Chapter 3:
Atom & Molecule
❖All matter was composed of atoms,
uncreatable, indivisible and indestructible~
❖While all atoms of an element were identical,
different elements had atoms of differing property, size
and mass~
❖All compounds were composed of combinations of these
atoms in defined ratio~
❖ Molecules are the smallest particle which retain the
chemical properties of a matter.
❖ Different types of molecules have different properties.
❖ Molecules are always in mobile state.
❖ There must be a gap between particles of molecules.
❑An atom is made of ‘subatomic particles’
❑At the centre of the atom is the ‘nucleus’
❑‘Nucleus’ is made from two kinds of ‘subatomic particles’ :
1.) proton (positive electrical charge)
2.) neutron (no electrical charge )
* Both of them have the same mass!
❑Around the nucleus are ‘subatomic particles’ called ‘electron’
(negative electrical charge)
➢In an atom, number of electron = number of proton
➢So, an atom is electrically neutral~
➢The electrons are arranged in ‘shells’ at different distances from the
nucleus~
shell
➢For example: carbon atom
1) two electrons making an inner shell
2) four electrons making an outer shell
➢Other atoms may have more shells than this~
- the actual mass of an atom basically depends
on the numbers of protons and neutrons in its nucleus.
• (mass of proton 1.672 x 10-27kg ≈ mass of neutron 1.674 x 10-27kg)
• (mass of electron is too too small so can be neglected)
→ Mass for atom H = 0.000000000000000000000000001674 kg!!!
❑ ratio of actual mass of the atom respect to the
mass of a ‘standard’ atom~
❑ at first, hydrogen atom was used as the standard :
→ Mass of H atom = 1.674 x 10-27 kg (the lightest)
→ Mass of C atom = 1.995 x 10-26 kg
H
H
H
H
H
H
H
H
H
H
H
H
C
Ar of C = 12
❑ Later, scientist found out that the ratio using hydrogen
atom as standard is not so accurate ~
❑ Carbon atom is used as standard nowadays~
H
H
H
H
H
H
H
H
H
H
H
H
C
H
Ar of C = 12
𝟏
Ar of C = 1
𝟏𝟐
H
Ar of
O
Na
Atom
Oxygen
Sodium
Relative
Atomic Mass
16
23
𝟏
C =1
𝟏𝟐
Relative Atomic Mass (Ar)：
-The calculated Ar is not the mass of exact atom.
-It is a ratio of actual mass of the atom respect to the
1/12 of the mass of a carbon-12 atom
*(there are many types of carbon atom, carbon-12 is one of it)
-In order to calculate Ar, first is to calculate the
1/12 of the mass of a carbon-12 atom!
• → mass of a carbon-12 atom = 1.993 x 10-26kg
• → 1.993x10-26/12 = 1.661x10-27 kg !!!
- then, compare this value with any other atom and
the obtained ratio is relative atomic mass for that atom
Element
Oxygen
Mass of atom
1/12 of mass of a
carbon-12 atom
Relative Atomic Mass
Ar of Oxygen
= 16
❖ also can be known as “ Formula Mass ”
❖ the sum of all the relative atomic masses of the
atoms in a molecule or compound~
Mr = Total sum of Ar for all the atoms in a formula
➢Can be divided into three types：
i. Calculate Mr of a specific substance
For example:
1. Calculate the relative molecular mass of NaCl.
(Ar : Na = 23, Cl = 35.5)
2. Calculate the relative molecular mass of CaF2.
(Ar : Ca = 40, F = 19)
ii. Calculate the mass ratio of each element in a substance
For example: (Ar : H = 1, C = 12, N = 14, O = 16)
1. Calculate the mass ratio of carbon & oxygen in CO2
2.Calculate the mass of nitrogen & hydrogen in ratio for
NH3.
iii Calculate the mass percentage of certain
element in a substance
Example: (Ar : H = 1, C = 12, O = 16)
1.Calculate the mass percentage of carbon in CH4
2.Calculate the mass of oxygen in percentage for CO2
❖In chemistry, the unit ‘mole’ is used to represent the
amount of substance containing 6.02 x 10 23 particles~
❖ 1 mol = 602 x 10 21
= 6.02 x 10 23
❖ For example :
-1 mol of sodium atoms
-2 mol of oxygen molecules
❖The NA is defined as the number of particles in
one mole of a substance~
❖NA = 6.02 x 1023 mol-1
( means there are 6.02 x 1023 particles per mole)
❖Particles in the substances can be atoms, molecules or
ions~
6.02 x 1023 = 六千零二十万亿亿个
O
C
H
Relative atomic
Mass, Ar
Assumption
Real
1
12
16
1 billion of H atom
= 0.02g
1 billion of C atoms
=0.02 x 12
=0.24 g
1 billion of O atoms
=0.02 x 16
=0.32 g
6.02 x 1023 of H
atoms
=1g
6.02 x 1023 of C atoms 6.02 x 1023 of O atoms
= 12 g
= 16 g
❖ 1 atom of carbon-12 = 1.993 x 10-23g
❖ 1 mole of carbon-12 = 6.02 x 1023 atoms
❖ Mass of 1 mole carbon-12
= (1.993 x 10-23g) x (6.02 x 1023) = 11.99 ≈ 12g
❖ So, the mass for 1 mole of carbon-12 atoms is 12g ;
2 mole of carbon-12 atoms is 24g ,
etc…
❑ The mass of one mole of the substance in grams.
❑ It has a unit of gram per mole (g mol-1)
❑ For example,
1 mol of carbon atom = 12g
2 mol of carbon atom = 24g
2.5 mol of carbon atom = 30 g
(So, the molar mass for carbon atom is 12g mol-1)
❑ The molar mass of any substance is numerically equal
to its relative atomic, molecular or formula mass~
(Given Ar : H = 1, He = 4, C = 12, Mg = 24)
63.5
63.5
31.75
❑ The volume of one mole of any gas~~~
-At STP (OoC,1 atm) → 22.4 L mol-1 ;
-At room temperature (~25oC,1 atm) → 24 L mol-1
1 L = 1000ml
1ml = 1cm3
1L = 1 dm3
❑ A chemical formula tells us the number of atoms of each
element in a compound.
❑ It contains the symbols of the atoms of the elements present
in the compound as well as how many there are for each
element in the form of subscripts.
❑ There are three types of chemical formula.
❑ Sometimes known as the true formula, shows
the actual number of the different elements in one molecule
of a compound.
❑ Each element is written as their symbols in the periodic table,
and the number of atoms for each element is shown by the
subscript (the small no. to the lower right of element)
❑ The simplest ratio of whole numbers of elements that make up
a compound.
❑ This type of formula is derived from experimental data.
❑ The chemical formula for all the ionic compounds is actually
their empirical formula.
❑ For example: NaCl is empirical formula for sodium chloride~
❑ NaCl does not exist as a molecule, it is composed of many Na+ and
Cl- ions arranged into a structure~
❑ This NaCl formula only indicates that it is made of an equal
sodium and chloride ions. Its ratio of cation and anion is 1:1~~~
2 Chemical Formulas:
1 Chemical Formula:
Empirical Formula =
NaCl
Molecular Formula =
N2O4
Empirical Formula =
NO2
❑ Both the actual number of atoms of elements in a compound,
how the atoms are arranged and which atoms are bonded to
one another are shown.
❑There are three ways in total:
A) Using the valency of elements
- the valency (化合价) of an element is
a measure of its combining power with
other atoms when it forms chemical compounds.
3+
Al 2
2-
O3
Al2O3
+1
Na 1
-1
Cl 1
Ca1
-1
Cl 2
CaCl2
Mg 2
-2
O2
MgO
NaCl
+2
+2
+3
Al 2
-2
O3
Al2O3
+1
Na 1
-1
NO3 1
NaNO3
+2
Mg 1
-1
NO3 2
Mg(NO3)2
+1
NH42
-2
SO4 1
(NH4)2SO4
+2
Mg 1
-1
NO3 2
Mg(NO3)2
Metal
Element
Symbol
Valency
Non-Metal
Element
Potassium
Hydrogen
Sodium
Fluorine
Silver
Oxygen
Calcium
Chlorine
Magnesium
Bromine
Barium
Iodine
Zinc
Nitrogen
Aluminium
Phosphorus
Copper
Sulfur
Iron
Carbon
Manganese
Silicon
Symbol
Valency
Polyatomic Ion
ammonium
hydroxide
nitrate
hydrogen carbonate
sulphate/sulfate
carbonate
phosphate
Symbol
Valency
Common
molecule
H2
O2
N2
F2
Cl2
Br2
I2
Name
hydrogen
oxygen
nitrogen
fluorine
chlorine
bromine
iodine
Common
molecule
NH3
O3
CO
CO2
HCl
HNO3
H2SO4
Name
ammonia
ozone
carbon monoxide
carbon dioxide
hydrochloric acid
nitric acid
sulfuric acid
B) Using the mass of each element in a substance
Al2O3
C) Using the percentage composition of a substance
Empirical formula: H3PO4
❑ The formula we get through the ways above
is mostly empirical formula.
❑ Molecular formula of a compound can be
determined if the following data is known:
a) Its empirical formula
b) Its relative molecular mass Mr
Mr of Molecular formula = n (Mr of Empirical formula)
*n = positive integer!
❑8.5g of hydrogen peroxide contains 0.5g of
hydrogen. If the Mr of hydrogen peroxide is 34.
Find its molecular formula. (Ar : H = 1, O = 16)
❖Two different chemicals can react and form a new compound~
❖The total mass of the reactants is the same as
the total mass of the products~
➔Matter is neither created nor destroyed during a chemical reaction~
H
H
Cl Cl
H
Cl
H
Cl
❖ Regardless of the amount, the elements in a pure compound are
always the same and having fixed proportion, no matter how the
compound is made~
❖ Compounds with the same mass proportions must be the
same compound~
❖ For example:
The mass percentage of carbon in CO2
will be always 27.27% !!!
If you have 100g of CO2 , 27.27g is carbon…
If you have 200g of CO2 , 54.55g is carbon…
Example 1
Cadmium sulphide is a yellow compound that is
used as a pigment in artist’s oil colors.
A sample of this compound is composed of
1.25g of cadmium and 0.357g of sulphur.
If a second sample of the same compound
contains 3.50g of sulphur,
how many grams of cadmium does it contain?
Example 2
A sample of chloroform is found to contain
12.0g of carbon, 106.4g of chlorine and 1.01g
of hydrogen.
If a second sample of chloroform is found to
contain 30.0 g of carbon,
how many g of chlorine and hydrogen does it
contain?
➢ If two or more different compounds are composed of
the same two elements,
then the ratio of the masses of the second element
combined with a certain mass of the first elements is
always a ratio of small whole numbers.
Example: H2O and H2O2 , SO2 and SO3 , CO and CO2
Example: nitrogen react with oxygen to form
→
If,
mass of 1 N
= 14
mass of 1 O
= 16
1:2
Exercise:
Which of the following statements indicates that law of multiple proportion
is being followed?
A.Sample of carbon dioxide taken from any source will always have carbon
and oxygen in the ratio 1:2
B.Carbon forms two oxides namely CO2 and CO, where masses of oxygen
which combine with fixed mass of carbon are in the simple ratio 2:1
C.when magnesium burns in oxygen, the amount of magnesium taken for
the reaction is equal to the amount of magnesium in magnesium oxide
formed
D.At constant temperature and pressure 200ml of hydrogen will combine
with 100ml oxygen to produce 200ml of water vapour
Exercise:
Which of the following groups of substances verify the Law of Multiple
Proportions?
A.S2, S6, S8
B.LiF, NaCl, KBr
C.CH4, CCl4, CBr4
D.C2H6, C3H6, C3H8
Exercise:
1. Compound A and compound B are formed by element X and element Y.
Composition of compound A and B is shown in the table below.
According to law of multiple proportions, state the chemical formula for
compound B if the chemical formula of compound A is XY4 .
A. XY2
2.
B. XY3
Element X
Element Y
Compound A
75%
25%
Compound B
80%
20%
C. X2Y3
D. X3Y2