from one mole of A, 0.6 mole of B and
0.72 mole of C?
(A)
(C)
25.
15.58
5.58
0.25
0.24
(B)
(D)
0.3
2.32
Sulphur trioxide is prepared by the following two
reactions:
S 8(s) + 8O 2(g) → 8SO 2(g)
2SO 2(g) + O 2(g) → 2SO 3(g)
How many grams of SO 3 are produced from 1 mole S 8 ?
(At. wt. S = 32, O = 16)
(A)
(C)
1280
960
(B)
(D)
640
320
XY
X3Y4
What is/are true about calcium carbonate? (At. Wt.
Ca = 40)
(a)
(b)
(c)
(d)
(A)
(C)
22.
(B)
(D)
24.
Two elements X (atomic mass = 50) and Y (atomic
mass = 16) combine to give a compound having 32 %
Y. The formula of the compound is
(A)
(C)
21.
CH 3
CH 2
A metal oxide has the formula X 2 O 3 . It can be
reduced by hydrogen to give free metal and water.
0.1596 g of metal oxide requires 6 mg of hydrogen
for complete reduction. The atomic mass of metal in
amu is
(A)
(C)
A solution containing 0.2 mole of ferric chloride is
allowed to react with 0.24 mole of sodium hydroxide.
The correct statement for this reaction is that
(A)
(B)
(C)
(D)
20.
40g MgO + 20g O 2
45g MgO + 15g O 2
50g MgO + 10g O 2
60g MgO only
3.0 g of a hydrocarbon on combustion gives 8.8 g of
CO 2 and 5.4 g of water. The empirical formula of the
compound is
(A)
(C)
23.
The percentage of carbon atom is 20 %
The percentage of oxygen is four times the
percentage of carbon.
10 g of calcium carbonate contains 4 g of
calcium.
10 g of calcium carbonate contains 6.02 × 1022
CO 32− ions.
All are true
a, c and d
(B)
(D)
b, c and d
c and d
254 g of iodine and 142 g of chlorine are made to
react completely to give a mixture of ICl and ICl 3 .
The moles of each one formed is
(At. wt. I = 127, Cl = 35.5)
(A)
(B)
(C)
(D)
0.1 ICl and 0.1 mole ICl 3
1.0 mol ICl and 1.0 mol ICl 3
0.5 ICl and 0.1 mol ICl 3
0.5 mol ICl and 1.0 mol ICl 3
Chemistry -1(Some Basic Concepts of Chemistry)
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XI + XII + Entrance - 2015
C-24
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8.
EXERCISE : HOMEWORK
1.
A pure substance can only be
(A)
x+y
(B)
A symbol not only represents the name of the
element but also its
x
y
y
(C) xy
(D)
x
At room temperature and pressure, two flasks of equal
volumes are filled with H2 and SO2 separately. Particles
which are equal in number, in the two flasks are
(A)
(C)
(A)
(C)
atoms
molecules
(B)
(D)
(A)
(B)
(C)
(D)
2.
3.
(C)
(D)
atomic number
atomic volume
All elements are homogeneous
Compounds always contain two or more
different elements
A mixture is not always heterogeneous
Air is a heterogeneous mixture
10.
11.
kg m2 s−2
kg m2 s−1
(B)
(D)
12.
reciprocal proportions
multiple proportions
constant proportions
conservation of mass
elements consist of impurities
these are mixtures of allotropes
these are mixtures of isobars
these are mixtures of isotopes
If the weight of 5.6 litres of a gas at S.T.P is 11 g, the
gas is
(Given at.wt. of P = 31, C = 12, N = 14, O = 16,
Cl = 35.5)
(A)
(C)
phosphine
nitric oxide
(B)
(D)
phosgene
nitrous oxide
290
34.4
(B)
(D)
180
10.4
XY
XY 2
(B)
(D)
X2Y
X2Y3
5 mL, 10 mL
5 mL, 15 mL
(B)
(D)
10 mL, 5 mL
10 mL, 10 mL
If 6 litre of H 2 and 5.6 litre of Cl 2 are mixed and
exploded in an eudiometer, the volume of HCl
formed is
(A)
(C)
15.
315 g
342 g
If 5 mL of methane is completely burnt, the volume
of oxygen required and the volume of CO 2 formed
under the same conditions are
(A)
(C)
14.
(B)
(D)
The simplest formula of a compound containing 50%
of element X (At. mass = 10) and 50% of the element
Y (At. mass = 20) is
(A)
(C)
13.
150 g
327 g
4.6 × 1022 atoms of an element weigh 13.8 g.
The atomic mass of the element is
(A)
(C)
kg m−1 s2
kg m2 s2
electrons
neutrons
What is the mass of one mole of aluminium sulphate
Al 2 (SO 4 ) 3 ?
(A)
(C)
The atomic masses of the elements are usually
fractional because
(A)
(B)
(C)
(D)
7.
(B)
(D)
Different proportions of oxygen in the various oxides
of nitrogen, prove the law of
(A)
(B)
(C)
(D)
6.
atomic mass
atomicity
9.
SI unit of energy is
(A)
(C)
5.
a compound
an element
an element or a compound
a heterogeneous mixture
Which one of the following statements is incorrect?
(A)
(B)
4.
The atomicity of a species is x and its atomic weight
is y. The molecular weight of the species is
6.0 litre
11.2 litre
(B)
(D)
5.6 litre
11.6 litre
23 g if sodium will react with ethyl alcohol to give
(A)
(B)
(C)
(D)
one mole of hydrogen
one mole of oxygen
one mole of NaOH
1/2 mole of hydrogen
ANSWERKEY
1.
(C)
2.
(A)
3.
(D)
4.
(A)
5.
(B)
6.
(D)
7.
(D)
8.
(C)
9.
(C)
10.
(D)
11.
(B)
12.
(B)
13.
(B)
14.
(C)
15.
(D)
Chemistry -1(Some Basic Concepts of Chemistry)
We teach success
XI + XII + Entrance - 2015