9.4.8 The Mole and Molar Mass
This is the eighth lecture from Chapter 4: “Stoichiometry” in the new Class 9 Chemistry book (Punjab Board – PCTB). It discusses the concept of mole and molar mass. It also explains the use of mole and molar mass in chemical calculations. The lecture also includes a multiple-choice quiz, short question and long question notes.
MCQs Based Quiz
Short Questions
What is meant by mole in chemistry?
A mole is the amount of a substance that contains $6.022 \times 10^{23}$ particles.
How many atoms are present in one mole of carbon?
One mole of carbon contains exactly $6.022 \times 10^{23}$ atoms.
How many molecules are present in two moles of oxygen (O2)?
Two moles of oxygen (O2) contains $12.044 \times 10^{23}$ molecules.
How many atoms are present in 12g of carbon?
12g of carbon is equal to its one mole and one mole is equal to $6.022 \times 10^{23}$ atoms.
How many formula units are there in 58.5g of NaCl?
58.5g of NaCl contains $6.022 \times 10^{23}$ formula units.
What is meant by molar mass?
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.
What is the molar mass of hydrogen (H2)?
Molar mass of hydrogen (H2) is 2.016g.
Calculate the molar masses of the following compounds:
(a) H3PO4
$$\begin{align}
\text{Atomic mass of H} &= 1\\
\text{Atomic mass of P} &= 31\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } H_3PO_4 &= 3(1) + (31) + 4(16)\\
&= 98 gmol^{-1}
\end{align}$$
(b) SiO2
$$\begin{align}
\text{Atomic mass of Si} &= 28\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } SiO_2 &= (28) + 2(16)\\
&= 60 gmol^{-1}
\end{align}$$
(c) C12H22O11
$$\begin{align}
\text{Atomic mass of C} &= 12\\
\text{Atomic mass of H} &= 1\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } C_{12}H_{22}O_{4} &= 12(12) + 22(1) + 11(16)\\
&= 342 gmol^{-1}
\end{align}$$
(d) N2O4
$$\begin{align}
\text{Atomic mass of N} &= 14\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } N_2O_4 &= 2(14) + 4(16)\\
&= 92 gmol^{-1}
\end{align}$$
(d) MgCO3
$$\begin{align}
\text{Atomic mass of Mg} &= 24\\
\text{Atomic mass of C} &= 12\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } C_{12}H_{22}O_{4} &= (24) + (12) + 3(16)\\
&= 84 gmol^{-1}
\end{align}$$
(e) H2SO4
$$\begin{align}
\text{Atomic mass of H} &= 1\\
\text{Atomic mass of S} &= 32\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } C_{12}H_{22}O_{4} &= 2(1) + (32) + 4(16)\\
&= 98 gmol^{-1}
\end{align}$$
(f) C6H12O6
$$\begin{align}
\text{Atomic mass of C} &= 12\\
\text{Atomic mass of H} &= 1\\
\text{Atomic mass of O} &= 16\\
\\
\text{Molar mass of } C_{12}H_{22}O_{4} &= 6(12) + 12(1) + 6(16)\\
&= 180 gmol^{-1}
\end{align}$$
Descriptive Question
Write a note on concept of mole and molar mass.
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.
$$\underset{\substack{
\text{Two atoms} \\
\text{Two moles of C} \\
2 \times 6.022 \times 10^{23}\ \text{atoms} \\
\text{24 g}}}{2\mathrm{C}}\ +
\underset{ \substack{\text{One molecule} \\\text{One mole of O}_2 \\
6.022 \times 10^{23}\ \text{molecules} \\
\text{32 g}}
}{\mathrm{O}_2}\ \rightarrow \underset{\substack{\text{Two molecules} \\
\text{Two moles of CO} \\
2 \times 6.022 \times 10^{23}\ \text{molecules} \\
\text{56 g}}}{2\mathrm{CO}}$$