CONSTITUENTS OF AN ATOM
The atom as explained earlier is made up of three sub-particles.
i. Protons,
ii. Electrons and
iii. Neutrons.
the Protons are positively charged.
the Electrons are negatively charged while
the Neutrons have no charge.
S/n sub-atomic particles charge mass
1. Protons + (positive) 1 (unit)
2. Electrons - (negative) 0.00005
3. Neutrons 0 (neutral) 1 (unit)
Atomic number: - Of an element is the number of protons in the nucleus of the atom of the element.
atomic numbers are whole numbers and NO two elements in the world have the same atomic number.
Mass number or atomic mass: - of an element is the sum of the protons and neutrons in the nucleus of its atom.
that is,
Atomic Mass = Number of protons + Number of neutrons
In chemistry an element X can be represented as A
ZX
where A = Atomic mass or mass number
Z = Atomic number
20 Ca atomic no = 20 (number of protons)
Elements Num. of P Num. of E Num. of N
40
1. 20 Ca 20 20 20
14
2. 7 N 7 7 14
39
3. 19 K 19 19 20
Examples
1. State the constituents of an atom.
2. What is the number of protons, neutrons and electrons in the following elements
11 12 23 32
(a) 5B (b) 6C (c) 11Na (d) 16S
ISOTOPES
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 37Cl
Carbon- 12 C 13 C and 14 C6
Potassium – 39K19 and 41K19
Oxygen - 16O16 and 18O16
EVALUATION
1. Define isotopy.
2. Write the isotopes of chlorine.
CALCULATION OF RELATIVE ATOMIC MASS
RELATIVE ABUNDANCE: this is the amount that a particular isotope occurs in nature, and it is expressed in percentage.
For example, Chlorine always occurs in two isotopic mixtures 35Cl and 37Cl with percentages of 75% and 25% respectively. This percentages are known as the relative abundance of each isotope
The following is an example of calculation of relative atomic mass of an element from percentage abundance of its isotopes.
1. X is an element which exists as an isotopic mixture containing 90% of 39X19 and 10% of 41X19
a. How many neutrons are present in the isotope 41X
b. Calculate the mean relative atomic mass of X
Solution
a. Neutrons in 41X19 : mass number - atomic number
= 41-19 = 22
b. R.A.M = 90 x 39 + 10 x 41
100 100
= 90 x 39 + 41 x 10
100
= 3920 = 39.20
100
1. How many neutrons are present in the isotope 37Cl17 ?
2. A given quantity of chlorine contains 75% 35Cl17, and 25% 37Cl17, determine the relative atomic mass of chlorine.
Example 2.
- The following are more examples on 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?
SOLUTION
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
= 39.4
3. 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?
SOLUTION
16.2 = 90 x 16 + 10 x m
100
16.2 = 9
x 16 + m
10 10
16.2 = 144 +
m
10 10
16.2 = 144 + m
10
16.2 x 10 = 144 + m
162 = 144 + m
162 – 144 = m
18 = m
The value of m is 18
OBJECTIVE QUESTIONS
1. 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
(d) the number of atoms in an element
2. 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
(d) the smallest particle of a substance that can take part in a chemical reaction
3. The mass number is
(a) proton number + neutron number
(b) electron number + proton number
(c) neutron number + electron number
(d) atomic number + electron number
4. 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
(d) 109
5. An element X has two isotopes 18.8X 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
(d) 17
THEORY
1. Consider the atoms represented below:
qX rX
p p
a. State the relationship between the two atoms.
b. What is the difference between them?
c. Give two examples of other elements which exhibit the phenomenon illustrated. (waec)
2. State the number of electrons, protons and neutrons present in the following atoms/ions
a) Ca b) S2- c) Al3+ d) P
b). If an element R has isotopes 60% 12R6 and 40% xR6 and the relative atomic mass of R is 12.4, find x.
3.(a) Define the term isotopy.
(b) Determine the number of electrons, protons and neutrons in each of the following: 39K19, 63.5Cu29
c) If an element R has isotopes 60% of 12R6 and 40% xR6 and the relative atomic mass is 12.4, find x.
4. How many neutrons are present in the isotope 37Cl17 ?
5. A given quantity of chlorine contains 75% 35Cl17, and 25% 37Cl17, determine the relative atomic mass of chlorine.
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