9.4.7 Empirical Formula
This is the seventh lecture from Chapter 4: “Stoichiometry” in the new Class 9 Chemistry book (Punjab Board – PCTB). It discusses Avogadro’s number and how it solved the problem of measuring chemical species. The lecture also includes a multiple-choice quiz, short question and long question notes.
MCQs Based Quiz
Short Questions
Q1. What was the need for Avogadro’s number (NA)?
- The know the mass ratio of reactants
- To express the masses of reactants in grams
Q2. What is Avogadro’s number?
Avogadro’s number (NA) is a special number which is equal to 6.022 x 1023. It is the number of units in one mole of a substance.
Q3. What is the relation between grams and amu?
1.00g = 6.022 x 1023 amu.
Q4. How many atoms are present in 24g of carbon.
24g of carbon contains 2 x 6.022 x 1023 atoms, which is 12.044 x 1023 atoms.
Q5. How many molecules are present in 32g of oxygen (O2)?
32 of oxygen (O2) contains 6.022 x 1023 molecules.
Q6. How many molecules are present in 56g of carbon monoxide?
56g of carbon monoxide contains 2 x 6.022 x 1023 molecules, which is 12.044 x 1023 molecules.
Descriptive Question
Q1. Write a comprehensive note on Avogadro’s number and how it is used to measure chemical species.
Need for Avogadro’s Number (NA):
- Chemical reactions involve a very large number of atoms and molecules of reactants and products.
- The mass ratio of these reactants and products must be known.
- The masses of reactants and products must also be expressed in grams.
Development of Avogadro’s Number (NA):
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.
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.
$$\underset{\substack{
2 \times 100 \text{ atoms} \\
2 \times 10000 \text{ atoms}}
}{2\mathrm{C}}+\underset{
\substack{
100 \text{ molecules} \\
10000 \text{ molecules}}
}{\mathrm{O}_2}\rightarrow\underset{\substack{
2 \times 100 \text{ molecules} \\
2 \times 10000 \text{ molecules}}
}{2\mathrm{CO}}$$
Increasing the number of reactants will not change the mass ratio but the problem is still not solved, because the number is still very small.
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:
$$\begin{align}
2 \times 6.022 \times 10^{23} \times \text{12 amu} &= 24.00 \text{ g Carbon atoms}\\
6.022 \times 10^{23} \times \text{32 amu} &= 32.00 \text{ g Oxygen molecules} \ \mathrm{(O_2)}\\
2 \times 6.022 \times 10^{23} \times \text{28 amu} &= 2 \times 28.00 \text{ g Carbon monoxide molecules} \ \mathrm{(CO)}
\end{align}$$
The mass ratio between reactants and products will then become:
$\ce{
\underset{\text{24g}}{2C} + \underset{\text{32g}}{O_2} -> \underset{\text{56g}}{2CO}
}$