Chemistry Basics – Study Notes
(For Mizoram PSC – General Awareness & Current Affairs)
Goal: Understand the core concepts of chemistry that are frequently asked in the PSC General Awareness paper.
Format: Simple explanation → Key points → Worked example → Quick revision summary.
1. Simple Explanation
Chemistry is the science that studies matter (anything that has mass and occupies space) and the changes it can undergo.
- Matter is made of atoms (the smallest units that retain an element’s identity).
- Atoms combine to form molecules (compounds) or remain as elements.
- Chemical reactions transform substances, releasing or absorbing energy.
2. Key Points
| Theme | What to Know | Why It Matters for PSC |
|---|---|---|
| Atomic Structure | - Nucleus (protons + neutrons) - Electrons orbit in shells - Atomic number = protons - Mass number = protons + neutrons |
Helps explain periodic trends and element symbols. |
| Periodic Table | - 7 periods, 18 groups - Metals, non‑metals, metalloids - Trends: electronegativity ↑, atomic radius ↓ across a period; ↑ down a group |
Many PSC questions test element symbols, group properties, and trends. |
| Chemical Bonding | - Ionic (electron transfer) - Covalent (electron sharing) - Metallic (electron “sea”) |
Explains why salts are soluble, why metals conduct electricity, etc. |
| States of Matter | - Solid, liquid, gas, plasma - Phase changes: fusion, vaporisation, sublimation, condensation, deposition |
Useful for questions on temperature, pressure, and everyday phenomena. |
| Stoichiometry | - Mole concept (Avogadro’s number) - Molar mass - Balancing equations - Limiting reagent, percent yield |
Core of quantitative chemistry questions. |
| Acids & Bases | - pH scale (0–14) - Strong vs. weak acids/bases - Indicators (litmus, phenolphthalein) |
Often asked in current‑affair stories about water pollution, acid rain, etc. |
| Redox Reactions | - Oxidation numbers - Oxidation vs. reduction - Electrochemistry (galvanic cells, electrolytic cells) |
Relevance to batteries, corrosion, and industrial processes. |
| General Equations | - Combustion: ( \text{C}_x\text{H}_y + O_2 \rightarrow CO_2 + H_2O ) - Acid‑base neutralisation: ( H_3O^+ + OH^- \rightarrow H_2O ) - Precipitation: ( AB + CD \rightarrow AD + CB ) |
Quick reference for balancing and answering “what is produced” questions. |
3. Worked Example
Problem
Balance the following combustion reaction and calculate the percent yield.
[ \text{C}_3\text{H}_8 + O_2 \rightarrow CO_2 + H_2O ]
a) Balanced equation.
b) Given 30.0 g of (C_3H_8) (molar mass = 44.11 g mol⁻¹), how many grams of (CO_2) (molar mass = 44.01 g mol⁻¹) are theoretically produced?
c) If the actual yield is 23.5 g of (CO_2), what is the percent yield?
Solution
a) Balancing the equation
- Carbon (C)
[ 3;C \rightarrow 3;CO_2 ] - Hydrogen (H)
[ 8;H \rightarrow 4;H_2O ] - Oxygen (O) – Count O atoms on right side:
(3 \times 2 = 6) (from (CO_2)) + (4 \times 1 = 4) (from (H_2O)) = 10 O atoms.
Since (O_2) provides 2 O atoms per molecule, we need 5 (O_2).
Balanced equation:
[ \boxed{C_3H_8 + 5,O_2 ;\longrightarrow; 3,CO_2 + 4,H_2O} ]
b) Theoretical yield of (CO_2)
- Moles of (C_3H_8)
[ n = \frac{30.0;\text{g}}{44.11;\text{g mol}^{-1}} = 0.680;\text{mol} ] - Stoichiometry – 1 mol (C_3H_8) gives 3 mol (CO_2).
[ n_{CO_2} = 0.680 \times 3 = 2.04;\text{mol} ] - Mass of (CO_2)
[ m = n_{CO_2} \times 44.01;\text{g mol}^{-1} = 2.04 \times 44.01 = 89.7;\text{g} ]
Theoretical yield ≈ 89.7 g (CO_2).
c) Percent yield
[ \text{Percent yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100 = \frac{23.5}{89.7} \times 100 \approx 26.2%