Summary of Chemistry Textbook: Section 4.1
–
The Mole Concept and Avogadro’s
Constant

A relative scale
is one in which all measurements are compared to one standard or reference measure 
Relative scales have no units
–
they are merely comparisons of one quantity with another 
Advantage of relative scales is that very large or very small numbers can be compared more easily 
To generate the relative scale of atomic masses, chemists chose the most abundant isotope of the element carbon, the carbon12 isotope (
12
C), relative mass of exactly 12 units 
Reasons for choosing carbon:
Carbon is very cheap and is widely available
It is relatively easy to isolate and purify this isotope
Carbon is not toxic in any way 
Mass of 12 units mirrors the mass number of the isotope 
Relative atomic mass will closely parallel the number of these fundamental particles in the nucleus of the element 
Thus, the lightest of all elements, hydrogen, will have a relative atomic mass of close to 1 
A mole
is defined as the amount of substance that contains as many elementary particles as there are atoms in precisely 12 grams of the carbon12 isotope 
Actual number of atoms determined as approximately 6.02 x 10
23

Number is known as
Avogadro’s constant
, (symbol,
L
) 
It is important to recognise that a mole is simply a number, just as a dozen is equal to 12 or a gross is equal to 144
L
= 6.02 x 10
23
mol
1

Atoms are so small that it is impossible to see an individual atom
Calculating numbers of particles

If you are asked to calculate a
number
of particles, the you are calculating the quantity represented by
N
and you will obtain a number with no units 
Asked to calculate an
amount
of substance, then you are trying to find a
number
of mole
of the substance 
This is the quantity represented by
n
and wi
ll have the units ‘mol’
Where n = number of mole (amount), N = number of particles and L = Avogadro’s constant
SECTION 4.1 EXERCISES 1.
Calculate the number of mole of the stated particles in 1.00 mole of sulphuric acid. a)
Atoms of sulphur
–
1
b)
Atoms of hydrogen
–
2
c)
Atoms of oxygen
–
4
d)
Total number of atoms
–
7
2.
Calculate the number of mole of the stated particles in each of the following. a)
Magnesium ions in 1.25 mole of magnesium nitrate
–
1.25
b)
Sodium ions in 3.25 mole of sodium sulfate
–
2 x 3.25 = 6.50 mole of sodium ions
3.
Calculate the number of mole of the states particles in each of the following. a)
Carbon atoms in 2.00 mole of propane
–
2.00 x 3 = 6 mole of carbon atoms
b)
Carbon and hydrogen atoms in 1.25 mole of pentane
–
1.25 x 17 = 21.25
4.
Determine the number of particles, present in the following: a)
3.50 mol of ammonia molecules
–
n = 3.50 L = 6.02 x 10
23
Therefore; 3.50 x 6.02 x 10
23
= 2.10
24
b)
3.22 x 10
3
mol of manganese nitrate
n = 3.22 x 10
3
L = 6.02 x 10
23
Therefore; 3.22 x 10
3
x 6.02 x 10
23
= 1.94 x 10
21
5.
Calculate the amount of substance in mol present in each of the following. a)
1.44 x 10
24
atoms of neon
= 2.39 mol of atoms
b)
1000 atoms of mercury
= 1.661 x 10
21
mol of atoms