Course Content
CHAPTER 3: CLASSIFICATION OF ELEMENTS
Section Name Topic Name 3 Classification of Elements and Periodicity in Properties 3.1 Why do we Need to Classify Elements ? 3.2 Genesis of Periodic Classification 3.3 Modern Periodic Law and the present form of the Periodic Table 3.4 Nomenclature of Elements with Atomic Numbers > 100 3.5 Electronic Configurations of Elements and the Periodic Table 3.6 Electronic Configurations and Types of Elements: s-, p-, d-, f – Blocks 3.7 Periodic Trends in Properties of Elements
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CHAPTER 7: EQUILIBRIUM
Section Name Topic Name 7 Equilibrium 7.1 Equilibrium in Physical Processes 7.2 Equilibrium in Chemical Processes – Dynamic Equilibrium 7.3 Law of Chemical Equilibrium and Equilibrium Constant 7.4 Homogeneous Equilibria 7.5 Heterogeneous Equilibria 7.6 Applications of Equilibrium Constants 7.7 Relationship between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G 7.8 Factors Affecting Equilibria 7.9 Ionic Equilibrium in Solution 7.10 Acids, Bases and Salts 7.11 Ionization of Acids and Bases 7.12 Buffer Solutions 7.13 Solubility Equilibria of Sparingly Soluble Salts
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CHAPTER 10: S-BLOCK ELEMENTS
Section Name Topic Name 10 The s-Block Elements 10.1 Group 1 Elements: Alkali Metals 10.2 General Characteristics of the Compounds of the Alkali Metals 10.3 Anomalous Properties of Lithium 10.4 Some Important Compounds of Sodium 10.5 Biological Importance of Sodium and Potassium 10.6 Group 2 Elements : Alkaline Earth Metals 10.7 General Characteristics of Compounds of the Alkaline Earth Metals 10.8 Anomalous Behaviour of Beryllium 10.9 Some Important Compounds of Calcium 10.10 Biological Importance of Magnesium and Calcium
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CHAPTER 12: CHARACTERIZATION OF ORGANIC COMPOUND
Section Name Topic Name 12 Organic Chemistry – Some Basic Principles and Techniques 12.1 General Introduction 12.2 Tetravalence of Carbon: Shapes of Organic Compounds 12.3 Structural Representations of Organic Compounds 12.4 Classification of Organic Compounds 12.5 Nomenclature of Organic Compounds 12.6 Isomerism 12.7 Fundamental Concepts in Organic Reaction Mechanism 12.8 Methods of Purification of Organic Compounds 12.9 Qualitative Analysis of Organic Compounds 12.10 Quantitative Analysis
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Class 11th Chemistry Online Class For 100% Result
About Lesson

Mole Concept and Molar Masses

  • Atoms and molecules are extremely small in size but their numbers are very large even in a small amount of any substance.
  • In order to handle such large numbers, a unit of similar magnitude is required.
  • We use unit dozen to denote 12 items; score for 20 items and so one. To count the entities at microscopic level mole concept was introduced.
  • In SI system, mole was introduced as seventh base quantity for the amount of a substance.
  • One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.
  • The mass of a carbon–12 atom was determined by a mass spectrometer and found to be equal to (1.992648 × 10–23) g.
  • We know that one mole of carbon weighs 12g,the number of atoms in it is equal to:-
  • (12 g / mol 12C)/( 1.992648 × 10–23 g/12C atom)
  • = 6.0221367 ×1023 atoms/mol.
  • This number of entities in 1 mol is so important that it is given a separate name and symbol.
  • It is known as ‘Avogadro constant’, denoted by NA in honour of Amedeo Avogadro.
  • If a number is written without using the powers of ten 602213670000000000000000, so many entities (atoms, molecules or any other particle) constitute one mole of a particular substance.
  • 1 mol of hydrogen atoms = 6.022 x 1023
  • 1 mol of water molecule = 6.022 x 1023
  • 1 mol of sodium chloride = 6.022 × 1023 units of NaCl.

1 mole of various substances

Class_11_Concepts_Of_Chemistry_1_Mole_of_Various_Substances
Concept of Chemistry

Molar mass

  • The mass of one mole of a substance in grams is called its molar mass.
  • The molar mass in grams is numerically equal to atomic /molecular/formula mass in u.
  • Molar mass of oxygen (O) =16.02g.
  • Molar mass of water (H2O) = 18.02g.
  • Molar mass of carbon (C) = 12g.
  • Molar mass of sodium chloride (NaCl) = 58.5g.

Percentage Composition

Percentage Composition

  • The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.
    • For example:-Consider H2O molecule. Molar mass of Hydrogen (H) = 2g, and Molar mass of Oxygen (O) = 16g.
      • Consider 18g of H2O it contains 2g of H and 16g of O.
      • Mass % of H = (2×1.008×100)/(18.02) = 11.18%
      • Mass % of O = (16.00×100)/(18.02)= 88.79%
    • Note: – By using information from percentage composition we can calculate empirical formula.

Empirical Formula for Molecular formula

  • An empirical formula represents the simplest whole number ratio of various atoms present in a compound.
  • Molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
  • If the mass percent of various elements present in a compound is known, then its empirical formula can be determined.
  • Molecular formula can further be obtained if the molar mass is known.

Problem:-

A compound contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?

Answer:-

Step 1:- Conversion of mass per cent to grams.

Since we are having mass per cent, it is convenient to use 100 g of the compound as the starting material. Thus, in the

100 g sample of the above compound,

4.07g hydrogen is present, 24.27g carbon is present and 71.65 g chlorine is present.

Step 2:- Convert into number moles of each element

Divide the masses obtained above by respective atomic masses of various elements.

Moles of hydrogen = (4.07 g/1.008 g) = 4.04

Moles of carbon = (24.27 g/12.01g) = 2.021

Moles of chlorine = (71.65 g/35.453 g) = 2.021

Step 3:- Divide the mole value obtained above by the smallest number

Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H: C: Cl.

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4:- Write empirical formula by mentioning the numbers after writing the symbols of respective elements.

CH2Cl is, thus, the empirical formula of the above compound.

Step 5:- Writing molecular formula

(a) Determine empirical formula mass

Add the atomic masses of various atoms present in the empirical formula.

For CH2Cl, empirical formula mass is

(12.01) + (2 × 1.008) + (35.453)

= 49.48 g

(b) Divide Molar mass by empirical formula mass

n = (Molar mass/Empirical formula)

n=2

(c) Multiply empirical formula by n obtained above to get the molecular formula

Empirical formula = CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.

Problem:-

Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).

Answer:-

The molecular formula of sodium sulphate is (Na2SO4)

Molar mass of (Na2SO4) = [(2 × 23.0) + (32.066) + 4 (16.00)]

= 142.066 g

Mass percent of an element:-

(Mass of that element in the compound/ Molar mass of the compound) x 100

Therefore, Mass percent of sodium:

= (46.0g/142.066g)/100

=32.379

=32.4%

Mass percent of sulphur:

= (32.066g)/ (142.066g) x 100

=22.57

=22.6%

Mass percent of oxygen:

= (64.0g)/ (142.066g) x 100

=45.049

=45.05%

Problem:-

Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.

Answer:-

% of iron by mass = 69.9 % [Given]

% of oxygen by mass = 30.1 % [Given] Relative moles of iron in iron oxide:

Relative moles of oxygen in iron oxide:

(% of iron by mass)/ (Atomic mass of iron)

= (69.9)/ (55.85)

Simplest molar ratio of iron to oxygen:

= 1.25: 1.88

= 1: 1.5

2: 3

Therefore, the empirical formula of the iron oxide is Fe2O3.

 

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