# Lesson Notes By Weeks and Term - Senior Secondary 1

Symbols, formulaes and equations I

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