This document provides an overview of some basic concepts in chemistry. It defines matter as anything that has mass and occupies space, and describes the three common states of matter as solids, liquids, and gases. It then discusses the characteristics of these different states, and introduces concepts like homogeneous and heterogeneous mixtures, pure substances, and the international system of units used in chemistry. Finally, it summarizes some important laws of chemical combination like the law of conservation of mass and Dalton's atomic theory.

1. SOME BASIC CONCEPTS OF
CHEMISTRY
KALPAJYOTI DIHINGIA
DEPARTMENT OF CHEMISTRY
DHSK COLLEGE, DIBRUGARH

2. MATTER
●
ANYTHING THAT HAS MASS AND OCCUPIES
SPACE IS CALLED MATTER
●
FOR EXAMPLE: BOOKS, PEN, PENCILL,
WATER AND ALL LIVING BEINGS.
●
STATES OF MATTER: SOLID, LIQUID AND
GAS

3. MATTER
●
ANYTHING THAT HAS MASS AND OCCUPIES
SPACE IS CALLED MATTER
●
FOR EXAMPLE: BOOKS, PEN, PENCILL,
WATER AND ALL LIVING BEINGS.
●
STATES OF MATTER: SOLID, LIQUID AND
GAS

4. REPRESENTATION OF SOLID
LIQUID AND GAS

5. CHARACTERISTICS
●
SOLIDS HAVE DEFINITE SHAPE AND
VOLUME
●
LIQUIDS HAVE DEFINITE VOLUME BUT NOT
THE DEFINITE SHAPE
●
GASES HAVE NEITHER DEFINITE VOLUME
NOR DEFINITE SHAPE.

6. CLASSIFICATION

7. Homogeneous and Heterogeneous
●
A mixture contains two or
more substances in it. Which
is called as components
●
Homogeneous mixture:
– Components are completely
mixed with each other.
●
Heterogeneous mixture:
– Composition is not uniform.
●
e.g: A mixture of salt, sugar,
and stone.

8. PURE SUBSTANCES
●
THEY HAVE FIXED COMPOSITION, WHEREAS
MIXTURE MAY CONTAIN THE COMPONENTS IN ANY
RATIO AND THEIR COMPOSITION IS VARIABLE.
●
COPPER, SILVER, GOLD, WATER, GLUCOSE are
some examples of Pure substances

10. International system of units (SI)
●
SI system has seven
base units and they are
listed in Table 1.1

11. Mass volume and density
●
Mass and Volume:
– Mass of a substance is
the amount of matter
present in it while
weight is the force
exerted by gravity on an
object.
– SI unit of mass is
kilogram.
●
Volume:
– Volume has unit of
(length)3
– So the S.I unit is m3
●
Density:
– Amount of mass per
unit volume.
– SI unit of density is
Kg/m3

12. Temperature
●
There are 3 common
scale to measure
temperature: °C, °F and
K.
●
Kelvin (K is the S.I unit
of temperature)
●
°F = (9/5)(°C) + 32
●
K = °C + 273.15

13. Scientific notation (Home work)
●
A Chemist has to deal with large number such as
602,200,000,000,000,000,000,000 or
0.000000000000000000000000000166 g mass of a
H atom
●
Similarly speed of light, plank’s constant, charges
on particles etc.

14. Scientific notation
●
N x 10^n,
Where n is an exponent having positive or negative values and N is a
number called as digit which varies between 1.000... to 9.999...

16. Multiplication and division

17. Practice

18. Division

19. Addition

20. Substraction

21. Substraction

22. Solve them:

23. Precision and Accuracy
●
Precision refers to the closeness of various
measurements for the same quantity.
●
Accuracy is the agreement of a particular value to
the true value of the result.

24. Example of Accuracy and
precision
●
Let us consider for an exeriment, the true value of the
result is 2.0 g. And three students viz. A, B, C
respectivly taking measurements in two sets.
– Student A reports results as 1.95 g and 1.93 g.
●
These values are precise as they are closed to each other but not
accurate.
– Student B reports results as 1.94 g and 2.05 g.
●
These obeservations are neither precise nor accurate.
– Student C reports results as 2.01 g and 1.99 g.
●
These values are both precise and accurate

25. Significant figures
●
2.0 Km = ?
(a) 2000m
(b) 2.0 x 103 m
(c) 0.2 x 104 m
(d)All of the above

26. Defination of Significant figure
●
Significant figures tells us the number of digits in a
measured value, we are confident of.
●
More the significant figures, more accurate
measurement.

27. Rules for significant figures
●
1. All non zero digits are significant.
●
Example: 1234 ------> 4 S.F
●
234 ------> 3 S.F
●
212 -------> 3 S.F
●
2. Trapped zeros, ie zeros between two nonzero digits
(or between two significant figures) are significant.
●
Example: 1.004 -------> 4 S.F
●
20.3 ------> 3 S.F
●
300.002 ----> 6 S.F

28. Rules for significant figure
●
3. Initial zeros/leading zeros are never
significant.
Example
●
0.001 ---------> 1 S.F
●
0.313 --------> 3 S.F
●
0.032 -------> 2 S.F
●
0.0204 ------> 3 S.F

29. ●
4. Ending Zeros/Trailing zeros are significant if
they appear after decimal
Example
●
2.00 --------> 3 S.F
●
3.120 -------> 4 S.F
●
0.003020 ---------> 4 S.F
●
200 ---------> 1 S.F
●
1400 -------> how many significant figure

30. Rules for significant figure
●
Numbers of Significant figure in
– 20.00 = ? , 1310.00 = ? , 13100 = ?
–
●
5. Order of magnitude is never significant.
– 2.1 x 103 -----------> 2 S.F
– 2.010 x 106 ---------> 4 S.F
– 200 = 2 x 102 -------> 1 S.F
– 1400 = 1.4 x 103-------> 2 S.F

31. Rules for significant figure
●
6. Pure numbers or constants have infinite
significant figure
– For example,
●
perimeter of a squre = 4a,
●
Area of a cube = 6a2
● Refractive index of water μw = 1.33
●
Speed of light

32. Calculating significant figures
a 212
b 2.120
c 2.0042
d 310.020
e 310.00
f 3100
g 4.20
h 0.004

33. While changing unit, No of S.F
remains same
●
1.5 m = 150 cm
●
2.0 Km = 2000 m
= 2.0 x 103 m ----------> 2 S.F

34. Rounding off
●
Round off to two significant figures:
– 3.72 =
– 3.76 =
– 3.752 =
– 4.653 =
– 3.750 =
– 3.650 =

35. Rounding off
●
Round off to two significant figures:
– 3.72 = 3.7 ( 2<5)
– 3.76 = 3.8 (6>5)
– 3.752 = 3.8 (5=5, followed by nonzero)
– 4.653 = 4.7
– 3.750 = 3.8 ( odd number is present before 5)
– 3.650 = 3.6 (even no is present before 5)

36. Calculations in S.F
●
Result of addition, substraction, multiplication and division
of measured values cannot be more accurate than least
accurate measurements.
– a+b+c = R
●
Addition & substraction:
– The result of addition & substraction must have the same number
of decimal places as present in the value with least decimal
places.

37. Questions
●
2.24 + 0.3 =
●
0.331 + 1.12 + 0.03 =
●
Box, mass = 1.4kg
●
Diamond, mass = 1.32 g
●
Total mass of the system = ?

38. Questions
●
2.24 + 0.3 = 2.54, Ans = 2.5 ( rounded off to 2 decimal
places)
●
0.331 + 1.12 + 0.03 = 1.481, Ans = 1.48
●
Box, mass = 1.4kg
●
Diamond, mass = 1.32 g
●
Note: Number of significant figure doesn’t change on changing units
●
Total mass of the system = (1.4 + 0.00132)kg
= 1.4 kg

39. Classwork
●
5.8 x 104 -1.5 x 103 = ?

40. LAWS OF CHEMICAL
COMBINATION
●
THE COMBINATION OF ELEMENTS TO
FORM COMPOUNDS IS GOVERNED BY
FOLLOWING 5 BASIC LAWS.
– LAW OF CONSERVATION OF MASS
– LAW OF DEFINITE PROPORTION
– LAW OF MULTIPLE PROPORTIONS
– GAY Lussac’s LAW OF GASEOUS VOLUMES
– AVOGADRO’S LAW

41. 1. LAW OF CONSERVATION
OF MASS
●
IT STATES THAT MATTER CAN NEITHER BE
CREATED NOR BE DESTROYED.
– THIS LAW WAS PROPOSED BY ANTONIE
LAVOISIER IN 1789. HE PERFORMED CAREFULL
EXPERIMENTAL STUDIES FOR COMBUSTION
REACTIONS FOR REACHING TO THE ABOVE
CONCLUSION.

42. 2. LAW OF DEFINITE
PROPORTION
●
A GIVEN COMPOUND ALWAYS CONTAINS
EXACTLY THE SAME PROPORTION OF
ELEMENTS BY ITS MASS.
●
He worked two samples of cupric carbonate
(CuCO3).
●
% of
Copper
% of
oxygen
% of
carbon
Natural
sample
51.35 9.74 38.91
Synthetic
sample
51.35 9.74 38.91

43. 3. LAW OF MULTIPLE
PROPORTION
●
IF TWO ELEMENTS COMBINE TO FORM MORE THAN
ONE COMPOUND, THE MASSES OF ONE ELEMENT
THAT COMBINE WITH A FIXED MASS OF OTHER
ELEMENTS ARE IN RATIO OF SMALL WHOLE
NUMBERS. This law was proposed by Dalton in 1803.

44. 4. Gay Lussac’s Law of Gaseous
Volumes
●
According to this law, in a gaseous reaction, the
reactants are always combined in a simple ratio by
volume and form products, at same temperature
and pressure.
A(g) +3B(g) 2C(g)
1V 3V 2V
1 : 3 : 2

45. Example
●
One volume of Hydrogen combines with one
volume of chlorine to produce 2 volume of
hydrogen chloride.
Simple Ratio = 1 : 1 : 2
H2(g) + Cl2(g) 2HCl(g)
1V 1V 2V
●
Ratio = 1 : 1 : 2

46. More examples
● N2(g) +3H2(g) 2NH3(g)

47. Avogadro’s Law
●
Avogadro proposed that equal volumes of all gases
at the same temperature and pressure should
contain equal number of molecules.
50 H2 molecules 50 O2 molecules 50 CO2 molecules

48. Under similar conditions
●
60 ml of CO ------------> x molecules
● 60 ml of H2 ------------> x molecules
● 60 ml of CO2 ------------> x molecules
● 60 ml of SO2 ------------> x molecules
●
30 ml of CO -------------> x/2 molecules
● 120 ml of N2 --------------> ?
● 240 ml of N2 ----------------> ?

49. ●
Volume of gases α Number of molecules
Temperature and pressure is constant
●
UNDER SAME CONDITION
– 50 l H2 Y molecules
– 100 l of CO2
– 200 l of CH4
– 75 l of N2

50. DALTON’S ATOMIC
THEORY,1808
●
Matter is made up of extremely small particles and
indestructible particles called as atoms
●
Atoms of same elements are identical in all
respects
●
Atoms of different elements are different in all
respects
●
Atom is the smallest unit that takes part in
chemical reactions.

51. ●
Atoms of two or more elements combine in a fix
ratio to form compounds.
●
Atoms can neither be created nor be destroyed
during any chemical and physical change.
●
Atoms of two elements combine in different ratio
to form more than one compound

52. Limitations
●
It could explain the laws of chemical composition by mass
but failed to explain the laws of gaseous volumes
●
Could not explain why atoms of different elements have
different size, mass and shape.
●
It could not explain why atoms of different elements combine
to form compounds or molecules
●
It failed to explain the nature of forces that bind atoms
together.
●
It makes no distinction between ultimate particles of elements
and compounds.

53. Atomic mass
●
Mass of one atom of an element is called as atomic
mass.
● 12C is assigned as a mass of exactly 12 atomic mass unit (amu)
and masses of all other atoms are given relative to this standard
12C mass= 12 amu
1 amu = (1/12) x mass of 12C
– One atomic mass is defined as a mass exactly equal to
one twelfth the mass of one Carbon-12 atom.
– And 1 amu = 1.66056 x 10-24 g

54. For H atom
●
Mass of an atom of hydrogen atom= 1.6736x10-24 g
Now, converting the mass of hydrogen atom to amu
We know that
1.66056 x 10-24 g = 1 amu
1.0 g = 1.0 /(1.66056 x 10-24) amu
1.6736x10-24 g = (1.6736x10-24 )/ (1.66056 x 10-24)
= 1.008 amu
●
At present, ‘amu’has been replaced by ‘u’, which is known as
unified mass

55. Average atomic mass
●
Most of the elements exist as more than one isotope.
●
Average atomic mass of that element can be computed considering
the existance of these isotopes and their relative
abundance( %occurance).
From the above data, the average atomic mass of carbon will come out to be:
(0.98892) (12 u) + (0.01108) (13.00335 u) + (2 × 10 –12 ) (14.00317 u) = 12.011 u

56. Molecular mass
●
Molecular mass is the sum of atomic masses of the
elements present in a molecule.
For example, molecular mass of methane,
CH4 = (12.001 u ) + 4 x (1.008 u) = 16.043 u
H2O = (2x1.008 u) + 16.00 = 18.02 u

57. Classwork
● Calculate the molecular mass of glucose (C6H12O6).
●
●
C = 12
●
H = 1
●
O = 16
12.011x6 + 1.008x12 + 16.00 x6

58. Classwork
● Calculate the molecular mass of glucose (C6H12O6).
Solution:
= 6(12.011 u) + 12(1.008 u) + 6(16.00 u)
= (72.066 u) + (12.096 u) + (96.00 u)
= 180.162 u

59. Formula mass
Some substances, such as sodium chloride, do not contain discrete
molecules as their constituent units. In such compounds, positive (sodium
ion) and negative (chloride ion) entities are arranged in a three-dimensional
structure, as shown in figure.
It may be noted that in sodium chloride, one Na + ion is surrounded by six
Cl – ion and vice-versa.

60. ●
The formula, such as NaCl, is used to calculate the
formula mass instead of molecular mass as in the
solid state sodium chloride does not exist as a
single entity.
Thus, the formula mass of sodium chloride is:
atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u

61. Mole concept
In SI system, mole (symbol, mol) was introduced as
seventh base quantity for the amount of a substance.
●
The mole, symbol mol, is the SI unit of amount of
substance. One mole contains exactly 6.02214076 ×1023
elementary entities. This number is the fixed numerical
value of the Avogadro constant, NA , when expressed in
the unit mol–1 and is called the Avogadro number.
●
1 mole = number = 6.022 x 1023 = Avogadro number
●
1 mole of H atoms = 6.022 x 1023 atoms
●
1 mol of water molecules = 6.022 x 1023 atoms

62. ●
6.022 x 1023 atom of C = 1 mol of carbon atom
Mass of 1 mol of C = 12 x 1.67 x10-24 x 6.022 x 1023
= 12 g
●
Mass of 1 mol of O atoms = ?
– Solution:
●
Mass of 1 O atom = 16 amu
= 16 x 1.67 x 10-24 g
mass of 1 mol O atoms
= 6.022 x 1023 x 16 x 1.67 x 10-24 g
= 16 g
The mass of one mole of a substance in
grams is called its molar mass.

63. ●
Find the number of moles of O atoms present in 64
g of O

64. ●
Number of moles = Given mass / molar mass
●

65. ●
Find the number of moles of atoms present in
– 28 g of N atoms
– 32 g of S atoms
– 14 g of H atoms
–
– Molar mass of N = 14
●
28/14 = 2 mol
– 32/32 = 1 mol
– 14/1 = 14 mol

66. PERCENTAGE COMPOSITION
●
Let us understand it by taking the example of water
(H2O). Since water contains hydrogen and oxygen,
the percentage composition of both these elements
can be calculated as follows:

67. ●
What is the percentage of carbon, hydrogen and
oxygen in ethanol?
– Formula of ethanol is C2H5OH

68. Empirical Formula for Molecular
Formula
●
The simplest formula of a substance which gives the
relative number of atoms of each element present in
the molecule present in the molecule of that
substance is called its empirical formula.
●
For example: consider a covalent compound with
molecular formula C6H6.
●
Lowest whole number ratio between the C and H is
1:1
●
Therefore, empirical formula of a compound having
molecular formula of C6H6 => CH

69. ●
The symbolic representation of a molecule of
any substance describing the actual number of
atoms in it, is called its molecular formula.
– Note: If the mass per cent of various elements present
in a compound is known, its empirical formula can be
determined. Molecular formula can further be obtained
if the molar mass is known. The following example
illustrates this sequence.

74. STOICHIOMETRY AND
STOICHIOMETRIC CALCULATIONS
●
Stoichiometry deals with the calculation of masses
(sometimes volumes) of the reactants and the
products involved in a chemical reaction.
●
Let us consider the combustion of methane. A balanced
chemical equation :
– CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
– In this reaction CH4 (g) i.e, methane and O2 (g) are reactants
and Carbon dioxide and water are the products.
– Coefficient 2 for O2 and H2O &
– Coefficient 1 for CH4 and CO2 are called as stoichiometric
coefficients

75. ●
Thus according to the formula
– CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)
– 1 mol CH4 reacts with 2 moles of O2 to give 1 mol of CO2
and 2 moles of H2O
– One molecule of CH4(g) reacts with 2 molecules of O2(g)
to give one molecule of CO2 (g) and 2 molecules of
H2O(g)
– 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to give 44 g
of CO2 (g) and 2×18 g of H2O (g).

76. ●
Calculate the amount of water (g) produced by the
combustion of 16 g of methane.
●
● CH4 + 2O2 ---------> CO2 + 2H2O
● 1 mol CH4 gives 2 mols of H2O
● 1 mol CH4 = 16 g
● Convert 2 mol of H2O into grams
● Answer: 2x 18= 36g of H2O

77. Limiting Reagent
●
In chemical reactions reactants reacts to give products.
In some reactions one reactant may present in more
amount than the amount required by the reaction. The
reactant which present in the least amount gets
consumed after some time and no further reaction takes
place whatever may be the amount of other reactant.
●
Hence, the reactant, which gets consumed first, limits
the amount of product formed and is, therefore, called
the limiting reagent.

78. Reactions in Solutions
●
A majority of reactions in the laboratories are carried out
in solutions. Therefore, it is important to understand as
how the amount of substance is expressed when it is
present in the solution. The concentration of a solution or
the amount of substance present in its given volume can
be expressed in any of the following ways.
– 1. Mass per cent or weight per cent (w/w %)
– 2. Mole fraction
– 3. Molarity
– 4. Molality

79. 1. Mass per cent
●
It is obtained by using the following relation:
Q. A solution is prepared by adding 2 g of a
substance A to 18 g of water. Calculate the mass
per cent of the solute.

80. 2. Mole fraction
●
It is the ratio of number of moles of a particular
component to the total number of moles of the
solution. If a substance ‘A’ dissolves in substance
‘B’ and their number of moles are nA and nB ,
respectively, then the mole fractions of A and B are
given as:

81. 3. Molarity
●
It is the most widely used unit and is denoted by
M. It is defined as the number of moles of the
solute in 1 litre of the solution. Thus,
●
Calculate the molarity of NaOH in the solution
prepared by dissolving its 4 g in enough water to
form 250 mL of the solution.

83. 4. Molality
●
It is defined as the number of moles of solute
present in 1 kg of solvent. It is denoted by m.
●
The density of 3 M solution of NaCl is 1.25 g/mL .
Calculate the molality of the solution.

85. THANK YOU