Atoms and Molecules
NCERT Solutions • Class 9 Science • Chapter 3Chapter Exercises
1. A $0.24 \text{ g}$ sample of compound of oxygen and boron was found by analysis to contain $0.096 \text{ g}$ of boron and $0.144 \text{ g}$ of oxygen. Calculate the percentage composition of the compound by weight.
Formula: $\text{Percentage} = \frac{\text{Mass of Element}}{\text{Mass of Compound}} \times 100$
For Boron:
$\text{Mass of Boron} = 0.096 \text{ g}$
$\% \text{ of Boron} = \frac{0.096}{0.24} \times 100 =$ $40\%$.
For Oxygen:
$\text{Mass of Oxygen} = 0.144 \text{ g}$
$\% \text{ of Oxygen} = \frac{0.144}{0.24} \times 100 =$ $60\%$.
For Boron:
$\text{Mass of Boron} = 0.096 \text{ g}$
$\% \text{ of Boron} = \frac{0.096}{0.24} \times 100 =$ $40\%$.
For Oxygen:
$\text{Mass of Oxygen} = 0.144 \text{ g}$
$\% \text{ of Oxygen} = \frac{0.144}{0.24} \times 100 =$ $60\%$.
2. When $3.0 \text{ g}$ of carbon is burnt in $8.00 \text{ g}$ oxygen, $11.00 \text{ g}$ of carbon dioxide is produced. What mass of carbon dioxide will be formed when $3.00 \text{ g}$ of carbon is burnt in $50.00 \text{ g}$ of oxygen? Which law of chemical combination will govern your answer?
Answer: $11.00 \text{ g}$ of Carbon Dioxide will be formed.
Explanation: Carbon and Oxygen react in a fixed ratio of $3:8$ by mass to form Carbon Dioxide.
Even if $50.00 \text{ g}$ of oxygen is available, only $8.00 \text{ g}$ will react with $3.00 \text{ g}$ of carbon. The remaining $42.00 \text{ g}$ of oxygen will remain unreacted.
Law: This is governed by the Law of Definite Proportions (or Constant Proportions).
Explanation: Carbon and Oxygen react in a fixed ratio of $3:8$ by mass to form Carbon Dioxide.
Even if $50.00 \text{ g}$ of oxygen is available, only $8.00 \text{ g}$ will react with $3.00 \text{ g}$ of carbon. The remaining $42.00 \text{ g}$ of oxygen will remain unreacted.
Law: This is governed by the Law of Definite Proportions (or Constant Proportions).
3. What are polyatomic ions? Give examples.
Definition: A group of atoms carrying a charge is known as a polyatomic ion. These atoms act as a single unit with a fixed net charge.
Examples:
• Nitrate ($NO_3^-$)
• Hydroxide ($OH^-$)
• Sulphate ($SO_4^{2-}$)
• Ammonium ($NH_4^+$)
Examples:
• Nitrate ($NO_3^-$)
• Hydroxide ($OH^-$)
• Sulphate ($SO_4^{2-}$)
• Ammonium ($NH_4^+$)
4. Write the chemical formulae of the following:
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate.
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate.
- (a) Magnesium chloride: $MgCl_2$
- (b) Calcium oxide: $CaO$
- (c) Copper nitrate: $Cu(NO_3)_2$
- (d) Aluminium chloride: $AlCl_3$
- (e) Calcium carbonate: $CaCO_3$
5. Give the names of the elements present in the following compounds:
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.
- (a) Quick lime ($CaO$): Calcium, Oxygen
- (b) Hydrogen bromide ($HBr$): Hydrogen, Bromine
- (c) Baking powder ($NaHCO_3$): Sodium, Hydrogen, Carbon, Oxygen
- (d) Potassium sulphate ($K_2SO_4$): Potassium, Sulphur, Oxygen
6. Calculate the molar mass of the following substances.
(a) Ethyne, $C_2H_2$:
$= 2 \times 12 + 2 \times 1 = 24 + 2 =$ $26 \text{ g/mol}$
(b) Sulphur molecule, $S_8$:
$= 8 \times 32 =$ $256 \text{ g/mol}$
(c) Phosphorus molecule, $P_4$:
$= 4 \times 31 =$ $124 \text{ g/mol}$
(d) Hydrochloric acid, $HCl$:
$= 1 + 35.5 =$ $36.5 \text{ g/mol}$
(e) Nitric acid, $HNO_3$:
$= 1 + 14 + (3 \times 16) = 1 + 14 + 48 =$ $63 \text{ g/mol}$
$= 2 \times 12 + 2 \times 1 = 24 + 2 =$ $26 \text{ g/mol}$
(b) Sulphur molecule, $S_8$:
$= 8 \times 32 =$ $256 \text{ g/mol}$
(c) Phosphorus molecule, $P_4$:
$= 4 \times 31 =$ $124 \text{ g/mol}$
(d) Hydrochloric acid, $HCl$:
$= 1 + 35.5 =$ $36.5 \text{ g/mol}$
(e) Nitric acid, $HNO_3$:
$= 1 + 14 + (3 \times 16) = 1 + 14 + 48 =$ $63 \text{ g/mol}$
Group Activity: Writing Formulae
Example 1:
Using placards: Right hand (Symbol), Left hand (Valency).
Criss-cross valencies to find the formula.
Example 2 (Model):
Sodium Sulphate: $Na^+$ has valency 1, $SO_4^{2-}$ has valency 2.
Two $Na^+$ ions fix onto one $SO_4^{2-}$ ion.
Formula: $Na_2SO_4$.
Do It Yourself: Sodium Phosphate ($Na_3PO_4$)
Sodium ($Na^+$) valency = 1. Phosphate ($PO_4^{3-}$) valency = 3.
Requires 3 Sodium ions for 1 Phosphate ion.
Example 1:
Using placards: Right hand (Symbol), Left hand (Valency).
Criss-cross valencies to find the formula.
Example 2 (Model):
Sodium Sulphate: $Na^+$ has valency 1, $SO_4^{2-}$ has valency 2.
Two $Na^+$ ions fix onto one $SO_4^{2-}$ ion.
Formula: $Na_2SO_4$.
Do It Yourself: Sodium Phosphate ($Na_3PO_4$)
Sodium ($Na^+$) valency = 1. Phosphate ($PO_4^{3-}$) valency = 3.
Requires 3 Sodium ions for 1 Phosphate ion.