Atoms and Molecules

Given,

Mass of the sample compound = 0.24 g,

mass of boron = 0.096 g,

mass of oxygen = 0.144 g

To calculate the percentage composition of the compound,

Percentage of boron :

= $\dfrac{\text{mass of boron}}{\text{mass of the compound}}$ x 100

= $\dfrac{0.096}{0.24}$ x 100

= 40%

Percentage of oxygen

= $\dfrac{\text{mass of oxygen}}{\text{mass of the compound}}$ x 100

= $\dfrac{0.144}{0.24}$ x 100

= 60%

Hence, percentage of boron = 40% and percentage of oxygen = 60% in the compound.

Given,

$\underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 8 g}}{\text{O}_2} \longrightarrow \underset{\text{11 g}}{\text{CO}_2}$

When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.

Now,

$\underset{\text{ 3 g}}{\text{C}} + \underset{\text{ 50 g}}{\text{O}_2} \longrightarrow \underset{\text{?}}{\text{CO}_2}$

Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO₂ and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.

The above answer is governed by the law of constant proportions.

Cluster of atoms that act as an ion are called polyatomic ions. They carry a fixed charge on them.

Example: Hydroxide [OH^-], Cyanide [CN^-]

(a) Magnesium chloride

Step 1 — Write each symbol with its valency

$\text{Mg}^{2+} \phantom{\nearrow} \text{Cl}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{Cl}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Magnesium chloride : $\bold{MgCl_2}$

(b) Calcium oxide

Step 1 — Write each symbol with its valency

$\text{Ca}^{2+} \phantom{\nearrow} \text{O}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Calcium oxide : $\bold{CaO}$

(c) Copper nitrate

Step 1 — Write each symbol with its valency

$\text{Cu}^{2+} \phantom{\nearrow} \text{NO}_3^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{NO}_3} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of copper nitrate : $\bold{Cu(NO_3)_2}$

(d) Aluminium chloride

Step 1 — Write each symbol with its valency

$\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Aluminium chloride : $\bold{AlCl_3}$

(e) Calcium carbonate –

Step 1 — Write each symbol with its valency

$\text{Ca}^{2+} \phantom{\nearrow} \text{CO}_3^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{CO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{CO}_3} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of copper nitrate : $\bold{Ca(CO_3)_2}$

(a) Quick lime (CaO) — Calcium and oxygen

(b) Hydrogen bromide (HBr) — Hydrogen and bromine

(c) Baking powder (NaHCO₃) — Sodium, hydrogen, carbon and oxygen

(d) Potassium sulphate (K₂SO₄) — Potassium, sulphur and oxygen,

(a) Molar mass of Ethyne C₂H₂ = (2 x Mass of C) + (2 x Mass of H) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g

(b) Molar mass of Sulphur molecule S₈ = (8 x Mass of S) = 8 x 32 = 256 g

(c) Molar mass of Phosphorus molecule, P₄ = 4 x Mass of P = 4 x 31 = 124 g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl = 1 + 35.5 = 36.5 g

(e) Molar mass of Nitric acid, HNO₃ = Mass of H + Mass of N + (3 x Mass of O) = 1 + 14 + (3 × 16) = 15 + 48 = 63 g

As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of products.

Sodium carbonate + acetic acid ⟶ Sodium acetate + carbon dioxide + water

5.3 g + 6 g ⟶ 8.2 g + 2.2 g + 0.9 g

⇒ 11.3 g ⟶ 11.3 g

As per the above reaction, L.H.S. = R.H.S. = 11.3 g

Hence, the observations are in agreement with the law of conservation of mass.

Given,

hydrogen : water = 1 : 8

So, for every 1 g of hydrogen, 8 g of oxygen is required

∴ For 3 g of hydrogen, oxygen required is $\dfrac{8}{1}$ x 3 = 24 g

Hence, 24 g of oxygen would be required for the complete reaction with 3 g of hydrogen gas.

One of the postulates of Dalton's atomic theory states that : 'Atoms cannot be created nor be destroyed in a chemical reaction'. This law is the result of law of conservation of mass.

The postulate of Dalton's atomic theory that can explain the law of definite proportions is : 'Relative number and kinds of atoms are equal in given compounds.'

Atomic mass unit is defined as 1⁄12 ^th the mass of an carbon atom C-12.

It is not possible to see an atom with naked eyes because:

1. Atoms are very small in size, measured in nanometres.
2. Except for atoms of noble gases, they do not exist independently.

(i) Sodium oxide:

Step 1 — Write each symbol with its valency

$\text{Na}^{1+} \phantom{\nearrow} \text{O}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{O}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Sodium oxide : $\bold{Na_2O}$

(ii) Aluminium chloride

Step 1 — Write each symbol with its valency

$\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Aluminium chloride : $\bold{AlCl_3}$

(iii) sodium sulphide

Step 1 — Write each symbol with its valency

$\text{Na}^{1+} \phantom{\nearrow} \text{S}^{2-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{S}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Sodium sulphide : $\bold{Na_2S}$

(iv) Magnesium hydroxide

Step 1 — Write each symbol with its valency

$\text{Mg}^{2+} \phantom{\nearrow} \text{OH}^{1-}$

Step 2 — Interchange the valencies

$\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \\[0.5em]$

Step 3 — Write the interchanged number, ignore equal numbers & hence the formula

Therefore, we get

Formula of Magnesium hydroxide : $\bold{Mg(OH)_2}$

(i) Al₂(SO₄)₃ — Aluminium sulphate

(ii) CaCl₂ — Calcium chloride

(iii) K₂SO₄ — Potassium sulphate

(iv) KNO₃ — Potassium nitrate

(v) CaCO₃ — Calcium carbonate

The chemical formula of a compound is a symbolic representation of its composition. It denotes in a compound, the number of atoms of each element present.

(i) In one molecule of H₂S, 2 atoms of hydrogen and 1 atom of sulphur are present. Hence, 3 atoms in total are present.

(ii) In one ion [PO₄^3-], 1 atom of phosphorus and 4 atoms of oxygen are present. Hence, 5 atoms in total are present.

1. The molecular mass of H₂ = 2 x atoms atomic mass of H = 2 x 1u = 2u

2. The molecular mass of O₂ = 2 x atoms atomic mass of O = 2 x 16u = 32u

3. The molecular mass of Cl₂ = 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

4. The molecular mass of CO₂ = atomic mass of C + (2 x atomic mass of O) = 12 + (2 × 16)u = 12 + 32 = 44u

5. The molecular mass of CH₄ = atomic mass of C + (4 x atomic mass of H) = 12 + (4 x 1) = 16u

6. The molecular mass of C₂H₆ = (2 x atomic mass of C) + (6 x atomic mass of H) = (2 x 12) + (6 x 1) = 24 + 6 = 30u

7. The molecular mass of C₂H₄ = (2 x atomic mass of C) + (4 x atomic mass of H) = (2 x 12) + (4 x 1) = 24 + 4 = 28u

8. The molecular mass of NH₃ = atomic mass of N + (3 x atomic mass of H) = 14 + (3 x 1) = 17u

9. The molecular mass of CH₃OH = atomic mass of C + (3 x atomic mass of H) + atomic mass of O + atomic mass of H = [12 + (3 × 1) + 16 + 1] = (12 + 3 + 17) = 32u

Given:

Atomic masses of Zn = 65u, Na = 23u, K = 39u, C = 12u, and O = 16u

The formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O = 65u + 16u = 81u

The formula unit mass of Na₂O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K₂CO₃ = 2 x Atomic mass of K + Atomic mass of C + 3 x Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u