Back to: CHEMISTRY SS1
Welcome to class!
In today’s class, we will be talking about atomic number, mass number, isotopes and calculations. Enjoy the class!
ATOMIC NUMBER, MASS NUMBER, ISOTOPES AND CALCULATIONS
CONSTITUENTS OF AN ATOM
Atoms are made up of sub-particles. Protons, electrons and neutrons. Proton has a positive charge, an 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
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 – 12C, 13Cl and 13Cl
Potassium – 39K 19 and 41K19
Oxygen – 16O16 and 18O16
EVALUATION
- Define isotopy.
- Write the isotopes of chlorine.
CALCULATION OF RELATIVE ATOMIC MASS
The following is an example of calculation of the 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
Solution
a.) Neutrons in 41X19
= 41-19
= 22
b.) R.A.M =
= (90 x 39 + 41 x 10)/ 100
= 3920/ 100 = 39.20
EVALUATION
- 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.
CALCULATIONS
- 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?
SOLUTION
2. 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)/10 + (m/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
GENERAL EVALUATION/REVISION
- 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.
2. State the number of electrons, protons and neutrons present in the following atoms/ions
a.) Ca b) S2- c) Al3+ d) P
3. If an element R has isotopes 60% 12R6 and 40% xR6 and the relative atomic mass of R is 12.4, find x.
READING ASSIGNMENT
- New School Chemistry for SSS by O.Y Ababio. Pg 48-49
WEEKEND ASSIGNMENT
- 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
THEORY
- (a) Define the term isotope. (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.
In our next class, we will be talking about the Structure of the Atom: Orbitals and Electronic Structure of the Atom. We hope you enjoyed the class.
Should you have any further question, feel free to ask in the comment section below and trust us to respond as soon as possible.
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Thanks for the lecture. But for the weekend assignment question 4&5. I could not find the answer to it.
Answer to number 4
RAM = 1/2 × 107 + 1/2 × 109
RAM = 53.5 + 54.5
RAM =108
how did you get your answer
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Pls under calculation. Example 1. Where did 5 come from?
They added both ratios together to get 5
5 was gotten from the sum of the ratio,I.e 2+3=5
I need a solving for molecular mass
Thanks for the lecture but the solution one is kind of confusing.
Hi Amarachi,
Kindly provide in details how we can help further.
Thanks.
Number 3 objective and number 5 theory I can’t get the answer
Ok
i don’t understand number on in the theory the calculation, please kindly teach me how to solve it
please don’t you use essential textbook