The document covers fundamental concepts of chemistry for Class 11, including the importance, applications, and calculations related to atoms, molecules, and stoichiometry. It highlights chemistry's role in agriculture, health, and environmental protection, along with detailed explanations of atomic and molecular masses, the mole concept, and practical applications in industry. Key topics such as percentage composition, empirical and molecular formulas, as well as stoichiometric calculations, are also addressed, providing essential knowledge for students.

1. • Chapter:1
• Some basic concepts of chemistry
• Class-11 CBSE
• SHORT NOTES
BY
S.ARUNKUMAR

2. Some Basic Concepts of Chemistry
General Introduction: Importance and scope
of Chemistry.
• Atomic and molecular masses.
• mole concept and molar mass,
• percentage composition,
• Empirical and molecular formula,
• chemical reactions,
• stoichiometry and calculations based on
stoichiometry.

3. Importance of chemistry
• Chemistry has a direct impact on our life and has
wide range of applications in different fields.
• (A) In Agriculture and Food:
(i) It has provided chemical fertilizers such as
urea, calcium phosphate, sodium nitrate,
ammonium phosphate etc.
(ii) It has helped to protect the crops from insects
and harmful bacteria, by the use of certain
effective insecticides, fungicides and pesticides.
(iii) The use of preservatives has helped to
preserve food products like jam, butter, squashes
etc. for longer periods.

4. • (B) In Health and Sanitation:
(i) It has provided mankind with a large number
of life-saving drugs. Today, dysentery and
pneumonia are curable due to discovery of
sulpha drugs and penicillin life-saving drugs.
Cisplatin and taxol have been found to be very
effective for cancer therapy and AZT
(Azidothymidine) is used for AIDS victims.
(ii) Disinfectants such as phenol are used to kill
the micro-organisms present in drains, toilet,
floors etc.
(iii) A low concentration of chlorine i.e., 0.2 to 0.4
parts per million (ppm) is used ’ for sterilization
of water to make it fit for drinking purposes.

5. • Saving the Environment:
The rapid industrialisation all over the world
has resulted in lot of pollution.
Poisonous gases and chemicals are being
constantly released in the atmosphere. They
are polluting environment at an alarming rate.
Scientists are working day and night to develop
substitutes which may cause lower pollution.
For example, CNG (Compressed Natural Gas), a
substitute of petrol, is very effective in
checking pollution caused by automobiles.

6. • (D) Application in Industry:
Chemistry has played an important role in
developing many industrially manufactured
fertilizers, alkalis, acids, salts, dyes, polymers,
drugs, soaps,
detergents, metal alloys and other inorganic
and organic chemicals including new
materials contribute in a big way to the
national economy.

7. Atoms and Molecules
• The smallest particle of an element, which
may or may not have independent
existence is called an atom.
• The smallest particle of a substance which is
capable of independent existence is called a
molecule.
• Homoatomic molecules are made up of the
atoms of the same element and heteroatomic
molecules are made up of the atoms of the
different element have different atomicity.

8. Atomic Mass Unit(amu)
• It is the mass of an atom.
• One atomic mass unit is defined as a
mass exactly equal to one twelfth the
mass of one carbon-12atom.
• 1amu=1.66056×10–24g. g.
Today, ‘amu’ has been replaced by
‘u’ which is known as unified mass.

9. Atomic Mass
• Atomic mass of an element is defined as
the average relative mass of an atom of
an element as compared to the mass of
an atom of carbon -12 .

10. Gram Atomic Mass
• Atomic mass of an element
expressed in grams is the gram
atomic mass or gram atom
• For example, the atomic mass of
oxygen = 16 amu Therefore gram
atomic mass of oxygen = 16 g

11. Average Atomic Mass

12. • The average atomic mass of an element
is the sum of the masses of its isotopes,
each multiplied by its natural abundance.
• Average mass= mass1 x Relative
abudance1 + mass2 x Relative abundance2
=12 x 0.9893 + 13 x 0.0107
=12.01amu

13. Molecular Mass
• It is the sum of atomic masses of all the
elements present in the molecule.

14. Formula Mass
• Formula mass of Sodium Chloride
• Atomic mass of sodium+ Atomic mass of
chlorine
• =23.0u+35.5u=58.5u

15. Mole Concept
• One mole is the amount of a substance that
contains as many particles or entities as there
are atoms in exactly 12g of the 12C isotopes.
• 1 mol of water molecules=6.022x𝟏𝟎 𝟐𝟑
• 1 mole of any substances is equal to
6.022x𝟏𝟎 𝟐𝟑
• The mass of one mole of a substance in grams
is called its molar mass.

16. Percentage Composition
• The mass percentage of each constituent
element present in any compound is called
its percentage composition.

17. • H2 o

18. What is the percentage of carbon,
hydrogen and oxygen in ethanol?
• C2H5OH
• Molar mass of ethanol is 46g
• Mass percent of carbon=
𝟐𝟒
𝟒𝟔
𝒙𝟏𝟎𝟎 =52%
• Mass percent of hydrogen=
𝟔
𝟒𝟔
x100=13%
• Mass percent of oxygen=
𝟏𝟔
𝟒𝟔
x100=34%

19. • An empirical formula represents the simplest
whole number ratio of various atoms present
in a compound.
Step:1
Conversion of mass percent to grams:
4.07% - 4.07g
24.7%- 24.7g
71.65%- 71.65g
98.96% - 98.96g

20. • Step:2 Convert into number moles of each
element.
• Mass of hydrogen=
4.07𝑔
1.008𝑔
= 4.04
• Mass of carbon=
24.27𝑔
12.01𝑔
=2.021
• Mass of chlorine=
• Step:3 Divide each of the mole values obtained
above by the smallest number amongst them
• Since 2.021 is smallest value, division by it gives a
ratio of 2:1:1 for H:C:Cl
𝟕𝟏. 𝟎𝟓𝒈
𝟑𝟓. 𝟒𝟑𝒈
= 𝟐. 𝟎𝟐𝟏

21. • Step:4 Write down the empirical formula by mentioning the
numbers after writing the symbols of respective elements
C𝐻2 𝐶𝑙 𝑖𝑠 𝑡ℎ𝑒 𝑒𝑚𝑝𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑏𝑜𝑣𝑒 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑.
Step:5 Writing Molecular Formula:
C𝐻2 𝐶𝑙, Empirical formula mass is 12.01+(2x1.008)+35.453
=49.48g
b) Divide Molar mass by empirical formula
=
𝑴𝒐𝒍𝒂𝒓 𝑴𝒂𝒔𝒔
𝑬𝒎𝒑𝒊𝒓𝒊𝒄𝒂𝒍 𝒇𝒐𝒓𝒎𝒖𝒍𝒂 𝒎𝒂𝒔𝒔
98.96𝑔
49.48𝑔
=2=(n)
Step:6 Multiply empirical formula by n obtained above to get the
molecular formula
Empirical formula= C𝐻2 𝐶𝑙x2 = 𝑪 𝟐 𝑯 𝟒 𝑪𝒍 𝟐

22. Stoichiometry and Stoichiometric
calculations
• Calculate the amount of water(g) produced by
the combustion of 16g of methane.
• C𝑯 𝟒 𝒈 + 𝟐𝑶 𝟐 𝒈 −→ 𝑪𝑶 𝟐 + 𝟐𝑯 𝟐 𝑶 𝒈
• 16g of methane corresponds to one mole.
• 1 mol of methane gives 2 mol of water
• 2 mol of water=2x(2+16)=2x18=36g
• The co-efficients 2 for O2 and H2O are called
stoichiometric coefficients. Similarly the
coefficient for CH4 and CO2 is one in each case.

23. How many moles of methane are
required to produce 22g of CO2(g) after
combustion
• According to the chemical equation
• C𝑯 𝟒 𝒈 + 𝟐𝑶 𝟐 𝒈 −→ 𝑪𝑶 𝟐 𝒈 + 𝟐𝑯 𝟐 𝑶 𝒈
44g C𝑶 𝟐 𝒈 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒇𝒓𝒐𝒎 𝟏𝟔𝒈 𝑪𝑯 𝟒 𝒈
Since 1mol of 𝑪𝑶 𝟐(g) is obtained from 1 mole
of 𝑪𝑯 𝟒(g).
Moles of 𝑪𝑶 𝟐=22g C𝑶 𝟐 𝒈 𝒙
𝟏 𝒎𝒐𝒍 𝑪𝑶 𝟐(𝒈)
𝟒𝟒𝒈 𝑪𝑶 𝟐(𝒈)
=0.5mol 𝑪𝑶 𝟐 𝒈
0.5 mol of 𝑪𝑯 𝟒 is required to produce 22g of
𝑪𝑶 𝟐.

24. Limiting reagent
• The reactant which is present in the lesser
amount gets consumed after sometime and
after that no further reaction takes place
whatever be the amount of other reactant
present.

25. • N2​+3H2⟶2NH3
• Molecular mass of Nitrogen =28g/mol=0.028kg/mol
• Molecular mass of Hydrogen =6g/mol=0.006kg/mol
• Molecular mass of Ammonia =17g/mol=0.017kg/mol
• Now, according to the balanced chemical equation,
• 0.028 kg of Nitrogen reacts with 0.006 kg of Hydrogen.
∴ 50 kg of Nitrogen reacts with[
(𝟎.𝟎𝟎𝟔𝒙𝟓𝟎)
𝟎.𝟎𝟐𝟖
] =10.71kg of
Hydrogen.
• The amount of Hydrogen (given 10 kg) is less than
the amount required (i.e., 10.71 kg) for 50 kg of
Nitrogen.
• Therefore, Hydrogen is the limiting reagent.

26. • Hence, the formation of Ammonia will
depend on the amount of Hydrogen
available for reaction.
• ∵ Amount of ammonia produced by 0.006
kg of hydrogen =2×0.017=0.034kg
• ∴ Amount of ammonia produced by 10 kg
of Hydrogen =
0.034𝑥10
0.006
=56.67kg
• Hence, 56.67 kg of ammonia gas will be
formed and hydrogen will be limiting
reagent.

27. Reactions in solutions
• Mass Percent:
• Mass percent=
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒆
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑺𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝒙𝟏𝟎𝟎
• Mass percent of A=
𝑴𝒂𝒔𝒔 𝒐𝒇 𝑨
𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏
𝒙𝟏𝟎𝟎
• =
𝟐𝒈
𝟐𝒈+𝟏𝟖𝒈
𝒙𝟏𝟎𝟎
• =
𝟐𝒈
𝟐𝟎𝒈
𝒙𝟏𝟎𝟎
• =10%

28. Mole Fraction
• It is the ratio of number of moles of a particular
component to the total number of moles of the
solution.
Mole fraction of A=
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐴
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑛 𝐴
𝑛 𝐴 + 𝑛 𝐵
Mole fraction of B=
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐵
𝑁𝑜 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
𝑛 𝐵
𝑛 𝐴 + 𝑛 𝐵

29. Molarity
• It is defined as the number of moles of the
solute in 1 litre of the solution.
• Molarity(M)=
𝑵𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒍𝒊𝒕𝒓𝒆𝒔
• Calculate the molarity of NaOH in the solution
prepared by dissolving its 4g in enough water
to form 250mL of the solution.
• Molarity(M)=
𝑵𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒍𝒊𝒕𝒓𝒆𝒔

30. • =
𝒈𝒊𝒗𝒆𝒏 𝒎𝒂𝒔𝒔
𝒎𝒐𝒍𝒂𝒓 𝒎𝒂𝒔𝒔
𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒍𝒊𝒕𝒓𝒆
• =
𝟒𝒈
𝟒𝟎𝒈
𝟎.𝟐𝟓𝟎𝑳
• =
𝟎.𝟏𝒎𝒐𝒍𝒆
𝟎.𝟐𝟓𝟎𝑳
• =0.4
𝒎𝒐𝒍
𝒍𝒊𝒕𝒓𝒆
• =0.4M

31. • Find the volume of NaOH to prepare 0.2M from
1M of NaOH solution. 1st-method
• If 1M NaOH means 1 mol of NaOH present in 1
litre of the solution.
• If 1 mol is present in 1L or 1000mL then 0.2 mol
is present in
•
𝟏𝟎𝟎𝟎𝒎𝒍
𝟏𝒎𝒐𝒍
𝒙 0.2 mol
• 200mL
• Thus 200ml of 1M NaOH are taken and added to
1 litre.

32. • M1 V1 = M2 V2 (2nd-method)
• 0.2M x1000ml = 1.0M x V2
• V2=
𝑀1𝑥𝑉1
𝑀2
• V2 =
0.2𝑀 𝑥1000𝑚𝑙
1.0𝑀
• =200ml.

33. Molality
• It is defined as the number of moles of solute present
in 1kg of solvent. It is denoted by m.
• Molality(m)=
𝑵𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝑴𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒌𝒈
• The density of 3M solution of NaCl is 1.25g /ml.
Calculate molality of the solution.
• density=
𝒎𝒂𝒔𝒔
𝒗𝒐𝒍𝒖𝒎𝒆(𝒍)

34. • Molarity = no. of moles of solute / volume of
solution in L
Given, no. of moles of NaCl = 3
Density = Mass/ Volume
Mass of the solution = density x volume
= 1.25g/ml x 1000ml
= 1250 g
Mass of solute = mass of 3 moles of NaCl
= 3 x (23 +35.5) = 175.5 g
∴ Mass of solvent = mass of solution - mass of solute
= 1250 - 175.5
= 1074.5 g =1.074 kg
∴ Molality = no. of moles of solute/ mass of solvent
= 3/1.074 = 2.79m

35. • m=
𝑵𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒗𝒆𝒏𝒕 𝒊𝒏 𝒌𝒈
• m=
𝟑
𝟏.𝟎𝟕𝟒𝟓𝒌𝒈
• =2.79m