Basic Concepts in Chemistry Formula Sheet
All key Basic Concepts of Chemistry formulas: mole concept, stoichiometry, concentration terms, normality, empirical formula. JEE Main & Advanced quick revision.
Last-minute revision sheet for the entire Basic Concepts chapter: mole concept, atomic theory, percentage composition, stoichiometry, equivalent concept, and concentration terms. Every formula here is pulled straight from the chapter topics.
Constants and Units to Memorize
| Quantity | Value | Notes |
|---|---|---|
| Avogadro’s number ($N_A$) | $6.022 \times 10^{23}\ \text{mol}^{-1}$ | Particles per mole |
| 1 amu | $1.66 \times 10^{-24}\ \text{g}$ | $\frac{1}{12}$ mass of $^{12}\text{C}$ atom |
| Molar volume (old STP) | $22.4\ \text{L/mol}$ | 273.15 K, 1 atm (used in JEE) |
| Molar volume (modern STP) | $22.7\ \text{L/mol}$ | 273.15 K, 1 bar (IUPAC) |
Mole Concept
The single most-used relation in the whole chapter:
$$\boxed{n = \frac{m}{M} = \frac{N}{N_A} = \frac{V}{22.4}\ \text{(gas at STP)}}$$| Conversion | Formula |
|---|---|
| Mass to moles | $n = m / M$ |
| Moles to mass | $m = n \times M$ |
| Moles to particles | $N = n \times N_A$ |
| Particles to moles | $n = N / N_A$ |
| Moles to volume (STP) | $V = n \times 22.4\ \text{L}$ |
Where $n$ = moles, $m$ = mass (g), $M$ = molar mass (g/mol), $N$ = number of particles, $V$ = gas volume at STP (L).
Molar Mass
$$M = \frac{\text{Mass in grams}}{\text{Number of moles}}$$- For elements: molar mass = atomic mass from periodic table.
- For compounds: sum of atomic masses of all atoms (mind the subscripts).
Mole Fraction and Mole Percent
$$\boxed{\chi_A = \frac{n_A}{n_\text{total}} = \frac{n_A}{n_A + n_B + n_C + \ldots}}$$$$\text{Mole \% of A} = \chi_A \times 100$$Properties: dimensionless, $0 \le \chi \le 1$, and $\sum \chi = 1$. For a binary solution, $\chi_\text{solute} + \chi_\text{solvent} = 1$.
Atomic Theory and Atomic Mass
$$\boxed{Z = \text{number of protons} = \text{number of electrons (neutral atom)}}$$$$\boxed{A = Z + n \quad (\text{Mass number} = \text{protons} + \text{neutrons})}$$Atomic notation: $^{A}_{Z}X^{\text{charge}}$. Electrons in an ion: $e^- = Z - \text{charge}$ (subtract for cations, add for anions).
Average Atomic Mass
$$\boxed{M_\text{avg} = \frac{M_1 a_1 + M_2 a_2 + \ldots}{100}}$$where $M_i$ = isotope mass and $a_i$ = percentage abundance.
Isotopes, Isobars, Isotones
| Term | Same | Different |
|---|---|---|
| Isotopes | Atomic number $Z$ | Mass number $A$ |
| Isobars | Mass number $A$ | Atomic number $Z$ |
| Isotones | Number of neutrons $n$ | $Z$ and $A$ |
Fundamental Particles
| Particle | Charge | Mass (amu) |
|---|---|---|
| Proton ($p^+$) | $+1$ | $1.00728$ |
| Neutron ($n^0$) | $0$ | $1.00866$ |
| Electron ($e^-$) | $-1$ | $0.000549$ |
Percentage Composition and Formulae
$$\boxed{\% \text{ of element} = \frac{\text{Mass of element in formula}}{\text{Molecular mass } M} \times 100}$$Check: all percentages must sum to $100\%$.
Empirical and Molecular Formula
$$\boxed{\text{Molecular formula} = (\text{Empirical formula})_n}$$$$\boxed{n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}}}$$Molecular mass from vapour density: $M = 2 \times \text{VD}$.
Empirical formula steps: assume 100 g (% becomes grams) → divide by atomic mass to get moles → divide all by the smallest → multiply to clear decimals.
Combustion Analysis (C, H, O compounds)
$$\boxed{\text{Mass of C} = \frac{12}{44} \times m_{CO_2}}$$$$\boxed{\text{Mass of H} = \frac{2}{18} \times m_{H_2O}}$$$$\boxed{\text{Mass of O} = m_\text{compound} - (m_C + m_H)}$$Stoichiometry
Conservation of mass: $\sum m_\text{reactants} = \sum m_\text{products}$.
The coefficients in a balanced equation give the mole ratio:
$$\boxed{\frac{n_A}{a} = \frac{n_B}{b}}$$where $a$, $b$ are stoichiometric coefficients of A and B.
Limiting Reagent
Divide moles of each reactant by its coefficient; the smallest value is the limiting reagent. Calculate products from the limiting reagent only. Excess = initial - consumed.
Percentage Yield
$$\boxed{\% \text{ Yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100}$$Actual yield is always less than theoretical (incomplete/side reactions, separation losses, impure reactants); yield can never exceed 100%.
graph LR
A["Given amount of A"] -->|"÷ M"| B["Moles of A"]
B -->|"mole ratio (from balanced eqn)"| C["Moles of B"]
C -->|"× M or × Nₐ"| D["Desired quantity of B"]Equivalent Concept and Normality
$$\boxed{E = \frac{M}{\text{n-factor}}}$$where $E$ = equivalent weight, $M$ = molecular/atomic weight.
n-factor Quick Reference
| Type | n-factor |
|---|---|
| Acid | Basicity (replaceable $H^+$) |
| Base | Acidity (replaceable $OH^-$) |
| Salt | Total charge on cation/anion |
| Redox agent | Electrons transferred per molecule |
| Element | Valency |
Gram Equivalents
$$\boxed{\text{Gram equivalents} = \frac{\text{Mass}}{E} = \frac{\text{Mass} \times \text{n-factor}}{M} = \text{Moles} \times \text{n-factor}}$$Law of equivalents: in any reaction, gram equivalents of A = gram equivalents of B.
Normality
$$\boxed{N = \frac{\text{Gram equivalents}}{V(\text{L})} = M \times \text{n-factor}}$$For reaction or dilution: $\boxed{N_1 V_1 = N_2 V_2}$
For mixing same solute: $N_\text{mix} = \dfrac{N_1 V_1 + N_2 V_2}{V_1 + V_2}$
Concentration Terms
| Term | Formula | Units | Temperature dependent? |
|---|---|---|---|
| Molarity ($M$) | $\dfrac{n}{V(\text{L})}$ | mol/L | Yes |
| Molality ($m$) | $\dfrac{n}{\text{mass solvent (kg)}}$ | mol/kg | No |
| Mole fraction ($\chi$) | $\dfrac{n_i}{n_\text{total}}$ | dimensionless | No |
| Mass % (w/w) | $\dfrac{m_\text{solute}}{m_\text{solution}} \times 100$ | % | No |
| Volume % (v/v) | $\dfrac{V_\text{solute}}{V_\text{solution}} \times 100$ | % | Yes |
| Mass/volume % (w/v) | $\dfrac{m_\text{solute (g)}}{V_\text{solution (mL)}} \times 100$ | % | Yes |
| ppm | $\dfrac{m_\text{solute}}{m_\text{solution}} \times 10^6$ | ppm | No |
| ppb | $\dfrac{m_\text{solute}}{m_\text{solution}} \times 10^9$ | ppb | No |
ppm / ppb Relations
$$1\ \text{ppm} = 1000\ \text{ppb} = 10^{-4}\% \qquad \text{For dilute aqueous: } \text{ppm} = \frac{\text{mg solute}}{\text{L solution}}$$Interconversion Formulae
Molarity to molality (d = density g/mL, $M_s$ = solute molar mass):
$$\boxed{m = \frac{1000\,M}{1000d - M\,M_s}}$$Molality to molarity:
$$\boxed{M = \frac{1000\,d\,m}{1000 + m\,M_s}}$$Mass % (w/w) to molarity:
$$\boxed{M = \frac{10 \times d \times \%(\text{w/w})}{M_\text{solute}}}$$Molality to mole fraction of solute ($M_\text{solvent}$ usually 18 for water):
$$\boxed{\chi_\text{solute} = \frac{m\,M_\text{solvent}}{1000 + m\,M_\text{solvent}}}$$Mole fraction to molality:
$$\boxed{m = \frac{1000 \times \chi_\text{solute}}{M_\text{solvent}(1 - \chi_\text{solute})}}$$Dilution and Mixing
$$\boxed{M_1 V_1 = M_2 V_2 \quad (\text{dilution})}$$$$\boxed{M_\text{final} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2} \quad (\text{mixing same solute, additive volumes})}$$One-Glance Memory Table
| Need | Use |
|---|---|
| Count atoms/molecules from mass | $n = m/M$, then $N = n N_A$ |
| Gas volume at STP | $V = n \times 22.4$ |
| Reaction calculation | balance, then $n_A/a = n_B/b$ |
| Which reactant runs out | divide moles by coefficient, smallest wins |
| Colligative property concentration | molality (T-independent) |
| Titration calculation | normality, $N_1V_1 = N_2V_2$ |
| % composition to formula | assume 100 g, moles, simplest ratio |