Some Basic Concepts of Chemistry

Memory aid 'C-D-MM-RP-G': (1) Law of Conservation of Mass (Lavoisier) - matter is neither created nor destroyed. (2) Law of Definite/Constant Proportions (Proust) - a pure compound always has the same elements in the same fixed mass ratio. (3) Law of Multiple Proportions (Dalton) - when two elements form more than one compound, the masses of one combining with a fixed mass of the other are in small whole-number ratios (e.g., CO and CO2: O ratio 1:2). (4) Gay-Lussac's Law of Gaseous Volumes - gases combine in simple whole-number volume ratios at same T,P. (5) Avogadro's Law - equal volumes of gases at same T,P contain equal number of molecules. Avogadro's law resolved the atom-molecule confusion of Dalton's theory.

Avogadro number NA = 6.022 x 10^23. Number of moles can be found four ways: n = given mass/molar mass = number of particles/NA = volume at STP/22.4 L (gases) = molarity x volume(L). At STP (273.15 K, 1 bar) molar volume = 22.7 L; at 273.15 K and 1 atm it is 22.4 L. Mass of one atom = molar mass/NA. Number of molecules = n x NA; number of atoms = n x NA x (atomicity). Shortcut: 1 mole = molar mass in grams = 6.022 x 10^23 particles = 22.4 L gas at STP(1 atm). Remember total moles of atoms in a molecule = moles of molecules x number of atoms per molecule.

Carbon forms two oxides. In CO, 12 g C combines with 16 g O. In CO2, 12 g C combines with 32 g O. Fixing carbon mass at 12 g, the oxygen masses are 16 g and 32 g, giving ratio 16:32 = 1:2, a simple whole-number ratio - illustrating the Law of Multiple Proportions. Another example: H2O (2 g H : 16 g O) and H2O2 (2 g H : 32 g O) gives oxygen ratio 1:2. Tip: always fix the mass of one element, then take the ratio of the other element's masses to verify the law quickly in exams.

One atomic mass unit (amu / u) = 1/12th the mass of one C-12 atom = 1.66 x 10^-24 g. Average atomic mass accounts for isotopic abundance: Average mass = Sum(isotopic mass x fractional abundance). Example for chlorine: (35 x 0.7577) + (37 x 0.2423) = 35.45 u. This is why atomic masses on the periodic table are usually non-integers. Memory aid: 'AVERAGE = mass-weighted by abundance'. The unified mass unit 'u' has replaced 'amu' in modern usage but both are accepted. Gram atomic mass = atomic mass expressed in grams = mass of 1 mole of atoms.

Molecular mass = sum of atomic masses of all atoms in a molecule (for covalent/molecular substances). Example: H2O = 2(1.008) + 16.00 = 18.02 u. Formula mass is used for IONIC compounds which do not exist as discrete molecules, e.g., NaCl formula mass = 23 + 35.5 = 58.5 u (we say formula unit, not molecule). Gram molecular mass = molecular mass in grams = mass of 1 mole of molecules. Shortcut: just add up atomic masses x number of atoms. Distinguish molecular mass (real molecules like CO2, H2O) from formula mass (ionic lattices like NaCl, CaCl2).

Boron has two isotopes: B-10 (abundance 19.9%, mass 10.013 u) and B-11 (abundance 80.1%, mass 11.009 u). Average atomic mass = (10.013 x 0.199) + (11.009 x 0.801) = 1.992 + 8.818 = 10.81 u. This matches the periodic table value of boron (10.81). Exam shortcut: if abundances are given as percentages, convert to fractions (divide by 100) before multiplying, or multiply by mass and divide the sum by 100. Always check the answer lies between the two isotopic masses, weighted toward the more abundant isotope.

Mass percent of an element = (mass of that element in 1 mole of compound / molar mass of compound) x 100. Example: in H2O, mass % of H = (2/18) x 100 = 11.11%; mass % of O = (16/18) x 100 = 88.89%. Always confirm the percentages add up to 100. Memory aid 'PART over WHOLE times hundred'. This data is the starting point for finding empirical formula. Mass of element = mass % x molar mass / 100. For hydrates and complex salts, count every atom carefully using the full formula including water of crystallisation.

Empirical formula = simplest whole-number ratio of atoms. Molecular formula = actual number of atoms = n x empirical formula, where n = molecular mass / empirical formula mass. STEPS (memory 'MMD-R-W'): (1) Take Mass % (assume 100 g sample). (2) Divide by atomic Mass to get moles. (3) Divide all by the smallest mole value to get Ratio. (4) round to Whole numbers (multiply if needed, e.g., 1.5 x2 = 3). That gives the empirical formula. Then find n and multiply. Example: empirical CH2O (mass 30); if molecular mass = 180, n = 180/30 = 6, molecular formula = C6H12O6 (glucose).

A compound contains 40% C, 6.7% H, 53.3% O; molar mass = 60 g/mol. Assume 100 g: C = 40/12 = 3.33 mol; H = 6.7/1 = 6.7 mol; O = 53.3/16 = 3.33 mol. Divide by smallest (3.33): C = 1, H = 2, O = 1. Empirical formula = CH2O, empirical mass = 30. n = 60/30 = 2. Molecular formula = C2H4O2 (acetic acid). Always double-check by computing the molar mass of your final formula. Tip: ratios near .33 or .67 hint multiply by 3; near .5 multiply by 2.

Molarity (M) = moles of solute / volume of solution in litres (temperature dependent). Molality (m) = moles of solute / mass of solvent in kg (temperature independent - preferred for accuracy). Mole fraction (x) = moles of component / total moles (xA + xB = 1, dimensionless). Mass percent = (mass of solute / mass of solution) x 100. ppm = (mass of solute / mass of solution) x 10^6. Memory aid: 'Molarity = Litres of Solution; molality = kg of Solvent'. Useful relation: Molarity = (10 x density x mass%)/molar mass. Dilution: M1V1 = M2V2.

The limiting reagent is the reactant that is completely consumed first and thus determines (limits) the amount of product formed; the other reactant is in excess. STEPS: (1) Write the balanced equation. (2) Convert all given masses/volumes to moles. (3) Divide moles of each reactant by its stoichiometric coefficient. (4) The smallest value identifies the limiting reagent. (5) Calculate product using the limiting reagent's moles. Memory aid: 'Least ratio LIMITS'. Always base product calculations on the limiting reagent, never on the excess reagent. Excess reagent leftover = initial moles - moles reacted.

Reaction: N2 + 3H2 -> 2NH3. Given 28 g N2 and 6 g H2. Moles N2 = 28/28 = 1; moles H2 = 6/2 = 3. Divide by coefficients: N2 = 1/1 = 1; H2 = 3/3 = 1. Both ratios are equal, so neither is in excess - they react completely. NH3 formed = 2 x 1 = 2 mol = 2 x 17 = 34 g. If instead H2 were 4 g (2 mol), then H2/3 = 0.67 < N2/1 = 1, so H2 is limiting and NH3 = 2 x (2/3) = 1.33 mol. Always recompute the ratio when quantities change.