Form 3 Chemistry – MOLE CONCEPT

What is mole?

Mole is a unit of measurements as any other units,

Example Pair, dozen, gross etc

1mole = 6.02 x 10²³ particles

Definition;

A mole is the amount of a substance as many as particles of elementary entities contained in 12g of carbon-12 isotopes. The particles can be atoms, molecules, electrons or ions. This is a unit mole which was introduced by Avogadro
as 6.02 x 10²³ particles               

Example; 1 mole of water (H₂O) contains 6.02 x 10²³ molecules

               1 mole of sodium (Na) contains 6.02 x 10²³ atoms

               1 mole of CuCl₂ contains 6.02 x 10²³ ions

               1 mole of Fe contains 6.02 x 10²³ electrons

MOLAR MASS OF A SUBSTANCE

Molar mass: is the mass of I mole of any element or compound

Its SI unit is g/mol. It is denoted by M

For example; Na = its molar mass = 23g/mol

       NaCl = its molar mass = 58.5g /mole

       Na₂SO₄ = its molar mass = (23 x 2) + (32) + (16 x 4) = 142g/mol

NB: Molar masses and relative molecular mass of a substance are calculated from the formula of that substance.

QUESTION;

Calculate the molar of the following

1. Na₂CO₃. 10 H₂O

11. Al₂ (SO₄)₃

Al = 27, S = 32, O = 16, Na = 23, C = 12, H = 1

2. Calculate the number of molecules in   i. 8g of oxygen gas

                                                              ii. 11g of CO₂

3. If the number of ions in CuCl₂ is 3.02 x 10 ²³, what was the mass of CuCl₂

 
 

ANSWERS:

1. Na₂CO₃.10H₂O
   

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   (23 x 2) + 12 + (16 x 3) + 10 [(1 x 2) + 16]

   46 + 12 + 48 + 180

            106 + 180        = 286g/mol

 
 

ii) Al₂ (SO₄)₃

         (27 x 2) + 3 (32 + (16 x 4))

            54 + 3(96)

            54 + 288          = 342g/mol

 
 

Iii). 8g of oxygen gas

           O₂ = 16 x 2 = 32 = 6.02 x 10²³ 
 
                           8   =?

                        

                      = 1.505 x 10 ²³ molecules

IV). 11g of CO2

            CO₂ = 12 + 32 = 44 = 6.02 x 10²³ 

                               11 =?

                                   

                                    = 1.505 x 10²³ molecules

CuCl₂

            64 + (35.5 x 2) = 64 + 71 = 135 = 6.02 x 10²³

                                                         ?   = 10²³

                                                                  

                                    = 67.5 g/mol

 RELATIVE MOLAR MASS (M_r)

This is the mass of one molecule of a compound compared with the mass of one atom of carbon – 12.

It has no units

Molar mass (M) = Relative molecular mass (M_r)

Molar mass of an element = relative atomic mass (A_r)

Example; Na    M of Na = 23gmol^-1

                        A_r of Na = 23

                        M or CH₄ = 16gmol^-1

                        Mr of CH₄ = 16

                        M of H₂O = 18gmole

                        Mr of H₂O = 18

 Example;

1. Calculating the molar mass of (NH₄)₂CO₃ given that N = 14, H= 1, C = 12, O = 16.

Solution

(NH₄)₂CO₃

(14g x 2) + (1g x 8) + 12g + (16g x 3)

            28g/mol + 8g/mol + 12g/mole + 48g/mol

            = 96g/mole

Molar mass of a compound

This is the sum of constituent atoms (elements)

AMOUNT OF A SUBSTANCE OR NUMBER OF MOLES (n)

Number of moles, n is the mass of the sample of a substance divided by the molar mass of that substance

                               M = Molar mass of substance

                                 N = number of moles (amount of substance)

Example;

What is the amount of substance in

a) 180g of carbon

            m = 180g

            M = 12gmol^-1

                  

            n = 15mol

b) 180g of CO₂

      

 n = 4.09 mol

QUESTIONS:

Finding the mass of each of the following substance

a) 2.4 mol of NaOH

b) 3.2 x 10^-3 mol of N

c) 0.780 mol of Ca(CN)₂

d) 7.00 mol of H₂O₂

How many moles are in each of the following?

a) 0.800g of Ca

b) 79.3g of Cl₂

c) 5.96g of KOH

d) 937g of Ca(C₂H₃O₂)₂

                        NUMBER OF PARTICLES (N) IN A GIVEN AMOUNT OF SUBSTANCE (n)

To find the number of particles in a given amount of substance we use the expression

N = n.L

Where:  n = number of moles

            N = number of particles in a substance

            L = Avogadro’s no = 6.02 x 1023 mol^-1

QUESTION:

1. How many particles are there in 20g of Ca?

Solution

Find n in 20g of Ca

  

N = 0.5 mol

From the expression

N = nL

= 0.5 mol x 6.02 x 10²³ mol^-1

= 3.01 x 10²³

2. How many molecules are there in 80g of the NaOH?

  MOLAR VOLUME (V_m)

MOLAR GAS VOLUME

This is the volume occupied by one mole of a gas at standard temperature and pressure and is 22.4 dm³ or 22.4l

            1dm³ = 1l = 1000cm³

            Standard temp = 0⁰c (273K)

Standard pressure = 1 atm (760 mm Hg)

Avogadro’s law states “At the same temperature and pressure volumes of all gases contain the same number of particles”

In calculating the amount of substance n, using molar volume (v_m), the expression used is

Example; Find the amount of substance present in

a) 18.8 dm³ of CO₂ at s.t.p

b) 48.8 dm³ of O₂ at s.t.p

Solution

Where; V = volume of a gas

                       V_m = molar volume of a gas

                        n = amount of substance

                        V = 18. 8 dm3

                        V_m = 22.4 dm3/ mol

                        n =?  

Substituting the volume in the expression

                       

                        n = 0.839 mol

            n = 2.18 mol

 THE MOLE IN STOICHIOMETRIC CALCULATIONS

A balance chemical equation tells us a great of quantitative information

Consider the following equation us a great deal of quantitative information

Consider the following balanced equation

      

Stoichiometry- is the quantitative relationship of reacting substances

Stoichiometric coefficients = moles

A balanced equation is used in calculating different quantities

     

From the above balanced equation, calculate the weight of CaO and Volume of CO₂ at s.t.p which will be produced by heating 75g of CaCO₃

(Ca = 40, C = 12, O = 16)

Solution

Finding the molecule mass

  

Molecular mass of CaCO₃ = 100g

                                CaO = 56g

 From the equation:

  

1 mol        1mol   1mol

100g           56g

15g             x?

75g    produces   ?

x=42g

If 100g/mol = 22.4dm³

     75g/mol =?

            = 16.8 dm³

QUESTION

a) How many ions are in 10g of Al₂(SO₄)₃?

b) How many fluorides are in 1.46mols of AIF₃?

Solution; 

342 g/mol = 5 mol of ions

       10 g     =     ?

If 342 g = 1 mol

   10 g   =  x

            x = 0.0292 mol of Al₂ (SO₄)₃

If 1 mol of Al₂ (SO₄)₃   =     5mol of ions

0.0292mol of Al₂ (SO₄) ₃  =     ?                  

           

= 0.0146 mol of ions

N = nL

   = 0.046 mol x 6.02 x 1023 mol-1

   = 8.79 x 10²³

     SOLUTION AND MOLAR CONCENTRATION
Solution is a mixture of solvent and solute.
   Solute – Is a solid crystal component which dissolved into solvent.
   Solvent – Is a liquid component which dissolve solute. (water is a common universal solvent).
So that when we prepare a standard solution, we must measure both two component. A solute measured in grams of weight and solvent in liter, dm3  or cm3 of its volume. The quantity of solute that may dissolved by 1litre of solvent make a standard solution.

^{STANDARD SOLUTION
Is a solution of known concentration. Standard solution has been prepared by accurate measurement and the concentration of solute in the solution is described into two ways.
        i. Weight concentration in grams.
        ii. Particles concentration in mole.}

i. WEIGHT CONCENTRATION
This concentration abbreviated as conc. of a grams dissolved in one litre (g/dm3 or g/L).
e.g 30g/L or 15g/dm3

Example;
Allen dissolved two spoons of sugar into the tea drink. If capacity of cup is 200cm3 and a spoon carry 25g, what is the concentration of sugar in the tea?

Solution
carrying capacity = 25g
               25 x 2 = 50g
Therefore, weight of solute = 50g
Volume of solvent = 200cm3
But,
    
MOLARITY (MOLAR CONCENTRATION)
This is a concentration of particles contained in one litre of the solution. (mol/L or M)

Example;
The Allen spin a common salt in the bowl filled by 320cm3 of water to make a solution. If the same spoon used, determine a molarity of salt in the solution.

Solution
Volume of water = 320cm3
Weigh of salt = 25g
But, Moles = weight of component/Molar mass (W/Mr)
        n = 25g/58.5g
        n = 0.43moles
Therefore,
       
RELATIONSHIP OF WEIGHT AND MOLAR CONCENTRATION
From,
        Molarity = moles/Volume in litre
        M = n/V in L
        But,
             n = W/Mr
        Therefore,
        M = W/Mr X 1/V in L
        But,
        W/V in L = Conc
        Therefore,
         M = Conc/Mr

So that,  
             Molarity = Conc/ Molar mas
Where by, 
              n = no of moles
              V = Volume in liter
              Mr = Molar mass
              W = Weight of component in grams

Assignment

Find the molarity of each component if both had a concentration of 72g/dm3 .
   (a) H₂^SO₄
   ^{(b) Na}₂^CO₃
   ^{(c) NaHCO}₃
   ^{(d) KCl}