ATOM AND
MOLECULES:
Atom: Atom is the ultimate
electrically neutral, made up of fundamental particle (Electron, neutron,
Proton) which shows the characteristic properties of the element and exist
freely in a chemical reaction.
Molecules: A molecule is the smallest
particle made up of one or more than one atom in a definite ratio having stable
and independent existence e.g.H2 ,O2 He HCl
, CaCO3 etc.
(1)
Atomic Mass: As
atoms are very tiny particles, their absolute masses are difficult to measure.
However it is possible to determine the relative masses of different atoms if
small unit of mass is taken as standard there is three conventions given as.
(1) This
standard was mass of one atom of hydrogen and taken as unity.
(2) And
also mass of one atom is 1/16th part of oxygen atom
(3) And now
mass of one atom is 1/12th part of C-12 atom.
The atomic
mass of an element can be defined as the number which indicates how many times
the mass of one atom of the element is heavier in comparison to the mass of one
atom of hydrogen.
(2)
Atomic Mass Unit:
(1) The
quantity 1/12Th mass of an atom of carbon-12 is known as the atomic mass unit
and is abbreviated as amu.
(2) The
actual mass of one atom of carbon-12 is –
Determination
of atomic mass:
(1) Applying Dulong and Petit’s
law.
(2) Cannizzaro’s methods
(3) By mitscherlich’s law of
isomorphism.
(4) By measurement of V.D. of
volatile chloride or bromide.
(1) Dulong
& Petits Law: The
product of specific heat of pure element and atomic mass of the element is
equal to 6.4.
NOTE:
(1) But this law is not applicable
to lighter element like boron, carbon, silicon. To obtain correct atomic mass
of element first of all equivalent mass of the element is known by any other
method and their atomic mass = eq. weight ´ valency
(2) In which valency has whole number
value which can be deduced by dividing approximate by equivalent
mass. Dulongs and Petit’s Law: Atomic mass ´ specific
heat = 6.4
ILLUSTRATIVE
EXAMPLE (1): The
specific heat of a metal of atomic mass 32 is likely to be:
(A) 0.25
(B) 0.24
(C) 0.20
(D) 0.1
SOLUTION:
ILLUSTRATIVE
EXAMPLE (2): On
dissolving 6 gm of metal in sulphuric acid, 13.53g of the metal sulphate was
formed. The specific heat of metal is 0.057 Cal/g. What is equivalent mass of
metal, valency and exact atomic mass?
SOLUTION:
(2) Cannizzaro’s
methods:
If an
element has several compound with other same or different elements of known
atomic mass then the compound that has minimum presence of former element
indicate the atomic mass of former element.
Procedure:
(1) First
of all the molecular mass of all compounds known by applying
V.D ´ 2
= mol. weight
(2) By
analysis the presence of the desired element in each compound is known.
(3) The
mass that is lowest among all the compound indicate the atomic mass
of that
element
ILLUSTRATIVE
EXAMPLE (3): Estimate
the atomic mass of nitrogen given that vapour density of NH3 =
8.5, Nitrous oxide = 22, Nitric oxide = 15, Nitrogen peroxide = 23, Nitrogen
trioxide = 38.
SOLUTION:
(3) Law
of Isomorphism
When two or
more compound forms similar type of crystals or able to form mixed crystals,
they are known as isomorphs. For examples: MgSO4.7H2O, ZnSO4.7H2O and FeSO4.7H2O are
isomorphs of each other as their crystals posses same shape.
According to
Mitscherlich [year 1819]. The valency of elements that are similarly placed to
that of other elements in their isomorphs are always same.
In the above
example Fe, Zn and Mg have same valency [2] and equal ratio of water molecule
in each isomorphs.
If
equivalent mass of one element is known then atomic mass can be calculated by
knowing the valency of other isomorphs key element.
ILLUSTRATIVE
EXAMPLE (4): Which
pair of the following substances is said to be isomorphous?
(A) White
vitriol and blue vitriol
(B) Epsom
salt and Glauber’s salt
(C) Blue
vitriol and Glauber’s salt
(D) White
vitriol and Epsom salt
SOLUTION: Epsom salt (MgSO4.7H2O)
and White vitriol (ZnSO4.7H2O) contains divalent cation
Mg2+ and Zn2+ and same number of water
molecules as water of crystallization which hold criteria for isomorphism.
Hence (D) is correct.
(4) Atomic
mass from vapour density of a chloride:
The
following steps are involved in this method
(1) Vapour density of chloride of
the element is determined
(2) Equivalent mass of the element
is determined
(3) Let the valency of the element
be x. The formula of its chloride will be MClX
ILLUSTRATIVE
EXAMPLE (5): One
gram of chloride was found to contain 0.835g of chlorine. Its vapour density is
85. Calculate its molecular formula
SOLUTION:
TRY
YOURSELF:
Exercise
(1): Two oxides
of a metal contain 63.2% and 69.62% of the metal. The specific heat of the
metal is 0.117. What is the formula of the two oxides?
Exercise
(2): 1 g of a
metal which has specific heat of 0.06 combines with oxygen to form 1.08 g of
oxide. What is the mass of M?
Exercise
(3): The
chloride of a solid metallic element contains 57.89% by mass of the element.
The specific heat of the element is 0.0324 cal deg-1g-1.
Calculate the exact atomic mass of the element.
Exercise
(4): White
vitriol (hydrated sulphite) is isomorphous with MgSO4.7H2O.
White vitriol contains 22.95% zinc and 43.9% of water of crystallization. Find
the atomic mass of zinc.
Exercise
(5): Two oxides
of metals A and B are isomorphous. The metal A whose atomic mass is 52, forms a
chloride whose vapour density is 79. The oxide of the metal B contains 47.1%
oxygen. Calculate the atomic mass of B.
ANSWER
KEYS:
Exercise
(1): MO2 and
M2O3
Exercise
(2): Molar mass
of metal 100 g
Exercise
(3): 195.21
Exercise
(4): 65.3
(atomic mass of zinc)
Exercise
(5): Atomic
mass of B = 27 amu
