Mole concept it's explanation and calculations on mole concept including concentration.

1. MOLE CONCEPT

2. HOW WE MEASURE
ENTITIES?
MASS- GRAMS
VOLUME – LITRES
DISTANCE -METRES

3. How to measure atoms or molecules of such
smaller magnitude?
• Just as we denote 1 Dozen for 12 items, it can be anything
12 eggs-1 Dozen
12 banana-1 Dozen
12 bangles- 1 Dozen
Similarly mole is used to count entities on the microscopic level
( i.e. atoms/molecules/ions/ particles etc.)

4. Define moles
One mole is the amount of substance that contains as many entities or particles
as there are present in 12 g of C-12 atom which is exactly equal to 6.02 x 1023
particles.
Here 6.02 x 1023 is the Avogadro number or Avogadro constant

5. Why C-12 is used as relative atomic
mass(atomic weight)
• Atomic weights are an average of the relative masses of all of the
isotopes of the given element.
The number of C-12 atoms in exactly 12.00 g of C-12 is 6.02 X 1023.
So C-12 is used as relative atomic mass in calculation of Avogadro
number

6. Relative atomic mass
• The ratio of average of mass of an atom of the element to the 1/12 th mass of the
carbon atom is defined as the relative atomic mass of any element.
• Example
• Mass of one N atom=14x1/12 (mass of 1 carbon -12 atom)
• Mass of one O atom=16x1/12(mass of 1 carbon-12 atom)
Atomic mass unit: 1/12 x mass of 1 carbon-12 atom

7. What is Avogadro number /Avogadro
constant?
• Avogadro number is the number of constituent particles usually
atoms/molecules present in 1 mole of the substance which is equals to the
6.02 x 1023
Example- 1 mol of hydrogen atoms = 6.02 x 1023 atoms
1 mol of water molecules= 6.02 x 1023 molecules
1 mol of sodium chloride = 6.02 x 1023 formula units

8. How we calculated Avogadro constant?
• Mass of a C-12 atom=1.99x 10-23 g ( determined by mass spectrometer)
• Knowing that one mole of the carbon atom weighs 12g, the number of atoms in it is
equal to
12g /mol
1.99x10-23 g/atom
 6.02 x 1023 atoms

9. The mole concept for (A) elements, (B) compounds, and (C) molecular
substances.
A mole contains
6.02 X 1023 particles. Since every mole contains the same number of
particles, the ratio of the mass of any two moles is the same as the ratio of
the masses of individual particles making up the two moles.

10. Molar mass
Molar mass is defined as the mass of 1 mol of substance in grams is
numerically equal to atomic mass.
molar mass of water= 18.02g/mol
Molar mass of sodiumm chloride =58.5g/mol
Molar mass in grams is numerically equal to atomic/molecular/formula
mass in u.

11. Molecular mass
• Molecular mass is defined as the sum of the atomic masses of all the
elements present in a compound.
• Example
• H2O= 2x1+1x16= 18
H3PO4 = 3X1+1X31+4X16= 98
Problems:
Calculate the molecular mass of glucose and hydrogen peroxide.

12. How to calculate moles?
Calculate the number of sodium atoms in 0.120 mol
Na?
0.120 mol Na x 6.02 x 1023 atoms Na = 7.22 x 1022 1
mol Na

13. Calculate the number of moles of potassium in 1.25 x 1021 atoms of K.
• 1.25 x 1021 atoms K x 1 mol K = 2.08 x 10-3 mol K
6.02 x 1023 atoms K

14. Problems
• What is the mass in grams of 2.01 x 1022 atoms of sulfur?
• How many O2 molecules are present in 0.470g of oxygen gas?

15. Mole calculations
• Calculate the number of Magnesium and Chlorine ions present in 0.450 mol of
MgCl2.
• # Mg2+ ions:
• 0.450 mol MgCl2 x 6.02 x 1023 formula units x 1 Mg2+
1 mole 1 formula unit
• # Mg2+ ions = 2.71 x 1023
• # Cl- ions:
• 0.450 mol MgCl2 x 6.02 x 1023 formula units x 2 Cl-
1 mole 1 formula unit
• # Cl- ions = 5.42 x 1023

16. Percentage composition
• Percentage composition is defined as the percentage by mass of each element contributed in the
compound.
• Mass % of an element=
Mass of an element in the compound x100
Molar mass of the compound
Example:
Calculate the percentage of Nitrogen, by mass in Ca(NO3)2:
Formula Weight = 1 x Ca + 2 x N + 6 x O
= 1 x 40.1 + 2 x 14.0 + 6 x 16.0
= 164.1 amu
% N = (2)(14.0) = 17.1%
164.1

17. Empirical formula
• Empirical formula represents the simplest whole number ratio of all the atoms
present in a compound
• What is different from the molecular formula?
• Molecular formula represents the actual number of different atoms present in a
compound.
• For example glucose molecular formula is C6H12O6 but the empirical formula
deduced from it after taking its whole number ratio is CH2O

18. Calculate the empirical formula of a compound composed
of 38.67 % C, 16.22 % H, and 45.11 %N.
• Assume 100 g so
• 38.67 g C x 1mol C = 3.220 mole C
12.01 gC
• 16.22 g H x 1mol H = 16.09 mole H
1.01 gH
• 45.11 g N x 1mol N = 3.219 mole N
14.01 gN
The ratio is 3.220 mol C = 1 mol C
3.219 mol N 1 mol N
The ratio is 16.09 mol H = 5 mol H
3.219 mol N 1 mol N
C1H5N1

19. Problems
• A compound is 43.64 % P and 56.36 % O.
What is the empirical formula?
• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its
empirical formula?
• The percent composition of a solid is known to be 68.4% Ba, 10.3% P, and
21.3% O. What is the empirical formula of the compound.

20. Q. What is the percent composition of sodium
hydrogen carbonate?
• The formula for sodium hydrogen carbonate is NaHCO3. The Na:H:C:O mole
ratio is 1:1:1:3. Convert this mole ratio into a mass ratio by assuming there is a 1
mole sample present. Answer = 23g Na, 1.0g H, 12g C, 48 g O
• To determine the percent composition, divide the mass of each element present
by the total mass of the compound and multiply by 100.
• Total mass of 1 mole of NaHCO3 = 84 g
• Answer = 27% Na, 1.2% H, 14% C, 57% O

21. A compound contains 4.07% hydrogen,
24.27%carbon,71.65%chlorine, its molar mass is 98.9g. What are its
empirical formula and molecular formula.
• Steps involved in the solution
• Conversion of mass per cent to grams
• Convert into number of moles of each element.
• Divide the mole value obtained above by the smallest number
• Write empirical formula by mentioning the numbers after writing the
symbols of respective elements
• Writing molecular formula

22. STOICHIOMETRY AND STOICHIOMETRIC
CALCULATIONS
• Stoichiometry deals with the calculation of masses of the reactant and the products involved in a chemical
reaction.
• But before calculation we need to know the exact amount of all the masses of reactant and product taking
part in any chemical reaction for which we need to balance the chemical equation.
For example a balanced equation for this reaction is:
CH4(g)+2O2(g) CO2 (g)+ 2H2O(g)
The coefficients 2 for O2 and H2O are called stoichiometric coefficients

23. Calculate the amount of water (g) produced by the combustion of
16g of methane.
• The balanced equation for combustion of the methane is :
• CH4(g) +202(g) CO2(g)+2H2O(g)
• 1. 16 g of CH4 corresponds to one mole
• From above equation , 1mol of CH4 gives 2 mol of H2O
2 mol of water H2O = 2x(2+16)
2x18= 36 g
1 mol of H2O =18 g H20= 18g H20= 1
1mol H20
Hence 2 mol H20X 18g H20
1 mol H2O
2X18 g H2O= 36 g H2O

24. Limiting reagent
• The reactants which get consumed and limits the amount of product formed is
therefore called the limiting reagent.
• For example in the reaction
• N2 (g)+ 3H2 (g) 2NH3 (g)
• In the reaction dihydrogen is the limiting reagent because the amount of product
formed ammonia is mostly depends on the availability of the dihydrogen

25. THANKS

26. QUERIES?????