Lesson Notes By Weeks and Term - Senior Secondary 1

Symbols, formulaes and equations I

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Term: 2nd Term

Week: 4

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:       

Subject:      Chemistry

Topic:-       Symbols, formulaes and equations I

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

  1. State the chemical symbols of elements and their valencies
  2. Derive the empirical and molecular formulae of compounds
  3. State the law of conservation of mass
  4. Solve exercises on the conservation of mass

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

PRESENTATION

TEACHER’S ACTIVITY

STUDENT’S ACTIVITY

STEP 1

INTRODUCTION

The teacher reviews the previous lesson on the sates of mater and kinetic theory

Students pay attention

STEP 2

EXPLANATION

He reviews the first 20 elements on the periodic table and their symbols. He also guides the learners on how to derive the empirical and molecular formulae of compounds and solves some examples

 

Students pay attention and participates

STEP 3

DEMONSTRATION

He states and explains the law of conservation of mass. He solves some exercises on the conservation of mass

Students pay attention and participate

STEP 4

NOTE TAKING

The teacher writes a summarized note on the board

The students copy the note in their books

 

NOTE

SYMBOLS, FORMULAES AND EQUATIONS

Brezelius in 1814 proposed the modern symbols being used on the periodic table. This involved the use of the letter or letters of the name of the respective elements; which could be English name, Latin, Greek, or as obtained. This also involved using capital letter for the first letter of the alphabets for the name of the element or the first letter and any other letter within the letters of the name of the element

 

NAMES OF THE FIRST 30 ELEMENTS ON THE PERIODIC TABLE AND THEIR SYMBOLS

 

 

 

Hydrogen

H

 

 

Helium

He

 

 

 

 

Lithium

Li

 

 

Beryllium

Be

 

 

 

 

Boron

B

 

 

Carbon

C

 

 

 

 

Nitrogen

N

 

 

Oxygen

O

 

 

 

 

Fluorine

F

 

 

Neon

Ne

 

 

 

 

Sodium

Na

 

 

Magnesium

Mg

 

 

 

Aluminum

Al

 

 

Silicon

Si

 

 

 

Phosphorus

P

 

 

Sulphur

S

 

 

 

Chlorine

Cl

 

 

Argon

Ar

 

 

 

Potassium

K

 

 

Calcium

Ca

 

Sc

 

 

 

 

Scandium

 

 

 

 

 

Titanium

Ti

 

 

Vanadium

V

 

 

Chromium

Cr

 

 

Manganese

Mn

 

 

Iron

Fe

 

 

Cobalt

Co

 

 

Nickel

Ni

Copper

Cu

Zinc

Zn

 

Valences for an element are the combining power of an element: This is the combining power of an element during a chemical reaction. In a simple term it is the number of electron an element is willing to accept or donate during a chemical reaction. Some elements are capable of exhibiting more than one valency e.g Iron (Fe), Cu, Sn, Pb etc.

 

EXAMPLES OF ELEMENT AND THEIR VALENCIES

 

Element

Valency

Element

Valency

Element

Valency

 

H

1

Mg

2

Zn

2

 

Na

1

Ca

2

Pb

2,4

 

Cl

-1

Al

3

Mn

2,4,5,7

 

O

-2

Sn

2,4

Cu

1,2

 

COMPOUND NAMES AND FORMULAE

Name of compound

Chemical

Name of

Chemical

 

Formula

Compound

formula

Sodium chloride

NaCl

Water

H2O

Carbon Iv oxide

CO2

Magnesium

hydroxide

Mg(OH)2

 

 

Calciumtrioxocarbonate (Iv)

CaCO3

Aluminum

sulphate

Al2(SO4)3

Zinc (II) 0xide

ZnO

Tin (Iv) oxide

SnO2

 

DIFFERENCES BETWEEN A MIXTURE AND A COMPOUND

 

 

MIXTURES

COMPOUNDS

1

It is either homogenous or heterogeneous

It is always homogenous

 

2

Constituents are physically combined together

Constituents are chemically joined together

3

Constituents can be combined in any ratio

Constituents have fixed ratio for combination

4

Their properties are the sum of the individual constituents

 

Their properties differ from that of the constituents

 

 

 

CALCULATIONS

Calculation of relative atomic mass of isotopic elements from % abundance

  1. Given that C-12 has % abundance of 98.9% and C-13 has 1.1% calculates its relative atomic mass
  2. Chlorine 35 =75%, and chlorine 37=25% calculate its relative atomic mass

 

Solutions :

  1. relative atomic mass of carbon = (Atomic mass X % abundance of C-12) + (atomic mass X % abundance of C-13)

= 12 x 98.9                   +       13 x 1.1

          100                                  100

= 11.868    +       0.143

= 12.011

  1. relative atomic mass of chlorine =(Atomic mass X % abundance of Cl-35) + (atomic mass X % abundance of C-37)

= 35 x 75                       +       37 x 25

      100                                     100

= 26.25 +   9.25

= 35.5

 

CALCULATION INVOLVING PERCENTAGE COMPOSITION OF ELEMENTS IN A COMPOUND

Examples: Calculate the molecular mass of each of the following compounds the percentage composition by mass of each of the elements in their respective compounds

  1. NaOH
  2. H2SO4
  3. Ca(OH)2
  4. Al2(SO4)3

 

Na =23, O=16, H=1, S=32, Ca=40, Al=23, Solution

i. Mr of NaOH = Na + O +H

= 23+16+1

= 40gmol-1

 

ii. Mr of H2SO4 = 2H + S + 4O

= (2x1) + 32 + (4X16)

= 2+32+64

= 98gmol-1

 

iii. Mr of Ca(OH)2 = Ca + 2O + 2H

= 40 + (2x16) + (2x1)

= 40 +32 +2

= 74gmol-1

 

EMPIRICAL FORMULA

It shows the simplest whole number ratio of the atoms present, e.g.

C2H6; C:H ratio is 1:3. and could be written simply as CH3

 

Calculation the Empirical Formula of a Compound

Find the empirical formula of an oxide of magnesium consisting of 0.32g of oxygen and 0.96g of magnesium. (Mg=24, O=16)

Step 1: find the number of moles of the 2 elements. n(Mg) =

Mole =mass /atomic mass

Mg =0.96g/24

=0.04mol.

O = 0.32g/16

=0.02mol.

 

Step 2: Divide the moles by the smallest number.

Mg = 0.04/0.02 =2

=0.02/0.02 =1

Therefore, the empirical formula is Mg2O

 

Calculating the Empirical Formula from Percentage Composition

An oxide of sulphur consists of 40% sulphur and 60% oxygen. (S=32, O=16)

Note the total composition is usually taking as 100%.

Step 1: find the number of moles of the 2 elements.

S =% composition/atomic mass

= 40/32

= 25mol.

 

O = 60/16

= 75mol.

 

Step 2: Divide the moles by the smallest number

S =1.25/1.25

= 1

O =3.75/1.25

=3

Therefore, the empirical formula is SO3

 

Molecular formula

It is the formula indicating the actual number of atoms of element present in a compound.

Being given the molecular mass of a molecule or a compound, the molecular formula could be calculated using its empirical formula

Example

A compound has an empirical formula of CH2 and a molecular mass of 42gmol.-1 find its molecular formula

Solution

Empirical formula X n = Molecular formula

Where n = is the multiplying factor (CH2) n = 42gmol.-1

[12 + (1x2)] x n=42gmol.-1

(12+2) x n =42gmol.-1

14n =42

n=42/14

n=3

Therefore molecular formula is empirical formula multiplied by 3 CH2 x3 = C3H6

 

LAW OF CONSERVATION OF MASS

The law of conservation of mass states that

“The mass in an isolated system can neither be created nor be destroyed but can be transformed from one form to another”.

According to the law of conservation of mass, the mass of the reactants must be equal to the mass of the products for a low energy thermodynamic process.

It is believed that there are a few assumptions from classical mechanics which define mass conservation. Later the law of conservation of mass was modified with the help of quantum mechanics and special relativity that energy and mass are one conserved quantity. In 1789, Antoine Laurent Lavoisier discovered the law of conservation of mass.

 

Formula of Law of Conservation of Mass

Law of conservation of mass can be expressed in the differential form using the continuity equation in fluid mechanics and continuum mechanics as:

∂ρ +▽(ρv)=0

∂t

Where,

  • ρ is the density
  • t is the time
  • v is the velocity
  • ▽ is the divergence

 

Law of Conservation of Mass Examples

  • Combustion process: Burning of wood is a conservation of mass as the burning of wood involves Oxygen, Carbon dioxide, water vapor and ashes.
  • Chemical reactions: To get one molecule of H2O (water) with the molecular weight of 10, Hydrogen with molecular weight 2 is added with Oxygen whose molecular weight is 8, thereby conserving the mass.

Law of Conservation of Mass Problems

Q1. 

10 grams of calcium carbonate (CaCO3) produces 3.8 grams of carbon dioxide (CO2) and 6.2 grams of calcium oxide (CaO). Represent this reaction in terms of law of conservation of mass.

Solution

According to the law of conservation of mass:
Mass of reactants = Mass of products
∴ 10 gram of CaCO3 = 3.8 grams of CO2 + 6.2 grams of CaO
10 grams of reactant = 10 grams of products

Hence, it is proved that the law of conservation of mass is followed by the above reaction.

 

Q2.

4 grams of hydrogen reacts with some oxygen to make 36 grams of water. Figure out how much oxygen must have been used by applying the law of conservation of mass?

 

Solution

 Hydrogen + Oxygen → Water

Mass given- Hydrogen = 4g, Oxygen = x g, Water = 36 g.

According to the Law of conservation of Mass-

Mass of reactants = Mass of products

4 g+ x g = 36 g

x g = 36 g – 4 g

x = 32 g.

Hence, we can say that 32 g of oxygen was used.

 

Q3.

On heating, 10.0 grams of sodium carbonate (Na2CO3), 4.4 g of carbon dioxide (CO2) and 5.6 g of sodium oxide (Na2O) is produced. Show that this reaction obeys the law of conservation of mass.

Solution 

Sodium Carbonate → Carbon dioxide + Sodium oxide

Na2CO3 → CO2 +Na2O

Mass given- Sodium carbonate = 10 g, Carbon dioxide = 4.4 g, sodium oxide = 5.6 g.

According to the Law of conservation of Mass-

Mass of reactants = Mass of products

Mass of reactant = 10 g

Mass of product = 4.4 + 5.6 = 10 g.

Hence, this reaction obeys the law of conservation of mass.

 

Q4.

How much sodium carbonate is produced when 224.4 g of NaOH reacts with 88 g of CO2? The reaction produces 36 g of water.

Solution

Sodium Hydroxide + Carbon dioxide → Sodium carbonate + Water

NaOH + CO2 → Na2CO3 + H2O

Mass given- Sodium Hydroxide = 222.4 g, Carbon dioxide = 88g, Water = 36g.

Let the mass of sodium carbonate be x grams.

According to the Law of conservation of Mass-

Mass of reactants = Mass of products

Mass of Sodium Hydroxide + Mass of Carbon dioxide = Mass of Sodium carbonate + Mass of Water

222.4 g + 88 g = x g + 36 g

x g = 276.4 g

276.4 g of sodium carbonate is produced when 224.4 g of NaOH reacts with 88 g of CO2 along with 36 g of water.

 

EVALUATION:    1. State the first 30 elements and their symbols

  1. What are the valencies of ferrous and ferric?
  2. What is the empirical and molecular formula for sucralose? The percent composition is C= 36.25%, H= 4.82%, Cl=26.75% and O= 32.19%. The molar mass is 397.63 g/mole.
  3. Differentiate between mixtures and compounds
  4. If the law of conservation of mass holds true, how much sodium chloride will react with 34.0 g of silver nitrate to produce 17 g of sodium nitrate and 28.70g of silver chloride?        
  5. Silicon dioxide, which is composed of the elements silicon and oxygen, contains 46.7 percent silicon by mass. How much oxygen will 10 g of silicon combine with?
  6. 0.1618 g of magnesium oxide (MgO) was produced when 0.0976 g of magnesium was heated in the air. How much oxygen is required to produce 0.1618 g MgO?
  7. State the law of conservation of mass

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively