Almost every numerical problem in chemistry — whether it appears in physical, inorganic or organic sections — eventually reduces to counting particles and relating them by mass or volume. This chapter builds that counting machinery, and NEET reliably draws 1–2 questions directly from it, plus dozens that depend on it indirectly.
Matter, Measurement and the Reliability of Numbers
Chemistry deals with matter, which occupies space and has mass. Matter can be classified physically (solid, liquid, gas — differing in how tightly particles are packed and how freely they move) or by composition:
- Pure substances — fixed composition, further split into elements (one kind of atom) and compounds (two or more elements in a fixed mass ratio).
- Mixtures — variable composition; homogeneous (uniform throughout, e.g. sugar solution, air) or heterogeneous (visibly distinct regions, e.g. sand in water).
Measurements in chemistry use SI units. The seven base quantities are length (m), mass (kg), time (s), temperature (K), amount of substance (mol), electric current (A) and luminous intensity (cd). Two conversions are worth memorising because they appear in gas-law and density problems: 1 L = 1 dm³ = 10⁻³ m³, and 1 atm = 760 mmHg = 101325 Pa = 1.01325 bar. Temperature conversions: K = °C + 273.15 and °F = (9/5)°C + 32.
Because every instrument has a limit, results are reported with significant figures — the digits that carry real information.
- All non-zero digits count. Zeros between non-zero digits count.
- Leading zeros never count (0.0032 has 2 s.f.).
- Trailing zeros count only if a decimal point is present (2.50 has 3 s.f.; 250 is ambiguous, so write 2.5 × 10² or 2.50 × 10²).
- Exact counted numbers and defined conversion factors have infinite significant figures.
For addition/subtraction, the answer keeps the smallest number of decimal places; for multiplication/division, the smallest number of significant figures. Note the distinction between precision (how close repeated readings are to each other) and accuracy (how close a reading is to the true value).
Dimensional analysis (factor-label method) is the safest way to convert units: multiply by ratios equal to 1 so that unwanted units cancel. For instance, converting 5.0 g cm⁻³ to kg m⁻³: 5.0 g cm⁻³ × (1 kg/1000 g) × (10⁶ cm³/1 m³) = 5.0 × 10³ kg m⁻³.
The Laws of Chemical Combination and Dalton's Theory
Five empirical laws, established before atomic theory was accepted, constrain how elements combine.
- Law of conservation of mass (Lavoisier): in a chemical change, total mass of reactants equals total mass of products; matter is neither created nor destroyed.
- Law of definite (constant) proportions (Proust): a given compound always contains the same elements in the same fixed mass ratio, regardless of its source or method of preparation. Water from any origin is 1 : 8 hydrogen to oxygen by mass.
- Law of multiple proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other bear a simple whole-number ratio. With 14 g nitrogen, oxygen present in N₂O, NO and NO₂ is 8 g, 16 g and 32 g — a ratio of 1 : 2 : 4.
- Gay Lussac's law of gaseous volumes: gases react and form products in volume ratios of simple whole numbers, at the same temperature and pressure. H₂ + Cl₂ → 2HCl is 1 : 1 : 2 by volume.
- Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Dalton's atomic theory rationalised these: matter is made of indivisible atoms; atoms of an element are identical in mass and properties; atoms combine in small whole-number ratios to form compounds; atoms are neither created nor destroyed in reactions. Its limitations — isotopes (atoms of one element with different masses), isobars, and its inability to explain why particular combining ratios occur — were addressed only much later.
Atomic Mass, Molecular Mass and the Mole
Atomic masses are expressed relative to the carbon-12 standard: one unified atomic mass unit (u), also written amu or Da, is exactly 1/12 the mass of a ¹²C atom, equal to 1.66 × 10⁻²⁴ g. Reported atomic masses are weighted averages over natural isotopes. Chlorine, for example, is 75% ³⁵Cl and 25% ³⁷Cl, giving (35 × 0.75) + (37 × 0.25) = 35.5 u — which is why no single chlorine atom weighs 35.5 u.
Molecular mass is the sum of atomic masses of all atoms in a molecule (H₂SO₄ = 2 + 32 + 64 = 98 u). For ionic compounds, which have no discrete molecules, we use formula mass (NaCl = 58.5 u).
A mole is the amount of substance containing as many entities as there are atoms in exactly 12 g of ¹²C — that is, 6.022 × 10²³ entities, the Avogadro constant (N_A). The crucial practical consequence: the mass of one mole of a substance in grams is numerically equal to its atomic/molecular mass in u. This is the molar mass (g mol⁻¹).
The three routes into a mole calculation:
$$n = \frac{\text{given mass (g)}}{\text{molar mass (g mol}^{-1})} = \frac{\text{number of particles}}{6.022\times10^{23}} = \frac{\text{volume of gas at STP (L)}}{22.4}$$
The last relation applies only to gases at STP (273.15 K, 1 bar, where molar volume is 22.7 L; the older 1 atm standard gives 22.4 L — NEET questions overwhelmingly use 22.4 L, so read the data given). Always identify the entity being counted: 1 mol of O₂ contains 6.022 × 10²³ molecules but 1.204 × 10²⁴ atoms; 1 mol of CaCl₂ furnishes 1 mol Ca²⁺ and 2 mol Cl⁻ ions.
Percentage Composition, Empirical and Molecular Formulae
The mass per cent of an element in a compound is (mass of that element in one mole ÷ molar mass) × 100. In ethanol, C₂H₅OH (46 g mol⁻¹), carbon is (24/46) × 100 = 52.2%.
The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual numbers. They are linked by
$$\text{Molecular formula} = n \times \text{Empirical formula}, \qquad n = \frac{\text{molecular mass}}{\text{empirical formula mass}}$$
Standard procedure from percentage data:
- Take 100 g of compound, so percentages become grams.
- Divide each mass by the element's atomic mass to get moles.
- Divide all mole values by the smallest one.
- If the resulting numbers are close to simple fractions (e.g. 1.5, 1.33, 2.5), multiply all of them by a common factor (2, 3, 2 respectively) to clear them.
- Write the empirical formula; use molecular mass to scale up if asked.
Example: a compound is 40.0% C, 6.7% H, 53.3% O with molecular mass 180 u. Moles: 40.0/12 = 3.33; 6.7/1 = 6.7; 53.3/16 = 3.33. Dividing by 3.33 gives 1 : 2 : 1, so the empirical formula is CH₂O (mass 30). Then n = 180/30 = 6, and the molecular formula is C₆H₁₂O₆.
Stoichiometry and the Limiting Reagent
A balanced equation is a molar recipe. In N₂ + 3H₂ → 2NH₃, the coefficients say 1 mol N₂ reacts with 3 mol H₂ to give 2 mol NH₃ — and, for gases at the same T and P, also 1 volume : 3 volumes : 2 volumes.
A general stoichiometry problem follows one path: given quantity → moles → mole ratio from the equation → moles of required species → required quantity (mass, volume or particles).
When amounts of more than one reactant are specified, one of them runs out first and caps the yield: it is the limiting reagent. To find it, divide the moles of each reactant by its stoichiometric coefficient; the smallest quotient identifies the limiting reagent. All product calculations must then be based on that reactant, and the excess reagent's leftover amount is found by subtraction.
Example: 28 g N₂ (1 mol) is mixed with 4 g H₂ (2 mol). Quotients: N₂ → 1/1 = 1; H₂ → 2/3 = 0.67. Hydrogen limits. NH₃ formed = 2/3 × 2 = 1.33 mol = 22.7 g, and 1 – 0.67 = 0.33 mol N₂ (9.3 g) remains unreacted.
Related ideas: theoretical yield is the stoichiometric maximum, actual yield is what is obtained, and percentage yield = (actual/theoretical) × 100.
Expressing Concentration of Solutions
A solution has a solute dissolved in a solvent. Several concentration scales appear in NEET problems:
- Mass per cent (w/w) = (mass of solute / mass of solution) × 100.
- Mole fraction x_A = n_A/(n_A + n_B); mole fractions of all components sum to 1. Unitless.
- Molarity (M) = moles of solute per litre of solution (mol L⁻¹). Temperature-dependent, because volume expands on heating.
- Molality (m) = moles of solute per kilogram of solvent (mol kg⁻¹). Independent of temperature — hence preferred in colligative-property work.
- Parts per million (ppm) = (mass of solute / mass of solution) × 10⁶, used for trace amounts.
Two frequently needed bridges:
- Molarity from mass percentage and density: M = (10 × mass % × density in g mL⁻¹) / molar mass.
- Dilution: on adding solvent, moles of solute stay constant, so M₁V₁ = M₂V₂.
Example: commercial HCl is 36.5% by mass with density 1.2 g mL⁻¹. M = (10 × 36.5 × 1.2)/36.5 = 12 mol L⁻¹. Diluting 10 mL of it to 250 mL gives M₂ = (12 × 10)/250 = 0.48 M.
Common Mistakes and Exam Traps
Mixing up molarity's denominator. Molarity uses the volume of the final solution, not the volume of solvent added; molality uses the mass of solvent, not of solution. Questions that give the density of a solution and quietly expect you to switch between the two scales are common — always find the mass of the solution first (from volume × density), subtract the solute's mass to isolate the solvent's mass, and only then compute molality.
Treating percentage yield as something to apply mid-calculation. Actual yield is always given or measured, never derived; only the theoretical yield comes from stoichiometry. Apply a stated yield percentage as the very last step of a problem, not partway through a multi-step mole calculation, otherwise rounding errors compound in the wrong place.
Rounding intermediate results instead of only the final answer. Carrying an already-rounded intermediate value into the next step of a calculation is a frequent reason a technically correct method still produces a numerically 'off' final answer relative to the expected one; round once, at the very end.
Forgetting to reduce the empirical formula to its lowest terms before finding n. n = molecular mass ÷ empirical formula mass assumes the empirical formula is already in its simplest whole-number ratio; skipping that reduction (using, say, C₂H₄O₂ instead of CH₂O) gives a fractional or wrong value of n and, in turn, a wrong molecular formula.