Chapter 1
Some Basic Concepts
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 of 12C atom
C → 12
CH4 → 16
It is unitless
,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 Concepts 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
Some Basic Concepts
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 of 12C atom
C → 12
CH4 → 16
It is unitless
,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 Concepts 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
