Chapter 4: Stoichiometry (Solved Exercise Notes)

Q2. Explain the concepts of Avogadro’s number and mole.

Avogadro’s Number:

Need for Avogadro’s Number (N_A) and the concept of mole:

1. Chemical reactions involve a very large number of atoms and molecules of reactants and products.
2. The mass ratio of these reactants and products must be known.
3. The masses of reactants and products must also be expressed in grams.

Development of Avogadro’s Number (N_A):

Amedeo Avogadro was an Italian scientist after whom this constant was named. To understand its development, lets consider the following reaction:

$\ce{
\underset{\text{2 Atoms}}{2C} + \underset{\text{1 Molecule}}{O_2} -> \underset{\text{2 Molecules}}{2CO}
}$

Even though it is a very simple reaction, we cannot perform this reaction with only 2 carbon atoms and 1 oxygen molecule because we cannot measure the mass of particles as small as atoms or molecules.

One thing that we can do is increase the number of reactants.

We should increase this number to such a value that is easy for us to calculate the masses of chemical species.

Thus,

$\ce{
\underset{2 \times 6.022 \times 10^{23}}{2C} + \underset{6.022 \times 10^{23}}{O_2} -> \underset{2 \times 6.022 \times 10^{23}}{2CO}
}$

$6.022 \times 10^{23}$ is a huge number and it is selected because

$1.00 \text{g} = 6.022 \times 10^{23} \text{amu}$

Now, the amounts of reactants and products can be written as follows:

$$\begin{align}
2 \times 6.022 \times 10^{23} \times \text{12 amu} &= 24.00 \text{ g Carbon atoms}\\
6.022 \times 10^{23} \times \text{32 amu} &= 32.00 \text{ g Oxygen molecules}\\
2 \times 6.022 \times 10^{23} \times \text{28 amu} &= 2 \times 28.00 \text{ g Carbon monoxide molecules}
\end{align}$$

The mass ratio between reactants and products will then become:

$\ce{
\underset{\text{24g}}{2C} + \underset{\text{32g}}{O_2} -> \underset{\text{56g}}{2CO}
}$

The number $6.022 \times 10^{23}$ is called Avogadro’s number (NA) which is also the number of units present in one mole of a substance.
Concept of Mole:
A mole is the amount of a substance that contains $6.022 \times 10^{23}$ particles.
These particles can be atoms, molecules or ions. Therefore, when we use the term mole of a substance, we must also explain what type of particles are present in that substance.
The mass of one mole of a substance is called molar mass. For example, molar mass of one hydrogen is 1.008g and it contains $6.022 \times 10^{23}$ atoms.

$$\begin{align}
\text{1 Mole of carbon atoms} &= 6.022 \times 10^{23} \text{ atoms} = \text{12g}\\
\text{1 Mole of hydrogen atoms} &= 6.022 \times 10^{23} \text{ atoms} = \text{1.008g}\\
\text{1 Mole of hydrogen molecules} &= 6.022 \times 10^{23} \text{ molecules} = \text{2.016g}\\
\text{1 Mole of oxygen molecules} (O_2) &= 6.022 \times 10^{23} \text{ molecules} = \text{32g}\\
\text{1 Mole of NaCl formula units} &= 6.022 \times 10^{23} \text{ formula units} = \text{58.5g}\\
\end{align}$$

So, the chemical equation of production of carbon monoxide can be understood in the following way.

$$\underset{\substack{
\text{Two atoms} \\
\text{Two moles of C} \\
2 \times 6.022 \times 10^{23}\ \text{atoms} \\
\text{24 g}
}
}{2\mathrm{C}}
\ +
\underset{
\substack{
\text{One molecule} \\
\text{One mole of O}_2 \\
6.022 \times 10^{23}\ \text{molecules} \\
\text{32 g}
}
}{\mathrm{O}_2}
\ \rightarrow
\underset{
\substack{
\text{Two molecules} \\
\text{Two moles of CO} \\
2 \times 6.022 \times 10^{23}\ \text{molecules} \\
\text{56 g}
}
}{2\mathrm{CO}}
$$