CHEMCA
EXAM MASTER FORMULA SHEET
Some Basic Concepts of Chemistry (Mole Concept)
Optimized for JEE Main, Advanced & NEET
1. Fundamental Mole Relations
Connecting the microscopic world of atoms to the macroscopic world of grams.
Calculating Number of Moles (\(n\)):
\[ n = \frac{\text{Mass (g)}}{\text{Molar Mass (M)}} = \frac{\text{Particles (N)}}{6.022 \times 10^{23}} = \frac{V_{\text{STP}} \text{ (L)}}{22.4} \]
Average Atomic Mass:
\[ A_{\text{avg}} = \frac{\sum (\% \text{abun.} \times \text{Mass})}{100} \]
Vapour Density (V.D.):
\[ \text{Molar Mass} = 2 \times \text{V.D.} \]
2. Composition & Formulas
Mass Percentage of Element:
\[ \% \text{ of X} = \frac{n \times \text{Atomic Mass of X}}{\text{Molar Mass of Compound}} \times 100 \]
Molecular Formula Relation:
\(\text{Molecular Formula} = n \times (\text{Empirical Formula})\)
Where \(n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}\)3. Concentration of Solutions
- Molarity (M): Moles of solute / Liters of solution
- Molality (m): Moles of solute / kg of solvent
- Mole Fraction (\(\chi\)): \(\frac{n_i}{n_{total}}\)
Chemca Master Relations (Conversions):
\[ M = \frac{10 \times d \times (\% w/w)}{M_{\text{solute}}} \]
\[ m = \frac{1000 \times M}{1000 \times d - (M \times M_{\text{solute}})} \]
4. Redox & Equivalents
Equivalent Weight (\(E\)) = \(\frac{\text{Molar Mass}}{n\text{-factor}}\)
Normality (\(N\)) = \(M \times n\text{-factor}\)
Law of Equivalence:
\[ N_1 V_1 = N_2 V_2 \]
At equivalence point, Eq. of Oxidant = Eq. of Reductant
5. Eudiometry
\[ C_xH_y + (x + y/4)O_2 \longrightarrow xCO_2 + (y/2)H_2O \]
Volume of liquid water is considered zero in gas analysis problems at room temperature.
