SYMBOLS AND FORMULAE - II
Week: Date: Time:
Period:
Duration: Average
age of learners:
Subject: CHEMISTRY
Class: SS ONE
Topic: SYMBOLS AND FORMULAE - I
Sub topic: Empirical and Molecular
formulae
Reference materials:
(1) ESSENTIAL CHEMISTRY, TONALD PUBLISHERS, I. O ODESINA
(2) NEW SCHOOL CHEMISTRY, AFRICAN FIRST PUBLISHERS, OSEI YAW
ABABIO
(3) INTERNET
Instructional materials:
Entry behavior: The students have
been familiar with some common element.
Behavioural objective: At
the end of the lesson the students should be able to:
Content
Empirical formula and Molecular formulae
The empirical
formula of a compound is the simplest whole number ratio of each type of atom
in a compound. It can be the same as the compound’s molecular formula - but not
always. An empirical formula can be calculated from information about the mass
of each element in a compound or from the percentage composition.
The molecular
formula of
a substance is the actual number of each type of atom in one unit of the
substance. For example, the molecular formula of ethane is C2H6 because
each ethane molecule contains 2 carbon atoms and 6 hydrogen atoms.
An empirical
formula is
the simplest
whole number ratio of
each type of atom in a compound. For example, the empirical formula of ethane
is CH3because
both numbers in its molecular formula can be simplified by dividing by 2.
The empirical formula and molecular formula
can be the same for some compounds. For example, they are the same for carbon
dioxide CO2 and
methane CH4 because
the numbers in their molecular formulae are already in their simplest whole
number ratios.
percentage mass
If you have the formula of a compound, you should be
able to work out the percentage
composition by mass of
an element in
it.
For example, the formula for sodium hydroxide
is NaOH. It contains three different elements - Na, O and H. But the percentage
by mass of each element is not simply 33.3%, because each element has a
different relative atomic mass.
Using known masses
You should be able to calculate the percentage
by mass of an element in a compound if you know the masses of the other
elements in it.
Example
12 g of magnesium reacts with oxygen to
produce 20 g of magnesium oxide. What is the percentage by mass of oxygen in
magnesium oxide?
mass of oxygen = 20 – 12 = 8 g
percentage of oxygen = 8/20 × 100
=
800/20
= 40%
You should be able to calculate the empirical formula of
a compound if
you are given the mass of
each element it
contains. Here is an example:
Example
3.2g of sulfur reacts with oxygen to produce
6.4g of sulfur oxide. What is the formula of the oxide?
(Ar of
S = 32 and Ar of
O = 16)
Step |
Action |
Result |
|
1 |
Write the element symbols |
S |
O |
2 |
Write the masses |
3.2 g |
6.4 – 3.2 = 3.2 g |
3 |
Write the Ar values |
32 |
16 |
4 |
Divide mass by Ar |
3.2 ÷ 32 = 0.1 |
3.2 ÷ 16 = 0.2 |
5 |
Divide by the smallest number |
0.1 ÷ 0.1 = 1 |
0.2 ÷ 0.1 = 2 |
6 |
Write the formula |
SO2 |
The action at Step 5 usually gives you the simplest
whole number ratio straightaway.
Sometimes it does not, so you might get 1 and 1.5. In this example, you would
multiply both numbers by 2, giving 2 and 3 (instead of rounding 1.5 up to 2).
You should be able to calculate the empirical
formula of a compound if you are given its percentage
composition by mass.
Presentation of teacher activity:
i. The
teacher defines element and list the first- thirty elements.
ii. The teacher defines a compound with
relevant examples.
iii. The teacher explains mixture and their
types
Students’ activities:
i. The students chorus the definition of
element
ii. The students mention example of compound
iii. The students differentiate between the
types of mixture
Evaluation: The teacher evaluates the lessons
by asking the following questions:-
define element
what are the constituents of calcium hydroxide
and potassium iodide
Briefly explain a mixture
Assignment
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