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
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
