Mole Concepts and Calculations

Mole is the amount of substance that contains the same number of particles as the number of carbon atoms in exactly 12g of the carbon-12 isotope.

If 1 atom of carbon has a mass of 1.992x10-23, let us find out how many atoms are there in 12g of carbon

Let the number of atoms be W 

1 atom of carbon = 1.992x10-23 g.

 W atoms of carbon =12g
 
 → 1 atom x 12g = W atoms x 1.992 x 10-23 g

 W atoms = 1 atom x 12g
           1.992 x 10-23g
         = 6.02 x 1023 atoms.

The number of atoms present in 12g of carbon-12 isotope is 6.02 x 23 atoms.

Thus 1 mole of a substance contains 6.02 x 1023 particles. This number (6.02 x 1023) is called Avogadro’s constant, and it is usually denoted by either L or NA.

Avogadro’s constant is the number of atoms in exactly 12g of carbon -12 isotope. It is also known as Avogadro’s number.

The Molar Mass

Molar mass is the mass of one mole of a particular substance in grams. Its unit is g/mol.

The molar mass of a substance is numerically equal to its relative atomic mass/relative molecular mass/relative formula mass

The molar mass of a single element is equal to its relative atomic mass. The molar mass of molecular substance is equal to its relative molecular mass of that molecule. Lastly, the molar mass of ionic substance is equal to its relative formula mass of that ionic compound.

The differences between molar mass and the stated equivalents is in its units. The units for molar mass is g/mol whereas relative formula/atomic/molecular masses have no units

Substance Molar massRMM/RFM/RAM
CaO 56g/mol 56
Al 27g/mol 27
CO2 44g/mol 44
C 12g/mol 12

Worked Examples on Molar Mass

  1. Calculate the molar mass of the following compounds:
    1. Al2(SO4)3

      2. The relative atomic masses: Al=27, S=32, O=16

      Soln.

      Al2(SO4)3
      (27 X 2) + (32 X 3) + (16 X 4 X 3)
      =342g/mol
      The molar mass of Al2(SO4)3 is 342g/mol.
    2. Na2CO3.10H2O

      3. Ar for Na = 23, C= 12, O= 16, H=1.

      Soln.

      Na2CO3.10H2O
      (23 x 2) + 12 + (16 x 3) + (10 x 1 x 2) + (10 x 16) = 286g/mol
      The molar mass of Na2CO3.10H2O is 286g/mol.
  2. Calculate the molar mass of CH3COOH.
    3. Answer, 60g/mol.

Exercise

Calculate the molar mass of the following substances. Relative atomic masses are Na=23, C= 12, O=16, N=14, H=1, S=32.
  1. (NH4)2SO4
  2. Na2CO3

The Relationship between Mass and Mole

In order to get the number of moles from the given mass, you will have to divide mass of the substance by its molar mass.

Number of moles = mass of the substance
	          molar mass of the substance

Worked Examples

Calculate the number of moles of present in
(a)180g of carbon
(b)180g of carbon dioxide

Soln.

(a) Mass of carbon = 180g
Molar mass of carbon =12g/mol.
Number of moles = mass of the substance
	          molar mass of the substance
 = 180g
  12g/mol
=15mol. Therefore, the molar mass of carbon in 180g is 15mol. (b) 180g of carbon dioxide

Soln.

Mass of carbon dioxide = 180g
Molar mass of carbon dioxide =44g/mol. 
Number of moles(n) = Mass of the substance
                    Molar mass of the substance     
		  = 180g
		    44g/mol
		 =4.09mol
The number of moles present is 4.09mol.

Molar Volume of Gases

NOTE

For some calculations, you will have to convert the units and the following relationship among volume units is helpful in such demand

1 litre = 1000cm3 = 1dm3

With exception to noble gases, most gases are made up of particles called molecules.

1 mole of any gas contains 6.02 x 1023 molecules.

Avogadro's law states:

"At the same temperature and pressure, equal volumes of different gases contain an equal number of molecules."

Or more simply:

"Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules."
Number of moles of a gas (n) = volume of a gas 
                               molar volume
    Therefore at STP n= volume of a gas
                            22.4dm3
     At r.t. p n= volume of a gas
                      24dm3
  1. What volume at STP will be occupied by 0.5mol of oxygen gas?

    Solutions:

    Method 1:

    From the definition
    1 mole of the gas occupies 22.4dm3 at STP.
    0.5mole will occupy ?dm3

    Volume occupied = (0.5mol x 22.4dm3)
                        1mol
                      = 11.2dm3
    The volume occupied by 0.5mol of oxygen is 11.2dm3

    Method 2: 

     Number of moles (n) = volume of a gas
                             22.4dm3
Thus, volume of a gas at STP = 22.4dm3/mol x 0.5mol
                              = 11.2dm3
  2. What volume at room temperature and pressure will be occupied by 0.2 mol of sulphur dioxide?

    Soln.

    1 mole of sulphur dioxide  =    24dm3
0.2 mole of suphur dioxide       ?dm3
    Volume of sulphur dioxide = (0.2mol × 24dm3)
                                  1mol
                                 = 4.8dm3
  3. What amount of substance in moles is present in 9.98dm3 of oxygen at room temperature and pressure?
  4. Calculate the volume which will be occupied by 3.65g of hydrogen chloride gas at STP.

Number of Particles in a Given Number of Moles of the Substance

The number of particles in one mole of any substance is 6.02 x 1023.

To get the number of particles in any given number of moles, the following relation can be used:

Number of particles (N) = number of moles (n) x Avogadro Constant(L)

Let us calculate the number of particles in:

Number of particles in elements

  1. How many atoms are there in 48g of magnesium element?

    2. Solution

    Molar mass of magnesium element=24g/mol 
    (i)Number moles = 48g 
               24g/mol
	     = 2moles
		 
(ii)Number of atoms in 2moles

Number particles (N) = number of moles (n) x Avogadro’s constant (L)
                     = 2mol x 6.02 x 1023atoms/mol
  		  = 1.204 x 1024atoms.
  2. Find the number atoms present in 0.45mol of Calcium element.

    3. Solution

    Number of particles = number of moles x Avogadro's constant 
Number of atoms = 0.45mol x 6.02 x 1023
                = 2.7 x 1023
Therefore, the number of atoms in 0.45mol 
of calcium are 2.7 x 1023.

Number of Particles in Gaseous Substances

Gases can be measured in terms of mass or volume. Most gases exist as molecules. What is molecule?

Worked Examples

  1. How many molecules are there in
    (a) 112cm3 CO2
    (b) 200cm3 H2 at STP?

    2. Solution

    Data given

    Volume of carbon dioxide =112cm3 = 0.112dm3
    From the formula

    Number of molecules = Number of moles × Avogadro’s constant

    →Number of moles of gas = volume of a gas
                             22.4dm3/mol
Number of moles of gas= 0.112dm3
                        22.4dm3/mol
                      = 0.005mol
    Number of molecules = 0.005mol × 6.02 ×1023molecules/mol
    
                    = 3.01 ×10 21molecules.
There are 3.01 × 1021 molecules of CO2 in 112cm3
  2. How many molecules are there in 2g of oxygen gas at STP?
  3. How many molecules of oxygen are in 3.2dm3 of oxygen gas at STP?

Atomicity in gaseous substances

Atomicity is the number of atoms present in one molecule of an element or compound.

Number of Atoms in Molecules

Number of atoms = number of atoms in a chemical formula x number of molecules

  1. How many (a) oxygen molecules and (b) oxygen atoms are contained in 20g of oxygen gas?

    2. Soln.

    Data given:
    Mass of oxygen gas = 20g
    From N = nL
    Where 
    n = mass
    molar mass
    Molar mass of oxygen gas (O2) = 16 × 2 = 32g/mol 
       n = 20g
      32g/mol
      =0.625mol
    Therefore, number of molecules = 0.625mol × 6.02×10 23 = 3.7625×10 23molecules.
    (b) Oxygen gas (O2) contains two atoms of oxygen, therefore the total oxygen atoms is
    =2×3.7625×10 23molecules.
    = 7.525 × 10 23 atoms of oxygen gas.
  2. Calculate the number of atoms in 25 grams of carbon dioxide (CO2). The relative atomic mmasses of each elements are C = 12 and O = 16. (i)Molar mass of carbon dioxide (CO2) 
    Molar mass of CO2 = (12  + (2 × 16 ))g/mol
                  = 44 g/mol
    (ii) Number of moles of carbon dioxide present in 25g: 
    Number of moles = Mass 
                 Molar mass
                = 25 g 
		 44 g/mol
                = 0.568 mol
    (iii) Calculate the number of molecules: 
    Number of molecules = Number of moles × Avogadro's number
                    = 0.568 mol × 6.02×  1023 molecules/mol
                    = 3.42 × 1023 molecules
    (iv) Calculate the number of atoms:
    There are 3.42 × 1023 molecules of carbon dioxide (CO2) each has 3 atoms (1 atom of carbon and 2 atoms of oxygen). We asked to tell the total of atoms irrespective of their type. 
    Number of atoms = Number of molecules × 3
               = 3.42× 1023 molecules × 3
               = 1.027 × 1024 atoms
    Therefore, there are approximately 1.027 × 1024 atoms present in 25 grams of carbon dioxide (CO2).

  3. Calculate the number of atoms in 30 grams of water (H2O). Use these constants, Ar H=1, O = 16.

    (i). Determine the molar mass of water (H2O).

Molar mass of H2O = 2 × Molar mass of H + Molar mass of O
                 = 2 × 1 g/mol + 16 g/mol
                 = 18 g/mol
    (ii). Number of moles of water. 
    Number of moles = Mass 
                  Molar mass
                = 30 g 
		18 g/mol
                = 1.667 mol
    (iii). Calculate the number of molecules
    Multiply the number of moles by Avogadro's number. 
    Number of molecules = Number of moles × Avogadro's number
                   = 1.667 mol × 6.02 × 1023 molecules/mol
                   = 1.004 × 1024 molecules
    (iv). Calculate the number of atoms:
    Since each water molecule (H2O) consists of 3 atoms (2 hydrogen atoms and 1 oxygen atom), multiply the number of molecules by 3. 
    Number of atoms = Number of molecules × 3
               = 1.004 × 1024 molecules × 3
               = 3.012 × 10 24 atoms
    Therefore, there are approximately 3.012 × 1024 atoms present in 30 grams of water (H2O).

  4. Calculate the number of atoms in 50 grams of methane (CH4). i. Determine the molar mass of methane (CH4). 
    Molar mass of CH4 = Molar mass of C + 4 × Molar mass of H
                 = 12 g/mol + 4 × 1 g/mol
                 = 16 g/mol
    ii. Number of moles of methane in 50grams. 
    Number of moles = Mass 
                  Molar mass
                = 50 g 
		16 g/mol
                = 3.125 mol
    iii. Calculate the number of molecules:
    Multiply the number of moles by Avogadro's number. 
    Number of molecules = Number of moles × Avogadro's number
                    = 3.116 mol × 6.02 × 1023 molecules/mol
                    = 1.876 × 1024 molecules
    iv. Calculate the number of atoms: Multiply the number of molecules by the number of atoms in each molecule. Since each methane molecule (CH4) consists of 5 atoms (1 carbon atom and 4 hydrogen atoms), multiply the number of molecules by 5. 
    Number of atoms = Number of molecules × 5
                = 1.876 ×1024 molecules × 5
                = 9.38 ×  1024 atoms
    Therefore, there are approximately 9.38 × 1024 atoms present in 50 grams of methane (CH4).

    5. Exercise

    1. Calculate the number of molecules in 0.50moles of carbon dioxide. Then calculate the number of atoms contained in these moles.
    2. How many oxygen atoms are present in 0.2moles of NO2?
    3. How many atoms are there in 10g of sulphur dioxide H2S?

    Number of Particles in Ionic Compounds

    Ionic compounds are made of ions, that is, the particles that make ionic compounds are ions. Therefore, calculations of ions in a given ionic compounds should consider the number of ions a given ionic compound it can produce when dissociated into its constituents.
    1. How many ions are there in 386g of Pb(NO3)2?
      2. (Pb=207, N= 14, O = 16)
      Soln
      i. Calculate the molar mass of Pb(NO3)2:
       Molar mass = (207 + 2x14 + 2x3x16)g/mol 
              = 331g/mol
      ii. Find the number of moles: 
      Number of moles = 386g
                  331g/mol
	        = 1.166mol
      iii. Write the dissociation equation for Pb(NO3)2 
      Pb(NO3)2 → Pb2+ + 2NO3-
      
Therefore each Pb(NO3)2 compound has 
      1 mole of ion of Pb2+ and  2 moles of ions of NO3- 

1 mole of Pb(NO3)2   =    3 moles of ions 
1.166moles will produce = ? moles of ions
                         3. 5 moles of ions.
Number of ions = 3.5mol x 6.02 x 1023ions/mol
               = 2.11 x 10 24 ions.
    2. How many ions are there in 10g of Al2(SO4)3

      3. (Use Ar for Al = 27, S=32, O = 16)
      Molar mass = (27x2) + (3x32) + (3x4x16)
           = 342g/mol
Number of moles of Al2(SO4)3 
       = 10g
	 342g/mol
      =0.029mol
Number of ionic compound (Al2(SO4)3)
         = 0.029mol x 6.02 x 1023
         = 1.75 x 1022		 
Dissociation eqn:
Al2(SO4)3 → 2Al2+ + 3SO42-
From this equation each Al2(SO4)3 has ability of producing 5 ions.
Therefore the number of ions that can be produced by 10g of Al2(SO4)3 
          = 1.75 x 1022 x 5

       Answer: 8.75 x 1022 ions.
    3. How many ions are present in 25g of CaCl2?
      4. (Ca = 40, Cl = 35.5)
      4 x 1023
    4. How many ions are present in 100g of (NH4)2SO4?
      1.37 X 10 24 ions.
    5. How many ions are present in 150g of MgCl2?
      6. 2.8 x 1024ions.

    Molar Solution

    A molar solution is a solution that contains one mole of a solute dissolved in one liter of the solution.

    Concentration is the amount of a substance (solute) that is dissolved in a given volume of solution.

    There are various ways of expressing the concentration, for this level we will only discuss two ways:
    (a) Molarity/ molar concentration
    (b) Mass concentration or concentration in g/dm3 

    Molarity = number of moles(of solute)
           volume of solution in dm3
OR:

Concentration(g/dm3) = mass of solute(g)
                     volume of solution(dm3)
					 
The relationship between molar concentration and concentration(g/dm3):

Molarity = Concentration(g/dm3)
           Molar mass
    Example:
    Calculate the molar concentration of a solution of sodium hydroxide which was made by dissolving 10g of solid sodium hydroxide in water and making up to 250cm3
    (Na = 23, O = 16, H = 1)
    Soln
    Number of moles = mass of substance
                  molar mass 
		
		= 10g
		40g/mol
		
		= 0.25mol
Molarity = number of moles
          volume of solution (dm3)
		
	=  0.25mol
 	 0.25dm3
		 
       = 1mol/dm3 or 1M

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