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Language of Chemistry
ICSE Class 8 Chemistry • Chapter 5 (Detailed Master Notes)
Chapter Overview
Just as English needs an alphabet (A, B, C) to form words, and words to form sentences, Chemistry needs its own language to describe the millions of reactions happening in nature. In this chapter, we learn how to write chemical symbols, formulas, and full chemical equations.
5.1 Chemical Symbols
In the early days, alchemists used strange drawings to represent elements. Today, we use a standard, internationally accepted system devised by J.J. Berzelius.
Symbol: A short, universally accepted shorthand abbreviation that represents a specific element or one atom of that element.
Rules for Writing Symbols:
- Single Letter: The first letter of the English name, always written in CAPITALS. (e.g., Hydrogen $\rightarrow$ H, Oxygen $\rightarrow$ O, Carbon $\rightarrow$ C).
- Two Letters: If two elements start with the same letter, we use the first letter (capital) followed by another prominent letter (small) from its name. (e.g., Calcium $\rightarrow$ Ca, Chlorine $\rightarrow$ Cl, Cobalt $\rightarrow$ Co).
- Latin Names: A few elements take symbols strictly from their ancient Latin names.
| Element (English) | Latin Name | Symbol |
|---|---|---|
| Sodium | Natrium | Na |
| Potassium | Kalium | K |
| Iron | Ferrum | Fe |
| Copper | Cuprum | Cu |
| Silver | Argentum | Ag |
| Gold | Aurum | Au |
| Lead | Plumbum | Pb |
5.2 Valency & Radicals
Before we can construct words (formulas), we must understand how elements link together.
Valency: The combining capacity of an element or a radical. It is numerically equal to the number of electrons an atom needs to lose, gain, or share to become stable.
Radical: An atom or a tightly bound group of atoms that carries a positive or negative charge and acts as a single chemical unit in reactions (e.g., Sulphate $SO_4^{2-}$, Ammonium $NH_4^+$).
Variable Valency: Certain transition metals show more than one valency depending on the reaction conditions. We use Roman numerals to indicate this.
- Iron (Fe): Ferrous ($Fe^{2+}$) and Ferric ($Fe^{3+}$)
- Copper (Cu): Cuprous ($Cu^+$) and Cupric ($Cu^{2+}$)
5.3 Chemical Formula (Criss-Cross Method)
A formula represents the composition of a molecule. Writing a formula is easy if you know the symbols and valencies. We use the Criss-Cross Method.
Steps:
- Write the symbol of the positive radical (metal) on the left, and negative radical (non-metal) on the right.
- Write their respective valencies at the top right corner.
- Criss-cross (swap) the valencies and write them as subscripts at the bottom right.
- Ignore the (+/-) signs. If subscripts are $1$, omit them. If they can be simplified by a common factor, divide them.
Example: Aluminum Oxide
- Symbols: $Al$ and $O$
- Valencies: $Al^{3+}$ and $O^{2-}$
- Criss-cross: $Al \rightarrow 2$ and $O \rightarrow 3$
- Result: $Al_2O_3$
5.4 Chemical Equations
A chemical equation is the shorthand representation of a full chemical reaction using symbols and formulas.
Example: Burning of Magnesium in Air.
Word Equation: Magnesium + Oxygen $\rightarrow$ Magnesium Oxide
Skeletal Equation: $Mg + O_2 \rightarrow MgO$ (This is unbalanced!)
Balancing an Equation
The Law of Conservation of Mass states that atoms can neither be created nor destroyed. The number of atoms of each element on the Reactant side (left) MUST equal the Product side (right).
Balanced Equation: $2Mg + O_2 \rightarrow 2MgO$
(Now we have 2 Mgs and 2 Os on both sides. The equation is structurally perfect).
AI Image Prompt: A visual scale (balance). On the left pan (Reactants): 2 grey spheres (Mg atoms) and 1 pair of red spheres (O2 molecule). On the right pan (Products): 2 combined units, each a grey sphere attached to a red sphere (MgO). The scale is perfectly balanced horizontally.
Q1. Write the chemical formula for Calcium Carbonate.
Answer: The symbol for Calcium is $Ca$ (valency $+2$). The formula for Carbonate radical is $CO_3$ (valency $-2$). By criss-crossing, we get $Ca_2(CO_3)_2$. Since 2 is common, we divide by 2. The final formula is $CaCO_3$.
Q2. Balance the following skeletal equation: $H_2 + N_2 \rightarrow NH_3$
Answer: There are 2 Nitrogen atoms on the left, but only 1 on the right. Multiply $NH_3$ by $2$, yielding $H_2 + N_2 \rightarrow 2NH_3$. Now we have $6$ Hydrogen atoms on the right (2x3), but only $2$ on the left. Multiply $H_2$ by $3$. The final balanced equation is $3H_2 + N_2 \rightarrow 2NH_3$.