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Mole Concept Formulas For JEE 2027
Mole Concept and Stoichiometry is a foundational chapter in JEE Chemistry that explains the relationship between mass, moles, volume, and the number of particles. It helps JEE 2027 aspirants use balanced chemical equations to calculate the quantities of reactants required and products formed. These concepts support numerical problem-solving across Physical and Inorganic Chemistry.
This chapter covers limiting reagents, percentage yield, percentage composition, empirical formula, and molecular formula. Students should understand each concept, practise calculations, and revise key relationships using a JEE Mains formula sheet. Taking a JEE Mains mock test can help aspirants assess their understanding, identify calculation errors, and improve speed and accuracy under exam conditions.
Mole and Avogadro's Number
Definition: Mole
A mole is the amount of substance that contains exactly $$6.022 \times 10^{23}$$ particles (atoms, molecules, ions, etc.). This number is called Avogadro's number ($$N_A$$).
Avogadro's Number
$$$N_A = 6.022 \times 10^{23} \text{ particles/mol}$$$
- 1 mole of carbon atoms = $$6.022 \times 10^{23}$$ carbon atoms
- 1 mole of water molecules = $$6.022 \times 10^{23}$$ water molecules
- 1 mole of electrons = $$6.022 \times 10^{23}$$ electrons
Worked Example: Counting Molecules
How many molecules are in 3 moles of $$\text{H}_2\text{O}$$?
Number of molecules $$= n \times N_A = 3 \times 6.022 \times 10^{23} = \textbf{1.807} \times \textbf{10}^{\textbf{24}}$$ molecules
Atomic Mass, Molecular Mass, and Molar Mass
Definition: Atomic Mass Unit (amu or u)
A unit of mass defined as exactly $$\frac{1}{12}$$ the mass of a carbon-12 atom. 1 amu $$= 1.66 \times 10^{-24}$$ g.
Definition: Molecular Mass
The sum of atomic masses of all atoms in a molecule.
Worked Example: Molecular Mass
Find the molecular mass of $$\text{H}_2\text{SO}_4$$.
$$= 2 \times (\text{H}) + 1 \times (\text{S}) + 4 \times (\text{O})$$
$$= 2 \times 1 + 1 \times 32 + 4 \times 16 = 2 + 32 + 64 = \textbf{98 u}$$
Molar Mass
$$$M = \text{Atomic or Molecular mass in grams per mole (g/mol)}$$$
- Molar mass of C = 12 g/mol (1 mole of carbon weighs 12 g)
- Molar mass of $$\text{H}_2\text{O}$$ = 18 g/mol
- Molar mass of $$\text{H}_2\text{SO}_4$$ = 98 g/mol
Tip: The molar mass in g/mol is numerically the same as the molecular mass in amu. This is a key shortcut — just look up the periodic table values and write g/mol as the unit.
The Mole Relationships
The mole connects three quantities: number of particles, mass, and volume (for gases). These relationships are the most important formulas in this chapter.
Mole ↔ Mass
$$$n = \frac{w}{M}$$$
where $$n$$ = number of moles, $$w$$ = given mass (g), $$M$$ = molar mass (g/mol).
Worked Example: Mass to Moles
How many moles are in 49 g of $$\text{H}_2\text{SO}_4$$?
$$n = \frac{w}{M} = \frac{49}{98} = \textbf{0.5 mol}$$
Mole ↔ Number of Particles
$$$n = \frac{N}{N_A} \quad \text{or} \quad N = n \times N_A$$$
where $$N$$ = number of particles, $$N_A = 6.022 \times 10^{23}$$.
Worked Example: Mass to Atoms
How many atoms are in 4 g of helium (He)?
Molar mass of He = 4 g/mol
$$n = \frac{4}{4} = 1$$ mol
Number of atoms $$= 1 \times 6.022 \times 10^{23} = \textbf{6.022} \times \textbf{10}^{\textbf{23}}$$ atoms
Mole ↔ Volume (Gases at STP)
At STP (Standard Temperature and Pressure: $$0°\text{C}$$, 1 atm):
$$$\text{1 mole of any gas} = 22.4 \text{ litres}$$$
$$$n = \frac{V}{22.4}$$$
where $$V$$ = volume in litres at STP.
Worked Example: Moles to Volume
What volume does 2 moles of $$\text{O}_2$$ occupy at STP?
$$V = n \times 22.4 = 2 \times 22.4 = \textbf{44.8 L}$$
Important Note
The 22.4 L/mol value applies only at STP. At other conditions, use the ideal gas equation $$PV = nRT$$.
Percentage Composition
Percentage of an Element
$$$\% \text{ of element} = \frac{\text{Total mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100$$$
Worked Example: Percentage Composition
Find the percentage of oxygen in $$\text{H}_2\text{O}$$.
Molar mass of $$\text{H}_2\text{O} = 2(1) + 16 = 18$$ g/mol
Mass of O in 1 mole = 16 g
$$\% \text{ O} = \frac{16}{18} \times 100 = \textbf{88.9\%}$$
Empirical and Molecular Formula
Definition: Empirical Formula
The simplest whole-number ratio of atoms in a compound. E.g., for glucose ($$\text{C}_6\text{H}_{12}\text{O}_6$$), the empirical formula is $$\text{CH}_2\text{O}$$.
Finding Empirical Formula
- Step 1: Find the mass (or %) of each element.
- Step 2: Divide each by its atomic mass → gives moles of each element.
- Step 3: Divide all by the smallest mole value → gives the simplest ratio.
- Step 4: If ratios are not whole numbers, multiply all by the smallest integer to make them whole.
Molecular Formula from Empirical Formula
$$$n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}$$$
$$$\text{Molecular formula} = n \times \text{Empirical formula}$$$
Worked Example: Finding Molecular Formula
A compound contains 40% C, 6.7% H, and 53.3% O. Its molecular mass is 60. Find its molecular formula.
Step 1: Moles of each element (assuming 100 g sample):
C: $$\frac{40}{12} = 3.33$$ H: $$\frac{6.7}{1} = 6.7$$ O: $$\frac{53.3}{16} = 3.33$$
Step 2: Divide by smallest (3.33):
C : H : O $$= 1 : 2 : 1$$
Step 3: Empirical formula $$= \text{CH}_2\text{O}$$ (mass = 30)
Step 4: $$n = \frac{60}{30} = 2$$
Molecular formula $$= 2 \times \text{CH}_2\text{O} = \textbf{C}_2\textbf{H}_4\textbf{O}_2$$ (acetic acid)
Stoichiometry and Limiting Reagent
Reading a Balanced Equation
Consider: $$\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$$
This tells us:
- 1 mole of $$\text{N}_2$$ reacts with 3 moles of $$\text{H}_2$$ to produce 2 moles of $$\text{NH}_3$$
- Mole ratio: $$\text{N}_2 : \text{H}_2 : \text{NH}_3 = 1 : 3 : 2$$
- Mass is conserved: total mass of reactants = total mass of products
Worked Example: Stoichiometry
How many grams of $$\text{NH}_3$$ are produced from 14 g of $$\text{N}_2$$?
$$\text{N}_2 + 3\text{H}_2 \to 2\text{NH}_3$$
Moles of $$\text{N}_2 = \frac{14}{28} = 0.5$$ mol
From the equation: 1 mol $$\text{N}_2$$ produces 2 mol $$\text{NH}_3$$
So 0.5 mol $$\text{N}_2$$ produces $$0.5 \times 2 = 1$$ mol $$\text{NH}_3$$
Mass of $$\text{NH}_3 = 1 \times 17 = \textbf{17 g}$$
Finding the Limiting Reagent
- Step 1: Calculate moles of each reactant.
- Step 2: Divide each by its coefficient in the balanced equation.
- Step 3: The reactant with the smallest value is the limiting reagent.
- Step 4: Use the limiting reagent to calculate the amount of product.
Worked Example: Limiting Reagent
5 g of $$\text{H}_2$$ reacts with 32 g of $$\text{O}_2$$. How many grams of $$\text{H}_2\text{O}$$ are formed?
$$2\text{H}_2 + \text{O}_2 \to 2\text{H}_2\text{O}$$
Moles: $$\text{H}_2 = \frac{5}{2} = 2.5$$ mol, $$\text{O}_2 = \frac{32}{32} = 1$$ mol
Divide by coefficients: $$\text{H}_2: \frac{2.5}{2} = 1.25$$, $$\text{O}_2: \frac{1}{1} = 1$$
$$\text{O}_2$$ has the smaller value → $$\text{O}_2$$ is the limiting reagent.
From 1 mol $$\text{O}_2$$: produces 2 mol $$\text{H}_2\text{O} = 2 \times 18 = \textbf{36 g}$$
Tip: In stoichiometry problems, always: (1) write and balance the equation first, (2) convert everything to moles, (3) check for limiting reagent if two reactant amounts are given.
Concentration Terms
| Term | Formula | Unit |
|---|---|---|
| Molarity ($$M$$) | $$\dfrac{\text{moles of solute}}{\text{volume of solution (L)}}$$ | mol/L |
| Molality ($$m$$) | $$\dfrac{\text{moles of solute}}{\text{mass of solvent (kg)}}$$ | mol/kg |
| Mole fraction ($$\chi$$) | $$\dfrac{n_A}{n_A + n_B}$$ | (no unit) |
| Mass % ($$w/w$$) | $$\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100$$ | % |
Worked Example: Molarity
Find the molarity of a solution made by dissolving 4 g of NaOH in water to make 500 mL of solution.
Molar mass of NaOH $$= 23 + 16 + 1 = 40$$ g/mol
Moles of NaOH $$= \frac{4}{40} = 0.1$$ mol
Volume $$= 500 \text{ mL} = 0.5$$ L
Molarity $$= \frac{0.1}{0.5} = \textbf{0.2 M}$$
Tip: Molarity changes with temperature (because volume changes), but molality does not (because it depends on mass). JEE prefers molality in colligative property problems.
Mole Concept Formulas For JEE 2027: Conclusion
Mole Concept formulas for JEE 2027 connect mass, moles, particles, and concentration, helping students solve questions on stoichiometry and limiting reagents. Revising these relationships with a JEE Main Chemistry formula sheet and practising numerical problems can improve calculation speed and accuracy for JEE Main and Advanced.
Alongside revision, reviewing JEE Mains Chemistry Chapter-wise Weightage can help students plan their study time and prioritise practice. Build a strong understanding of the mole concept, check units carefully, and use balanced chemical equations to approach reaction-based calculations with confidence.
