TOPIC 3
ATOMIC MASS AND THE CONCEPT OF
MOLECULES
By:
Maisara Shahrom binti Raja Shahrom
Chemistry Lecturer
School of Allied Health Sciences
City University College of Science and Technology
RELATIVE ATOMIC MASS (R.A.M)

Relative mass of an atom =
Mass of an atom of the element
1/12 of the mass of one atom carbon-12

Why carbon-12 is used as a standard?

Its mass can be more easily measured with a mass
spectrometer
Mass of an atom of the element
1/12 of the mass of one atom carbon-12
Eg.
RAM of Mg =

RAM of Pb =
24
1/12 x 12
207
1/12 x 12
= 24
= 207
RELATIVE MOLECULAR MASS (R.M.M)

A molecule is made up of two or more atoms

Calculation of RMM
1.
2.
3.
Determined the molecular formula
Find the RAM of each element in the molecule
Add up of all the RAM of the element
Molecule
Molecular formula
RMM
Chlorine
Cl2
35.5 x 2 = 72
Nitrogen
N2
14 x 2 = 28
Ammonia
NH3
14 + (1 x 3) = 17
Ethanol
C2H5OH
Carbon
dioxide
CO2
(12 x 2) + (1 x 5) + 16 + 1 =
46
12 + (16 x 2) = 44
LET’S TRY!!
Find Relative molecular mass for:
(RAM: Cu=64 ; O=16 ; C=12 ; H=1 ; S=32 ; Cl=35.5 ;
Na=23)
i.
CuO
ii.
CH4
iii. SO2
iv. HCl
v.
Na2CO3
ANSWER
Find Relative molecular mass for:
(RAM: Cu=64 ; O=16 ; C=12 ; H=1 ; S=32 ; Cl=35.5 ;
Na=23)
i.
CuO = 80
ii.
CH4 = 16
iii. SO2 = 64
iv. HCl = 36.5
v.
Na2CO3 = 106
THE MOLE AND THE NUMBER OF
PARTICLES



In our daily lives, we need to count the amount of
objects
Pairs and dozen are examples of units that we used
In chemistry, we use the unit of mole to measure the
amount of substances.
1 pair of shoes = 2 shoes
1 dozen of eggs = 12 eggs
1 mole of carbon = 6.02 x 1023 atoms


1 mole of any element contains 6.02 x 1023 atoms.
The value of 6.02 x 1023 atoms is called as
Avogadro’s constant
AVOGADRO’S CONSTANT = 6.02 x 1023

Key:
1 mole of atom = 6.02 x 1023 particles
THE MOLE AND THE MASS OF
SUBSTANCE

Number of mole atoms
= Mass
RAM
MASS
MOLE
R.A.M
Problem solving
Determined the moles
How many moles of matter are there in:
1.
5 g of nitrogen, N
2.
10 g of sodium, Na
3.
80 g of carbon, C
(RAM: N=14 ; Na=23 ; C=12)
MASS
MOLE
R.A.M
Solution:
1.
2.
5 g of nitrogen, N
5 /14 =0.357 moles
10 g of sodium, Na
10/23 = 0.435 moles
MASS
3.
80 g of carbon, C
80/12 = 6.67 moles
MOLE
R.A.M
Problem solving
Determined the mass
How many grams of matter are there in:
1.
2 moles of nitrogen, N
2.
0.5 moles of sodium, Na
3.
6.0 moles of carbon, C
(RAM: N=14 ; Na=23 ; C=12)
MASS
MOLE R.A.M
Solution:
1.
2 moles of nitrogen, N
2 x 14 = 28 g
2.
3.
0.5 moles of sodium, Na
0.5 x 23 = 11.5 g
6.0 moles of carbon, C
6.0 x 12 =72 g
(RAM: N=14 ; Na=23 ; C=12)
MASS
MOLE
R.A.M
Determined The Number Of Particles
Number of particles = moles x 6.02 x 1023

1.
How many atoms are there in:
0.5 moles of carbon
0.5 x 6.02 x 1023 = 3.01 x 1023 atoms
2.
1.2 moles of sodium
1.2 x 6.02 x 1023 = 7.224 x 1023 atoms
Determined The Number Of Moles From
The Number Of Atoms
Number of moles of atoms = Number of atoms
6.02 x 1023

1.
How many moles are there in:
12.04 x 1023 atoms of chlorine
12.04 x 1023 / 6.02 x 1023 = 2 moles
2.
1.02 x 1046 atoms of sodium
1.02 x 1046 / 6.02 x 1023 = 1.69 x 1023 moles
÷ RAM
MASS
× NA
MOLE
× RAM
NUMBER OF PARTICLES
÷ NA
Number Avogadro (NA) = 6.02 x 1023 particles
CHEMICAL FORMULA




Cation : +ve charge
eg: Na+, Mg2+, Al3+…
Anion : -ve charge
eg: Cl-, O2-, N3-….
If the cation Xm+ and the anion is Yn-, then the formula
of the compound is XnYm
If m=n, then the formula XY
m+
X
nY
Xn
Ym
= XnYm
Example:
Chemical compound
Cation
Anion
Iron (II) chloride
Iron (III) chloride
Copper (I) sulphate
Copper (II) sulphate
Fe2+
Fe3+
Cu+
Cu2+
ClClSO2-4
SO2-4
Chemical
formula
FeCl2
FeCl3
Cu2SO4
CuSO4
Manganese (II) oxide
Manganese (III) oxide
Manganese (IV) oxide
Mn2+
Mn3+
Mn4+
O2O2O2-
MnO
Mn2O3
MnO2
Empirical formula

1.
2.
3.
4.
Step
Write the mass of element
Calculate the moles
Calculate the simplest mole ratio by dividing the
each number with the smallest number
Write the empirical formula
Example



Determine the empirical formula of this compound
Element
Ca
Cl
Mass
1.6 g
4.26 g
RAM
40
35.5
Element
Ca
Cl
Mass
1.6 g
4.26 g
Moles
1.6 g / 40 = 0.04
4.26 / 35.5 = 0.12
Simplest ration
0.04 / 0.04 = 1
0.12 / 0.04 = 3
Solution:
So, empirical formula = CaCl3
Molecular formula


Shows the actual number of atoms of elements
that combine to form the compound.
(Empirical formula)n = Relative molecular mass
Eg:
Relative molecular mass of CaCl3 is 293
(CaCl3) n = 293
(40+ (3x35.5)) n = 293
(146.5) n = 293
n=2
So, molecular formula is (CaCl3) n = (CaCl3) 2
= Ca2Cl6

Let’s try
1.
2.
0.19 g of aluminium and 0.79 g atom oxygen are
burnt to form aluminium oxide. What is the
empirical formula of aluminium oxide [RAM:
Al=27 ; O=16)
The emipirical formula of ethene is (CH3)n. Its
molecular formula mass is 30. Calculate the
molecular formula of this compound.
Solution
1. Element
Mass
RAM
Moles
Ratio
x2
Al
0.91 g
27
0.91 / 27 = 0.034
0.034 / 0.034 = 1
2
So, empirical formula is Al2O3
O
0.79 g
16
0.79 / 16 = 0.05
0.05 / 0.034 = 1.5
3
2.
The emipirical formula of ethene is (CH3)n. Its
molecular formula mass is 30. Calculate the
molecular formula of this compound.
(CH3)n = 30
(12 + 3)n = 30
(15)n = 30
n=2
So, molecular formula is C2H6
THANK YOU…