Physical ChemistryHigh Yield (1-2 Qs in JEE)

Mole Concept & Stoichiometry Formulas

Mole concept, molarity, molality, normality, mole fraction, limiting reagent, and empirical formula.

8 Formulas•471+ Questions in Bank

Core Formulas & Mathematical Identities

M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}}, \quad m = \frac{n_{\text{solute}}}{w_{\text{solvent (kg)}}}, \quad N = M \times n\text{-factor}

Fundamental quantitative measures of solution concentration.

Examiner Tip: Normality equals Molarity times valence n-factor.

X_A + X_B = 1, \quad m = \frac{X_B \times 1000}{(1 - X_B) M_A}

Converts mole fraction of solute X_B to molality m.

Examiner Tip: M_A is the molar mass of the solvent.

M_{\text{avg}} = \frac{\sum n_i M_i}{\sum n_i}, \quad \text{Molar Mass} = 2 \times \text{Vapor Density}

Molecular weight calculation for gas mixtures and unknown volatile compounds.

Examiner Tip: Vapor density is measured relative to Hydrogen gas.

E = \frac{\text{Molar Mass}}{n\text{-factor}}, \quad n\text{-factor}(\text{KMnO}_4): \begin{cases} 5 & \text{Acidic } (\text{Mn}^{2+}) \\ 3 & \text{Neutral } (\text{MnO}_2) \\ 1 & \text{Alkaline } (\text{MnO}_4^{2-}) \end{cases}

Calculates chemical equivalent weight. For acids: replaceable H⁺ count; bases: replaceable OH⁻ count; redox: change in oxidation state per molecule.

Examiner Tip: For K₂Cr₂O₇ in acidic medium: Cr goes from +6 to +3, giving n-factor = 2 × 3 = 6 (E = M / 6).

N_1 V_1 = N_2 V_2, \quad \text{mEq} = N \times V_{\text{mL}} = \frac{\text{mass (g)}}{E} \times 1000

In any complete stoichiometric reaction, the number of gram-equivalents of all reactants and products are identical.

Examiner Tip: Does NOT require balancing the stoichiometric reaction equation beforehand.

\text{MF} = (\text{EF})_n \quad \left(n = \frac{\text{Molar Mass}}{\text{EF Mass}}\right), \quad \text{Atomic Weight} \times \text{Specific Heat (cal/g)} \approx 6.4

Determination of true molecular formula from simplest atomic ratio and Dulong-Petit approximate atomic mass estimation for solid metals.

Examiner Tip: Valence of metal = (Approximate Atomic Weight) / (Equivalent Weight).

\text{LR: Smallest } \frac{\text{Moles}}{\text{Coefficient}}, \quad \% \text{ Yield} = \frac{\text{Actual}}{\text{Theoretical}} \times 100\%, \quad \% \text{ Purity} = \frac{m_{\text{pure}}}{m_{\text{sample}}} \times 100\%

Limiting reagent is consumed first and dictates the maximum theoretical amount of product formed.

Examiner Tip: Always balance the chemical reaction before dividing moles by stoichiometric coefficients.

\text{Volume Strength} = 11.2 \times M = 5.6 \times N, \quad \text{ppm} = \frac{w_{\text{solute}}}{w_{\text{solution}}} \times 10^6

Volume of O₂ liberated at STP from 1 volume of hydrogen peroxide solution upon decomposition, and ppm trace concentration unit.

Examiner Tip: 20 volume H₂O₂ has molarity M = 20 / 11.2 ≈ 1.78 M and normality N = 20 / 5.6 ≈ 3.57 N.

Frequently tested misconceptions in Mole Concept & Stoichiometry

•Molarity changes with temperature (due to volume expansion), but Molality and Mole Fraction are temperature INDEPENDENT.
•Limiting reagent is determined by dividing moles by stoichiometric coefficients (lowest ratio is limiting).