Transcript Matter

Matter
Classification
Matter
Metals
Metallic Bonds
Giant metallic strcuture
Atoms
Ions
Molecules
Periodic Table
Ionic Bonds
Covalent Bonds
Ionic compounds
Covalent compounds
Giant ionic strcuture
Giant covalent strcuture Simple molecular structure
Semi-metals
or Metalloids
Non-metals
van der Waals' forces
Atoms
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Z is the atomic number
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A
Z X
= number of protons
= number of electrons
A is the mass number
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= no. of protons + no. of neutrons
Atoms
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atomic number of Cl = 17
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35
17 Cl
= 17 protons
= 17 electrons
mass number = 35
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= no. of protons + no. of neutrons
= 17 + no. of neutrons
 no. of neutrons = 35 – 17 = 18
Electronic arrangement and
electronic diagram
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35
37
Electronic arrangement of
or
Cl
17
17 Cl
= 2, 8, 7
Electronic diagram
Cl
Isotopes
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35
17 Cl
37
17 Cl
Isotopes are atoms of the same element with equal
number of protons but different number of neutrons.
OR Isotopes are atoms of the same element with
equal atomic number but different mass number.
Isotopes have similar chemical properties because
they have the same number of outermost shell
electrons.
Isotopes have different physical properties because
their atomic masses are different.
Atomic mass

The actual mass of one
35
=
g
35
17 Cl
atom
6.02  1023
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It is meaningless to remember such value. Moreover,
it is found that mass of an electron is just 1 of the
1840
mass of a proton.
As a result, the mass of an atom is mainly due to
mass of nucleus, i.e. total mass of protons and
neutrons.
Relative isotopic mass
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12
Scientist assigned the mass of an C atom
= 12.000 atomic mass unit (a.m.u.)
Carbon-12 consists of 6 protons and 6
neutrons in the nucleus, 12 sub-atomic
particles totally.
As a result, relative mass of proton = relative
mass of neutron = 1 a.m.u.
Relative isotopic mass

Relative isotopic mass
= 1
12
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mass of
 mass of
isotope
carbon  12
79
Relative isotopic mass of 35
Br
= 79 a.m.u. (79 sub-atomic particles in the nucleus)
Relative isotopic mass
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Elements contain isotopes and each isotope has its
own abundance.
Usually, we mention the names of reagents required
in a reaction without specifying which isotopes.
E.g. 23g of sodium burns in chlorine to form sodium
chloride. What is the mass of chlorine gas required
for the complete reaction?
As a result, we have to find the weighted average of
chlorine isotopes.
Relative atomic mass
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Due to the presence of isotopes, atomic mass
of an element should not be an integer. It
should be the weighted means of the isotopes.

Relative atomic mass
=
weighted
average of
the mass of
1
 mass of
12
carbon  12
isotope
Isotopes
35
17 Cl
37
17 Cl
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Relative abundance = 75% : 25%
= 3 : 1
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Relative atomic mass of Cl
= 35 x 75% + 37 x 25%
= 35.5 a.m.u.
Isotopes
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69
Ga
71
Ga
If the relative atomic mass of Ga = 69.8, find
the relative abundance of each isotope.
69
71
Ga
Let the relative abundance of
and Ga
be x% and (100 – x)% respectively
69(x%) + 71(100-x)% = 69.8
x = 60
Isotopes
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Ga has two isotopes 69Ga and xGa . If the
relative abundance of 69Ga is 60% and the
relative atomic mass of Ga = 69.8. Find the
value of x.
69(60%) + x(40)% = 69.8
x = 71