Download SOME BASIC CONCEPTS OF CHEMISTRY NOTES

CHARAN’S PRE UNIVERSITY (PU) COLLEGE
(Recognized by the Govt of Karnataka)
No 326, Ist Main Road, Cambridge Layout, Halasuru, BENGALURU-560 008
1 PUC-CHEMISTRY
BASIC CONCEPTS OF CHEMISTRY-NOTES
MATTER: Anything which has mass and occupies space is called matter
Occurrence of Matter:
Matter exists in three physical states
 Solid
 Liquid
 Gas
Molecular Association:
Solid: In solids, the particles are held together very strongly in an orderly fashion and
there is not much freedom for the particles to move around. Hence, Solids have a
definite volume and a definite shape.
Liquid: In liquid, the particles are loosely packed compared to solids and liquids hence
take the shape of the container. Liquids have a definite volume but not a definite
shape.
Gas: The particles are far apart as compared to solids or liquids and their movement is
easy and fast. Hence Gases have neither definite volume nor a definite shape.
CLASSIFICATION OF MATTER:
MATTER
PURE
SUBSTANCES
ELEMENTS
COMPOUNDS
MIXTURES
HOMOGENEOUS
HETEROGENEOUS
Pure Substances:
Pure substances are made up of a single component with a fixed composition. Ex:
Copper, silver, gold etc. They are further classified into a) Elements and b) Compound
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Element:
An element is defined as any substance that cannot be broken down further into simpler
components. Ex: Sodium, potassium, gold etc. The constituent particles in an element are
either atoms or molecules.
Compound:
When two or more atoms of different elements combine, a compound is formed. Ex:
CaCl3, NaCl
MIXTURES:
A mixture consists of two or more substances present in any ratio. Ex: salt solution in
water, tea, mixture of salt and sugar etc
Mixtures are further classified into:
1. Homogeneous Mixtures:
In a homogeneous mixture, the components are completely mixed with each other
and its composition is uniform throughout. Ex: salt solution in water
2. Heterogeneous Mixture:
In heterogeneous mixtures, the composition is not uniform throughout and
sometimes the components can be seen separately in the mixture. Ex: misture of salt
and sugar
ATOM:
An atom is the smallest particle of an element, which may or may not be capable of
independent existence. Ex: atoms of iron, copper etc
MOLECULE:
A molecule is the smallest particle of an element or compound which can exist freely. Ex:
molecules of hydrogen (H2), water molecule (H2O)
PROPERTIES OF MATTER:
Physical Properties: They are those properties that can be measured or observed
without changing the identity or composition of the substance. Ex: colour, odour, melting
point, boiling point etc
Chemical Properties: They are those properties where a chemical change needs to occur
Ex: acidity, basicity, combustibility etc.
MEASUREMENT:
THE INTERNATIONAL SYSTEM OF UNITS (SI UNITS):
What are SI Units?
In order to standardize the unit of measurement across different countries, a new system
of units was adopted in 1971 at the General conference of Weights and Measures. This
system was called as the ‘International System of Units’ abbreviated as SI units.
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The SI system has seven base units as listed below:
NO
1
2
3
4
5
6
7
QUANTITY
Length
Mass
Time
Temperature
Amount of
substance
Electric
current
Luminous
intensity
UNIT
metre
kilogram
seconds
kelvin
mole
SYMBOL
m
kg
sec
K
mol
Ampere
A
candela
cd
OTHER IMPORTANT PROPERTIES:
MASS AND WEIGHT:
Mass of a substance is the amount of matter present in it. Weight of a substance is the
force exerted by gravity on an object.
If m is the mass of a substance, its weight w=mg, where g is acceleration due to gravity.
Thus weight of a substance may change from one place to another (as acceleration due to
gravity changes)
The SI unit of mass is kilogram (kg).
VOLUME:
The SI unit of volume is m3. One of the commonly used units for volume of liquids is
litre (L).
1L= 1000mL=1000cm3=1dm3
DENSITY:
Density of a substance is its mass per unit volume. The SI unit of density is kg/m3.
Density=
Mass
Volume
TEMPERATURE:
There are three common scales to measure temperature


°C (degree celsius)
°F (degree Fahrenheit)
kelvin(K), K is the SI unit
The formula for conversion between celsius and fahrenheit scales is:
𝟗
°F= (°C) + 32
𝟓
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The formula for conversion between kelvin and celsius scale is:
K=°C+273.15
SIGNIFICANT FIGURES:
Significant figures are meaningful digits which are known with certainty.
DIMENSIONAL ANALYSIS:
Any calculation involving the use of the dimensions of the different physical quantities
involved is called dimensional analysis.
LAWS OF CHEMICAL COMBINATION:
Law of conservation of Mass (Proposed by Antoine Lavoisier:
It states that matter can neither be created nor destroyed. It can only be changed from one
form to another during a chemical reaction.
Law of Definite Proportions (Proposed by Joseph Proust):
A given compound always contains exactly the same proportion of elements by weight.
Law of Multiple Proportions (Proposed by Dalton):
If two elements can combine to form more than one compound, the masses of one
element that combine with a fixed mass of the other element, are in the ratio of small
whole numbers
For example, hydrogen combines with oxygen to form two compounds namely water and
hydrogen peroxide.
Hydrogen + Oxygen Water
2g
16g
18g
Hydrogen + Oxygen  Hydrogen Peroxide
2g
32g
34g
Here the masses of oxygen( i.e 16g and 32g) which combine with a fixed mass of
hydrogen (2g) gives a simple ratio i.e 16/32= ½
GAY LUSSAC’S LAW OF GASEOUS VOLUMES:
When gases combine in a chemical reaction, they combine in a simple ratio by volume
provided that all the gases combine at the same temperature and pressure.
Hydrogen+ Oxygen  Water
100 mL
50mL
100mL
(100) / (50)--------Ratio of 2:1
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Thus the volume of hydrogen and oxygen which combine together (i.e 100 mL and
50mL) bear a simple ratio of 2:1
AVOGADRO LAW:
Equal volumes of all gases contain equal number of molecules under the same conditions
of temperature and pressure.
DALTON’S ATOMIC THEORY:
In 1808, Dalton proposed hi atomic theory. According to this theory,
1.
2.
3.
4.
5.
6.
All matter is made up of extremely small particles called atoms.
Atoms cannot be divided.
Atoms cannot be created or destroyed.
All atoms of a given element are identical in nature
Atoms of different elements differ in their mass, size and chemical properties.
Atoms of same element can combine in more than one ratio to form more than
one compound.
7. The number and kind of atoms in a given compound is fixed.
8. During a chemical combination, atoms of different elements combine in small
whole no ratios.
DRAWBACKS OF DALTON’S ATOMIC THEORY:
1. Dalton said that atoms cannot be divided but under some conditions, it can be
divided further into protons, neutrons and electrons.
2. Dalton said that atoms of the same element have the same mass but atoms of the
same element can have different masses also.
3. Dalton said that atoms of different elements have got different mass but during
certain conditions they can have the same mass.
ATOMIC AND MOLECULAR MASSES:
The atomic mass of an atom is extremely small. Earlier, scientists used to determine the
mass of one atom relative to another by experimental means. Hydrogen being the lightest
atom was assigned a mass of 1(without any units) and other elements were assigned
masses relative to it.
In the present system, the atomic mass of an element is given relative to the mass of
carbon-12 (12C).
One atomic mass unit (amu) is defined as a mass exactly equal to one-twelfth the
mass of one carbon-12 atom.
‘amu’ is now replaced by ‘u’ and is also known as unified atomic mass unit or
unified mass.
1 amu = 1.66056 x 10-24g
ISOTOPES:
Atoms having same atomic number with different mass number are called isotopes.
Isotopes of carbon are 12C, 13C, 14C
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AVERAGE ATOMIC MASS:
When an element exists in two or more stable isotope forms, its atomic mass can be given
as the average atomic mass by taking into account of the relative abundance of each of the
isotope and their respective atomic masses.
Example: Carbon exists in three isotope forms: carbon-12 (12C), carbon- 13 (13C) ,
carbon-14 (14C) . Their relative abundance is 98.9 %, 1.1 % and 2 x 10-10 %. The atomic
masses of the three forms are 12 u, 13 u and 14 u respectively.
Average atomic mass of C = (0. 989)(12u)+(0.011)(13u) + (2x 10-12) (14 u)
= 12.011 u
MOLECULAR MASS:
Molecular mass is the sum of atomic masses of all the elements present in a molecule.
Example: Molecular mass of water (H2O) = 2 (1.008u) + 16.00 u = 18.02 u
NUMERICAL:
1. Calculate the molecular mass of the following a) Ethane (C2H6)
b) Ammonia (NH3)
Ans: a) Ethane (C2H6)
Ethane = C2H6= 2 x atomic mass of carbon + 6 x atomic mass of Hydrogen
= 2 x (12.011u) + 6 x (1.008 u)
= 24.022u + 6.048u
= 30.070u
b) Ammonia (NH3)
Ammonia=(NH3) = 1x atomic mass of Nitrogen + 3 x atomic mass of Hydrogen
= 1x(14.01u) + 3 (1.008u)
= 14.01u + 3.024u = 17.034u
FORMULA MASS:
The formula mass of a molecule is the sum of the atomic masses of the atoms in its
empirical formula.
Example:
The formula mass of NaCl is the sum of atomic mass of sodium and atomic mass of
chlorine.
Formula mass of NaCl = 23.0 u + 35.5 u = 58.5 u
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MOLE CONCEPT AND MOLAR MASSES:
The mole is a unit of measurement used to express amounts of a chemical substance. It is
one of the fundamental SI units and its symbol is mol.
One mole is the amount of substance that contains as many particles or entities as
there are atoms in exactly 12 g of the 12C isotope.
This corresponds to a value of 6.023 x1023 elementary entities of the substance.
Ex: 1 mol of hydrogen atoms = 6.023 x 1023 atoms of hydrogen.
1 mol of water molecules= 6.023 x 1023 molecules of water
AVOGADRO NUMBER:
The number of entities in one mole (6.023 x 1023) is known as Avogadro number and is
denoted as N.
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.
Example:
Molar mass of water= 18.02 g mol-1
Molar mass of sodium chloride= 58.5 g mol-1
PERCENTAGE COMPOSITION:
The percentage composition of a compound is a relative measure of the mass of each
different element present in the compound.
METHOD TO CALCULATE THE PERCENT BY MASS OF A COMPOUND:
1. Calculate the molar mass of the compound.
2. Calculate the total mass of each element present in the formula of the compound.
3. Calculate the % composition:
% by weight(mass) of an element =
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑝𝑟𝑒𝑠𝑒𝑛𝑡
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
x 100
Example Problem:
1. Calculate the percent by mass of sodium(Na) and chlorine (Cl) in NaCl
a) Sol: Step 1- Calculate the molar mass of the compound
Molar mass of NaCl = Molar mass of Na+ Molar mass of Cl
= 23 + 35.5 = 58.5 g mol-1
b) Calculate the total mass of each element present:
Total mass of Na present = 23 g
Total mass of Cl present = 35.5 g
c) Calculate % composition
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% by mass of Na in NaCl = (23|58.5) x 100
= 39.32 %
% by mass of Cl in NaCl = (35.5|58.5) x 100
= 60.68%
2. Calculate the percent by mass of ethanol (C2H5OH)
Ans: Molecular formula of ethanol is: C2H5OH
a) Calculate the molar mass of the compound
Molar mass of ethanol is = Molar mass of C +Molar mass of H+ Molar mass of O +
= (2x 12.01 + 5x 1.008 + 1x16 + 1x 1.008) g
= 46.068 g
b) Calculate the total mass of each element present:
Total mass of C present = 2 x12 = 24 g
Total mass of H present= 5x 1.008 + 1x 1.008 = 6.048 g
Total mass of O present= 1 x 16 = 16 g
c) Calculate % composition
% by mass of C in C2H5OH is {24|46.068} x 100 = 52.14%
% by mass of H in C2H5OH is (6.068|46.068) x 100 = 13.1%
% by mass of O in C2H5OH is (16|46.068) x 100 = 34.73%
HOME ASSIGNMENT:
1. Calculate the percent by mass of each element present in water.
2. Calculate the percent by mass of each element present in sodium sulfate (Na2SO4)
Ans: 1) Mass % of hydrogen= 11.18, Mass % of oxygen = 88.79
2) Mass % of Na=32.38%, Mass % of S= 22.57%, Mass% of O= 45.05%
EMPIRICAL FORMULA:
An empirical formula gives the simplest whole number ratio of atoms of different
elements present in a compound.
The molecular formula shows the exact number of atoms of different types present in a
molecule of a compound.
Problems:
1. A compound contains 4.07% hydrogen, 24.25% carbon and 71.75% chlorine. Its
molar mass is 99.03g. What are its empirical and molecular formulae?
Ans:
Step 1: Conversion of mass percent to grams:
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Since the composition is given in percentage(4.07%,24.25% and 71.75%), 100g sample
of the above compound contains 4.07g hydrogen, 24.25g carbon and 71.75 g
chlorine.
Step 2: Convert into number moles of each element: Divide the masses obtained
above by respective atomic masses of various elements.
Atomic mass of hydrogen= 1, Atomic mass of carbon= 12, Atomic mass of
chlorine= 35.5
Therefore,
4.07g
Moles of hydrogen: 1.008g = 4.04
Moles of carbon=
24.25 g
Moles of chlorine=
12.0g
= 2.021
71.75𝑔
35.5𝑔
= 2.021
Step 3: Divide the mole value obtained above by the smallest number:
Since 2.021 is the smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl.
Step 4: Write empirical formula by mentioning the numbers after writing the symbols
of respective elements: CH2Cl, is, thus the empirical formula.
Step 5: Write the molecular formula: Determine empirical formula mass.
For CH2Cl, empirical formula mass is 12.0+2 x 1.008+35.5 =49.52 g
Divide Molar mass(Given in the sum) by empirical formula mass:
𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠
99.03
=
=2=𝑛
𝐸𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑚𝑎𝑠𝑠
49.52
Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH2Cl, n=2. Hence molecular formula is C2H4Cl2.
HOME ASSIGNMENT:
1. Find the empirical and molecular formulae for the compound which contains
51.28%, 9.40%,27.35% and 11.97% of carbon, hydrogen, oxygen and nitrogen
respectively. Molar mass of the compound is 234 g mol-1. Atomic masses: C=12,
H=1, O=16, N=14
Ans: Empirical formula: C5H11O2N1
Molecular formula: C10H22O4N2
2. An organic compound contain 57.14% of Carbon, 6.16% Hydrogen, 9.52%
Nitrogen, 27.18% oxygen. Calculate the empirical formula and molecular formula
if its molecular mass is 294.3 gm/mole.
Sol: Empirical formula: C7H9N1O
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Molecular formula: C14H18N2O4
STOICHIOMETRY AND STOICHIOMETRY CALCULATIONS:
Stoichiometry is a branch of chemistry that deals with the relative quantities of
reactants and products in chemical reactions.
A balanced chemical equation has the same number of atoms of each element on
both sides of the equation.
Let us consider the combustion of methane. A balanced equation for this reaction is
CH4(g) + 2O2(g) CO2(g) + 2H2O(g)
In this reaction methane and dioxygen are reactants and carbon dioxide and water are
products. All the reactants and products are gases in the above reaction and are denoted
by (g).
The coefficients 2 for O2 and H2O are called stoichiometric coefficients. Similarly the
coefficient for CH4 and CO2 is one in each case.
From this reaction we can conclude that one mole of CH4 (g) reacts with two
moles of O2(g) to give one mole of CO2(g) and two moles of H2O(g)
or
16g of CH4 (g) reacts with 64 g of O2 (g) to give 44g of CO2(g) and 36 g of H2O (g)
NUMERICAL PROBLEMS:
1. Calculate the amount of water(g) produced by the combustion of 44g of
propane
Ans: The balanced equation for the combustion of propane is
C3H8 + 5 O23CO2 + 4H2O
44g of propane corresponds to 1 mole(C3H8----3x 12+ 8x1=44)
From the above equation we can see that one mole of propane gives four moles of
water(g) on combustion.
I mole of H2O = (2x1+16=18g)
Therefore 4 moles of water = (4x18=72g)
This means that 72 g of water is produced by the combustion of 44g of
propane.
2. How many moles of methane are required to produce 88 gms of CO2 after
combustion?
Ans: Ans: Combustion reaction
CH4+2 O2  CO2+ 2H2O
10
(g)
(g)
(g)
(g)
16 gms of Methane = 1 mole = 44 gms of CO2
44 gms of CO2 is produced from 1 mole of methane
88
88 gms of CO2 is produced from 44 x 1 = 2 moles
HOME ASSIGNMENT:
1. How many moles of propane are required to produce 22g CO2(g) after
combustion
Ans: Hint: Equation of combustion--- C3H8+5O23CO2+4H2O
7.33 g of propane are required.
LIMITING REAGENT:
In a chemical equation, the reactant that is completely consumed is called the limiting
reagent.
NUMERICAL PROBLEMS:
1. 30g of magnesium and 30 g of oxygen reacted. Calculate the composition of the
residual mixture (Atomic mass of Mg = 24). What is the limiting reagent.
Ans: 2Mg + O2 2MgO
2x24g
2x16g
2(24+16) g
48g
32g
80g
30g
?
30 x 32
Mass of oxygen reacted = 48 = 20 g
Mass of oxygen left over = 30 – 20 = 10g
Mass of MgO formed =
30 𝑋 80
48
= 50 𝑔
Hence composition of the residual mixture is 10 g of oxygen and 50 g of MgO.
Magnesium is the limiting reagent.
HOME ASSIGNMENT:
1. 50 kg of N2 (g) and 10 kg of H2(g) are mixed to produce NH3(g). Calculate the
NH3 (g) formed. Identify the limiting reagent in the production of NH3 in this
situation.
Ans: 56.1 kg NH3 [ Refer Textbook Pg 19 for Solution]
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REACTIONS IN SOLUTIONS:
1. Mass percent or weight percent (w/w%):
Mass percent is calculated by the formula,
Mass of solute
Mass percent of a substance = Mass of solution x 100
2. Mole fraction:
It is the ratio of number of moles of a particular component to the total number of
moles in the solution.
If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB;
then the mole fractions of A and B are given as
No of moles of A
Mole fraction of A = No of moles of solution
=
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵
Mole fraction of B = 𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
3. Normality:
Normality can be defined as no of gram equivalent mass of the substance present in
one litre of the solution.
Normality =
𝑁𝑜 𝑜𝑓 𝑔𝑟𝑎𝑚 𝑒𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑉𝑜𝑙 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠
4. Molarity:
Molarity is defined as the number of moles of solute in 1 litre of solution.
No of moles of solute
Molarity (M) = Volume of solution in litres
4. Molality:
It is defined as the number of moles of solute present in 1 kg of solvent.
No of moles of solute
Molality (m) = Mass of solvent in kg
NUMERICAL PROBLEMS:
1. 4g of sodium chloride was dissolved in 46 g of water. Calcualte mass percentage of
the substance in the solution.
Ans: Mass % of substance =
Mass of substance
Mass of solution
x 100
12
4g
= 4 g of NaCl+46g of water x 100
=8%
2. Calculate the molarity of NaOH in the solution prepared by dissolving its 4 g in
enough water to form 250 mL of the solution.
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
Sol: Molarity (M)= 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑙𝑖𝑡𝑟𝑒𝑠
=
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻⁄𝑀𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
4 𝑔⁄40 𝑔
0.25 𝐿
= 0.250 𝐿
=
0.1 𝑚𝑜𝑙
0.250 𝐿
= 0.4 mol L-1 = 0.4 M
3. The density of 3M solution of NaCl is 1.25 g mL-1. Calculate molality of the
solution.
Ans: M = 3 mol L-1
Mass of NaCl in 1 L of solution = 3x 58.5[NaCl=23+35.5] = 175.5 g
We know that Density = Mass/ Volume
Therefore,
Mass of 1L solution = Volume x Density = 1000 x 1.25 = 1250 g
(since density = 1.25 g mL-1)
Mass of water in solution = Total mass of 1L solution – Mass of NaCl in 1L solution
= 1250 – 175.5
= 1074.5 g
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
3 𝑚𝑜𝑙
Molality = 𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡 𝑖𝑛 1 𝑘𝑔 = 1.0745 𝑘𝑔= 2.79 m.
HOME ASSIGNMENT:
1. 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.
Sol: 10 %
2. Calculate the molarity of sodium Hydroxide (NaOH) in the solution prepared by
dissolving 4 gm in 500ml of the solution.
Sol: 0.2 mol/litre
3. The density of 2M solution of sodium chloride (NaCl) 1.13 g mol/litre. Calculate
molality of the solution. M= 2 mol per litre.
Sol: 1.9743 m
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