Introduction to Mole Concept, Study notes of Chemistry

An introduction to the mole concept in chemistry. It covers the definition of a mole, avogadro's number, and how to perform calculations involving moles, molar masses, and molar volumes of gases. Learning objectives, study questions, key words, and a detailed lesson plan with examples. It is intended for use in a year 10 chemistry class during the 2023/2024 academic year.

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2023/2024

Uploaded on 08/27/2024

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WEEK: 1 TERM: 2ND SESSION: 2023/2024
SUBJECT: CHEMISTRY TOPIC: INTRODUCTION TO MOLE
CONCEPT
CLASS: YEAR 10 DATE: 8/1/24 - 12/1/24 LESSON No.: 1
LEARNING OBJECTIVES: At the end of the
lesson, the students should be able to:
I. Define mole
II. Understand Avogadro’s number and the mole
concepts
III. Perform simple calculations involving number
of moles, molar masses and molar volumes of
gases
STUDY QUESTIONS:
What are the units of measurement you know?
List them.
KEY WORDS:
Mole, Molar mass, Avogadro’s
number, Atoms, Molecules,
Particles
PREVIOUS LEARNING :
Students have studied
separation techniques and
understand the principle behind
the used of each separation
technique.
LEARNING ACTIVITIES/CLASSROOM ORGANISATION:
STEP 1: - The Teacher asks the students to calculate the molar masses of
NaOH, KOH
STEP 2: Using a beam balance, the teacher weighs 1 mole of the above
substances, introduces the lesson and describes the mole as amount of a
substance.
STEP 3: - The teacher explains Avogadro’s number, performs simple
calculations involving moles and leads the students to do same
STEP 4: - Students do some exercises as given by the teacher
STEP 5: - The teacher explains the concept of molar volume of gases and
does some classwork with the students .
DIFFERENTIATION:
Level 1: All students should be able to review step 1
Level 2 : Most students should be able to review steps 1, 2 and 3
Level 3: Some students should be able to review step 4
IT:
BBC Bitesize:
The Mole
EVALUATION FOR LESSON TAUGHT:
a. What is the number of molecules in 6.4g of SO2 (NA = 6.02 X 1023)?
b. What is the volume in cm3 of 2.2g of CO2 at s.t.p ( C=12, O=16)?
c. Determine the percentage by mass of oxygen in Al2(SO4).2H2O.( Al=27,
S=32, O=16, H=1)
d.
RESOURCES:
Cambridge IGCSE
Chemistry 2nd Edition by
KEY SKILLS:
Analytical, Problem
Solving
HOMEWORK:
Standout Chemistry Workbook
For Senior Secondary 1
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WEEK: 1 TERM: 2ND SESSION: 2023/

SUBJECT: CHEMISTRY TOPIC: INTRODUCTION TO MOLE

CONCEPT

CLASS: YEAR 10 DATE: 8/1/24 - 12/1/24 LESSON No.: 1 LEARNING OBJECTIVES: At the end of the lesson, the students should be able to: I. Define mole II. Understand Avogadro’s number and the mole concepts III. Perform simple calculations involving number of moles, molar masses and molar volumes of gases STUDY QUESTIONS: What are the units of measurement you know? List them.

KEY WORDS:

Mole, Molar mass, Avogadro’s number, Atoms, Molecules, Particles PREVIOUS LEARNING : Students have studied separation techniques and understand the principle behind the used of each separation technique. LEARNING ACTIVITIES/CLASSROOM ORGANISATION: STEP 1: - The Teacher asks the students to calculate the molar masses of NaOH, KOH STEP 2: Using a beam balance, the teacher weighs 1 mole of the above substances, introduces the lesson and describes the mole as amount of a substance. STEP 3 : - The teacher explains Avogadro’s number, performs simple calculations involving moles and leads the students to do same STEP 4: - Students do some exercises as given by the teacher STEP 5: - The teacher explains the concept of molar volume of gases and does some classwork with the students. DIFFERENTIATION: Level 1: All students should be able to review step 1 Level 2 : Most students should be able to review steps 1, 2 and 3 Level 3: Some students should be able to review step 4

IT:

BBC Bitesize: The Mole EVALUATION FOR LESSON TAUGHT: a. What is the number of molecules in 6.4g of SO 2 (NA = 6.02 X 10^23 )? b. What is the volume in cm^3 of 2.2g of CO 2 at s.t.p ( C=12, O=16)? c. Determine the percentage by mass of oxygen in Al 2 (SO 4 ).2H 2 O.( Al=27, S=32, O=16, H=1) d. RESOURCES: Cambridge IGCSE Chemistry 2 nd^ Edition by

KEY SKILLS:

Analytical, Problem Solving

HOMEWORK:

Standout Chemistry Workbook For Senior Secondary 1

Roger Norris New School Chemistry by Ose Yaw Ababio LESSON CONTENT INTRODUCTION TO MOLE CONCEPT The mole A mole is a number of particles of a substance which may be atoms, ions, molecules or electrons. This number of particles is approximately 6.02 x 10^23 in magnitude and is known as Avogadro’s number of particles. The mole is defined as the amount of a substance which contains as many elementary units as there are atoms in 12g of Carbon-12. A. MOLE IN TERMS OF THE FORMULA The mole can be expressed as: Element: Mole = mass of the element

  • Relative Atomic Mass of Element OR Compound: = Mole mass in grams Or molecule R.M.M Example: Calculate the number of moles of atoms, present in 40g of calcium carbonate or calcium trioxocarbonate (IV) (CaC0 3 ) Solution: Mass of CaC0 3 = 40g R.M.M. CaC0 3 = (40 + 12 +16) = 100glmol Mole (n) = Mass is g R.M.M CaC0 3 N = 40g 100g/mole = 0.4 mole LESSON 2
  1. Calculate the number of particles i. 44g of iron (II) sulphide (Fes) ii. 5.5g of manganese (Mn) iii. 8g of oxygen molecule (0 2 ) iv. 8g of oxygen atom (0) (Mn – 55, 0-16, Fe =56, S=32) Mole = mass of given substance (g) (n) Gram atomic/molar mass ii. Mole of Mn = 5.5g 55 g/mole Number of particles = NA^ x n : .1 mole of Mn = 6.023 x 10^23 atoms 0.1mole Mn = 0.1 x 6.023 x 10^23 atoms = 6.023 x 10 22 atoms ii. Mole of Fe = massing of Fes n G|.M.M = 44g = 44g (56 +32) 88g1mole = 0.5 mole 1 mole fess = 6.023 x 10^23 molecules :. 0.5 mole= 6.023 x 10^23 x. = 3.012 x 10^23 molecules of Fes iii. Mole of 0 2 = massing G.M.M Gmm of 0 2 = 16 x 2 = 32/gmole :. Mole = 8 g 32/gmole = 0.25 mole 1 mole of 0 2 = 6.023 x 10^23 mole of 0 2 :. 0.25 moles of 0 2 = 6.023 x 10^23 x 0.25 molecules = 1.506 x 10^23 molecules of 0 2 iv. G.m.m of oxygen = 16g/mole Mole of 0 = massing of 0 G.a.m = 8g 16g/mole = 0.5mole 1 mole of 0 = 6.023 x 10^23 atoms of 0 :. 0.5 mole of 0

= 0.5x 6.023 x 10^23 atoms = 3. 012 x10^23 atoms of 0 LESSON 3: MOLAR VOLUME OF GASES The volume occupied by 1 mole of a gas at standard conditions of temperature and pressure (s.t.p.) is 22.4 dm^3. NOTE: 1 mole of any gaseous substance = molar volume of a gas at s.t.p. i.e. 1 mole of 0 2 (32g) = 22.4dm^3 at s.t.p 1 mole of C0 2 (44g) = 22.4dm^3 at s.t.p 1 mole of N 2 (28g) = 22.4dm^3 at s.t.p 1 mole of S0 2 (64g) = 22.4dm^3 at s.t.p 1 mole of C1 2 (71) 22.4dm3 at s.t.p Examples:

  1. Calculate the volume occupied by 5 moles of carbon dioxide (carbon (iv) oxide) at s.t.p SOLUTION: 1 mole of gas at s.t.p = 22.4dm^3 1 mole of C0 2 = 22.4dm3 at s..t.p 5 mole of CO 2 =5 x 22.4 at s.t.p = 112.0dm^3 at s.t.p
  2. Determine the number of mole present in 11.2dm^3 of nitrogen (IV) oxide (nitrogen dioxide) = N0 2 (g) at s.t.p Mole = volume G.m.v 1 mole of N0 2 (g) = 22.4dm^3 at s.t.p :. 22.4dm of N0 (g) = 1 mole of N0 2 at s.t.p 1 dm^3 of N0 2 (g) = 1 ofN0 2 (g) at s.t.p 22.4dm^3 :.11.2dm^3 of N0 2 (g) = 1 x 11.2dm3 of N0 2 (g) at s.t.p 22.4dm^3 = 11.22m^3 of N0 2 (g) at s.t.p = 0.5mole of N0 2 at s.t.p
  • How many grammes of gas are present in 5600cm^3 of chlorine gas at s.t.p? (Cl=35.5) SOLUTION: