Download COMPOUNDS Chapter 2 : Preliminary course Chemistry 2

COMPOUNDS
Chapter 2 : Preliminary course
Summary
What is:
Law of Constant proportions?
Law of Multiple proportions?
Key Points
Key Terms
CLASS NOTES
What is a compound?
Pure substance - has definite physical
properties (eg. MP, BP)
Can be decomposed into elements or
simpler compounds (eg. )
Are composed of elements combined
in fixed ratios:

of numbers of atoms

of weights
For example NH3 (ammonia)
1 nitrogen : 3 hydrogen (by number)
14:3 (by weight)
Keeping the weight of one element
fixed, the ratios of the other elements,
simple whole number ratios.
Compound
Number
Weight
NO
1:1
1:1.14
NO2
1:2
1:2.29
N2O
2:1
1:0.57
N2O4
1:1
1:2.29
Keeping N fixed and comparing masses of
O in different compounds
1.14
: 2.29 : 0.57 : 2.29 =
2
:4
:1
:4
This is the law of multiple proportions
A compound always has the same
properties, regardless where found
A compound is made up of either:

the same discrete molecules of same size, do not join, eg.
CO2 is a discrete molecule

Crystal lattice - all joined, can
be of different sizes, eg. salt
crystals are a crystal lattice, as
is diamond.
Law of Constant Proportions
when compared with each other are in
The law of constant (fixed) proportions
states that the same compound,
however formed, contains the same
Chemistry 2: Compounds Page 1 Sat 12 February 2000
elements chemically combined in a
constant proportion by weight.
Example: Water
Always made from same elements

the elements are always the
same weight
2 : 16
1:8
Compounds
are
consistent
everywhere.
Always the same,
regardless of how they are formed
NO2
14 : 32
1 : 2.29
0.57 : 1.14 : 1.7 : 2.29
1
:2
:3
:4
When the weight of one element is
fixed, in different compounds, the
other element is present in a simple
whole number ratio.
Law of Multiple proportions
The law of multiple proportions states
that when two elements combine to
form more than one compound, the
weights of one element that combine
with a fixed weight of the other are in a
simple whole number ratio.
Example:
N20
NO
N2O3
By weight
28 : 16
14 : 16
28 : 48
1 : 0.57
1 : 1.14
1 : 1.7
2.4 Properties of compounds and
their component elements: A
comparison
.......(Basically means that a compound
can be very different from the
elements that it is made up from.)
Example 1: Copper with Sulphur
Copper Sulfide
Cu(s) + 2HNO3(aq) --> NO2(g) + H2O
Brown gas
Valency
Metalloid
4
4±
Valency is the measure of the ability of
an element to combine with other
elements to form compounds i.e.
combining power.
Non-meta
ls
5
3-
6
2-
7
1-
8
0
Inert gas
Valency is directly related to the
number of outer shell electrons:
No. of outer
Type
Valency
shell electrons
Metals
1
1+
2
2+
3
3+
Chemistry 2: Compounds Page 2 Sat 12 February 2000
Metals have positive valencies
Non-metals have negative valencies
(as a general rule)
Radicals have valencies that equal the
sum of the valencies of the component
atoms. (generally)
= 20.2
example: ammonium ion
NH4+
made up of N3- and 4 x H+
Some elements have only one isotope.
Some elements have more than one
valency.
Example: Iron
Fe2+ Fe3+
Copper
Cu+ Cu2+
To
distinguish
between
different
valencies of an element we use
Roman numerals.
Example:
FeCl2
FeCl3
Iron(II) Chloride
Iron(III)Chloride
Isotopes
Atomic number (Z) is the number of
protons in the nucleus.
Mass number (A) is the number of
protons and neutrons in the nucleus.
A
zX
Example:
12 C
6
1 H
1
23 Na
11
Number of neutrons N = A - Z
The atomic weights on the periodic
table
represent the average weights of all
the isotopes of the element, based on
their relative abundance.
Example: Neon has two isotopes
Ne-20 @ 90%, Ne-22 @ 10%
Average weight = (0.9 x 20) + (0.1 x
22)
Chemistry 2: Compounds Page 3 Sat 12 February 2000
Relative atomic mass (Ar)
A more modern (and more correct)
name for atomic weight is relative
atomic mass.
Relative mass is more important to
chemists that actual mass e.g. it is
more important to know that a sulphur
atom is 32x the mass of a Hydrogen
atom than to know that the actual
mass is 5.31 x 10-27 kg.
The standard element against which
the mass of all atoms is compared is
carbon-12 . 1 atom = 12 atomic mass
units.
Empirical Formula and Molecular
Formula
Empirical formula shows the simplest
ratio of atoms in a substance
Molecular formula shows the actual
number of atoms in the molecule.
Example:
Molecular formula
Empirical formula
H2O
H2O
H2O2
HO
O3
O
C6H6
CH
Relative Molecular Mass (Mr)
(Mass of the molecule)
The relative molecular mass is the
sum of the relative atomic masses of
all the elements present in a molecule.
Example: Mr H2O = 2 + 16 = 18 amu
1. Find mass of E.F.
N = 14, H2 = 2
14 + 2 = 16
2. See how many times it goes into the
relative atomic weight (Mr).
Chemistry 2: Compounds Page 4 Sat 12 February 2000
Mr CO2 = 12 + 2(16) = 44 amu
(amu = atomic mass units)
Percentage Composition
If the Empirical formula (EF)
or
molecular formula (MF) of a substance
is known, it is possible to calculate the
percentage composition by mass.
Example: Sulpur Dioxide SO2
Ar S = 32
Ar O = 16
Mr = 32 + 2(16) = 64
%S in SO2 = 32/64 = 50%
%O in SO2 = 32/64 = 50%
Example 2: Sodium Sulphate Na2SO4
Ar Na2 = 2(23) = 46
Ar S = 32
Ar O4 = 4(16) = 64
Total = 142
Na = 46/142 = 32.4%
S = 32/142 = 22.5%
O = 64/142 = 45.1%
If the EF and Mr are known the MF
can be calculated
Example: EF is CH3 and Mr = 30
relative mass of CH3 = 12 + 3 = 15
30 = 15 x 2
Therefore MF is (CH3)2 = C2H6
Example 2: EF is NH2 and Mr = 32
32/16 = 2
Therefore MF is N2H4
Example 3: Three elements are
present in a compound in different
weights. Mg = 16.4%, N = 18.9%, O =
2. Divide by smallest number to work
64.7%. Find the Molecular formula.
1. Divide percentages by relative
atomic weight.
Mg = 24.3
N = 14O = 16
Mg: 16.4/24.3 = 0.67
N:
18.9/14 = 1.35
O:
64.7/16 = 4
out simplest whole number ratio.
0.67 1.35 4
To find Molecular Formula given
percentage weights
1. Divide percentages by
relative atomic mass (Mr).
2. Divide by smallest number to
work out simplest whole number
ratio.
The Mole
Chemists have defined an entirely
different and distinct physical quantity
called amount of substance. The unit
chosen for this quantity is the mole.
0.67
=1:2:6
Therefore
Molecular
formula
is
MgN2O6
1 mol of an element can be found by
looking at the Ar.
Example:
1 mol of sodium = 23 g (Mr Na = 23)
1 mol of zinc = 65.4 g (Mr Zn = 65.4)
2 mol of sodium = 2 x 23 = 46 g
Amount of substance (n)
Amount of substance =
mass of substance(m) / molar mass
(M) (Molar mass = Mr - for elements)
n=
m
M
A mole is the amount of substance
that contains as many elementary
entities as 12 g of Carbon-12.
Elementary entities include atoms,
molecules, ions and electrons.
m = mass of substance
M = molar mass (Mr)
The symbol for mole is mol.
Amount of substance is represented
by the symbol n.
= 1.74 mol
Example 2: Calculate the amount of
substance in 20 g of calcium
carbonate, CaCO3?
1 mol of all substance contains 6.02 x
1023 particles. This is known as
Avogadro’s number.
Mr CaCO3 = 40 + 12 + 3(16)
= 100 g/mol
n = 20 g / 100 g/mol
= 0.2 mol
Chemistry 2: Compounds Page 5 Sat 12 February 2000
Example 1: 40 g of Vanadium, how
many mol?
n = 40 g / 23
Example 3: Amount of substance in
To
Molecular
formula
using
4.0 g He.
percentage weight,
percentage by the Ar.
% on top
divide
each
Percentage Composition
To find percentage composition divide
weights of each atom by total weight.
Example:




Elements and compounds have
fixed physical and chemical
properties. They are pure
substances.
compounds are made of 2 or
more elements combined in
fixed mass ratios and are
ionicly or covalently bonded
together.
Mixtures
are
not
pure
substances they aren’t bonded
together and are present in
varying ratios in the mixture.
Metals bonding - Metals
cannot bond together because
they both have small numbers
of outer shell electrons and
hence a positive valency
Chemistry 2: Compounds Page 6 Sat 12 February 2000
find
Ar on bottom
Elements and Compounds
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