Learn the essential concepts of chemistry with our comprehensive and free Some Basic Concepts Of Chemistry notes pdf. This article at Sidclasses covers everything you need to know, from atoms and molecules to chemical equations and stoichiometry. The below notes are based on the NCERT book.
Class 11 Chemistry Chapter 1 Notes
Matter
The matter is defined as anything that occupies space and possesses mass.
Classification of Matter
(1) Physical classification
- (a) Solid
- (b) Liquid
- (c) Gases
| Solid | Liquid | Gases |
| (1) They have fixed shape and volume | They have a moderate density | They do not have fixed shapes and volume |
| (2) They cannot be compressed | They cannot be compressed | They can be compressed easily |
| (3) They have high density | They do not have fixed shapes and volume | They have low density |
| (4) They do not flow | They flow easily | They flow easily |
| (5) They do not fill their container | They do not fill their container | They fill their container |
| (6) The forces of attraction are strong | The forces of attraction are less strong than solids | The forces of attraction are weak. |
| Particles are not close as solids | Kinetic energy is more than solids | Kinetic energy is maximum. |
| (8) Particles are closely packed | Particles are not close as in solids | Particles are much farther apart from one another. |
| (7) Kinetic energy is the least | For Ex: Water, petrol, cold drinks | For Ex: Oxygen, nitrogen, helium |
- Vapors represent a gaseous state of a substance that is liquid at room temperature.
- A substance that is in a gaseous state at room temperature is called a gas.
- For Ex: Ammonia is a gas but on heating water forms vapors.
(2) Chemical classification
All kinds of matter are classified into two types:
(a) Homogeneous – Material is said to be homogeneous if it has uniform composition and identical properties throughout a material is said to be homogeneous if it consists of only one phase.
(b) Heterogeneous– A material is said to be heterogeneous if it consists of a number of phases. The different phases are separated from each other by distinct boundaries.
Pure Substance
A pure substance is one which is made up of only one kind of particles (atoms or molecules).
Pure substances are of two types:
(1) Elements
(2) Compounds
Mixture
A mixture is one which contains two or more different kinds of particles(atoms or molecules).
Mixtures are of two types:
(1) Homogenous mixture
(2) Heterogeneous mixture
Elements
An element is usually defined as the simplest form of a pure substance with definite physical and chemical properties and which can neither be broken into nor built from simpler substances by any chemical or physical method.
An element contains only one kind of particle. These particles may be atoms or molecules.
For example carbon, sulfur, iron, gold, hydrogen, oxygen, nitrogen, silver, etc.
The number of elements known to date is 114 .out of these 92 occur in nature in the earth’s crust and the remaining have been prepared artificially in the laboratory through reactions.
Types of Elements
(1) Metals
(2) Non-metals
(3) Metalloids
| Metals | Non-metal |
| (1) They are malleable | They are not malleable |
| (2) They are ductile | They are not ductile. |
| (3) They are good conductors of heat and electricity. | They are bad conductors of heat and electricity. |
| (4) They are lustrous | They are not lustrous. |
| They have a low tensile strength | They have low tensile strength |
| (6) They are sonorous | They are not sonorous |
| (7) They are hard | They are soft |
Metalloids
The elements whose properties are intermediate between those of metals and non-metals are known as metalloids.
For Example Silicon, Germanium, Arsenic, and Tellurium.
Compounds
Compounds are pure substances containing more than one kind of element or atom.
In compounds, the two elements are in a fixed proportion by mass and can be decomposed into their constituent elements by suitable chemical methods. The properties of a compound are completely different from those of its constituent elements.
For example, Water is a compound containing hydrogen and oxygen combined together in a fixed proportion of 1:8 by weight.
It can be decomposed into its constituent elements hydrogen and oxygen by passing electricity through water. Water is completely different from its constituent hydrogen and oxygen.
For Example, Carbon dioxide, Sulphur Dioxide, hydrochloric acid, nitric acid, washing soda, and common salt are examples of compounds.
Types of compounds
All the compounds may be divided into two categories :
(1) Organic compounds
Organic Compounds are compounds containing carbon and a few other elements like carbon, Hydrogen, oxygen, nitrogen, sulfur, and halogens. They were originally obtained only from plants and animals.
(2) Inorganic compounds
Inorganic Compounds are compounds containing any two or more elements out of more than one 114 elements known.
Mixtures
A material containing two or more substances in any proportion is called a mixture. The properties of a mixture are the properties of its constituents. A mixture can be separated into its constituent by a simple physical method.
Types of mixture
(1) Homogeneous
(2) Heterogeneous
| Homogeneous Mixture | Heterogeneous Mixture |
| (1) Those mixtures in which the substances are completely mixed together and are indistinguishable from one another. | Those mixtures in which the substances remain separate and one substance is spread throughout the other. |
| (2) They have uniform composition throughout its mass. | They do not have uniform composition. |
| (3) It has no visible boundaries of separation between various constituents. | It has visible boundaries. |
| (2) They have uniform composition throughout their mass. | For Ex: Sugar and sand, Salt and Sand, Milk, Soil, Blood, Starch, Muddy water etc. |
Difference between Mixture and Compound
| Mixture | Compounds |
| (1) A mixture can be separated into its constituents by the physical processes. | A compound can not be separated into its constituents by the physical processes. |
| (2) A mixture shows the properties of its constituents. | A compound can not be separated into its constituents by physical processes. |
| (3) Energy is usually neither given out nor absorbed in the preparation of a mixture. | A compound does not show the properties of its constituents. |
| (4)The composition of a mixture is variable, the constituents can be present in any proportion by mass. | Energy is usually given out or absorbed during the preparation of a compound. |
| (5) A mixture does not have a fixed melting and boiling point. | A compound has a fixed melting and boiling point. |
Atom
An atom is the smallest particle of an element that may or may not be capable of independent existence.
For example: Atoms of iron, copper, Silver, and Gold can exist freely whereas atoms of hydrogen, oxygen, and nitrogen cannot exist freely.
Molecule
A molecule is the smallest particle of an element or a compound that can exist freely.
Molecules may be classified into two categories:
(1) Molecules of elements: They are made up of only one kind of atom, they are called homoatomic or homonuclear molecules.
The atomicity of metal elements like sodium, magnesium, aluminum, copper, iron, etc. is taken as 1. They are called Monoatomic molecules.
(2) Hydrogen, Nitrogen, Oxygen, Chlorine, bromine, etc. have two atoms in their molecules. So the atomicity is 2. They are called diatomic molecules.
(3) Ozone has 3 atoms in its molecules, so the atomicity is 3. They are called triatomic molecules.
(4) Phosphorus has 4 atoms in its molecules, so the atomicity is 4. They are called tetra-atomic molecules.
Molecules of compounds: They are made up of atoms of different elements and hence are called heteroatomic or heteronuclear molecules.
They may be diatomic, or triatomic depending upon the number of atoms present in a molecule of the compound.
For Example HCl, water, carbon dioxide, ammonia, methane, etc.
Some Basic Concepts Of Chemistry Notes pdf
Physical Quantities
Physical Properties
Physical Properties are those which can be measured or observed without changing the identity or composition of the substance.
For example, Mass, volume, melting point, and boiling point.
Chemical properties
Chemical properties are those in which a chemical change in the substance occurs.
The measurement of any physical quantity consists of two parts:
(a)The number
(b)The unit
For example: If an object weighs 4.5 kg it involves two parts:
4.5 is the number and kg is the unit.
A unit is defined as the standard of reference chosen to measure any physical quantity.
French Academy of Science in 1791 introduced a new system of measurement called metric system in which the different units of a physical quantity are related to each other as multiples of powers of 10.
For example, The improved System of units has been accepted internationally and is called the International
System of units or in short SI units (Systeme Internationale in French).
Seven base units
The seven basic physical quantities on which the International System of Units is based, their symbols, the name of their units, and the symbols of these units are given as:
Measurement of Temperature
There are 3 scales of temperature:
(1) Degree Celsius
(2) Degrees Fahrenheit
(3) Kelvin
The SI unit of Temperature is Kelvin
Relation between Degree kelvin and degree Celsius —- Kelvin = 0C + 273.15
Relation between degree Celsius and Degrees Fahrenheit —— 0F = 9/5 0C +32
Measurement of Volume
The SI unit of volume is m³
1 L =1000 mL =1000 cm³
1 L = 1 dm³
1 m³ = 100 (cm)³ = 10³ L
Measurement of Mass
Mass is the quantity of matter contained in the sample and for the given sample it is constant and it does not depend upon the place.
weight is the force with which the body is attracted toward the earth. It depends upon the acceleration due to gravity which varies from place to place. The mass of a substance can be determined very accurately in the laboratory by using an analytical balance or electrical balance.
S.I. unit of mass is the kilogram
1 Kg =1000 g
Units of Length
The S. I. unit of length is a meter. It is also expressed in angstroms or nanometres or picometres.
Commonly used physical quantities and their derived units
Commonly used prefixes with the base units
Significant Figures
Accuracy of Measurement
The accuracy of any measurements depends upon the
(a) accuracy of the measuring device used
(b) the skill of its operator.
If the average value of different measurements is close to the correct value, the measurement is said to be accurate.
If the values of different measurements are close to each other and hence close to the average value, the measurement is said to be precise.
A measurement can have good accuracy but poor precision because different measurements may give a correct average.
Errors in measurements, when the same mistake is made repeatedly, are called systematic errors. They do not affect the Precision but they often affect the accuracy of a measurement.
Significant Figures
The total number of digits in a number including the last digit whose value is uncertain is called the number of significant figures.
Rules for determining the number of significant figures
(1) All non-zero digits as well as the zeros between the non-zero digit are significant.
576 cm has three significant figures
5004 has 4 significant figures
0.48 has two significant figures
2.05 Three significant figures
(2) Zeros to the left of the first nonzero digit in a number are not significant.
0.05 m has 1 significant figure
0.0045 has two significant figures
(3) If a number ends in zeros but these zeros are to the right of the decimal point then these zeros are significant.
5.0 meters has two significant figures
2.500 g has 4 significant figures
0.0200 has 3 significant figures
(4) If a number ends in zero but these zeros are not to the right of a decimal point, these zeros may or may not be significant.
1.05 * 10³ has three significant figures
1.050 * 10³ has four significant figures
1.0500 * 10³ has five significant figures
Mathematical operation on numbers expressed in scientific notation
1) 4683.507 will be written as 4.683507 * 10³
Decimal has moved three places towards the left so that only one non-zero digit is left and the number of places moved is the exponent of 10 in scientific notation.
2) 0.000256 will be written as 2.56 * 10-4
Decimal has moved four places towards the right so that there is only one non-zero digit before the decimal point and the exponent of 10 is -4.
Calculations involving multiplication and division
In multiplication the coefficients i.e. the numbers before the factor 10n are multiplied and the exponents of 10 are added up.
For Example: ( 5.7 × 106) × ( 4.2 × 105)
( 5.7 × 4.2 ) (10 6+5 )
23.94 × 1011
In division, the factor N is divided, and exponents are subtracted.
For Example: (5.7 × 106 ) ÷ ( 4.2 × 103 )
( 5.7 ÷ 4.2 ) × (106-3 )
1.357 × 103
In addition and subtraction, first, the numbers are written in such a way that they have the same exponents. Taking out 10n common, the coefficients are added or subtracted.
For Example: 4.56 × 103 + 2.62 × 102
45.6 × 102 + 2.62 × 102
( 45.6 + 2.62 ) × 102
58.22 × 102
For Example 4.5 × 10-3 – 2.6 × 10-4
4.5 × 10-3 -0.26 ×10-3
4.24 × 10-3
Rules for determining the number of significant figures
Rule 1 The result of an addition or subtraction should be reported to the same number of decimal places as that of the term with the least number of decimal places.
For Example: 4.523 + 2.3 + 6.24
Actual sum= 13.063
Reported sum = 13.1
4.523 has 3 decimal places
2.3 has 1 decimal place
6.24 has 2 decimal places
The answer should be reported only up to one decimal place.
Example 2: 18.4215 – 6.01
Actual difference = 12.4115
Reported difference= 12.41
As the second number has 2 decimal places only the answer is reported up to 2 decimal places.
Rule 2: The result of a multiplication or division should be reported to the same number of significant figures as is possessed by the least precise term used in the calculation.
Example 1: 4.327 × 2.8
Actual product =12.1156
Reported product= 12
The first number has 4 significant figures while the second has 2. The actual product has been rounded off to give a reported product containing two significant figures.
Example 2: 0.46 ÷ 15.734
Actual quotient = 0.029236
Reported quotient = 0.029
It should contain only 2 significant figures because the least precise term in calculation has only 2 significant figures.
Rule 3 If a calculation involves a number of steps, the result should contain the same number of significant figures as that of the least precise number involved, other than the exact numbers.
(42.967 * 0.02435) ÷ (0.34 * 4)
=0.7692988
0.77
Laws of Chemical Combination
Law of Conservation of Mass
This Law was studied by French chemist Antoine Lavoisier in 1789
This law states that
In all physical and chemical changes, the total mass of the reactants is equal to that of the products.
Or
Mass can neither be created nor destroyed.
This law is also called the law of indestructibility of matter.
The mass and energy are interconvertible but the total sum of the mass and energy during any physical or chemical change is the names constant.
Law of constant composition or definite proportions
This law was discovered by French chemist J.L. Proust in 1799.
It states that
A chemical compound is always found to be made up of the same elements combined together in the same fixed proportion by mass.
For example A pure water obtained from whatever source or any country will always be made up of only hydrogen and oxygen elements combined together in the same fixed ratio of 1: 8 by mass.
A sample of carbon dioxide may be prepared in the laboratory (a) by heating limestone (b) by burning coal in air (c) by the action of dilute hydrochloric acid on marble (d) by heating sodium carbonate.
It is found that carbondioxide is made up of same element ie. carbon and oxygen combine together in the same fixed ratio of 3:8 by mass.
Limitation
1) It Is not applicable if an element exists in different isotopes which may be involved in the formation of a compound.
2) The elements may combine in the same ratio but the compounds formed may be different.
Law of multiple proportion
When two elements combine to form two or more chemical compounds ,then the masses of one of the elements which combined with a fixed mass of the other, bear a simple ratio to one another.
For Ex: Compounds of carbon and oxygen : Carbon combines with oxygen to form two compounds namely carbon dioxide and carbon monoxide.
In carbon dioxide, 12 parts by mass of carbon combine with 32 parts by mass of oxygen while in carbon monoxide, 12 parts by mass of carbon combine with 16 parts by mass of oxygen.
The masses of oxygen which combined with a fixed mass of carbon in carbon monoxide and carbon dioxide are 16 and 32. These masses of oxygen bear a simple ratio of 16:32 or 1:2 to each others.
For Ex: Compound of Sulphur and oxygen : The element sulphur also form two oxides Sulphur dioxide and sulphur trioxide.
In sulphur dioxide ,32 parts by mass of Sulphur combine with 32 parts by mass of oxygen but in sulphur trioxide 32 parts by mass of Sulphur combined with 48 parts by mass of oxygen. The masses of oxygen which combined with the fixed mass of sulphur in the two oxides are 32 and 48. These bear a simple ratio of 32 : 48 or 2 : 3 to each other.
Law of reciprocal proportion
This law was put forward by Richter in 1792.
It states that The ratio of masses of 2 elements A and B which combines separately with a fixed mass of the third element C is either the same or some multiple of the ratio of the masses in which A and B combine directly with each other.
For example: The elements carbon and oxygen combine separately with the third element hydrogen to form methane and water and they combine directly with each other to form carbon dioxide.
In methane, 12 parts by mass of carbon combine with 4 parts by mass of hydrogen. In water, 2 parts by mass of hydrogen combine with 16 parts by mass of oxygen. The masses of carbon and oxygen which combine with fixed mass of hydrogen are 12 and 32 i.e… they are in the ratio 12 : 32 or 3 : 8.
In carbon dioxide 12 parts by mass of carbon combine directly with 32 parts by mass of oxygen ie they combined directly in the ratio of 12 : 32 or 3 : 8 which is the same as the first ratio.
Gay Lussac’s law of gaseous volume
When gases react together they always do so in volumes which bear a simple ratio to one another and to the volume of the products, if these are also gases, provided all measurements of volumes are done under similar conditions of temperature and pressure.
For Example : 1) Combination between hydrogen and chlorine: 1 volume of hydrogen and 1 volume of chlorine always combine to form two volumes of hydrochloric acid gas. The ratio between the volumes of the reactants and the product in this reaction is simple ie. 1 : 1 : 2.
2) Combination between nitrogen and hydrogen:One volume of Nitrogen always combines with 3 volume of hydrogen to form two volumes of ammonia. This reaction also indicate a simple ratio of 1: 3: 2 between the volume of the reactants and products.
Dalton’s Atomic Theory
John Dalton in 1808 put forward a theory known as Dalton’s atomic theory
The main points of this theory are:
(1) Matter is made up of extremely small indivisible particles called atoms.
(2) Atoms of the same element are identical in all respect i.e. shape, size and mass.
(3) Atoms of different elements have different masses, sizes and also possess different chemical properties.
(4) Atoms of the same or different elements combine together to form molecules.
(5) When atoms combine with one another to form molecules, they do so in simple whole number ratios.
(6) Atoms of two elements may combine in different ratios to form more than one compound.
(7) An atom is the smallest particle that take parts in a chemical reaction.
(8) An atom can neither be created nor destroyed.
Limitations of Dalton’s Atomic Theory
(1) It could explain the law of chemical combination by Mass but failed to explain the law of gaseous volume.
(2) It could not explain why atoms of different elements have different masses, size, valencies.
(3) It could not explain the nature of binding forces between atoms and molecules which accounts for the existence of three States of matter.
Modern Atomic Theory
(1) Atom is no longer considered to be indivisible: It is made up of electrons, protons and neutrons.
(2) Atoms of the same element may have different atomic masses: Atoms of the same element which possess different atomic masses are called isotopes.
For Example:
(1) Isotopes of Hydrogen
1H1 1H2 1H3
(2) Isotopes of Carbon
14C 12C 13C
(3) Atoms of different elements may have same atomic masses: Atoms of the different elements which have the same atomic masses are called isobars.
For Example : Calcium and Argon have atomic masses of 40 with atomic number as 20 and 18.
(4) The ratio in which the different atoms combine with one another may be fixed and integral but may not be simple :
For example : In sugarcane the ratio in which the elements carbon, hydrogen and oxygen combine together 12 : 22 : 11 which is not simple.
(5) Atom is the smallest particle that take part in a chemical reaction.
(6) Atom is no longer Indestructible: By carrying out nuclear reactions atom of an element may be changed into another.
Avogadro’s Hypothesis
It states that an equal volume of all gases under similar conditions of temperature and pressure contains an equal number of molecules.
Berzelius a Swedish chemist, gave a hypothesis called the Berzelius hypothesis which states:
An equal volume of all gases under similar conditions of temperature and pressure contains an equal number of atoms.
Applications of Avogadro’s Law
1) In the calculation of the atomicity of Elementary gases
The atomicity of an elementary substance is defined as the number of atoms of the element present in 1 molecule of the substance.
For example, The atomicity of oxygen is 2 while that of ozone is 3.
Hydrogen + oxygen → Water Vapours
2n molecules n molecules 2n molecules
On dividing throughout by 2n
1 molecule ½ molecule 1 molecule
Thus one molecule of water contains ½ molecule of oxygen. But 1 molecule of water contains 1 atom of oxygen.
Hence ½ molecule = 1 molecule of oxygen
1 molecule of oxygen= 2 atoms of oxygen = 1 atom of oxygen= 2
2) To find the relationship between molecular mass and vapor density of a gas
Molecular mass = 2 × vapour density
3) To find the relationship between the mass and volume of a gas
22.4 liters of any gas at STP weigh equal to the molecular mass of the gas expressed in grams. This is called gram molecular volume law.
Atomic and Molecular mass
As an atom is so small a particle that it cannot be seen or isolated, therefore it is impossible to determine the actual mass of a single atom by weighing it.
The problem was finally solved by Avogadro’s hypothesis. If equal volumes of two different gases are taken under similar conditions of temperature and pressure and then weighted, the ratio of their masses will be equal to the ratio of their single molecules. Thus, though the actual masses of the atoms could not be determined their relative masses could be determined. If the atomic mass of the hydrogen is taken is 1 , the relative atomic mass of oxygen is 16.
Initially, the atomic masses of all the elements were obtained by comparing with the mass of hydrogen taken as 1 but by doing so, the atomic masses of most of the elements came out to be fractional.
Therefore carbon is taken reference for the determination of atomic masses.
The atomic mass of an element is the number of times an atom of that element is heavier than an atom of carbon taken as 12.
One atomic mass unit is equal to one-twelfth of the mass of an atom of carbon 12 isotope.
The atomic mass of an element is the average relative mass of its atoms as compared with an atom of carbon 12 taken as 12.
The fractional abundance of an isotope is the fraction of the total number of atoms that is comprised of that particular isotope.
Gram Atomic mass
The atomic mass of an element expressed in grams is called gram atomic mass.
For Ex: The atomic mass of oxygen is 16 amu.
Gram atomic mass of oxygen = 16 g.
Molecular mass
The molecular mass of a substance is the number of times the molecule of the substance is heavier than one-twelfth the mass of an atom of carbon -12.
or
The molecular mass is equal to the sum of the atomic masses of all the atoms present in one molecule of a substance.
For Example Mass of H= 1 u
The atomic mass of O =16 u
Molecular mass of water =2 × atomic mass of H+1 × atomic mass of O
=2 × 1 + 16 × 1
= 18 u
Gram molecular mass
The molecular mass of a substance expressed in grams is called gram molecular mass.
For Example Molecular mass of oxygen = 32u
Gram molecular mass of oxygen=32 g
Dimensional Analysis
Any calculations involving the use of the dimensions of the different physical quantities involved is called dimensional analysis.
It is used for any one of the following purpose:
(1) To convert a physical quantity in one type of unit into some other units: The method used is called the factor label method or unit factor method.
(a) First determine the unit conversion factor
(b) Multiply the given physical quantity with the unit conversion factor, retaining the units of the physical quantity as well as that of the unit conversion factor in such a way that all units cancel out leaving behind only the required units.
For Example Conversion of pounds (lb) into kg.
1 kg =2.205 lb
Question: A man weigh 175 lb. Express his weight in kg.
Answer: 1 kg =2.205 lb
= 79.4 kg
(2) In solving problems
Question: What is the mass (in grams) of an aluminum block whose dimensions are 2.0 in? × 3.0 in. × 4.0 in. and whose density is 2.7 g/cm3? Given that 1 inch = 2.54 cm
Answer : Unit conversion factors are 1 = 2.54cm/1 inch = 1 inch/2.54 cm
Required mass (in g) = 2.0 in. x 3.0 in. x 4.0 in. x 2.54 cm/1 in. x 2.54 cm/1 in. x 2.54 cm/1 in. x 2.7/cm3
= 1.1 × 103 g
Mole Concept
Avogadro’s number or Avogadro’s constant (NA)
One gram atom of any element contains the same number of atoms and one gram molecule of any substance contains the same number of molecules.
The value was found to be 6.022137 × 1023
The value generally used is 6.022 × 1023 .
This is called Avogadro’s number or Avogadro’s constant (NA)
Avogadro’s number may be defined as the number of atoms present in one gram atom of the element or the number of molecules present in one gram molecule of the substance.
A mole is a chemist unit of counting particles such as atom, molecules, ions, electrons, protons which represent a value of 6.022 × 1023
A mole of hydrogen atom means 6.022 × 1023 atoms of hydrogen whereas a mole of hydrogen molecule means 6.022 × 1023 molecules of hydrogen or 2 × 6.022 × 1023 atoms of hydrogen.
A mole of oxygen molecule means 6.022 × 1023 molecules of oxygen or 2 × 6.022 × 1023 atoms of oxygen.
A mole is defined as that amount of substance which has a mass equal to a gram atomic mass if the substance is atomic or gram molecular mass if the substance is molecular.
1 mole of carbon atoms =12 grams
1 mole of sodium atoms = 23 grams
1 mole of Oxygen atom = 16 grams
1 mole of Oxygen molecule = 32 grams
1 mole of water molecule = 18 grams
1 mole of carbon dioxide molecule = 44 grams
Mole
A mole is defined as the amount of substance that contains Avogadro’s number of atoms if the substance is atomic or Avogadro’s number of molecules if the substance is molecular.
1 mole of carbon atoms = 6.022 ×1023 atoms of carbon.
1 mole of sodium atom = 6.022 ×1023 atoms of sodium
1 mole of Oxygen atom = 6.022 ×1023 atoms of oxygen
1 mole of Oxygen molecule = 6.022 ×1023 molecules of oxygen
1 mole of water = 6.022 ×1023 molecules of water
In case of gases, a mole is defined as that amount of the gas which has a volume of 22.4 litres at STP.
1 mole of Oxygen gas = 22.4 litres of oxygen at STP
one mole of carbon dioxide gas = 22.4 litres of carbon dioxide at STP
A mole of an ionic compound is defined as that amount of the substance which has mass equal to gram formula mass or which contains Avogadro’s number of formula unit.
1 mole of NaCl = 58.5 grams of NaCl
1 mole of NaCl =6.022 ×1023 formula units of NaCl = 6.022 ×1023 Na+ ion and 6.022 ×1023 Cl– ion.
Importance of Avogadro’s number and Mole Concept
- In the calculation of actual mass of a single atom of an element or a single molecule of a substance.
- In the calculation of the number of atoms or molecules in a given mass of the element of the compound.
- In the calculation of the number of molecules present in a given volume of the gas under given conditions.
- In the calculation of the size of the individual atoms and molecules assuming them to the spherical.
- In the calculation of actual masses of 1 amu or 1 u.
1 amu = 1.6606 × 10-24 g
1 amu = 1.6606 × 10-27 kg
Mole Concept in Solutions
A solution is defined as a homogeneous mixture of two or more chemically non reacting substances, the relative amount of which can be varied up to a certain limit.
If a solution consist of only two components it is called a binary solution.
The component present in smaller amount is called solute while the other present in larger amount is called solvent
The concentration of the solution can be expressed in number of ways:
(1) Mass percent or weight percent :Mass percent of a solute in a solution is the mass of the solute in grams present in 100 grams of solution.
(2) Strength: Strength of a solution is defined as the amount of the solute in grams present per litre of the solution.
(3) Molarity: Molarity of a solution is defined as the number of moles of the solute present per litre of the solution.It is represented by the symbol, M.
(4) Molality :The molality of a solution is defined as number of moles of the solute dissolved in 1 kg of the solvent.
It is represented by the symbol m.
(5) Normality : Normality of a solution is defined as the number of gram equivalent of the solute present per litre of the solution.
It is represented by the symbol N.
Basicity is the number of displaceable H+ ions present in 1 molecule of the acid.
Acidity is the number of displaceable OH– ions present in 1 molecule of the base.
(6) Mole fraction : Mole fraction of any component in the solution is equal to the number of moles of that component divided by the total number of moles of all the components.
For a solution containing n2 moles of the solute dissolved in n1 moles of the solvent,
(7) Parts per million: The concentration of the very dilute solution is expressed in terms of parts of the solute by mass present in million parts by mass of the solution.
Molarity equation
If a solution having molarity M1 and volume V1 is diluted to volume V2 so that the new molarity is M2 then as the total number of moles in the solution remains the same, we have :
Normality Equation
Relation between molarity and normality of the solution of an acid or base
Normality of solution of an acid = Molarity × Basicity
Normality of solution of a base = Molarity × Acidity
Calculation of Molarity and Normality of the solution obtained on mixing two or more solution
If the solutions mixed are of the same substance
If V1 mL of a solution of molarity M1 are mixed with V2 mL solution of the same substance with molarity M2 Or V1 mL of M1 HCL solution are mixed with V2 mL of M2 , then the molarity of the solution obtained is calculates as:
M1 V1 + M2 V2 = M3 (V1 +V2)
N1V1 + N2V2 = N3 (V1 +V2)
some basic concepts of chemistry notes class 11 pdf
Empirical and Molecular Formula
Calculation of Percentage Composition
The percentage of any element or constituent in a compound is the number of parts by mass of that element or constituent present in 100 parts by mass of the compound.
Step 1: Calculate the molecular mass of the compound from its formula by adding the atomic masses of the element present.
Step 2: Calculate the percentage of elements or constituents by applying the relation:
Question: Calculate the percentage of Carbon, Hydrogen, and oxygen in Ethanol.
Answer
Empirical and Molecular Formula
The empirical formula of a compound is the chemical formula which expresses the simplest whole number ratio of the atoms of the various elements present in one molecule of the compound.
For Example : The empirical formula of benzene is CH, hydrogen peroxide is HO, Glucose is CH2O.
The empirical formula represents only the atomic ratio of the various elements present in its molecule.
The molecular Formula of a compound is the chemical formula which represents the true formula of its molecules. It expresses the actual number of atoms of various elements present in one molecule of the compound.
The molecular formula of benzene is C6H6 , hydrogen peroxide is H2O2 , Glucose is C6H6.
Molecular Formula = n × Empirical formula
where n is any integer such as 1, 2, 3 ,4 etc.
The value of n can be obtained from the relation:
Molecular mass = 2 × Vapour Density
Calculation of Empirical Formula
Step 1 : Convert the mass percentage into grams.
Step 2 : Calculate the number of moles.
Step 3 : Calculate the simplest molar ratio: Divide the moles obtained in step 1 by the smallest quotient or the least value from amongst the values obtained for each element.
Step 4 : Calculate the simplest whole number ratio.
Step 5 : Write the empirical formula.
The Empirical Formula is C2H4O.
Calculation of molecular Formula
The molecular formula of a compound can be deduced from:
1) Empirical Formula
2) Molecular mass
Step 1 Calculation of the empirical formula from the percentage composition.
Step 2 Calculation of empirical formula mass by adding the atomic mass of all the atoms present in the empirical formula.
Step 3 Determination of the molecular mass of the compound from the given data.
Step 4 Determination of the value of n.
Step 5 Determination of the molecular formula
Question 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
| Element | Symbol | Percentage of Element | Atomic Mass | Moles of Element | Simplest Molar Ratio | Simplest whole no. Molar ratio |
| Carbon | C | 24.27 | 12 | 2.02 | 1 | 1 |
| Hydrogen | H | 4.07 | 1 | 4.07 | 2 | 2 |
| Chlorine | Cl | 71.65 | 35.5 | 2.02 | 2 | 1 |
The empirical formula of the compound is CH2ClEmpirical mass of CH2Cl = 12 + 2 × 1 +35.5 = 49.5
n=2
Molecular formula = n × E. F.
= 2 × CH2Cl
=C2H4Cl2
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Balancing Of A Chemical Equation
Balancing of a chemical equation means making the number of atoms of each element equal on both sides of the equation.
The methods of balancing equation are:
1) Hit and Trial Method:
The simplest method to balance a chemical equation is by hit and trial method.
Step 1 : Write down the correct formula of the reactants and products with plus sign in between with an arrow pointing from reactants to Products. This is called skeletal equation.
Step 2: Select the biggest formula from the Skeleton equation and equalise the number of atoms of each of its constituent elements on both sides of the chemical equation by suitable multiplication .
Step 3: When an elementary gas appear as a reactant or a product ,the equation is balanced more easily by keeping the elementary gas in the atomic state. The balanced atomic equation is then made molecular by multiplying the whole equation by 2.
Question : Magnetic oxide when heated with hydrogen is reduced to iron and water is also produced. Write balanced equation for the reaction?
Answer : Magnetic oxide + Hydrogen ———-> Iron + Water
Fe3O4 + H2 ——–> Fe + H2O
To equalise the number of atoms of Fe on both sides, multiply Fe by 3 , we have
Fe3O4 + H2 ——–> 3Fe + H2O
The above equation has 4 atoms of O on L.H.S. and 1 atoms of O on R.H.S.
To equalise, multiply water by 4.
Fe3O4 + H2 ——–> 3Fe + 4H2O
The above equation has 8 H atoms on R.H.S. and 2 H atoms on L.H.S.
To equalise ,multiply H2 on L.H.S. by 4
Fe3O4 + 4H2 ——–> 3Fe + 4H2O
This a balanced chemical equation
2) Partial Equation method
Step 1 The chemical reaction represented by the equation is supposed to proceed in two or more steps.
Step 2 The Skeleton equation representing each step are written and then balanced by hit and trial method. These equations are known as partial equations.
Step 3 If necessary the partial equation are multiplied by suitable integers so as to cancel those intermediate products which do not occur in the final reaction .
Step 4 The partial equations are added up to get the final balanced equation.
Question: The skeleton equation to represent the action of chlorine on a hot solution of sodium hydroxide is
NaOH + Cl2 ————> NaCl + NaClO3 + H20
Balance this equation by the method of partial equations
Answer : Cl2 + H2O ———-> H2O + HClO
NaOH + HCl ————> NaCl + H2O
NaOH + HClO ———–> NaClO + H2O
3 NaClO ———–> 2 NaCl + NaClO3
3 Cl2 + 6 NaOH ————> 5 NaCl + NaClO3 + 3 H2O
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Limiting Reagent
The reactant which reacts completely in the reaction is called limiting reactant or limiting reagent.
The reactant which is not consumed completely in the reaction is called excess reactant .
Question : 3 g of H2 react with 29 g of O2 to form H20.Which is the limiting reagent ?
Answer:
Thus O2 is present in excess.Hence H2 is the limiting reagent.