Mole Concept and Molar Masses: Definition, Common Mistakes to Avoid & Sample Questions

A mole is the amount of substance containing $6.022\times {10}^{23}$ entities (atoms, molecules, ions, etc.), equal to the number of atoms in 12 g of carbon-12, defined as Avogadro's number ( $N_A$ ). For example:

• 1 mole of ${\mathrm{N}}_2=6.022\times {10}^{23}$ molecules.
• 1 mole of ${\mathrm{C}\mathrm{l}}^{-}=6.022\times {10}^{23}$ ions.

Number of particles: $N=n\times N_A,\mathrm{}n=\frac{\text{number of particles}}{N_A}$ where $n$ is moles. JEE/NEET problems test conversions between moles and particles.

Molar mass is the mass of one mole of a substance ( $\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ), numerically equal to its atomic, molecular, or formula mass in atomic mass units (amu). Examples:

• Molar mass of $\mathrm{O}=16\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .
• Molar mass of ${\mathrm{C}\mathrm{O}}_2=12+2\times 16=44\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .
• Molar mass of ${\mathrm{C}\mathrm{a}\mathrm{C}\mathrm{O}}_3=40+12+3\times 16=100\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ .

Moles from mass: $n=\frac{\text{mass}\left(\mathrm{g}\right)}{\text{molar mass}(\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l})}$

Molar mass is key for stoichiometric calculations in exams.

At standard temperature and pressure (STP: $0^{\circ }\mathrm{C},1\mathrm{a}\mathrm{t}\mathrm{m}$ ), 1 mole of any gas occupies 22.4 L , containing $N_A$ molecules.

$n=\frac{\text{volume at STP}\left(\mathrm{L}\right)}{22.4}$

Example: Volume of $16\mathrm{g}{\mathrm{C}\mathrm{H}}_4$ at STP: $n=\frac{16}{16}=1\mathrm{m}\mathrm{o}\mathrm{l},\mathrm{}V=1\times 22.4=22.4\mathrm{L}$

NEET questions often involve gas volume calculations.

Percentage composition is the mass percentage of each element in a compound: $\mathrm{\%}\text{of element}=\frac{\text{mass of element in}1\text{mole of compound}}{\text{molar mass of compound}}\times 100$

Example: ${\mathrm{H}}_2\mathrm{O}$ (molar mass $=18\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ): $\mathrm{\%}\mathrm{H}=\frac{2\times 1}{18}\times 100=11.11\mathrm{\%},\mathrm{}\mathrm{\%}\mathrm{O}=\frac{16}{18}\times 100=88.89\mathrm{\%}$

JEE/NEET problems test percentage composition for compound analysis.

Empirical Formula: Simplest whole-number ratio of atoms in a compound.

Molecular Formula: Actual number of atoms, related by: $\text{Molecular formula}=n\times \text{empirical formula,}\mathrm{}n=\frac{\text{molecular}\text{mass}}{\text{empirical formula mass}}$

Example: A compound with 40 $\text{Moles}\mathrm{C}=\frac{40}{12}=3.33,\mathrm{}\mathrm{H}=\frac{6.67}{1}=6.67,\mathrm{}\mathrm{O}=\frac{53.33}{16}=3.33$

Ratio: 1:2:1. Empirical formula: ${\mathrm{C}\mathrm{H}}_2\mathrm{O}$ (mass $=30\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ).

$n=\frac{180}{30}=6,\mathrm{}\text{Molecular formula}={\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$

JEE Main exam and NEET exam questions frequently involve formula determination.

Concentration quantifies solute in a solution, used in JEE/NEET solution chemistry:

1. Molarity (M): Moles of solute per liter of solution: $M=\frac{\text{moles of solute}}{\text{volume of solution}\left(\mathrm{L}\right)}$

2. Molality (m): Moles of solute per kg of solvent: $m=\frac{\text{moles of solute}}{\text{mass of solvent}\left(\mathrm{k}\mathrm{g}\right)}$

3. Mass Percentage (% w/w): $\mathrm{\%}\mathrm{w}/\mathrm{w}=\frac{\text{mass of solute}}{\text{mass of solution}}\times 100$

4. Mole Fraction ( $x$ ): Ratio of moles of a component to total moles: $x_A=\frac{n_A}{n_A+n_B}$

Example: 18 g glucose in 100 g water. Molality: $\text{Moles glucose}=\frac{18}{180}=0.1,\mathrm{}\text{Molality}=\frac{0.1}{0.1}=1\mathrm{m}$

JEE/NEET problems test concentration conversions.

Example 1: Calculate the number of oxygen atoms in $4.4\mathrm{g}{\mathrm{C}\mathrm{O}}_2$ . (JEE Main/NEET) Molar mass ${\mathrm{C}\mathrm{O}}_2=44\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ . Moles: $n=\frac{4.4}{44}=0.1\mathrm{m}\mathrm{o}\mathrm{l}$

Molecules: $0.1\times 6.022\times {10}^{23}=6.022\times {10}^{22}$

Oxygen atoms (2 per molecule): $2\times 6.022\times {10}^{22}=1.2044\times {10}^{23}$

 

Example 2: A compound contains 52.17 Assume 100 g : $\text{Moles}\mathrm{C}=\frac{52.17}{12}=4.35,\mathrm{}\mathrm{H}=\frac{13.04}{1}=13.04,\mathrm{}\mathrm{O}=\frac{34.78}{16}=2.17$

Ratio: 2:6:1. Empirical formula: ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ (mass $=46\mathrm{g}/\mathrm{m}\mathrm{o}\mathrm{l}$ ).

$n=\frac{46}{46}=1,\mathrm{}\text{Molecular formula}={\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$