Atoms and Molecules
Atoms and Molecules
Atoms and Molecules
Prefix
Symbol
Prefix
Symbol
10
deci
102
centi
exa
103
milli
10
peta
106
micro
1012
24
10
yotta
1021
zetta
1018
15
Multiple
1
tera
109
nano
10
giga
1012
pico
106
mega
1015
femto
atto
zepto
18
10
kilo
10
102
hecto
1021
10
deca
24
da
10
yocto
Interpret
(b) 2
(d) 12.5
4.18 J
(d), Conversion factor =
1 cal
4.18 J
3 cal = 3 cal
1 cal
= 12.5 J
1 nm =1000 pm
Quantity
Area
Volume
Density
Velocity
Acceleration
Definition of quantity
Length squared
Length cubed
Mass per unit volume
Distance travelled per
unit time
Velocity changed per
unit time
Mass times
acceleration of object
Force per unit area
Expression in terms
of SI base units
m2
m3
kg /m 3 or kg m 3
m/s or ms1
m/s2 or ms2
kg m/s2 or kg ms2
(newton, N)
Pressure
kg /(ms2 ) or kg m 1 s2
(pascal, Pa)
Energy (work,
Force times distance
kg m 2 /s2 or kg m 2 s2
heat)
travelled
(joule, J)
Electric charge
Ampere times second A-s (coulomb, C)
Electric potential Energy per unit charge J/(A-s) potential
difference (volt, V)
Force
Scientific Notation
In scientific notation, all numbers (however large or small)
are expressed as a number between 1.000 and 9.999 multiplied or
divide by 10, ie, here a number is generally expressed in the form
N 10n
Here, N is called digit term. It is a number between 1.000
and 9.999.
n is called an exponent.
eg, 138.42 can be written as 1. 3842 102
or 0.013842 can be written as 1. 3842 102
Significant Figures
The digits in a properly recorded measurement are known as
significant figures or in other words we can say that
significant figures are the meaningful digits in a measured or
calculated quantity.
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A significant figure includes all those digits that are known
with certainty plus one more which is uncertain or estimated.
Always remember that greater the number of significant
figures in a reported result, smaller the uncertainty.
While determining significant figures
1. Read the number from left to right and count all the
digits, starting with the first digit that is not zero.
2. When adding or subtracting, the number of decimal
places in the answer should not exceed the number of
decimal places in either of the numbers. eg,
0.13
2 significant figures
1.5
2 significant figures
20.911
5 significant figures
22.541
Mixture
Substance
(b) Silica
(d) Plasma sulphur
12 6.022 1023
= 1.66 1024 g
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Molar Mass
Molar mass of an element is defined as mass of 1 mole of that
element, ie, mass of 6.023 1023 entities or particles of that
element. eg, molar mass of oxygen = 32 g/mol, that means
6.023 1023 molecules of oxygen weigh 32 g,
or molar mass of Na = 23 g/mol, that means 6.023 1023
monoatomic molecules of Na weigh 23 g.
Formula Mass
The formula mass of a substance is the sum of the atomic
masses of all atoms in the formula unit of the compound. It is
normally calculated for ionic compounds. eg, formula mass of
NH3 is 14 + 3 = 17 amu or u
or formula mass of NaCl is 23 + 35.5 = 58.5 amu or u.
mass of metal
8.0
mass of oxygen combined
mass of metal
35.5
mass of chlorine combined
98
= 49
2
eg,
58.5
= 58.5
1
106
Equivalent weight of Na 2CO 3 =
= 53
2
Equivalent weight of NaCl =
Interpret
(b) 28
(d) 112
atomic wt. 56
(a), Eq. wt. of metal =
=
= 18.6
3
valency
Mole Concept
The word mole was introduced around 1896 by W. Ostwald
who derived it from Latin word moles means a heap or pile.
In 1967 this word was accepted as a unit of chemical substances
under SI system. It is represented by the symbol mol.
One mole of any substance is defined as :
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(i) The amount which weighs exactly same as its formula
weight in gram or atomic mass in gram or molecular
mass in gram.
(ii) The amount which has same number of entities (atoms,
molecules or other particles) as there are atoms
in exactly 0.012 kg (or 12 g) of carbon-12 isotope
(ie, 6.023 1023 entities).
Here, 6. 023 1023 is called Avogadros number and
denoted by N A .
(iii) The amount which occupies 22.4 L at STP (if it is taken
for a gas).
The formulae used to convert amount of substance into
l
moles are :
amount of substance in gram
Number of moles =
molecular wt. in gram
number of particles at STP
=
Avogadro s number
volume of gas at STP (in L)
or
=
22.4
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Elaborations of above written formulae are given below, but
before elaborations it is better to understand concentration
terms used in them. These concentration terms are as
follows :
1. Normality (N) Which is defined as the number of
g-equivalents of solute per litre of solution or as the
number of mg-equivalents of a substance per millilitre of
solution. eg, 0.12 N H2SO 4 means a solution which
contains 0.12 g-equivalent of H2SO 4 per litre of solution.
This also means that each millilitre of this solution can
react, for example, with 0.12 mg-eq. of CaO or with 0.12
mg-eq. of Na 2CO 3.
specific gravity % strength 10
Normality =
equivalent weight
l
1
molecular weight of solute
= molarity
+
molality
1000
(a) % by weight of solute in solution
weight of solute (g) 100
=
weight of solution (g)
(b) % by volume of solute in solution
weight of solute (g) 100
=
volume of solution (mL)
18. Molecular
phase).
(for
weight
molecular weight
64
=2
Moles of O2 =
32
Also, Q 1 mole of O2 contains 6.023 1023 molecules
2 moles of O2 contains 6.023 1023 2
= 12.046 1023 molecules
1
500
= N
1000
10
1
2 1
N=
= = 0.20
10 1 5
18
106
Now,
Molarity =
= 55.56
18 103
or
or
Molarity = 55.56 M
(b) 0.1
(d) 0.25
Interpret (b),
20
0
0
20
20
= 0.1
200
(ii)
(a) 45.50 M
(c) 55.56 M
(i)
64 g O2 ?
(b) 2,12.046 1023
N
Mole fraction of benzene =
= 0.9
n + N
gaseous
(b) 3.205
(d) 1.86
0
20
Stoichiometry
The methods used to describe the atomic and molecular
masses constitute the generalised approach to chemical
calculations and quantitative methods called chemical
stoichiometry. (Stoichiometry from Greek words stoichion =
element; metron = measure). Solving of stoichiometric problems
require a firm grasp of mole concept, balancing chemical
equations and care in consistent use of units.
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The numerals used to balance a chemical equation are called
stoichiometric coefficients. For the stoichiometric
calculations, the mole relationships between different
reactants and products are required, as from them the
massmass, massvolume and volumevolume relationships
can be obtained between different reactants and products.
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The following steps are involved in stoichiometric
calculations :
1. Firstly a balanced chemical equation is written.
2. On the basis of this equation, the molar relationship is
written.
3. All the values are converted into desired parameters and
then, calculation is done by unitary method.
1
mol
2
Interpret (d), NH 3 + O2 NO + H 2 O
2NH 3 +
2 mol
2 17 g
1 mol
5
mol
2
5 6.8
6.8 g NH 3 will react with oxygen =
mol
2 2 17
= 0.5 mol
(c) C4 H 9 F
(d) C5 H11 F2
1 mol H
(11.92 g H)
= 11.8 mol H
1.008 g H
1 mol F
(24.97 g F)
= 1.31 mol`F
19.0 g F
Division by 1.31 yields the integer ratio 4 : 9 : 1 thus the empirical
formula is C4 H 9 F.
(c) C6 H12 O3
168
168
=
=3
12 3 + 4 + 16 56
= C9 H12 O3
(b) 40 g
(d) 4 g
= 0.05 mol
Instance 17 NH 3 + O2 NO + H 2 O
In the above equation, the moles of oxygen required for reacting with
6.8 g NH 3 is
(a) 5
(c) 1
1 mol
1 mol
(b) 2.5
(d) 0.5
5 mol
2
5 mol
2
5
O2 2NO + 3H 2 O (balanced equation)
2
Caution Points
1. While calculating the significant figures of numbers, it is
better to convert them into scientific notation because
exponential term does not contribute to the significant
figures.
2. Law of definite proportions and law of multiple
proportions do not hold good when same compound is
prepared by different isotopes of the same element, eg, H2O
16
18
Chapter Compendium
1. Units of length, mass, time, electric current, temperature,
amount of substance and luminous intensity are called
fundamental or basic units. All other units are derived from
these units, so called derived units.
2. All non-zero digits or zero present between two non-zero
digits are significant. Zero of the left side of a number are
never significant while that of right side, if present after a
number and also a decimal point, are significant. eg,
0.0245 contains 3 significant figures. 0.02450 contains 4
SF and 2000 contains only 1 SF.
3. A more precise result is obtained when expressed to greater
number of decimal points, eg, between 1.5 and 1.51, the
latter is more precise than the former.
4. Anything that occupies space and has mass is called matter.
On physical basis, it is classified into solid, liquid and gases
while on the basis of chemical composition, it is classified
as mixture and substance.
5. Daltons atomic theory has been modified and replaced by
modern atomic theory.
6. Law of conservation of mass state that in a chemical
change, mass is neither created nor destroyed. In any
chemical reaction,
Total masses of reactants = Total masses of products
7. Law of gaseous volumes is proposed by Gay-Lussac, which
state that gases react with each other in simple ratio of
their volumes and if product is also in gaseous state, its
volume also bears a simple ratio with the volumes of
gaseous reactants under similar conditions of temperature
and pressure.
8. Atomic and molecular weights are expressed with respect
to carbon-12 as reference in amu (or u).
1 u =1.66 1024 g
gas = 10 g
22.4
L
volume
will
be
occupied by hydrogen fluoride
10
gas =
22.4
5.6
= 40 g
Among the given molecular formulae, molecular mass of H 2 F2
is 40. Thus, the molecular formula of the gas is H 2 F2.
2.65 1022 atoms of carbon and 0.132 mol of oxygen atoms. Its
(b) Na2 C2 O4
(d) None of these
2.04 g
= 0.0887 mol
23 g mol 1
:
:
:
C
0.0440
1
:
:
:
0
0.132
3
(a) 781.25
empirical formula is
(a) NaCO2
(b) H 3 F3
(d) H 4 F4
(c) H 2 F2
1 vol
1 molecule
1 molecule
2 + 32n = 34
n =1
(c) 9.06 10
mass in gram
molecular mass in gram mol 1
2.5 104
32
= 781.25
Number of moles of N 2 = 3 781.25 = 2343.75
Mass of nitrogen in the cylinder = 2343.75 28
= 65625 g
= 6.5625 104 g
Total mass of the gas in the cylinder
= 2.5 104 + 6.5625 104
= 9.0625 104 g
=
Element
55.83
44.17
Atomic ratio
55.83
=1.1
52
44.17
= 2.76
16
V : O = 2: 5
Thus, empirical formula = V2 O5
x=7
On solving,
If the specific heat of the metal, M is 0.216 cal deg 1 g 1 , the molecular
formula of its oxide is
(a) MO
(b) M 2 O3
(c) M 2 O4
(d) M 2 O
=193.93 g
193.93
= 2 (nearest whole number)
100
So, accurate atomic mass = eq. mass valency
= 100 2 = 200 g
mol. mass 199.87
Atomicity =
=1
=
at. mass
200
Valency of M =