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## Chapter 3 Atoms and Molecules

Intext Questions
Question 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

Na₂CO₃ + CH₃COOH → C₂H₃NaO₂ + CO₂ + H₂O
(5.3 g)         (6 g)               (8.2 g)     (2.2 g) (0.9 g)
According the Law of conservation of mass, the mass can’t be created nor destroyed in a chemical reaction.
5.3 + 6 = 8.2 + 2.2 + 0.9
11.3 = 11.3
LHS = RHS
Hence, the above chemical reaction is in agreement with the law of conservation of mass.

Question 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?

Given
Hydrogen and oxygen combine in the ratio of 1:8. For every 1 g of hydrogen, 8 g of oxygen is required.
So, for 3 g of hydrogen, the mass of oxygen required will be 3 × 8 = 24 g.
Therefore, 24 g of oxygen gas would be required to react completely with 3 g of hydrogen gas.

Question 3: Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
The second postulate of Dalton’s atomic theory which states that, “Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction” is the result of law of conservation of mass.

Question 4: Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
The sixth postulate of Dalton’s atomic theory which states that, “The relative number and kinds of atoms are constant in a given compound” is the result of law of definite proportions.

Question 5: Define the atomic mass unit.
An atomic mass unit is a unit of mass used to express atomic and molecular weights; equal to 1/12th of the mass of an atom of carbon-12.

Question 6: Why is it not possible to see an atom with naked eyes?
Atoms are very small in size. The size of atoms is measured in nanometers. Also, the atoms of an element do not exist independently. Hence, it is not possible to see an atom with naked eye.

Question 7: Write down the formulae of
i) sodium oxide
ii) aluminium chloride
iii) sodium suphide
iv) magnesium hydroxide

Question 8: Write down the names of compounds represented by the following formulae:
i) Al₂(SO₄)₃
ii) CaCl₂
iii) K₂SO₄
iv) KNO₃
v) CaCO₃

i) Aluminium sulfate
ii) Calcium chloride
iii) Potassium sulfate
iv) Potassium nitrate
v) Calcium carbonate

Question 9: What is meant by the term chemical formula?
The chemical formula of a compound is a symbolic representation of its composition.

Question 10: How many atoms are present in a
i) H₂S molecule and
ii) PO₄³⁻ ion?

i) H₂S mole has 1 sulphur and 2 hydrogen atoms. So, there are 3 atoms in H₂S molecule.
ii) PO₄³⁻ ion has 4 oxygen atoms and 1 phosphorus atom. So, there are 5 atoms in PO₄³⁻ ion.

Question 11: Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

i) Molecular mass of H₂ = 2 × atomic mass of H
= 2 × 1 = 2 u

ii) Molecular mass of O₂ = 2 × atomic mass of O
= 2 × 16 = 32 u

iii) Molecular mass of Cl₂ = 2 × atomic mass of Cl
= 2 × 35.5 = 71 u

iv) Molecular mass of CO₂ = 1 × atomic mass of C + 2 × atomic mass of O
= (1 × 12) + (2 × 16)
= 12 + 32
= 44 u

v) Molecular mass of CH₄ = 1 × atomic mass of C + 4 × atomic mass of H
= (1 × 12) + (4 × 1)
= 12 + 4
= 16 u

vi) Molecular mass of C₂H₆ = 2 × atomic mass of C + 6 × atomic mass of H
= (2 × 12)  + (6 × 1)
= 24 + 6
= 30 u

vii) Molecular mass of C₂H₄  = 2 × atomic mass of C + 4 × atomic mass of H
= (2 × 12) + (4 × 1)
= 24 + 4
= 28 u

viii) Molecular mass of NH₃ = 1 × atomic mass of N + 3 × atomic mass of H
= (1 × 14) + (3 × 1)
= 14 + 3
= 17 u

ix) Molecular mass of CH₃OH = 1 × atomic mass of C + 4 × atomic mass of H + 1 × atomic mass of O
= (1 × 12) + (4 × 1) + (1 × 16)
= 12 + 4 + 16
= 32 u

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

Fomula unit mass of ZnO = 1 × Atomic mass of Zn + 1 × Atomic mass of O
= (1 × 65) + (1 × 16)
= 65 + 16
= 81 u

Formula unit mass of Na₂O = 2 × Atomic mass of Na + 1 × Atomic mass of O
= (2 × 23) + (1 × 16)
= 46 + 16
= 62 u

Formula unit mass of K₂CO₃ = 2 × Atomic mass of K + 1 × Atomic mass of C + 3 × Atomic mass of O
= (2 × 39) + (1 × 12) + (3 × 16)
= 78 + 12 + 48
= 138 u

Question 13: If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?

Given
1 mole of Carbon atoms = 12 g
1 mole of Carbon atoms = 6.022 × 10²³ atoms of C
The molecular mass of carbon atoms = an atom of carbon mass = 12 g
Mass of 1 carbon atom = 12/6.022 × 10²³
= 1.99 × 10⁻²³ g

Question 14: 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)?

In 100 g of Na
Given mass (m) = 100 g
Molar mass (M) = 23 u
Avogadro number (N₀) = 6.022 × 10²³
Number of particles = ?
N = m × N₀/M
N = 100 × 6.022 × 10²³/23
N = 26.18 × 10²³ atoms

In 100 g of Fe
Given mass (m) = 100 g
Molar mass (M) = 56 u
Avogadro number (N₀) = 6.022 × 10²³
Number of particles (N) = ?
N = m × N₀/M
N = 100 × 6.022 × 10²³/56
N = 10.75 × 10²³ atoms
Therefore, the number of atoms is more in 100 g of Na than 100 g of Fe.

Exercise Questions
Question 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
Weight of compound of oxygen and boron = 0.24 g
Weight of boron in sample = 0.096 g
Weight of oxygen in sample = 0.144 g
% of boron = Mass of boron/Mass of compound × 100
= 0.096/0.24 × 100
= 40%
% of oxygen = Mass of oxygen/Mass of compound × 100
= 0.144/0.24 × 100
= 60%

Question 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?

C      +    O₂ → CO₂
(3 g)    (8 g)     (11 g)

C      +    O₂ → CO₂
(3 g)    (50 g)     (53 g)
The Law of Constant Proportion, which states that “In a chemical substance the elements are always present in definite proportions by mass” governs this answer.

Question 3: What are polyatomic ions? Give examples.
A group of atoms carrying a charge is known as a polyatomic ion. Example: OH⁻ (hydroxide), NO₃⁻ (nitrate), HCO₃⁻ (Hydrogen Carbonate) etc.

Question 4: Write the chemical formulae of the following.
a) Magnesium chloride
b) Calcium oxide
c) Copper nitrate
d) Aluminium chloride
e) Calcium carbonate

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

Question 6: Calculate the molar mass of the following substances.
a) Ethyne, C₂H₂
b) Sulphur molecule, S₈
c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
d) Hydrochloric acid, HCl
e) Nitric acid, HNO₃

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

b) Molar mass of Sulphur molecule = 8 × Atomic mass of S
= 8 × 32
= 256 g

c) Molar mass of Phosphorus molecule = 4 × Atomic mass of P
= 4 × 31
= 124 g

d) Molar mass of Hydrochloric acid = 1 × Atomic mass of H + 1 × Atomic mass of Cl
= (1 × 1) + (1 × 35.5)
= 1 + 35.5
= 36.5 g

e) Molar mass of Nitric acid = 1 × Atomic mass of H + 1 × Atomic mass of N + 3 × Atomic mass of O
= (1 × 1) + (1 × 14) + (3 × 16)
= 1 + 14 + 48
= 63 g

Question 7: What is the mass of
a) 1 mole of nitrogen atoms?
b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
c) 10 moles of sodium sulphite (Na₂SO₃)?

a) Given
Number of moles = 1
Atomic mass of Nitrogen = 14 u
Mass (m) = molar mass (M) × number of moles (n)
= 14 × 1
= 14 g

b) Given
Number of moles = 4
Atomic mass of Nitrogen = 27 u
Mass (m) = molar mass (M) × number of moles (n)
= 27 × 4
= 108 g

c) Given
Number of moles = 10
Molecular mass of Na₂SO₃ = 2 × Atomic mass of Na + 1 × Atomic mass of S + 3 × Atomic mass of O
= (2 × 23) + (1 × 32) + (3 × 16)
= 46 + 32 + 48
= 126 g
Mass (m) = molar mass (M) × number of moles (n)
= 126 × 10
= 1260 g

Question 8: Convert into mole.
a) 12 g of oxygen gas
b) 20 g of water
c) 22 g of carbon dioxide.

a) Given
Molecular mass of O₂ = 2 × Atomic mass of O = 2 × 16 = 32 u
Given Mass of O₂ = 12 g
Number of moles (n) = given mass (m)/Molar mass(M)
= 12/32
= 0.375 moles

b) Given
Molecular mass of H₂O = 2 × Atomic mass of H + 1 × Atomic mass of O = (2 × 1) + (1 × 16) = 2 + 16 = 18 u
Given Mass of H₂O = 20 g
Number of moles (n) = given mass (m)/Molar mass(M)
= 20/18
= 1.11 moles

c) Given
Molecular mass of CO₂ = 1 × Atomic mass of C + 2 × Atomic mass of O = (1 × 12) + (2 × 16) = 12 + 32 = 44 u
Given Mass of CO₂ = 22 g
Number of moles (n) = given mass (m)/Molar mass(M)
= 22/44
= 0.5 moles

Question 9: What is the mass of
a) 0.2 mole of oxygen atoms?
b) 0.5 mole of water molecules?

a) Given
Number of moles = 0.2
Atomic mass of O = 16 u
Mass (m) = Number of moles (n) × Molar mass (M)
= 0.2 × 16
= 3.2 g

b) Given
Number of moles = 0.5
Molecular mass of H₂O = 2 × Atomic mass of H + 1 × Atomic mass of O = (2 × 1) + (1 × 16) = 2 + 16 = 18 u
Mass (m) = Number of moles (n) × Molar mass (M)
= 0.5 × 18
= 9 g

Question 10: Calculate the number of molecules of sulphur (S₈) present in 16 g of solid sulphur.

Given
Molecular mass of S₈ = atomic mass of S × 8 = 32 × 8 = 256 u
Number of moles = ?
Given mass = 16 g
Number of moles = given mass (m)/Molar mass (M)
= 16/256
= 0.0625 moles

Number of molecules of sulphur in 16 g of solid sulphur = ?
Number of molecules = Number of moles × Avogadro number
= 0.0625 × 6.022 × 10²³
= 0.3763 × 10²³
= 3.763 × 10²² molecules of S

Question 11: Calculate the number of aluminium ions present in 0.051 g of aluminium oxide. (Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)