Categories
Chemistry

Mass and volume relationship

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 CO= 6.023 x 10^23 molecules of CO2

1mole of O= 32g of O= 22.4dm^3 of O= 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. Chemical calculation based on mass volume relationship

Example:

Calculate the volume of carbon dioxide formed at STP in ‘ml’ by the complete thermal decomposition of 3.125 g of pure calcium carbonate (Relative atomic mass of Ca=40, C=12, O=16)

Solution:

As per the equation,

CaCO3 → CaCO + CO2

40 + 12 + (16 x 3) 1 mole

100g 22.4 litres

100 g 24400 ml

Volume of carbon dioxide formed from 100 g of CaCO3 = 22400 ml

Volume of carbon dioxide from 3.125 g of CaCO3 = ?

CaCO3 : CO2

100 g : 22400 ml

3.125 g : x

mass-example

Volume of carbon dioxide formed = 700 ml.

Example:

Calculate the volume of ammonia formed at STP in ‘ml’ by treating 2.675 g of ammonium chloride with excess of calcium hydroxide (Relative Atomic Mass of N=14, O=16, H=1, Cl=35.5, Ca=40).

Solution:

As per the equation,

NH4Cl + Ca(OH)2OH → CaCl2 + H2O + NH3

2 (14 + 4 + 35.5) 2 vols.(of NH3)

107 2 x 22.4 litres

107 g 44.8 litres or 44800 ml

NH4Cl : NH3

107 g : 44800 ml

2.675 g : x

x = (2.675 X 44800)/107

= 1120ml

Volume of ammonia formed at STP = 1120 ml.

We are interested in promoting FREE learning. Tell your friends about Stoplearn.com. Click the share button below!

Download our free Android Mobile application: Save your data when you use our free app. Click to Download StopLearn app.

Download free editable Resume/CV templates: Click here. Ask your question: Expert tutors will personally reply you usually within 24 hours.