1. The mole is the SI unit of the amount of substance.
2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles.
3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.
The Avogadros Constant contain 6.023 x10 23 particles. i.e.
1mole = 6.023 x10 23 particles = 6.023 x10 23
2 moles = 2 x 6.023 x10 23 particles = 1.205 x10 24
0.2 moles = 0.2 x 6.023 x10 23 particles = 1.205 x10 22
0.0065 moles = 0.0065 x 6.023 x10 23 particles = 3.914 x10 21
3. The mass of one mole of a substance is called molar mass. The molar mass of:
(i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g.
Molar mass of carbon(C)= relative atomic mass = 12.0g
6.023 x10 23 particles of carbon = 1 mole =12.0 g
Molar mass of sodium(Na) = relative atomic mass = 23.0g
6.023 x10 23 particles of sodium = 1 mole =23.0 g
Molar mass of Iron (Fe) = relative atomic mass = 56.0g
6.023 x10 23 particles of iron = 1 mole =56.0 g
(ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass is the sum of the relative atomic masses of the elements making the molecule.
The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.
Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g
6.023 x10 23 particles of Oxygen molecule = 1 mole = 32.0 g
Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g
6.023 x10 23 particles of chlorine molecule = 1 mole = 71.0 g
Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g
6.023 x10 23 particles of Nitrogen molecule = 1 mole = 28.0 g
(ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass is the sum of the relative atomic masses of the elements making the compound. e.g.
(i)Molar mass Water(H2O) = relative formular mass =[(1.0 x 2 ) + 16.0]g =18.0g
6.023 x10 23 particles of Water molecule = 1 mole = 18.0 g
6.023 x10 23 particles of Water molecule has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-1 x 6.023 x10 23 particles of Oxygen atoms
(ii)Molar mass sulphuric(VI)acid(H2SO4) = relative formular mass
=[(1.0 x 2 ) + 32.0 + (16.0 x 4)]g =98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) = 1 mole = 98.0g
6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
–1 x 6.023 x10 23 particles of Sulphur atoms
-4 x 6.023 x10 23 particles of Oxygen atoms
(iii)Molar mass sodium carbonate(IV)(Na2CO3) = relative formular mass
=[(23.0 x 2 ) + 12.0 + (16.0 x 3)]g =106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) = 1 mole = 106.0g
6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) has:
– 2 x 6.023 x10 23 particles of Sodium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(iv)Molar mass Calcium carbonate(IV)(CaCO3) = relative formular mass
=[(40.0+ 12.0 + (16.0 x 3)]g =100.0g.
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) = 1 mole = 100.0g
6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) has:
– 1 x 6.023 x10 23 particles of Calcium atoms
–1 x 6.023 x10 23 particles of Carbon atoms
-3 x 6.023 x10 23 particles of Oxygen atoms
(v)Molar mass Water(H2O) = relative formular mass
=[(2 x 1.0 )+ 16.0 ]g =18.0g
6.023 x10 23 particles of Water(H2O) = 1 mole = 18.0g
6.023 x10 23 particles of Water(H2O) has:
– 2 x 6.023 x10 23 particles of Hydrogen atoms
-2 x 6.023 x10 23 particles of Oxygen atoms
Practice
1. Calculate the number of moles present in:
(i)0.23 g of Sodium atoms
Molar mass of Sodium atoms = 23g
Moles = mass in grams = > 0.23g = 0.01moles
Molar mass 23
(ii) 0.23 g of Chlorine atoms
Molar mass of Chlorine atoms = 35.5 g
Moles = mass in grams = > 0.23g = 0.0065moles /6.5 x 10-3 moles
Molar mass 35.5
(iii) 0.23 g of Chlorine molecules
Molar mass of Chlorine molecules =( 35.5 x 2) = 71.0 g
Moles = mass in grams = > 0.23g = 0.0032moles /3.2 x 10-3 moles
Molar mass 71
(iv) 0.23 g of dilute sulphuric(VI)acid
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
2. Calculate the number of atoms present in:(Avogadros constant L = 6.0 x 10 23)
(i) 0.23 g of dilute sulphuric (VI)acid
Method I
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
Moles = mass in grams = > 0.23g = 0.0023moles /2.3 x 10-3 moles
Molar mass 98
1 mole has 6.0 x 10 23 atoms
2.3 x 10-3 moles has (2.3 x 10-3 x 6.0 x 10 23) = 1.38 x 10 21atoms
1
Method II
Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g
98.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 1.38 x 10 21atoms
98
(ii)0.23 g of sodium carbonate(IV)decahydrate
Molar mass of Na2CO3.10H2 O=
[(2 x 23) + 12 + (3 x16) + (10 x 1.0) + (10 x 16)] = 276.0g
Method I
Moles = mass in grams = > 0.23g = 0.00083moles /
Molar mass 276 8.3 x 10-4 moles
1 mole has 6.0 x 10 23 atoms
8.3 x 10-4 moles has (8.3 x 10-4 moles x 6.0 x 10 23) = 4.98 x 10 20atoms
1
Method II
276.0g = 1 mole has 6.0 x 10 23 atoms
0.23 g therefore has (0.23 g x 6.0 x 10 23 ) = 4.98 x 10 20atoms
276.0
(iii)0.23 g of Oxygen gas
Molar mass of O2 = (2 x16) = 32.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00718moles /
Molar mass 32 7.18 x 10-3 moles
1 mole has 2 x 6.0 x 10 23 atoms in O2
7.18 x 10-3moles has (7.18 x 10-3moles x 2 x 6.0 x 10 23) =8.616 x 10 21atoms
1
Method II
32.0g = 1 mole has 2 x 6.0 x 10 23 atoms in O2
0.23 g therefore has (0.23 g x 2 x 6.0 x 10 23 ) = 8.616 x 10 21atoms
32.0
(iv)0.23 g of Carbon(IV)oxide gas
Molar mass of CO2 = [12 + (2 x16)] = 44.0 g
Method I
Moles = mass in grams = > 0.23g = 0.00522moles /
Molar mass 44 5.22 x 10-3 moles
1 mole has 3 x 6.0 x 10 23 atoms in CO2
7.18 x 10-3moles has (5.22 x 10-3moles x 3 x 6.0 x 10 23) =9.396 x 10 21atoms
1
Method II
44.0g = 1 mole has 3 x 6.0 x 10 23 atoms in CO2
0.23 g therefore has (0.23 g x 3 x 6.0 x 10 23 ) = 9.409 x 10 21atoms
44.0
