Chapter 1
Some Basic Concept
of Chemistry
SOME USEFUL CONVERSION FACTORS
1 Å = 10–10m, 1nm = 10–9 m
1 pm = 10–12m
1 litre = 10–3 m3 = 1 dm3
1 atm = 760 mm or torr
= 101325 Pa or Nm–2
1 bar = 105 Nm–2 = 105 Pa
1 calorie = 4.184 J
1 electron volt(eV) = 1.6022 ×10–19 J
(1 J = 107 ergs)
(1 cal > 1 J > 1 erg > 1 eV)
ATOMIC MASS OR MOLECULAR MASS
Mass of one atom or molecule in a.m.u.
C → 12 amu
NH3 → 17 amu
Actual Mass
Mass of one atom or molecule in grams
C → 12 ×1.6 × 10–24 g
CH4 → 16 ×1.6 × 10–24 g
Relative Atomic Mass or Relative Molecular Mass
Mass of one atom or molecule w.r.t. 1/12th or 12C atom
C → 12
CH4 → 16
It is unitless
1
,GRAM ATOMIC MASS OR GRAM MOLECULAR MASS
Mass of one mole of atom or molecule
C → 12 g
CO2 → 44 g
It is also called molar mass
DEFINITION OF MOLE
One mole is a collection of that many entities as there are number of atoms
exactly in 12 gm of C-12 isotope.
The number of atoms present in exactly 12 gm of C-12 isotope is called
Avogadro’s number [NA = 6.022 × 1023]
1g
1u = 1amu = (1/12)th of mass of 1 atom of C12 =
NA
= 1.66 ×10–24 g
For Elements
● 1 g atom = 1 mole of atoms = NA atoms
● g atomic mass (GAM) = mass of NA atoms in g
Mass ( g )
● Mole of atoms =
GAM or molar mass
For Molecule
● 1g molecule = 1 mole of molecule = NA molecule
● g molecular mass (GMM) = mass of NA molecule in g.
Mass ( g )
● Mole of molecule =
GMM or molar mass
1 Mole of Substance
● Contains 6.022 × 1023 particles
● Weighs as much as molecular mass/ atomic mass/ionic mass in grams
● If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K
or 22.7 L at STP
For Ionic Compounds
● 1 g formula unit = 1 mole of formula unit = NA formula unit.
● g formula mass (GFM) = mass of NA formula unit in g.
Mass ( g )
● Mole of formula unit =
GMM or molar mass
Hand Book (Chemistry) 2
, VAPOUR DENSITY
Ratio of density of vapour to the density of hydrogen at similar pressure and
temperature.
Molar mass
Vapour density =
2
STOICHIOMETRY BASED CONCEPT
aA + bB → cC + dD
● a,b,c,d, represents the ratios of moles, volumes [for gaseous]
molecules in which the reactants react or products formed.
● a,b,c,d does not represent the ratio of masses.
● The stoichiometic amount of components may be related as
Moles of A reacted Moles of Breacted Moles of C reacted Moles of D reacted
= = =
a b c d
Concept of limiting reagent
If data of more than one reactant is given then first convert all the data into
moles then divide the moles of reactants with their respective stoichiometric
coefficient. The reactant having minimum ratio will be L.R. then find the
moles of product formed or excess reagent left by comparing it with L.R.
through stoichiometric concept.
Percentage Purity
The percentage of a specified compound or element in an impure sample
may be given as
Actual mass of compound
=%purity × 100
Total mass of sample
If impurity is unknown, it is always considered as inert (unreactive)
material.
3 Some Basic Concept of Chemistry
,EMPIRICAL AND MOLECULAR FORMULA
● Empirical formula: Formula depicting constituent atoms in their
simplest ratio.
● Molecular formula: Formula depicting actual number of atoms in
one molecule of the compound.
● The molecular formula is generally an integral multiple of the
empirical formula.
i.e. molecular formula = empirical formula × n
molecular formula mass
where n =
empirical formula mass
● For determination of atomic mass:
Dulong’s & Petit’s Law:
Atomic weight of metal × specific heat capacity (cal/gm-°C) ≈ 6.4.
It should be remembered that this law is an empirical observation
and this gives an approximate value of atomic weight. This law gives
better result for heavier solid elements, at high temperature conditions.
CONCENTRATION TERMS
Concentration
Mathematical Formula Concept
Type
Percentage by mass Mass of solute
w Mass of solute × 100
% = (in gm) present
w Mass of solution in 100 gm of
solution.
Volume percentage Volume of
v Volumeof solute × 100
% = solute (in cm3)
v Volumeof solution present in 100
cm3 of solution.
Mass-volume Mass of solute
w Mass of solute × 100
percentage % = (in gm) present
v Volumeof solution in 100 cm3 of
solution.
Parts per million Parts by mass
Mass of solute × 106
ppm = of solute per
Mass of solution million parts
by mass of the
solution
Hand Book (Chemistry) 4
, Concentration
Mathematical Formula Concept
Type
Mole fraction Moleof A Ratio of number
XA = of moles of one
Moleof A + Moleof B + Moleof C + ... component to
the total number
Moleof B
XB = of moles.
Moleof A + Moleof B + Moleof C + ...
Molarity Moleof solute Moles of solute
M= in one liter of
Volumeof solution (in L) solution.
Molality Moleof solute Moles of solute
m= in one kg of
Mass of solvent (Kg) solvent
MIXING OF SOLUTIONS
It is based on law of conservation of moles.
(i) Two solutions having same solute:
Total moles M V + M 2 V2
Final molarity = = 1 1
Total volume V1 + V2
M1V1
(ii) Dilution Effect: Final molarity, M 2 =
V1 + V2
Volume Strenght of H2O2 Solutions
Labelled as ‘volume H2O2’ means volume of O2 (in litre) at 1 bar & 273
K that can be obtained from 1 litre of such a sample when it decomposes
according to
2H2O2 → 2H2O + O2
● Volume Strenght of H2O2 solution = 11.35 × molarity
PERCENTAGE LABELLING OF OLEUM
Labelled as ‘% oleum’ means maximum amount of H2SO4 that can be
obtained from 100 gm of such oleum (mixture of H2SO4 and SO3) by adding
sufficient water. For example, 109% oleum sample means, with the addition
of sufficient water to 100 gm oleum sample 109 gm H2SO4 is obtained.
% labelling of oleum sample = (100 + x)%
x = mass of H2O required for the complete conversion of SO3 in H2SO4
40
● % of free SO3 in oleum = × x %
9
5 Some Basic Concept of Chemistry
,EUDIOMETRY
Some basic assumptions related with calculations are:
1. Gay-Lussac’s law of volume combination holds good. According to
this law, the volumes of gaseous reactants reacted and the volumes of
gaseous products formed, all measured at the same temperature and
pressure, bear a simple ratio.
N2(g) + 3H2(g) → 2NH3(g)
1 vol. 3 vol. 2 vol.
Problem may be solved directly is terms of volume, in place of mole.
The stoichiometric coefficients of a balacned chemical reactions gives
the ratio of volumes in which gaseous substances are reacting and
products are formed at same temperature and pressure.
2. The volumes of solids or liquids is considered to be negligible in
comparison to the volume of gas. It is due to the fact that the volume
occupied by any substance in gaseous state is even more than thousand
times the volume occupied by the same substance in solid or liquid
states.
2H2(g) + O2(g) → 2H2O(l)
2 mole 1 mole 2 mole
2 vol. 1 vol. 0 vol.
3. Air is considered as a mixture of oxygen and nitrogen gases only. It is
due to the fact that about 99% volume of air is composed of oxygen
and nitrogen gases only.
4. Nitrogen gas is considered as an non-reactive gas.
5. The volume of gases produced is often given by certain solvent which
absorb contain gases.
Solvent Gases absorb
KOH CO2, SO2, Cl2
Ammonical Cu2Cl2 CO
Turpentine oil O3
Alkaline pyrogallol O2
water NH3, HCl
CuSO4/CaCl2 H2O
qqq
Hand Book (Chemistry) 6
, Chapter 2
Atomic Structure
IMPORTANT DEFINITIONS
Proton (mP)/anode Neutron (mn) Electron (me)
rays /cathode rays
mass = 1.67 × 10–27 kg mass = 1.67 × 10–27 kg mass = 9.1 × 10–31 kg
mass = 1.67 × 10–24 g mass = 1.67 × 10–24 g mass = 9.1 × 10–28 g
mass = 1.00750 amu mass = 1.00850 amu mass = 0.000549 amu
e/m value is dependent e/m of electron is found
on the nature of gas to be independent
taken in discharge tube. of nature of gas &
electrode used.
REPRESENTATION OF AN ELEMENT
Mass number A Symbol
X of the
Atomic number Z element
Terms associated with elements:
● Atomic Number (Z): = No. of protons
Electron = Z–C (charge on atom)
● Mass number (A) = Total number of neutron and proton present
A = Number of proton + Number of Neutrons
● Isotopes: Same atomic number but different mass number
Example: 6C12, 6C13, 6C14
● Isobars: Same mass number but different atomic number
Example: 1H3, 2He3
● Isodiaphers: Same difference of number of Neutrons & protons
Example: 5B11, 6C13
7
Some Basic Concept
of Chemistry
SOME USEFUL CONVERSION FACTORS
1 Å = 10–10m, 1nm = 10–9 m
1 pm = 10–12m
1 litre = 10–3 m3 = 1 dm3
1 atm = 760 mm or torr
= 101325 Pa or Nm–2
1 bar = 105 Nm–2 = 105 Pa
1 calorie = 4.184 J
1 electron volt(eV) = 1.6022 ×10–19 J
(1 J = 107 ergs)
(1 cal > 1 J > 1 erg > 1 eV)
ATOMIC MASS OR MOLECULAR MASS
Mass of one atom or molecule in a.m.u.
C → 12 amu
NH3 → 17 amu
Actual Mass
Mass of one atom or molecule in grams
C → 12 ×1.6 × 10–24 g
CH4 → 16 ×1.6 × 10–24 g
Relative Atomic Mass or Relative Molecular Mass
Mass of one atom or molecule w.r.t. 1/12th or 12C atom
C → 12
CH4 → 16
It is unitless
1
,GRAM ATOMIC MASS OR GRAM MOLECULAR MASS
Mass of one mole of atom or molecule
C → 12 g
CO2 → 44 g
It is also called molar mass
DEFINITION OF MOLE
One mole is a collection of that many entities as there are number of atoms
exactly in 12 gm of C-12 isotope.
The number of atoms present in exactly 12 gm of C-12 isotope is called
Avogadro’s number [NA = 6.022 × 1023]
1g
1u = 1amu = (1/12)th of mass of 1 atom of C12 =
NA
= 1.66 ×10–24 g
For Elements
● 1 g atom = 1 mole of atoms = NA atoms
● g atomic mass (GAM) = mass of NA atoms in g
Mass ( g )
● Mole of atoms =
GAM or molar mass
For Molecule
● 1g molecule = 1 mole of molecule = NA molecule
● g molecular mass (GMM) = mass of NA molecule in g.
Mass ( g )
● Mole of molecule =
GMM or molar mass
1 Mole of Substance
● Contains 6.022 × 1023 particles
● Weighs as much as molecular mass/ atomic mass/ionic mass in grams
● If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K
or 22.7 L at STP
For Ionic Compounds
● 1 g formula unit = 1 mole of formula unit = NA formula unit.
● g formula mass (GFM) = mass of NA formula unit in g.
Mass ( g )
● Mole of formula unit =
GMM or molar mass
Hand Book (Chemistry) 2
, VAPOUR DENSITY
Ratio of density of vapour to the density of hydrogen at similar pressure and
temperature.
Molar mass
Vapour density =
2
STOICHIOMETRY BASED CONCEPT
aA + bB → cC + dD
● a,b,c,d, represents the ratios of moles, volumes [for gaseous]
molecules in which the reactants react or products formed.
● a,b,c,d does not represent the ratio of masses.
● The stoichiometic amount of components may be related as
Moles of A reacted Moles of Breacted Moles of C reacted Moles of D reacted
= = =
a b c d
Concept of limiting reagent
If data of more than one reactant is given then first convert all the data into
moles then divide the moles of reactants with their respective stoichiometric
coefficient. The reactant having minimum ratio will be L.R. then find the
moles of product formed or excess reagent left by comparing it with L.R.
through stoichiometric concept.
Percentage Purity
The percentage of a specified compound or element in an impure sample
may be given as
Actual mass of compound
=%purity × 100
Total mass of sample
If impurity is unknown, it is always considered as inert (unreactive)
material.
3 Some Basic Concept of Chemistry
,EMPIRICAL AND MOLECULAR FORMULA
● Empirical formula: Formula depicting constituent atoms in their
simplest ratio.
● Molecular formula: Formula depicting actual number of atoms in
one molecule of the compound.
● The molecular formula is generally an integral multiple of the
empirical formula.
i.e. molecular formula = empirical formula × n
molecular formula mass
where n =
empirical formula mass
● For determination of atomic mass:
Dulong’s & Petit’s Law:
Atomic weight of metal × specific heat capacity (cal/gm-°C) ≈ 6.4.
It should be remembered that this law is an empirical observation
and this gives an approximate value of atomic weight. This law gives
better result for heavier solid elements, at high temperature conditions.
CONCENTRATION TERMS
Concentration
Mathematical Formula Concept
Type
Percentage by mass Mass of solute
w Mass of solute × 100
% = (in gm) present
w Mass of solution in 100 gm of
solution.
Volume percentage Volume of
v Volumeof solute × 100
% = solute (in cm3)
v Volumeof solution present in 100
cm3 of solution.
Mass-volume Mass of solute
w Mass of solute × 100
percentage % = (in gm) present
v Volumeof solution in 100 cm3 of
solution.
Parts per million Parts by mass
Mass of solute × 106
ppm = of solute per
Mass of solution million parts
by mass of the
solution
Hand Book (Chemistry) 4
, Concentration
Mathematical Formula Concept
Type
Mole fraction Moleof A Ratio of number
XA = of moles of one
Moleof A + Moleof B + Moleof C + ... component to
the total number
Moleof B
XB = of moles.
Moleof A + Moleof B + Moleof C + ...
Molarity Moleof solute Moles of solute
M= in one liter of
Volumeof solution (in L) solution.
Molality Moleof solute Moles of solute
m= in one kg of
Mass of solvent (Kg) solvent
MIXING OF SOLUTIONS
It is based on law of conservation of moles.
(i) Two solutions having same solute:
Total moles M V + M 2 V2
Final molarity = = 1 1
Total volume V1 + V2
M1V1
(ii) Dilution Effect: Final molarity, M 2 =
V1 + V2
Volume Strenght of H2O2 Solutions
Labelled as ‘volume H2O2’ means volume of O2 (in litre) at 1 bar & 273
K that can be obtained from 1 litre of such a sample when it decomposes
according to
2H2O2 → 2H2O + O2
● Volume Strenght of H2O2 solution = 11.35 × molarity
PERCENTAGE LABELLING OF OLEUM
Labelled as ‘% oleum’ means maximum amount of H2SO4 that can be
obtained from 100 gm of such oleum (mixture of H2SO4 and SO3) by adding
sufficient water. For example, 109% oleum sample means, with the addition
of sufficient water to 100 gm oleum sample 109 gm H2SO4 is obtained.
% labelling of oleum sample = (100 + x)%
x = mass of H2O required for the complete conversion of SO3 in H2SO4
40
● % of free SO3 in oleum = × x %
9
5 Some Basic Concept of Chemistry
,EUDIOMETRY
Some basic assumptions related with calculations are:
1. Gay-Lussac’s law of volume combination holds good. According to
this law, the volumes of gaseous reactants reacted and the volumes of
gaseous products formed, all measured at the same temperature and
pressure, bear a simple ratio.
N2(g) + 3H2(g) → 2NH3(g)
1 vol. 3 vol. 2 vol.
Problem may be solved directly is terms of volume, in place of mole.
The stoichiometric coefficients of a balacned chemical reactions gives
the ratio of volumes in which gaseous substances are reacting and
products are formed at same temperature and pressure.
2. The volumes of solids or liquids is considered to be negligible in
comparison to the volume of gas. It is due to the fact that the volume
occupied by any substance in gaseous state is even more than thousand
times the volume occupied by the same substance in solid or liquid
states.
2H2(g) + O2(g) → 2H2O(l)
2 mole 1 mole 2 mole
2 vol. 1 vol. 0 vol.
3. Air is considered as a mixture of oxygen and nitrogen gases only. It is
due to the fact that about 99% volume of air is composed of oxygen
and nitrogen gases only.
4. Nitrogen gas is considered as an non-reactive gas.
5. The volume of gases produced is often given by certain solvent which
absorb contain gases.
Solvent Gases absorb
KOH CO2, SO2, Cl2
Ammonical Cu2Cl2 CO
Turpentine oil O3
Alkaline pyrogallol O2
water NH3, HCl
CuSO4/CaCl2 H2O
qqq
Hand Book (Chemistry) 6
, Chapter 2
Atomic Structure
IMPORTANT DEFINITIONS
Proton (mP)/anode Neutron (mn) Electron (me)
rays /cathode rays
mass = 1.67 × 10–27 kg mass = 1.67 × 10–27 kg mass = 9.1 × 10–31 kg
mass = 1.67 × 10–24 g mass = 1.67 × 10–24 g mass = 9.1 × 10–28 g
mass = 1.00750 amu mass = 1.00850 amu mass = 0.000549 amu
e/m value is dependent e/m of electron is found
on the nature of gas to be independent
taken in discharge tube. of nature of gas &
electrode used.
REPRESENTATION OF AN ELEMENT
Mass number A Symbol
X of the
Atomic number Z element
Terms associated with elements:
● Atomic Number (Z): = No. of protons
Electron = Z–C (charge on atom)
● Mass number (A) = Total number of neutron and proton present
A = Number of proton + Number of Neutrons
● Isotopes: Same atomic number but different mass number
Example: 6C12, 6C13, 6C14
● Isobars: Same mass number but different atomic number
Example: 1H3, 2He3
● Isodiaphers: Same difference of number of Neutrons & protons
Example: 5B11, 6C13
7
