CONSTITUENTS OF AN ATOM
Atoms are made up of sub-particles. Protons, electrons and neutrons. Proton has a positive charge, electron has a negative charge and neutron has no charge.
Atomic number and mass number
The atomic number of an element is the number of protons in the nucleus of its atom.
Mass number or atomic mass of an element is the sum of the number of protons and neutrons in the nucleus of its atom.
Mass Number = Number of proton + Number of neutron
An element X can be represented as
where A= Atomic mass or mass number
Z= Atomic number
e.g. 40 mass no = 40
- atomic no = 20
- State the constituents of an atom.
- What is the number of proton in the following elements?
(a) 11B (b) 12C
Isotopy is the occurrence of atoms of elements having the same atomic number but different mass numbers. This is due to the difference in the number of neutrons present in the atoms. The atoms that exhibit isotopy are called ISOTOPES.
Examples of atoms that exhibit isotopy are chlorine 35Cl and 37 Cl
Carbon- 12 C, 13 C and 14 C
Potassium – 39K19 and 41K19
Oxygen – 16O16 and 18O16
- Define isotopy.
- Write the isotopes of chlorine.
CALCULATION OF RELATIVE ATOMIC MASS
The following is an example of calculation of relative atomic mass of an element from percentage abundance of its isotopes.
- X is an element which exists as an isotopic mixture containing 90% of 39X19 and 10% of 41X19
- How many neutrons are present in the isotope 41X
- Calculate the mean relative atomic mass of X
- Neutrons in 41X19
= 41-19 = 22
- R.A.M = 90 x 39 + 10 x 41
= 90 x 39 + 41 x 10
= 3920 = 39.20
- How many neutrons are present in the isotope 37Cl17 ?
- A given quantity of chlorine contains 75% 35Cl17, and 25% 37Cl17, determine the relative atomic mass of chlorine.
- The following are more examples of calculations of relative atomic masses of elements.
- An element Y exist in two isotopic forms 39Y18 and 40Y18 in the ratio 3:2 respectively. What is the relative atomic mass of the element?
R.A.M of Y = 3 x 39 + 2 x 40
5 1 5 1
= 0.6 x 39 + 0.4 x 40
= 23.4 + 16
- An element with relative atomic mass 16.2 contains two isotopes 16P8 with relative abundance 90% and mP8 with relative abundance 10%. What is the value of m?
16.2 = 90 x 16 + 10 x m
16.2 = 9 x 16 + m
16.2 = 144 + m
16.2 = 144 + m
16.2 x 10 = 144 + m
162 = 144 + m
162 – 144 = m
18 = m
The value of m is 18
- Consider the atoms represented below:
- State the relationship between the two atoms.
- What is the difference between them?
- Give two examples of other elements which exhibit the phenomenon illustrated.
- State the number of electrons, protons and neutrons present in the following atoms/ions
- Ca b) S2- c) Al3+ d) P
- If an element R has isotopes 60% 12R6 and 40% xR6 and the relative atomic mass of R is 12.4, find x.
- The atomic number of an element is precisely
(a) the number of protons in the atom (b) the number of electrons in the atom (c) the number of neutrons in the atom
- An atom can be defined more accurately as (a) the smallest indivisible parts of an element that can take part in a chemical reaction (b) the smallest part of an element that can take part in a chemical reaction (c) a combination of protons, neutrons
- The mass number is (a) proton number + neutron number (b) electron number + proton number (c) neutron number + electron number
- Calculate the relative atomic mass of an element having two isotopes 107 Ag and 109Ag in the ratio 1:1 (a)106 (b)107 (c)108
- An element X has two isotopes 8X and 15.8X in the proportion of 1:9 respectively. Find the relative atomic mass of X (a)16.1 (b)13.6 (c)16.8
- (a) Define the term isotopy.
(b) Determine the number of electrons, protons and neutrons in each of the following: 39K19, 63.5Cu29
- If an element R has isotopes 60% of 12R6 and 40% xR6 and the relative atomic mass is 12.4, find x.
NOMENCLATURE OF CHEMICAL COMPOUNDS
STANDARD SEPARATION TECHNIQUES