Language of Chemistry

A. Berzelius' System of Symbols:

• A Symbol is a short abbreviation representing one atom of a specific chemical element.
• Derived from the first letter of its English name: $H$ (Hydrogen), $C$ (Carbon), $N$ (Nitrogen), $O$ (Oxygen), $S$ (Sulfur), $P$ (Phosphorus).
• When names share the same first letter, two letters are used (First capital, second lowercase): $Ca$ (Calcium), $Cl$ (Chlorine), $Co$ (Cobalt), $Cr$ (Chromium).

B. Essential Latin-Derived Symbols:

A. Definition of Valency:

Valency is the combining capacity of an atom of an element. It is measured by the number of hydrogen atoms, or chlorine atoms, or twice the number of oxygen atoms with which one atom of the element combines.

• For Metals (Valence electrons $1, 2, 3$): $\text{Valency} = \text{Number of valence electrons}$. $$\text{Na (2,8,1)} \implies 1; \quad \text{Mg (2,8,2)} \implies 2; \quad \text{Al (2,8,3)} \implies 3$$
• For Non-Metals (Valence electrons $4, 5, 6, 7$): $\text{Valency} = 8 - \text{Valence electrons}$. $$\text{C (2,4)} \implies 8 - 4 = 4; \quad \text{N (2,5)} \implies 8 - 5 = 3; \quad \text{O (2,6)} \implies 8 - 6 = 2; \quad \text{Cl (2,8,7)} \implies 8 - 7 = 1$$
• Noble Gases ($Z=2, 10, 18$): Complete octet/duplet $\implies \mathbf{\text{Valency} = 0}$.

B. Variable Valency:

Certain transition metals exhibit more than one valency by losing electrons from both their outermost and penultimate shells:

• Iron ($Fe$): Ferrous (lower valency $+2$) and Ferric (higher valency $+3$).
• Copper ($Cu$): Cuprous (lower valency $+1$) and Cupric (higher valency $+2$).
• Lead ($Pb$): Plumbous ($+2$) and Plumbic ($+4$).

A. Radicals (Polyatomic Ions):

A Radical is a group of atoms of different elements carrying a net positive or negative charge that behaves as a single chemical unit in reactions.

• Basic Radicals (Electropositive / Cations): Ammonium ($NH_4^+$ [valency 1]), Hydrogen ($H^+$ [1]), Sodium ($Na^+$ [1]), Calcium ($Ca^{2+}$ [2]), Aluminum ($Al^{3+}$ [3]).
• Acidic Radicals (Electronegative / Anions):
  • Monovalent (1): Hydroxide ($OH^-$), Nitrate ($NO_3^-$), Bicarbonate / Hydrogen carbonate ($HCO_3^-$), Chloride ($Cl^-$).
  • Divalent (2): Oxide ($O^{2-}$), Carbonate ($CO_3^{2-}$), Sulfate ($SO_4^{2-}$), Sulfide ($S^{2-}$).
  • Trivalent (3): Phosphate ($PO_4^{3-}$), Nitride ($N^{3-}$).

B. The Criss-Cross Method Algorithm:

1. Write the symbols of the positive radical (metal/cation) on the left and the negative radical (anion) on the right.
2. Write their respective valencies directly below each symbol.
3. Criss-cross (interchange) the valency numbers to become subscripts of the opposite radical.
4. Simplify the subscripts to the lowest whole-number ratio. Enclose polyatomic radicals in parentheses if subscript $> 1$.
5. Example 1: Aluminum Oxide: $Al$ (3) and $O$ (2) $\to$ Cross valencies $\to \mathbf{Al_2O_3}$.
6. Example 2: Calcium Hydroxide: $Ca$ (2) and $OH$ (1) $\to$ Cross valencies $\to \mathbf{Ca(OH)_2}$.
7. Example 3: Ammonium Sulfate: $NH_4$ (1) and $SO_4$ (2) $\to$ Cross valencies $\to \mathbf{(NH_4)_2SO_4}$.
8. Example 4: Magnesium Carbonate: $Mg$ (2) and $CO_3$ (2) $\to$ Ratio $2:2 = 1:1 \to \mathbf{MgCO_3}$.

A. Chemical Equation & Mass Conservation:

A Chemical Equation is the symbolic representation of a chemical reaction using chemical formulas of reactants (left) and products (right).

Law of Conservation of Mass: Matter can neither be created nor destroyed in a chemical reaction. Therefore, the total number of atoms of each element on the Reactant side (LHS) must be strictly equal to the number on the Product side (RHS).

B. Hit-and-Trial Balancing Technique:

1. Write the skeletal (unbalanced) equation: $$\text{Skeletal: } Fe + H_2O \to Fe_3O_4 + H_2$$
2. Balance the element with the maximum number of atoms first (Oxygen has 4 on RHS; balance LHS by multiplying $H_2O$ by 4): $$Fe + 4H_2O \to Fe_3O_4 + H_2$$
3. Balance Hydrogen (LHS has $4 \times 2 = 8$ H; balance RHS by multiplying $H_2$ by 4): $$Fe + 4H_2O \to Fe_3O_4 + 4H_2$$
4. Balance Iron ($Fe$ has 3 on RHS; balance LHS by multiplying $Fe$ by 3): $$\mathbf{3Fe + 4H_2O \to Fe_3O_4 + 4H_2}$$
5. Verify all atom counts: Fe (3 on both sides), H (8 on both sides), O (4 on both sides). The equation is perfectly balanced!