Atomic Foundations of Matter

1. Chapter at a glance

• Matter can neither be created nor destroyed in a chemical reaction (Law of Conservation of Mass).
• In a compound, elements are always combined in a fixed ratio by mass irrespective of source (Law of Constant Proportions).
• Atoms are indivisible particles that combine in simple whole-number ratios to form compounds (Dalton’s Atomic Theory).
• Atoms attain stable electronic configuration by sharing valence electrons (covalent bond) or by transfer of electrons forming cations and anions (ionic bond).
• Chemical formulae of covalent compounds show actual number of atoms; formulae of ionic compounds show simplest whole-number ratio of ions.
• Ionic compounds conduct electricity only when dissolved in water or in molten state; covalent compounds generally do not.
• Molecular mass of a covalent compound is the sum of atomic masses of all atoms in its molecule; formula unit mass is the sum of atomic masses in the simplest ratio of ions in an ionic compound.

2. Definitions and laws

• Law of Conservation of Mass (Lavoisier, 1789): “matter can neither be created nor destroyed in a chemical reaction.”
• Law of Constant Proportions (Proust’s Law / Law of Definite Proportions): “in any compound formed by two or more elements, the elements combine in a fixed ratio by mass.”
• Molecule: “an electrically neutral entity consisting of more than one atom that is capable of independent existence and shows all the properties of that substance.”
• Covalent bond: formed by sharing of a pair of electrons between atoms.
• Ionic bond: “the electrostatic force of attraction between oppositely charged ions that holds them together.”
• Cation: positively charged ion formed by loss of electron(s).
• Anion: negatively charged ion formed by gain of electron(s).
• Formula unit: “the collection of the simplest whole number ratio of ions” in an ionic compound.

Dalton’s postulates (exact): - All matter is made up of very tiny particles called atoms, which participate in chemical reactions. - Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. - Atoms of a given element are identical in mass and chemical properties. - Atoms of different elements have different masses and chemical properties. - Atoms combine in the ratio of simple whole numbers to form compounds. - The relative number and kinds of atoms are constant in a given compound.

3. Important diagrams and activities

• Activity 9.2 (closed balloon set-up): demonstrates that total mass remains constant before and after a chemical reaction when gas is not allowed to escape.
• Activity 9.3 (Na₂SO₄ + BaCl₂ precipitation): verifies Law of Conservation of Mass in an open system with no gaseous product.
• Fig. 9.6–9.10: electron-dot or line-bond diagrams showing formation of H₂, Cl₂, O₂, HCl and H₂O (demonstrates covalent bonding by sharing).
• Fig. 9.11–9.13: formation of Na⁺ and Cl⁻ ions and NaCl by electron transfer (demonstrates ionic bonding).
• Fig. 9.14: crystal lattice of NaCl (shows 3-D arrangement of ions in ionic compounds).
• Activity 9.4 (solubility & conductivity tests): distinguishes ionic compounds (soluble in water, conduct in solution) from covalent compounds (insoluble in water, non-conducting).

4. Common misconceptions and exam pitfalls

• Assuming mass is always conserved even when gas escapes in an open beaker (Law violated only if system is open and mass is not measured properly).
• Confusing “fixed ratio by mass” with ratio by number of atoms or volume.
• Writing molecular formulae for ionic compounds (e.g., writing NaCl as a molecule) or using brackets incorrectly with polyatomic ions.
• Believing all compounds containing metals conduct electricity in solid state (only molten or aqueous ionic compounds conduct).
• Mixing up molecular mass (covalent) with formula unit mass (ionic) in numericals.
• Forgetting to divide subscripts by common factor after criss-cross method.

5. Formula sheet

Molecular / formula unit mass examples

H₂O = 2×1 + 16 = 18 u
CO₂ = 12 + 2×16 = 44 u
Na₂O = 2×23 + 16 = 62 u
Ca(NO₃)₂ = 40 + 2×(14 + 3×16) = 164 u