11C01
Some Basic Concepts of Chemistry
Introduction to Chemistry
Chemistry
S
C
E
N
C
E
Importance of Chemistry
History of Chemistry
Introduction to Chemistry
History of Chemistry
The name Chemistry is derived from the word Al-Chemy
Philosopher’s stone (Paras)
Modern chemistry developed in Europe as result of those two quests of the Arabs
‘Elixir of life’ which would grant immortality
History of Indian Chemistry
Indians had their own alchemical traditions
That included much knowledge of
chemical processes and techniques
Chemistry –
Metallurgical Knowledge Chalcolithic cultures
Knowledge of Dyes Atharvaveda
Ancient Medicines Charaka Samhita
Cosmetic Products Varähmihir’s Brihat Samhita
Ancient Indian Chemistry Included
History of Indian Chemistry
Importance of Chemistry
Functioning of Brain
Life-saving Medicines
Eg. Cis - platin
Cancer treatment
Agriculture
Making Fertilisers
11C01.1
Nature of Matter
11C01.1 Nature of Matter
Learning Objectives
Matter and its Physical States
Classification of Matter
11C01.1
CV 1
Matter and its Physical States
Matter
A matter is defined as anything that occupies space, possesses mass and the presence of which can be felt by any one or more of our senses
Air in Football
Weight
Moving Hair in Wind
Not a Matter
Feel
Measure
Not Occupy Space
No Mass
Temperature
Matter
Physical states of Matter
Solid
Liquid
Gas
Properties of Solid, Liquid and Gas
Properties | Solid | Liquid | Gas |
Volume | Definite | Definite | Indefinite |
Shape | Definite | Indefinite | Indefinite |
Intermolecular force | Very high | Moderate | Negligible |
Intermolecular space | Very small | Slightly greater | Very large |
Compressibility | No | No | Very high |
Expansion on heating | Very little | Very little | Very high |
Rigidity | Highly rigid | Not rigid | Not rigid |
Fluidity | Can’t flow | Can flow | Can flow |
11C01.1
CV 2
Classification of Matter
Matter
Mixture
Pure Substance
Homogeneous
Heterogeneous
Elements
Compounds
Classification of Matter
Pure substances
Pure Ghee
Pure Milk
Pure Honey
Are they really Pure
?
Elements
Compounds
Atoms
Molecules
K
Fe
Ni
C
H
H
O
O
O
H
H
O
O
Molecules
Pure substances
C
Mixture
Contains particles of two or more pure substances which may be present in any ratio
Composition is variable
Pure substances forming mixture are called components
Can be separated by filtration, distillation, evaporation etc.
Examples - sugar solution in water, air, tea, dal etc.
Homogeneous Mixture
Example - Lemonade
Heterogeneous Mixture
Example - Daal
Cook
11C01.1
PSV 01
Q. Categorise the following as homogeneous and heterogeneous mixtures.
(i) sugar-water solution, (ii) air, (iii) mixture of pulses
Pause the video
Time duration : 1 minute
Q. Categorise the following as homogeneous and heterogeneous mixtures.
(i) sugar-water solution, (ii) air, (iii) mixture of pulses
Sol.
Water + Sugar
Effect of Air
Mixture of pulses
Homogeneous mixture of gases
Homogeneous mixture
Heterogeneous mixture
ConcepTest
Ready for Challenge
Pause the video
Time duration : 1 minute
Sol.
Fat, moisture and vitamin-
Mixture
Carbohydrate, sugar, water-
Mixture
Calcium, carbohydrate, fat, sugar, water-
Mixture
Summary
History of Chemistry
Importance of Chemistry
Matter and its physical states
Pure substance – atoms and molecules
Properties of solid, liquid and gas
Classification of matter
Mixture – homogeneous and heterogeneous
Workbook questions : Q9, Q10
11C01.1 Reference questions
11C01.2
Properties of Matter and their Measurement
11C01.2 Properties of Matter and their Measurement
Learning Objectives
Properties of Matter and measurement of Physical Properties
Mass, Volume, Density and Temperature
Scientific Notation and Uncertainty in Measurement
11C01.2
CV 1
Properties of Matter and measurement of Physical Properties
Properties of Matter
Physical properties
Chemical properties
Physical properties
Can be measured without changing the composition of the substance
Colour of Substance
Melting Point
Boiling point
Measurement does not require occurrence of a chemical change
Chemical properties
During measurement, there is a change in composition of the substance
Burning of substance
Acid-base nature
Chemical reactivity
Chemical change will occur during measurement
Measurement of Physical Properties
❓
Quantitative measurement of properties is required for scientific investigation
Quantitative physical properties
Length
Area
Volume
How to measure a Physical quantity
❓
Compare the physical quantity with a Reference Standard
Unit
Mass standard is the kilogram
Representation of Physical Properties
Magnitude
28
m
Unit
Measuring Length
Physical Quantity = Magnitude(n) × Unit(u) = n u
Systems of Measurement
Earlier Systems
New Systems
English System
Metric System
Originated in France. Based on decimal system thus, more convenient
The International System of Units (SI)
Units in SI System
Seven Fundamental or Base Units
Derived Units
Related to seven fundamental scientific quantities
Derived with the help of base units
Base Physical Quantity | Symbol of Quantity | Name of SI Unit | Symbol of SI Unit |
Length | l | metre | m |
Mass | m | kilogram | kg |
Time | t | second | s |
Electric current | I | ampere | A |
Thermodynamic temp. | T | kelvin | K |
Amount of substance | n | mole | mol |
Luminous intensity | Iv | candela | cd |
Seven Base Physical Quantities and their Units
11C01.2
CV 2
Mass, Volume, Density and Temperature
Mass
Amount of matter present in substance
❓
SI unit of mass is kilogram
Determined by analytical balance
Volume
Space occupied by a substance
SI unit of volume is m3
Relation between different units of volume -
1 L = 1000 mL = 1000 cm3 = 1 dm3 = 10-3 m3
Burette
Volumetric flask
Density
Mass per unit volume
Chemist often expresses density in g cm–3
SI unit of density
Temperature
°C (degree Celsius)
°F (degree Fahrenheit)
K (kelvin)
Freezing point of water
Boiling point of water
Temperature
Relation b/w Celsius and Fahrenheit
Relation b/w Celsius and Kelvin
11C01.2
CV 3
Scientific Notation and Uncertainty in Measurement
Scientific Notation
To simplify the calculation, any number can be represented in terms of exponential notation
Digit term
varies 1.000 to 9.999
Exponent
232.508 can be written as-
2.32508 ×102
0.00016 can be written as-
1.6 × 10–4
Q. Using scientific notation, solve the following
i. (6.65 × 104) + (8.95 × 103) ii. (2.5 × 10–2 ) – (4.8 × 10–3)
Pause the video
Time duration : 1 minute
Q. Using scientific notation, solve the following
i. (6.65 × 104) + (8.95 × 103) ii. (2.5 × 10–2 ) – (4.8 × 10–3)
Sol.
(6.65 × 104) + (8.95 × 103)
= (6.65 × 104) + (0.895 × 104)
= (6.65 + 0.895) × 104
= 7.545 × 104
(2.5 × 10–2 ) – (4.8 × 10–3)
= (2.5 × 10–2) – (0.48 × 10–2)
= (2.5 – 0.48) × 10–2
= 2.02 × 10–2
Uncertainty in Measurement
Experimental measurement or result has some amount of uncertainty
Uncertainty in experimental values is indicated by significant figures
Significant figures
Uncertainty is indicated by writing the certain digits and the last uncertain digit
Experimental result
11 . 2 ml
Certain
Uncertain
Rules for determining Significant figures
1- All non-zero digits are significant
2- Zeros preceding to 1st non-zero digit are not significant
285 cm
3
0.25 ml
2
0.0052
2
3- Zeros between 2 non-zero digits are significant
2.005
4
Rules for determining Significant figures
4- Zeros at the end or right of a number are
significant, provided they are on the right side of the decimal point
5- Exact numbers have infinite significant figures
0.200
3
2 balls
Infinite
30 eggs
Infinite
Precision and Accuracy
Precision
Accuracy
Closeness of our result in different attempts
Closeness of our result with actual result
Precision and Accuracy
Neither precise nor accurate
Precise but not accurate
Precise and
accurate
11C01.2
PSV 1
Q. Calculate the significant figure for the following addition-
12.11 + 18.0 + 1.012
Pause the video
Time duration : 1 minute
Q. Calculate the significant figure for the following addition-
12.11 + 18.0 + 1.012
Sol.
12.11 + 18.0 + 1.012 = 31.122
Result cannot have more digits to the right of the decimal point than either of the original numbers
Here, 18.0 has only one digit after decimal
Result should be reported only up to one digit after decimal
31.122
11C01.2
PSV 2
Q. A jug contains 2 L of milk. Calculate the volume of the milk in m3
Pause the video
Time duration : 1 minute
Q. A jug contains 2 L of milk. Calculate the volume of the milk in m3
Sol.
We know, 1 L = 10-3 m3, Thus
Called Unit Factors
The numerator should have that part which is required in the desired result
Dimensional Analysis
Summary
Physical properties
Chemical properties
Measurement of physical properties
Uncertainty in measurement – significant figures, precision and accuracy
SI units
Mass, volume, density and temperature
NCERT Exercise Questions: 1.15, 1.16, 1.18, 1.19, 1.20, 1.22, 1.27, 1.31
11C01.2 Reference questions
Workbook Questions: 1, 7, 8, 9, 10, 19
11C01.3
Laws of Chemical Combinations,
Atomic and Molecular Masses
11C01.3 Laws of Chemical Combinations,
Atomic and Molecular Masses
Learning Objectives
Laws of Chemical Combinations
Dalton’s Atomic Theory, Atomic and Molecular Masses
11C01.3
CV 1
Laws of Chemical Combinations
Laws of Chemical Combinations
Elements combine together chemically to form compounds
O
C
O
C
O
O
These chemical combinations are based on some laws
Laws of Chemical Combinations
Law of Conservation of Mass
Law of Definite Proportions
Law of Multiple Proportions
Gay Lussac’s Law of Gaseous Volumes
Avogadro’s Law
Law of Conservation of Mass
Antoine Lavoisier
In all physical and chemical changes, no net change in mass during the process
Matter
Neither be created
Nor be destroyed
Law of Definite Proportions or
Law of Definite Composition
Joseph Proust
Cupric Carbonate
Elements | % |
Cu | 51.35 |
C | 9.74 |
O | 38.91 |
Irrespective of the source, a compound always contains same elements combined together in the same proportion by mass
Natural
Synthetic
Law of Multiple Proportions
Dalton
If two elements combine to form more than one compounds, masses of one element that combine with fixed mass of the other element, are in ratio of small whole numbers
Reaction between Hydrogen and Oxygen
Hydrogen 2 g + Oxygen 16 g
Hydrogen 2 g + Oxygen 32 g
O
H
H
O
O
H
H
Water 18 g
Hydrogen Peroxide 34 g
Ratio b/w masses of Oxygen
16 : 32 = 1 : 2
Gay Lussac’s Law of Gaseous Volumes
At constant T and P, when gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume
Reaction between Hydrogen and Oxygen
Substance
O
H
H
H
H
O
O
Volume in ml
Simple Ratio
100
50
100
2
1
2
H
H
H
H
N
N
N
N
N
N
N
N
H
H
H
H
Equal volumes of all gases at same T and P should contain equal number of molecules
H
H
H
H
H
H
N
N
N
N
N
N
Much lighter gas
Heavier than Hydrogen
Number of Molecules
n
V ∝ n
Irrespective of mass of gas molecule
Volume = V
Volume = V
Avogadro’s Law
O
H
H
H
H
H
O
O
H
O
H
H
Avogadro’s Law
V ml
V ml
2V ml
n
n
2n
V ml
Reaction between Hydrogen and Oxygen
n
11C01.3
CV 2
Dalton’s Atomic Theory
Atomic & Molecular Masses
Dalton’s Atomic Theory
1 - Matter consists of indivisible atoms
2 - All atoms of a given element have identical
properties, including identical mass. Atoms of different elements differ in mass
3 - Compounds are formed when atoms of
different elements combine in a fixed ratio
4 - Chemical reactions involve reorganisation of atoms. These are neither created nor destroyed in a chemical reaction
O
H
H
H
H
O
O
H
H
O
H
H
Dalton’s Atomic Theory
Explain the laws of chemical combination
Could not explain the laws of gaseous volumes
Could not provide the reason for combining of atoms
Atomic Mass
The mass of an atom
An atom is very small
Mass of an atom is also extremely small
How to measure ?
Determination of Atomic Mass
In 19th century, scientists could determine the mass of one atom relative to another by experimental means
Lightest atom
Other elements were assigned masses relative to it
H
Arbitrarily assigned a mass of 1 (without any units)
But was not successful
Determination of Atomic Mass
Present system of atomic masses is based on
C-12
A mass of exactly 12 atomic mass unit (amu)
Masses of all other atoms are given relative to this standard
amu is one-twelfth of the mass of one carbon - 12 atom
amu = u (unified mass)
11C01.3
PSV 1
Pause the video
Time duration : 1 minute
Sol.
Average Atomic Mass
Generally, naturally occurring elements exist as more than one isotope
A
X
Y
B
X
Z
Same atomic number
Different atomic mases
Calculation of Average Atomic Mass
Avg. Atomic Mass
11C01.3
PSV 2
Q. Calculate average atomic mass of carbon using table given below.
Pause the video
Time duration : 1 minute
Isotope | Relative Abundance (%) | Atomic Mass (amu) |
C - 12 | 98.892 | 12 |
C - 13 | 1.108 | 13.00335 |
C - 14 | 2 ×10–10 | 14.00317 |
Sol.
Isotope | Relative Abundance (%) | Atomic Mass (amu) |
C - 12 | 98.892 | 12 |
C - 13 | 1.108 | 13.00335 |
C - 14 | 2 ×10–10 | 14.00317 |
Molecular Mass
It is the sum of atomic masses of all the elements present in a molecule
ConcepTest
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Pause the video
Time duration : 1 minute
Sol.
Molecular mass of glucose (C6H12O6) -
ConcepTest
Ready for Challenge
Q. Calculate the molecular mass of crystalline oxalic acid.
Pause the video
Time duration : 1 minute
Sol.
Q. Calculate the molecular mass of crystalline oxalic acid.
Formula Mass
Some substances do not contain discrete molecules as their constituent units
Positive and negative entities are arranged in a 3-D structure
Formula is used to calculate the formula mass instead of molecular mass
Summary
Laws of chemical combinations
Dalton’s atomic theory
Atomic mass
Average atomic mass
Molecular mass
NCERT Exercise Questions: 1.21, 1.32
11C01.3 Reference questions
Workbook Questions: 11, 14, 18, 20
11C01.4
Mole Concept and percentage Composition
Learning Objectives
Mole , Avogadro’s number & Molar Mass
Percentage Composition , Empirical & Molecular Formula
11C01.4 Mole Concept and Percentage Composition
11C01.4
CV 1
Mole, Avogadro’s number & Molar Mass
Why Mole?
To deal with such large numbers we invented mole.
≈ 1.67 x 1021 Water molecules
Mole is just a number as :
What is Mole??
C
C
Entities in Mole :
11C01.4
PSV 1
Pause the video
Time duration : 1 minute
Sol.
Molar Mass
The mass of one mole of a substance in grams is called its molar mass
O
H
Example :
H
ConcepTest
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Time duration : 1 minute
Sol.
ConcepTest
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Time duration : 1 minute
11C03.4
CV 2
Percentage Composition , Empirical & Molecular Formula
Cu
S
O
H
% composition of elements??
Percentage Composition
Percentage Composition
ConcepTest
Ready for Challenge
Q. Calculate the mass percentage of each element in ammonia.
Pause the video
Time duration : 1 minute
Empirical Formula
It represents the simplest whole number ratio of various atoms present in a compound.
Molecular formula
It shows the exact number of different types of atoms present in a molecule of a compound.
Molecular Formula = n × Empirical formula
11C01.4
PSV 2
Q. 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?
Pause the video
Time duration : 1 minute
Sol.
Tabular Method for calculation of Empirical formula
Elements in Compound | % Composition | | %Composition/ molar mass | Molar ratio | Simple whole number ratio |
| | | | | |
| | | | | |
| | | | | |
Summary
11C01.4 Reference questions
Workbook questions : Q2, Q4, Q6, Q12, Q17
NCERT Exercise questions : 1.1, 1.2 , 1.3, 1.8, 1.10, 1.28, 1.30, 1.33, 1.34
11C01.5
Balancing, Stoichiometry & limiting reagent of a Chemical Equation
Learning Objectives
Balancing of Chemical equations
Stoichiometry & Stoichiometric calculations
Limiting Reagent
11C01.5 Balancing, Stoichiometry & limiting reagent of a
Chemical Equation
11C01.5
CV 1
Balancing of Chemical equation
Balancing of a chemical Equation
What is Balanced chemical Reaction/Equation??
A balanced chemical equation has the same number of atoms of each element on both sides of the equation .
Why balancing is necessary??
The chemical equation needs to be balanced so that it follows the law of conservation of mass.
How to Balance a Chemical Equation
The best way to balance a chemical equation is by hit & trail method.
Carbon is balanced
Hydrogen is balanced
Oxygen is balanced
11C01.5
CV 2
Stoichiometry & Stoichiometric calculations
Stoichiometry
Quantitative analysis of a balanced chemical reaction
Example : Burning of Methane
Stoichiometric Observations :
11C01.5
PSV 1
Q. Calculate the amount of water (g) produced by the combustion of 48 g of methane.
Pause the video
Time duration : 1 minute
Q. Calculate the amount of water (g) produced by the combustion of 48 g of methane.
ConcepTest
Ready for Challenge
Q. Calculate the amount of carbon dioxide that could be produced when 2 mole of carbon is burnt in air.
C
Pause the video
Time duration : 1 minute
C
Sol.
Q. Calculate the amount of carbon dioxide that could be produced when 2 mole of carbon is burnt in air.
11C01.5
CV 3
Limiting Reagent
Limiting Reagent
The reactant which gets consumed first and limits the amount of product formed
Method to solve problems based on limiting Reagent :
Step 1. Write balanced chemical equation
Step 2. Calculate moles of each compound
Step 3. Calculate ratio of number of moles to the stoichiometry coeff.
“ Compound with minimum ratio will be the limiting reagent.”
Step 4. Do all calculations based on the availability of limiting reagent.
What is limiting Reagent ???
11C01.5
PSV 2
Q.
Pause the video
Time duration : 2 minute
Sol.
>
ConcepTest
Ready for Challenge
Q. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous hydrochloric acid according to the reaction 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g) How many grams of HCl react with 5.0 g of manganese dioxide?
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Time duration : 1 minute
Sol.
<
Summary
Stoichiometry & Stoichiometric Observations
Balancing of Chemical Reactions
Method to Determine Limiting Reagent
11C01.5 Reference questions
Workbook questions : Q13 .
NCERT Exercise questions : 1.4, 1.7, 1.23, 1.24, 1.36 .
11C01.6
Reactions in Solution
Learning Objectives
Introduction to Solutions
Mass % & Mole Fraction
Molarity
Molality
11C01.6 Reactions in Solution
11C01.6
CV 1
Introduction to Solutions
Solutions??
A solution is a homogeneous mixture of two or more substances
A solution may exist in any phase
Introduction to the world of solutions
Bronze
Cold drinks
Air
Solute
Solvent
Substance to be dissolved
Substance which dissolve other substances
Solute is usually present in a smaller amount than the Solvent
Solution = Solute + Solvent
Examples :
11C01.6
CV 2
Mass % & Mole Fraction
Concentration
The quantity of solute present in a given quantity of solution
Concentration can be measured in :
Mass Percent
Mole Fraction
Mole Fraction
Note : sum of mole fractions of all the compounds present in the solution is unity
A
B
A
A
A
A
A
A
B
B
B
B
B
B
11C01.6
PSV 1
Q. If 4 moles of alcohol and 6 moles of water are mixed then
calculate mole fraction of each component.
Pause the video
Time duration : 1 minute
Q. If 4 moles of alcohol and 6 moles of water are mixed then
calculate mole fraction of each component.
Sol.
11C01.6
CV 3
Molarity
Molarity
The number of moles of solute present per litre of solution. It is represented by “M”
For dilution of solutions
( On adding water )
11C01.6
PSV 2
Pause the video
Time duration : 1 minute
Sol.
ConcepTest
Ready for Challenge
Pause the video
Time duration : 1 minute
Sol.
Molarity of solution = 0.375 M
Volume of solution = 500mL = 0.5L
11C01.6
CV 4
Molality
Molality
11C01.6
PSV 3
Pause the video
Time duration : 1 minute
Sol.
ConcepTest
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Pause the video
Time duration : 1 minute
Sol.
Summary
11C01.6 Reference questions
Workbook questions : Q16, Q21
NCERT Exercise questions : 1.5, 1.6, 1.11, 1.12, 1.29, 1.35