Mole Concept: Overview

Mole concept is arguably the broadest topic in chemistry, as it cuts across every other branch of chemistry, as far as reactions are concerned. It is the foundation of calculations in chemistry. 

Chemical reactions, expressed as equations, are ways of confirming the Law of Conservation of Mass, which states that matter is neither created nor destroyed, but is transformed from one form to another. This is also what was paraphrased in one of the postulations of Dalton's Atomic Theory, which is, atoms are neither created nor destroyed (during a chemical reaction), but are changed from one form to another.

The above implies that if two substances A and B combine to give another substance C, i.e,

             A  +  B  --->  C

then, the total masses of A and B will equal the mass of C. In other words, all of A and B will be converted to C, assuming there is no loss. Meaning that if we know the masses of A and B, we can easily calculate what we should be expecting as C.

For instance, in the reaction between sodium and chlorine to form sodium chloride;

          2Na  +  Cl2  --->  2NaCl

        Before reacting (reactants):

            Na = 2*1 = 2 atoms
            Cl = 1*2 = 2 atoms

        After reacting (products):

            Na = 2*1 = 2 atoms
            Cl = 2*1 = 2 atoms

we can see that 2 atoms of Na, combined with 2 atoms of Cl to form 2 moles of NaCl, made up of 2 atoms of Na and 2 atoms of Cl each; and if we carry out atom counts, we will observe that no atom was created (added)and none was destroyed (lost).

From the above example, it means that whenever Na and Cl combine to form NaCl, they MUST do so in the ratio of 2:1:2 That is to say, 2 moles of Na molecule will always combine with 1 mole of Cl molecule to form 2 moles of NaCl molecule. This is known as the stoichiometry of the reaction, (i.e, the relationship between the amounts of the reactants and the products, in terms of the mole ratio and mass ratio of the species involved).

Definition of Terms

Mole
Since the substances that react contain different amounts of elementary particles (atoms, molecules, ions, electrons etc), which directly or indirectly contribute to their masses; it is always difficult to measure the exact amount of a particular substance that may be required to react with another. Thus, the need to take their masses and volumes (if gaseous), to "the same base". This "base" is known as the mole.

Hence, a mole can be defined as the amount of elementary entities or particles (atoms, molecules, ions, electrons etc) present in a substance, which is exactly as those present in 12 grammes of carbon-12.

Molar Mass
This is the mass of 1 mole of a substance expressed in grammes. Its is expressed in grammes per mole (g/mol).

Molar Volume
This is the volume occupied by 1 mole of a substance, usually gases, at standard temperature and pressure (s.t.p). It is also known as the molar gas volume (G.M.V), and has a constant value of 22.4dm^3 or 22400cm^3.

Avogadro Number
This is a constant value of 6.023 x 10^23, which represents the amount of elementary entities present in 1 mole of a substance. It can be likened to a dozen - 12, a score - 20, or a gross - 144.

Important Formulae Used in Mole Concept

1) Mole (n) = Reacting Mass (m)
                        ------------------
                          Molar Mass (M)
i.e,   n = m/M

This implies that,

Reacting Mass = Mole x Molar Mass

Molar Mass = Reacting Mass/Mole

2) Mole (n) = Reacting Volume (v)
                       --------------------
                        Molar Volume (V)

i.e,   n = v/V

This implies that,

Reacting Volume = Mole x Molar Volume

Molar Volume = Reacting Vol/Mole

3) Mole (n) = Number of Particles
                       --------------------
                             6.023 x 10^23

Relationship Between Mole & Mass (m), Volume (v) and Avogadro Number (L)

The relationships between the four quantities stated above can be expressed as shown below:

Molar Mass --- Mole --- Molar Volume
                                |
                 Avogadro Number

i.e, 1 mole = 6.023 x 10^23 = 22.4dm^3 = molar mass

1mole of CO2 = 44g of CO2 = 22.4dm^3 of CO2 = 6.023 x 10^23 molecules of CO2

1mole of O2 = 32g of O2 = 22.4dm^3 of O2 = 6.023 x 10^23 molecules of O2 = 2 x 6.023 x 10^23 atoms

1mole of Na = 23g of Na = 6.023 x 10^23 molecules of Na = 6.023 x 10^23 atoms of Na.

Example:

   2Na(s)  +  Cl2(g)  --->  2NaCl(s)

You will observe that we have introduced the state symbols (s - solid, g - gas), which enables us to know the physical states at which the reacting species and products are at room temperature.

Firstly, let's calculate the molar masses:

Molar mass of Na = 1* Relative Atomic Mass of Na
= 1* 23 = 23
Thus, 1mole of Na weighs 23 g/mol.

Molar mass of Cl2 = 2* Relative Atomic Mass of Cl
= 2 * 35.5 = 71
Hence, 1mole of Cl2 molecule weighs 71 g/mol.

Similarly, Molar mass of NaCl = (1* Relative Atomic Mass of  Na) + (1* Relative Atomic Mass of Cl)
= (1* 23) + (1 * 35.5)
= 23 + 35.5 = 58.5
Therefore, 1mole of NaCl molecule weighs 58.5 g/mol.

Now, let's fly:

                2Na(s)  +  Cl2(g)  --->  2NaCl(s)

Mole ratio    2       :      1          :       2

Molar mass   23         71             58.5

Molar vol         -      22.4dm^3         -

Rxting mass 2*23      1*71           2*58.5
                      46g          71g               117g
(*Rxting Mass = Reacting Mass)

From the above, it means that if we were to prepare sodium chloride in the lab, we would need 46g of sodium metal and 71g of chlorine gas, which would give us 117g of sodium chloride.

Also, remember that chlorine is a gas at room temperature and therefore, possesses molar volume. This implies that 46g of sodium metal can also react with 22.4dm^3 of chlorine gas to form 117g of sodium chloride at s.t.p.

The question now is what if we do not have up to this amounts of reagents in the lab, does it mean we will not be able to prepare sodium chloride? Of course, we will. Let's say, we have:
(i) 35g of sodium, and
(ii) 50g of chlorine in the lab, how much sodium chloride can we prepare from each?

According to the equation:
2Na(s)  +  Cl2(g)  --->  2NaCl(s)

(i) If 46g of Na produce 117g of NaCl
Then, 35g of Na will produce xg of NaCl

Solve for x by cross multiplying:
46 * x = 117 * 35
    46x = 4095
        x = 4095/46
           = 89.02g of NaCl

(ii) If 71g of Cl produce 117g of NaCl
Then, 50g of Cl will produce xg of NaCl
71 * x = 117 * 50
    71x = 5850
        x = 5850/71
           = 82.39g of NaCl

We can see from the above that we can achieve whatever we desire in the lab, irrespective of the amount of reagents available. All we need is the stoichiometry of the reaction, and every other thing will fall into place.

There is something l want us to note in our calculations above:

In (i), we ignored Cl, while in (ii), we ignored Na. I like calling species like these, 'distractions', because though, they are part of the equation, the question did not provide any information about them in each case. In other words, the questions were just silent about them.

Also, we made use of simple proportions in our calculations, by using the relationship between the two species involved in each question, i.e, Na and NaCl in question (i), and Cl and NaCl in question (ii)

Beginners Steps to Take When Solving Problems Involving Mole Concept

1) Write out the balanced chemical equation.
2) Get the stoichiometry or mole ratio of the reaction.
3) Calculate the molar masses of the substances involved.
4) Calculate the molar volumes of the gaseous substances involved (if any).
5) Calculate the reacting masses/volumes of the substances as the case may be.
6) Identity the 'distractions' in the equation and ignore them, based on the question(s).
7) Apply simple proportion in solving the problem using the reacting masses/volumes/no. of particles.

Examples

Question 1
Calculate the number of moles present in 8g of oxygen gas (O = 16)

Answer
Oxygen gas, O2 is a diatomic molecule, i.e, it is made up of two atoms.
Reacting mass = 8g
Molar mass = 2 x Relative Atomic Mass of O
                      = 2 x 16 = 32g/mol
but,
No of moles = Reacting mass
                         ---------------
                         Molar mass
                      = 8/32 = 0.25mole

Question 2
Calculate the number of
(i) atoms (ii) molecules
present in 35g of nitrogen gas [N = 14, Avogadro No = 6.02 x 10^23/mol]

Answer
Nitrogen gas, N2 is also a diatomic molecule. Before calculating the number of particles in a substance, first calculate the molar mass.

Molar mass of N2 = 2 x 14
                                = 28g/mol
Reacting mass = 35g
No of moles = Reacting mass
                         ---------------
                          Molar mass
                      = 35/28 = 1.25moles
remember,
1 mole of any substance = 6.02 x 10^23 entities or particles

(i) N2 contains 2 moles of atoms (diatomic)
Thus, 1mole of N2 = 2 x 6.02 x 10^23 atoms
                   = 1.204 x 10^24 atoms
Therefore,
If 1mole of N2 = 1.204 x 10^24 atoms
Then, 1.25 moles of N2 = x atoms
         x = 1.25 x 1.204 x 10^24
            = 1.505 x 10^24 atoms

(ii)N2 is 1 mole of a molecule of nitrogen, and 1 mole of N2 = 6.02 x 10^23 molecules. Therefore,

If 1mole of N2 = 6.02 x 10^23 molecules
Then, 1.25 moles of N2 = x molecules
x = 1.25 x 6.02 x 10^23
   = 7.525 x 10^23 molecules

Question 3
What volume in cm^3 will be occupied by 45g of ozone at s.t.p? [O = 16, G.M.V = 22.4dm^3]

Answer
Ozone, O3 is a triatomic molecule, i.e, it is made up of three atoms.
Reacting mass = 45g
Molar mass = 3 x Relative Atomic Mass of O
                      = 3 x 16 = 48g/mol
but,
No of moles = Reacting mass
                         ---------------
                         Molar mass
                      = 45/48 = 0.938mole

1 mole = 22.4dm^3 or 22400cm^3

(Since the answer is expected in cm^3, then we will use 22400cm^3)

So,
If 1mole of O3 gas = 22400cm^3
Then, 0.938mole of O3 = x cm^3
x = 0.938 x 22400
   = 21011.2cm^3
Therefore, 0.938mole of ozone will occupy 21011.2cm^3.

Question 4
C3H8(g) + 5O2(g) ---> 4H2O(g) + 3CO2(g)
From the equation above, calculate the volume of oxygen at s.t.p. required to burn 50cm^3 of propane. [C = 12, H = 1, O = 16, Molar gas volume at s.t.p. = 22.4dm^3]

Answer
From the equation and the accompanying question, all the species are in the gaseous phase, and as such, we will work with their volumes. In other words, we do not need to waste our time calculating their molar masses, so we make use of the relationship between mole and volume.

C3H8(g) + 5O2(g) ---> 4H2O(g) + 3CO2(g)

Mole ratio   1 :     5    :     4     :      3
Molar vol   22.4   22.4   22.4   22.4
Rxting vol  (1x22400)  (5x22400)  (4x22400)  (3x22400)
            = 22400  112000    89600   67200

Therefore, from the stoichiometry,

22400cm^3 of propane = 112000cm^3 of oxygen
Then, 50cm^3 of propane = x cm^3 of oxygen

Solve for x by cross multiplying
22400 * x = 50 * 112000
    22400x = 5600000
               x = 5600000/22400
                  = 250cm^3
Therefore, 250cm^3 of oxygen will be required to burn 50cm^3 of propane.

Question 5
2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g)
From the equation above, calculate the mass of sodium hydroxide produced by 2.3g of sodium [H = 1, O = 16, Na = 23].

Answer
2Na(s) + 2H2O(l) ---> 2NaOH(aq) + H2(g)
Mole ratio    2   :    2   :    2    :    1
Molar mass  23    18      40       2
Rxting mass  46    32      80      2

From the stoichiometry,
46g of Na produce 80g of NaOH
Then, 2.3g of Na will produce xg of NaOH
Solving for x gives,
46 * x = 2.3 * 80
    46x = 184
        x = 184/46
           = 4g
Therefore, 4g of sodium hydroxide will be produced by 2.3g of sodium.

Question 6
16.8g of sodium hydrogentrioxocarbonate (IV) is completely decomposed by heat. Calculate the volume of carbon (IV) oxide given off at s.t.p. [Na = 23, C = 12, O = 16, H = 1, Molar volume of a gas at s.t.p = 22.4dm^3]

Answer
Equation of reaction:
2NaHCO3(s) ---> Na2CO3(s) + H2O(l) + CO2(g)

(Note that when a hydrogentrioxocarbonate (IV) salt, [HCO3]- is heated, it decomposes to give a trioxocarbonate (IV) salt, [CO3--], water and carbon (IV) oxide)

Molar masses,
NaHCO3 = (1*Na) + (1*H) + (1*C) + (3*O)
             = (1*23) + (1*1) + (1*12) + (3*16)
             = 23 + 1 + 12 + 48
             = 84g/mol

    CO2 = (1*C) + (2*O)
            = (1*12) + (2*16)
            = 12 + 32
            = 44g/mol

(We will focus only on the two compounds, whose molar masses are calculated above. The others [Na2CO3 and H2O] are considered as 'distractions', because nothing was mentioned about them in the question)

Mole ratio       2   :   1   :    1   :    1
Molar mass    84     -        -        44
Molar vol          -      -        -     22.4
Rxting mass  168    -        -        44
Rxting vol         -      -        -     22.4

(Since there is a relationship between mass and volume using the mole, based on the question, we can directly compare the reacting mass of NaHCO3 to the reacting volume of CO2)

Hence, from the stoichiometry;
168g of NaHCO3 = 22.4dm^3 of CO2
16.8g of NaHCO3 = x dm^3 of CO2
168 * x = 16.8 * 22.4
    168x = 376.32
           x = 376.32/168
              = 2.24dm^3
Therefore, 2.24dm^3 of CO2 is liberated at s.t.p. when 16.8g of NaHCO3 decomposes on heating.

Application of Mole Concept

The following are some of the areas where mole concept is applied:

1) Gay-Lussac's Law
2) Avogadro's Law
3) Water of Crystallization
4) Law of Conservation of Mass
5) Calculation of Masses from Chemical Equations
6) Redox Reactions
7) Acid-Base Reactions
8) Organic Chemistry

Do These

Question 1
How many moles of limestone, CaCO3 will be required to produce 5.6g of quicklime, CaO?
CaCO3(s) ---> CaO(s) + CO2(g)
[Ca = 40, C = 12, O = 16]
A. 0.20  B. 0.10  C. 1.12  D. 0.56

Question 2
CaCO3(s) + 2HCl(aq) ---> CaCl2(aq) + H2O(l) + CO2(g)
What volume of gas is evolved at s.t.p. if 2g of calcium trioxocarbonate (IV) is added to a solution of hydrochloric acid? [Ca = 40, C = 12, O = 16, Cl = 35.5, H = 1, Molar volume of a gas at s.t.p. = 22.4dm^3]
A. 112cm^3  B. 224cm^3   C. 448cm^3   D. 2240cm^3

Question 3
If a solution contains 4.9g of tetraoxosulphate (VI) acid, calculate the amount of copper (II) oxide that will react with it. [Cu = 64, O = 16, S = 32, H = 1]
A.  0.8g   B.  4.0g   C.  40.0g   D. 
80g
(You have to write the equation of reaction by yourself)

Follow us on Twitter: @gmtacademy
WhatsApp: 07034776117
Facebook: www.facebook.com/greatermindstutors