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CC01 — Atoms, Molecules & Chemical Arithmetic

📖 CC01  ·  CDS General Science — Chemistry ★ High Priority

This chapter forms the quantitative backbone of all chemistry. Every reaction and compound you study in later chapters connects back to atoms, molecules, and the mole concept. CDS expects you to know Avogadro's number, mole relationships, basic laws of combination, and how to balance simple equations.

📌 CDS Focus: Avogadro's number (6.022 × 10²³), gram atomic/molecular mass as the mole bridge, Law of Conservation of Mass and Law of Constant Proportion, balancing equations by inspection, and percentage composition. Expect 1–2 direct factual questions.
PART 1 — BASIC DEFINITIONS

1. Atom, Molecule & Atomic Mass

An atom is the smallest particle of an element that retains its chemical identity. A molecule is the smallest particle of a compound (or some elements) that can exist independently.

Fig. 1 — Atoms, Molecules and the Mole: How They Connect
ATOMS → MOLECULES → MOLE → MASS: The Mole Bridge H ATOM Smallest unit of an element that retains its identity. Atomic mass measured in atomic mass units (amu). H = 1 amu, C = 12 amu, O = 16 amu, Na = 23 amu H H MOLECULE Two or more atoms bonded. Molecular mass = sum of atomic masses. H₂O = 2(1)+16 = 18 amu | CO₂ = 12+2(16) = 44 amu | NaCl = 23+35.5 = 58.5 amu ∞ MOLE 1 mole = 6.022 × 10²³ particles (Avogadro's Number Nₐ). Gram molecular mass of H₂O = 18 g/mol. At STP, 1 mol gas = 22.4 L. THE MOLE BRIDGE (most important formula) Moles = Mass (g) / Molar Mass (g/mol) = Particles / 6.022×10²³ = Volume (L) / 22.4 (at STP)
Key Formulae — Mole Concept:

● Moles = mass (g) / molar mass (g/mol)
● Number of particles = moles × 6.022 × 10²³
● Volume at STP (gas) = moles × 22.4 L
● Molar mass numerically equals molecular mass in g/mol
● Gram atomic mass: mass of 1 mole of atoms = atomic mass in grams
PART 2 — LAWS OF CHEMICAL COMBINATION

2. Laws of Chemical Combination

Fig. 2 — Five Laws of Chemical Combination with Examples
FIVE LAWS OF CHEMICAL COMBINATION Law & Scientist Statement Example 1. Conservation of Mass Lavoisier, 1789 Mass is neither created nor destroyed in any reaction. Reactant mass = Product mass. 2H₂ + O₂ → 2H₂O 4 + 32 = 36 g both sides 2. Constant Proportion Proust, 1808 A compound always contains elements in the same mass ratio, regardless of source. H₂O always has H:O = 1:8 by mass 3. Multiple Proportions Dalton, 1803 When two elements form more than one compound, masses of one are in a simple ratio. CO and CO₂: oxygen = 16 and 32 (ratio 1:2) 4. Reciprocal Proportions Richter, 1792 Masses combining with a fixed mass of a third element are simple multiples of each other. H₂S, H₂O and SO₂: related S and O ratios 5. Combining Volumes Gay-Lussac Gases react in simple whole-number ratios by volume at same T and P. H₂ + Cl₂ → 2 HCl 1 vol + 1 vol → 2 vol
PART 3 — CHEMICAL EQUATIONS

3. Balancing Chemical Equations

A balanced equation has equal numbers of each type of atom on both sides (Law of Conservation of Mass). Use the trial-and-balance (inspection) method for CDS level.

Fig. 3 — How to Balance a Chemical Equation Step by Step
BALANCING: Fe + O₂ → Fe₂O₃ — Step by Step STEP 1 — Write the unbalanced equation Fe + O₂ → Fe₂O₃ Count: Fe = 1 left, 2 right; O = 2 left, 3 right → NOT balanced STEP 2 — Balance by adjusting coefficients (never subscripts) 4Fe + 3O₂ → 2Fe₂O₃ LCM method: need 4 Fe and 6 O atoms on both sides (3 × O₂ = 6 O) STEP 3 — Verify: count atoms on both sides LHS: Fe = 4, O = 6   RHS: Fe = 4, O = 6   ✓ BALANCED RULES: Change coefficients only (never subscripts). Balance metals first, then non-metals, then H and O last. Verify atom count on both sides after each step. Coefficients must be whole numbers.

4. Percentage Composition & Empirical Formula

Percentage Composition:

% of element = (mass of element in 1 mole of compound / molar mass of compound) × 100

Example — H₂O: % H = (2/18) × 100 = 11.1%    % O = (16/18) × 100 = 88.9%

Empirical Formula: Simplest whole-number ratio of atoms. Molecular formula = n × Empirical formula.
Example: Glucose C₆H₁₂O₆ → Empirical formula CH₂O (n = 6)
✎ Worked Example — Mole Concept
How many molecules are there in 36 g of water (H₂O)? Molar mass of H₂O = 18 g/mol.
Moles of H₂O = 36/18 = 2 moles
Number of molecules = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules
✔ 1.204 × 10²⁴ molecules

📝 CDS PYQs — Atoms, Molecules & Chemical Arithmetic

Q1. Avogadro's number is: CDS PYQ
(a) 6.022 × 10²¹ (b) 6.022 × 10²² (c) 6.022 × 10²³ (d) 6.022 × 10²⁴
✔ Answer: (c) 6.022 × 10²³
Avogadro's number (Nₐ) = 6.022 × 10²³ — this is the number of atoms, molecules, or ions present in one mole of any substance. It was determined by Amedeo Avogadro's hypothesis that equal volumes of gases at the same T and P contain equal numbers of molecules. This is one of the most fundamental constants in chemistry and is directly tested in CDS.
Q2. Which law states that matter is neither created nor destroyed in a chemical reaction? CDS PYQ
(a) Law of Constant Proportion (b) Law of Multiple Proportions (c) Law of Conservation of Mass (d) Avogadro's Law
✔ Answer: (c) Law of Conservation of Mass
The Law of Conservation of Mass (Lavoisier, 1789) states that the total mass of reactants equals the total mass of products in a chemical reaction. This is why chemical equations must be balanced. Example: 2H₂ + O₂ → 2H₂O — 4g + 32g = 36g on both sides. This is the most fundamental law of chemistry.
Q3. The molecular mass of CO₂ is: CDS PYQ
(a) 28 g/mol (b) 40 g/mol (c) 44 g/mol (d) 48 g/mol
✔ Answer: (c) 44 g/mol
CO₂ = 1 carbon + 2 oxygens = 12 + 2(16) = 12 + 32 = 44 g/mol. Note: CO (carbon monoxide) = 12 + 16 = 28 g/mol (common CDS confusion). CO₂ is the gas exhaled during respiration and is a greenhouse gas. Its molecular mass of 44 is a frequently tested fact.
Q4. Which law explains why CO and CO₂ are different compounds? ⚡ Tricky
(a) Law of Conservation of Mass (b) Law of Constant Proportion (c) Law of Multiple Proportions (d) Avogadro's Law
✔ Answer: (c) Law of Multiple Proportions
CO and CO₂ are two compounds formed by the same two elements (C and O). In CO, oxygen = 16g per 12g carbon; in CO₂, oxygen = 32g per 12g carbon. The ratio of oxygen in both = 16:32 = 1:2 (a simple whole-number ratio). This is exactly what Dalton's Law of Multiple Proportions states: when two elements form more than one compound, the masses of one element relative to a fixed mass of the other are in a simple ratio.
Q5. How many grams is 1 mole of NaCl? (Na = 23, Cl = 35.5) CDS PYQ
(a) 23 g (b) 35.5 g (c) 58.5 g (d) 47 g
✔ Answer: (c) 58.5 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol. One mole of any substance has a mass equal to its molar mass in grams. So 1 mole of NaCl weighs 58.5 g and contains 6.022 × 10²³ formula units (Na⁺ and Cl⁻ ion pairs).

🧠 Quick Memory Chart — CC01

⚛ Mole Facts
  • 1 mole = 6.022 × 10²³ particles
  • 1 mole gas at STP = 22.4 L
  • Moles = mass/molar mass
  • Molar mass numerically = molecular mass
  • H₂O = 18 g/mol; CO₂ = 44 g/mol
⚖ 5 Laws
  • Conservation: mass neither created nor destroyed
  • Constant Proportion: same ratio by mass
  • Multiple Prop: simple whole-number ratios
  • Gay-Lussac: gases combine in simple volume ratios
  • Avogadro: equal volumes → equal molecules
⚡ Key Values
  • H = 1; C = 12; N = 14; O = 16
  • Na = 23; Mg = 24; Al = 27; S = 32
  • Cl = 35.5; Ca = 40; Fe = 56; Cu = 64
  • NaCl = 58.5; H₂SO₄ = 98; CaCO₃ = 100
  • Nₐ = 6.022 × 10²³

📝 Practice Exercise — Attempt Before Checking

E1. How many moles are there in 44 g of CO₂? (Molar mass = 44 g/mol)
(a) 0.5 mol(b) 1 mol(c) 2 mol(d) 44 mol
E2. A compound always has the same elements in the same mass ratio. This is:
(a) Law of Conservation of Mass(b) Law of Constant Proportion(c) Law of Multiple Proportions(d) Dalton's Atomic Theory
E3. The percentage of oxygen in H₂O is approximately:
(a) 11.1%(b) 66.7%(c) 88.9%(d) 50%
E4. 1 mole of oxygen gas (O₂) at STP occupies:
(a) 11.2 L(b) 22.4 L(c) 32 L(d) 44.8 L
Answers:
E1 → (b) 1 mol  [44/44 = 1 mol]  |  E2 → (b) Law of Constant Proportion (Proust's Law)  |  E3 → (c) 88.9%  [16/18 × 100]  |  E4 → (b) 22.4 L  [any gas at STP occupies 22.4 L per mole]
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