Atoms and Molecules

Given:
- Total mass of compound = 0.24 g
- Mass of boron = 0.096 g
- Mass of oxygen = 0.144 g

Formula used:
$$\text{Percentage by weight} = \frac{\text{Mass of element}}{\text{Total mass of compound}} \times 100$$

Step 1: Percentage of Boron
$$\% \text{ Boron} = \frac{0.096}{0.24} \times 100 = 40\%$$

Step 2: Percentage of Oxygen
$$\% \text{ Oxygen} = \frac{0.144}{0.24} \times 100 = 60\%$$

Verification: $40\% + 60\% = 100\%$ ✓

Answer: The compound contains 40% Boron and 60% Oxygen by weight.

Given (first case):
- Mass of carbon = 3.0 g
- Mass of oxygen = 8.00 g
- Mass of CO₂ produced = 11.00 g

Observation from first case:
Carbon and oxygen combine in the ratio:
$$\text{C : O} = 3.0 : 8.0 = 3 : 8$$
This means 3 g of carbon requires exactly 8 g of oxygen to produce 11 g of CO₂.

Second case:
- Mass of carbon = 3.00 g
- Mass of oxygen available = 50.00 g

Step 1: Find oxygen required to burn 3.00 g of carbon completely.
From the fixed ratio, oxygen required $= 8.00$ g.

Step 2: Since 50.00 g of oxygen is available but only 8.00 g is needed, oxygen is in excess. Carbon is the limiting reactant.

Step 3: Mass of CO₂ formed
$$= 3.00 + 8.00 = 11.00 \text{ g}$$

Answer: 11.00 g of carbon dioxide will be formed.

Law governing this: This is governed by the Law of Constant Proportions (Law of Definite Proportions), which states that in a chemical compound, elements are always present in definite proportions by mass, regardless of the source or method of preparation.

Definition:
A polyatomic ion is a group of atoms carrying a net electric charge (positive or negative). These atoms are covalently bonded together and the group as a whole acts as a single ion.

Examples:

| Ion | Formula | Charge |
|-----|---------|--------|
| Ammonium ion | $\text{NH}_4^+$ | $+1$ |
| Hydroxide ion | $\text{OH}^-$ | $-1$ |
| Carbonate ion | $\text{CO}_3^{2-}$ | $-2$ |
| Sulphate ion | $\text{SO}_4^{2-}$ | $-2$ |
| Nitrate ion | $\text{NO}_3^-$ | $-1$ |
| Phosphate ion | $\text{PO}_4^{3-}$ | $-3$ |

These ions participate in ionic bonding just like monoatomic ions.

Concept: Chemical formulae are written by criss-crossing the valencies of the combining ions.

(a) Magnesium chloride
- Magnesium ion: $\text{Mg}^{2+}$ (valency = 2)
- Chloride ion: $\text{Cl}^-$ (valency = 1)
- Criss-cross: Formula = $\mathbf{MgCl_2}$

(b) Calcium oxide
- Calcium ion: $\text{Ca}^{2+}$ (valency = 2)
- Oxide ion: $\text{O}^{2-}$ (valency = 2)
- Criss-cross and simplify (ratio 2:2 → 1:1): Formula = $\mathbf{CaO}$

(c) Copper nitrate
- Copper ion: $\text{Cu}^{2+}$ (valency = 2)
- Nitrate ion: $\text{NO}_3^-$ (valency = 1)
- Criss-cross: Formula = $\mathbf{Cu(NO_3)_2}$

(d) Aluminium chloride
- Aluminium ion: $\text{Al}^{3+}$ (valency = 3)
- Chloride ion: $\text{Cl}^-$ (valency = 1)
- Criss-cross: Formula = $\mathbf{AlCl_3}$

(e) Calcium carbonate
- Calcium ion: $\text{Ca}^{2+}$ (valency = 2)
- Carbonate ion: $\text{CO}_3^{2-}$ (valency = 2)
- Criss-cross and simplify (ratio 2:2 → 1:1): Formula = $\mathbf{CaCO_3}$

(a) Quick lime
- Chemical formula: $\text{CaO}$
- Elements present: Calcium (Ca) and Oxygen (O)

(b) Hydrogen bromide
- Chemical formula: $\text{HBr}$
- Elements present: Hydrogen (H) and Bromine (Br)

(c) Baking powder
- Baking powder contains Sodium hydrogen carbonate ($\text{NaHCO}_3$) as the main active ingredient.
- Elements present: Sodium (Na), Hydrogen (H), Carbon (C), and Oxygen (O)

(d) Potassium sulphate
- Chemical formula: $\text{K}_2\text{SO}_4$
- Elements present: Potassium (K), Sulphur (S), and Oxygen (O)

Concept: Molar mass = Sum of atomic masses of all atoms in the molecular formula.

Atomic masses used: H = 1 u, C = 12 u, N = 14 u, O = 16 u, S = 32 u, Cl = 35.5 u, P = 31 u

(a) Ethyne, $\text{C}_2\text{H}_2$
$$M = (2 \times 12) + (2 \times 1)$$
$$M = 24 + 2 = \mathbf{26 \text{ g mol}^{-1}}$$

(b) Sulphur molecule, $\text{S}_8$
$$M = 8 \times 32$$
$$M = \mathbf{256 \text{ g mol}^{-1}}$$

(c) Phosphorus molecule, $\text{P}_4$
$$M = 4 \times 31$$
$$M = \mathbf{124 \text{ g mol}^{-1}}$$

(d) Hydrochloric acid, $\text{HCl}$
$$M = 1 + 35.5$$
$$M = \mathbf{36.5 \text{ g mol}^{-1}}$$

(e) Nitric acid, $\text{HNO}_3$
$$M = 1 + 14 + (3 \times 16)$$
$$M = 1 + 14 + 48 = \mathbf{63 \text{ g mol}^{-1}}$$