# CHEMISTRY: THE MOLE-Introduction to the Mole, Molar Masses and Relative Atomic Masses

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