ATOMS AND MOLECULES - Q&A

Questions Page 27

1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water

Explanation:
The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. This means the total mass of the reactants must equal the total mass of the products.

Step 1: Calculate Mass of Reactants
Reactants are Sodium carbonate and Ethanoic acid.
Mass of reactants = Mass of Sodium carbonate + Mass of Ethanoic acid
Mass of reactants = 5.3 g + 6 g = 11.3 g

Step 2: Calculate Mass of Products
Products are Sodium ethanoate, Carbon dioxide, and Water.
Mass of products = Mass of Sodium ethanoate + Mass of Carbon dioxide + Mass of Water
Mass of products = 8.2 g + 2.2 g + 0.9 g = 11.3 g

Conclusion:
Since the Mass of Reactants (11.3 g) is equal to the Mass of Products (11.3 g), these observations are in agreement with the law of conservation of mass.

2. Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

Explanation:
The law of constant proportions states that hydrogen and oxygen always combine in a fixed ratio of 1:8 by mass.

Step 1: Understand the Ratio
1 g of Hydrogen requires 8 g of Oxygen.

Step 2: Calculate for 3 g of Hydrogen
If 1 g Hydrogen → 8 g Oxygen
Then 3 g Hydrogen → 3 × 8 g Oxygen
Mass of Oxygen required = 24 g

3. Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

The postulate is: "Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction."

4. Which postulate of Dalton's atomic theory can explain the law of definite proportions?

The postulate is: "The relative number and kinds of atoms are constant in a given compound."

Questions Page 34

1. Calculate the molecular mass of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

Atomic masses: H=1 u, O=16 u, Cl=35.5 u, C=12 u, N=14 u

1. H₂: 2 × 1 = 2 u
2. O₂: 2 × 16 = 32 u
3. Cl₂: 2 × 35.5 = 71 u
4. CO₂: 1 × 12 + 2 × 16 = 12 + 32 = 44 u
5. CH₄: 1 × 12 + 4 × 1 = 12 + 4 = 16 u
6. C₂H₆: 2 × 12 + 6 × 1 = 24 + 6 = 30 u
7. C₂H₄: 2 × 12 + 4 × 1 = 24 + 4 = 28 u
8. NH₃: 1 × 14 + 3 × 1 = 14 + 3 = 17 u
9. CH₃OH: 1 × 12 + 3 × 1 + 1 × 16 + 1 × 1 = 12 + 3 + 16 + 1 = 32 u

2. Calculate the formula unit mass of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

1. ZnO: 65 + 16 = 81 u
2. Na₂O: (2 × 23) + 16 = 46 + 16 = 62 u
3. K₂CO₃: (2 × 39) + 12 + (3 × 16) = 78 + 12 + 48 = 138 u

Questions Page 35

1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Explanation:
1 mole of carbon atoms = 6.022 × 10²³ atoms.
Mass of 1 mole (6.022 × 10²³ atoms) = 12 g.

Step 1: Calculate mass of 1 atom
Mass of 1 atom = Mass of 1 mole / Avogadro's number
Mass of 1 atom = 12 / (6.022 × 10²³) g
Mass of 1 atom ≈ 1.99 × 10^-23 g

2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?

Explanation:
To find the number of atoms, we first need to find the number of moles. The element with more moles will have more atoms.

Step 1: Calculate moles of Sodium (Na)
Moles of Na = Given Mass / Molar Mass
Moles of Na = 100 / 23 ≈ 4.35 moles

Step 2: Calculate moles of Iron (Fe)
Moles of Fe = Given Mass / Molar Mass
Moles of Fe = 100 / 56 ≈ 1.78 moles

Conclusion:
Since 4.35 > 1.78, 100 grams of sodium has more atoms.

Exercises

1. A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

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

Step 1: Calculate Percentage of Boron
% Boron = (Mass of boron / Mass of compound) × 100
% Boron = (0.096 / 0.24) × 100 = 0.4 × 100 = 40%

Step 2: Calculate Percentage of Oxygen
% Oxygen = (Mass of oxygen / Mass of compound) × 100
% Oxygen = (0.144 / 0.24) × 100 = 0.6 × 100 = 60%

Answer: Boron: 40%, Oxygen: 60%

2. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Explanation:
First reaction: 3 g Carbon + 8 g Oxygen → 11 g Carbon dioxide.
This shows Carbon and Oxygen combine in a fixed ratio of 3:8 by mass.

Step 1: Analyze the second scenario
We have 3 g of Carbon and 50 g of Oxygen.
According to the fixed ratio (3:8), 3 g of Carbon can only react with exactly 8 g of Oxygen.

Step 2: Calculate Product
The remaining Oxygen (50 - 8 = 42 g) will be left unreacted.
Mass of Carbon dioxide formed = Mass of Carbon + Mass of Reacted Oxygen
Mass of Carbon dioxide = 3 g + 8 g = 11 g

Governing Law:
This is governed by the Law of Constant Proportions (or Law of Definite Proportions), which states that elements always combine in fixed ratios by mass.

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 charged unit.

Examples:
1. Ammonium ion (NH₄⁺)
2. Nitrate ion (NO₃^-)
3. Sulphate ion (SO₄^2-)
4. Carbonate ion (CO₃^2-)

4. Write the chemical formulae of the following.

(a) Magnesium chloride: MgCl₂
(b) Calcium oxide: CaO
(c) Copper nitrate: Cu(NO₃)₂
(d) Aluminium chloride: AlCl₃
(e) Calcium carbonate: CaCO₃

5. Give the names of the elements present in the following compounds.

(a) Quick lime (CaO): Calcium, Oxygen
(b) Hydrogen bromide (HBr): Hydrogen, Bromine
(c) Baking powder (NaHCO₃): Sodium, Hydrogen, Carbon, Oxygen
(d) Potassium sulphate (K₂SO₄): Potassium, Sulphur, Oxygen

6. Calculate the molar mass of the following substances.

(a) Ethyne, C₂H₂: (2 × 12) + (2 × 1) = 24 + 2 = 26 g/mol
(b) Sulphur molecule, S₈: 8 × 32 = 256 g/mol
(c) Phosphorus molecule, P₄: 4 × 31 = 124 g/mol
(d) Hydrochloric acid, HCl: 1 + 35.5 = 36.5 g/mol
(e) Nitric acid, HNO₃: 1 + 14 + (3 × 16) = 1 + 14 + 48 = 63 g/mol