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Mole Concept (Avogadro Constant) , Molar mass, Percentage Composition,Emprical and Molecular Force





Mole Concept (Avogadro Constant) And Molar mass

1 Mole is defined as the quantity of substance comprising the same number of distinct entities (like atoms, ions, and molecules) as the count of atoms in a sample comprising pure 12C weighing accurately 12 g 1 mole = 6.022 × 1023 entities
This is also called Avogadro Constant
This is true for both atoms and Molecules ,so we say that
1 mole of hydrogen atom=6.022 × 1023 atoms
1 mole of water molecule=6.022 × 1023 molecule
It is denoted by letter NA
The mass of one mole of a substance in grams is called its molar mass
Molar Mass in gram is numerically equal to the atomic mass or Molecular mass or formula mass in u
Unit of Molar Mass is g/Mol

Molar Volume

The volume occupied by one mole of any substance is called its molar volume. It is denoted by Vm. One mole of all gaseous substances at 273 K and 1 atm pressure occupies a volume equal to 22.4 litre or 22,400 mL. The unit of molar volume is litre per mol or millilitre per mol

List of Molar Mass of common elements

Element Molar Mass (g/mol)
H 1.0079
He 4.0026
Li 6.941
Be 9.0122
B 10.811
C 12.0107
N 14.0067
O 15.9994
F 18.9984
Ne 20.1797
Na 22.9897
Mg 24.305
Al 26.9815
Si 28.0855
P 30.9738
S 32.065
Cl 35.453
K 39.0983
Ar 39.948
Ca 40.078
Sc 44.9559
Ti 47.867
V 50.9415
Cr 51.9961
Mn 54.938

List of Molar Mass of common Molecules

Chemical Compound Formula Molar Mass (gram/mol)
Ammonia NH2 17
Carbon Dioxide CO2 44
Ethanol C2H5OH 46.1
Hydrogen H20 2.02
Nitrogen N2 28
Oxygen O2 32
Water H20 18
Glucose C6 H12 O6 180
Sodium Chloride NaCl 58.5
ethane C2H6 30.07
Methane CH4 16.043
Carbon Monoxide CO 28
Hydrogen Chloride HCL 36.461

Gram Molecular Mass

Mass of NA molecules is Gram Molecular Mass.
Gram Molecular Mass in gram is numerically equal to Molecule mass/Formula mass in u (unified mass)

Gram Atomic Mass

It is the mass of NA atoms.
Gram Atomic Mass in gram is numerically equal to atomic mass in u (unified mass)


Solved Example


Question 1 Calculate the no. of moles present in 6 gm Carbon.
Solution
For Carbon
G. A. M. = 1 mole
12 gm = 1 mole
1 gm = 1/12mole
For 6 gm = 6 x (1/12)moles
=1/2 moles.

Question 2 Calculate the no. of moles in 2.8 gm N?
Solution
G. A. M. = 1 mole
14 gm = 1 mole
1 gm = 1/14 moles
2.8 gm = 2.8 *(1/14) moles
= 0.2 moles

Question 3Give the no. of atoms present in five moles N
Solution
Moles Na = 6.022 × 1023 atoms
5 moles Na =30.11 × 1023 atoms

Question 4Calculate the molesin 10 gm Carbon atom
Solution
1Mole = G. A. M.
1Mole = 12 gm
1 gm = 1/12 mole
10 gm= 10* (1/12) mole
=10/12 mole

Question 5Find the mass occupied by 0.5 mole N.
Solution
1 mole = Gram atomic mass
1 mole = 14 gm
0.5 mole= 14 *0.5
= 7 gm

Question 6 Give the no. of atoms present in 2.5 moles carbon
Solution
1 mole =NA atoms
1 mole = 6.022 × 1023 atoms
So, 2.5 moles = 6.022 *2.5 * 1023
= 15.0490 × 1023 atoms
= 15.049 × 1023 atoms

Question 7Give the no. of moles present in 3.01 × 1023 atoms of Sodium
Solution
1 mole =NA atoms
1 mole = 6.022 × 1023 atoms
1 atom = ( 1/ 6.022 × 1023 ) moles
So, in 3.01 × 1023 atoms =3.01× 1023 × ( 1/ 6.022 × 1023 ) = .5 Mole

Question 8Give the no. of moles present in 0.6 gm Carbon.
Solution
G. A. M. = 1 mole
12 gm = 1 mole
1 gm =1/12 mole
So, in 0.6 gm = .6 X (1/12) mole
= 0.05 mole

Question 9 Calculate the no. of atoms present in 2.3 gm Sodium.
Solution
NA atoms = G. A. M.
6.022* 1023 atoms = 23 gm
23 gm= 6.022* 1023 atoms
1 gm = (1/23) * 6.022* 1023 atoms
2.3 gm = (2.3/23) * 6.022* 1023 atoms
=6.022* 1022atoms

Question 10 Give the G. A. M. of an element occupying 1.21 volume at STP is 3.4 gm.
Solution
For 11.2 L = 3.4 gm
Multiplying it by ‘2’
22.4 L = 6.8 gm
So G.A.M =6.8 gm

Question 11 Give the no. of Molecules present in 0.18 gm of water.
Solution
Water Molecular mass =18
so 18 gm = 1moles
0.18 g = .18 * (1/18) moles
= 0.01 mole
=6.022 * 1021 molcules

Percentage composition: -


The mass percentage of each constituent element present in any compound is called its percentage composition
Percentage composition of elements in compound
Example
C6 H12 O6 =(Glucose)
% of C is given by

=$ \frac {6 * 12}{180} * 100$
= 40%
% of Hydrogen is
=$ \frac {12 *1}{180} * 100$
=6.67%
% of Oxygen
=$ \frac {6* 16}{180} * 100$
=53.33%

Empirical Formula and Molecular Formula
An empirical formula represents the simplest whole number ratio of various atoms present in a compound. E.g. CH is the empirical formula of benzene (C6H6)
The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. E.g. C6H6 is the molecular formula of benzene.
Relationship between empirical and molecular formulae
The two formulas are related as Molecular formula = n x empirical formula

Quiz Time

Question 1 Find the number of moles in $12.044 \times 10^{23}$ atoms of Sodium.
A. .5 moles
B. 1 moles
C. .25 moles
D. 2 moles
Question 2 The molar mass of $C_6 H_{12} O_6$ is
A. 190 g/mol
B. 180 g/mol
C. 100 g/mol
D. 180/kg/mol
Question 3 The number of Molecules present in 0.036 gm of water
A. 12.044 * 1020 molcules
B. 6.022 * 1021 molcules
C. 12.044 * 1022 molcules
D. 6.022 * 1023 molcules
Question 4 1 amu
A. 1.660056 × 10-24 g
B. 1.660056 × 10-20 g
C. 1.76 × 10-24 g
D. 1.0 × 10-24 g
Question 5 Find the mass occupied by 2.5 mole N
A. 14 gm
B. 35 gm
C. 28 gm
D. 21 gm
Question 6The unit of Molar mass is
A. moles/gm
B. g/moles
C. g x moles
D. None of the these


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