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Atoms and Molecules – CBSE NCERT Study Resources

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Atoms and Molecules

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9th

9th - Science

Atoms and Molecules

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Overview of the Chapter: Atoms and Molecules

This chapter introduces the fundamental concepts of atoms and molecules, which are the building blocks of matter. Students will learn about the laws of chemical combination, atomic theory, and the structure of atoms. The chapter also covers the concept of molecules, molecular masses, and the mole concept, which are essential for understanding chemical reactions and stoichiometry.

Atom: The smallest particle of an element that retains the chemical properties of that element.

Molecule: A group of two or more atoms chemically bonded together, representing the smallest unit of a compound.

Laws of Chemical Combination

The chapter begins with the laws governing chemical reactions:

  • Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction.
  • Law of Definite Proportions: A chemical compound always contains the same elements in the same proportion by mass.

Dalton's Atomic Theory

John Dalton proposed the atomic theory, which includes the following postulates:

  • All matter is composed of indivisible particles called atoms.
  • Atoms of the same element are identical in mass and properties.
  • Compounds are formed by the combination of atoms in simple whole-number ratios.

Structure of an Atom

Atoms consist of three subatomic particles:

  • Protons: Positively charged particles found in the nucleus.
  • Neutrons: Neutral particles found in the nucleus.
  • Electrons: Negatively charged particles orbiting the nucleus.

Molecules and Ions

Molecules can be formed by the combination of atoms of the same or different elements. Ions are charged particles formed when atoms gain or lose electrons.

Ion: An atom or molecule with a net electric charge due to the loss or gain of electrons.

Molecular Mass and Mole Concept

The molecular mass is the sum of the atomic masses of all atoms in a molecule. The mole concept helps in quantifying substances in chemical reactions.

Mole: A unit of measurement used to express amounts of a chemical substance, defined as 6.022 × 10²³ particles (Avogadro's number).

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define atom.
Answer:

Smallest particle of an element.

Question 2:
Name the symbol of gold.
Answer:

Au

Question 3:
What is the atomicity of oxygen gas (O2)?
Answer:

2

Question 4:
Give the chemical formula of water.
Answer:

H2O

Question 5:
What is the molecular mass of CO2?
Answer:

44 u

Question 6:
Name the law that states mass is conserved in reactions.
Answer:

Law of conservation of mass

Question 7:
What is the valency of chlorine?
Answer:

1

Question 8:
Give an example of a triatomic molecule.
Answer:

Ozone (O3)

Question 9:
What is the charge of an electron?
Answer:

Negative (-1)

Question 10:
Name the scientist who proposed atomic theory.
Answer:

John Dalton

Question 11:
What is the formula unit mass of NaCl?
Answer:

58.5 u

Question 12:
Define mole.
Answer:

SI unit for amount of substance.

Question 13:
Give the symbol of sodium.
Answer:

Na

Question 14:
What is the atomic number of carbon?
Answer:

6

Question 15:
Define atomic mass unit (amu).
Answer:

The atomic mass unit (amu) is a standard unit of mass used to express the masses of atoms and molecules.
1 amu is defined as one-twelfth the mass of a carbon-12 atom.

Question 16:
What is the law of constant proportions?
Answer:

The law of constant proportions states that a chemical compound always contains the same elements combined together in the same proportion by mass, regardless of its source or method of preparation.

Question 17:
Name the scientist who proposed the atomic theory.
Answer:

John Dalton proposed the atomic theory in 1808, which laid the foundation for modern chemistry.

Question 18:
What is the molecular mass of water (H2O)?
Answer:

The molecular mass of water (H2O) is calculated as:
Mass of 2 Hydrogen atoms = 2 × 1 = 2 amu
Mass of 1 Oxygen atom = 16 amu
Total molecular mass = 2 + 16 = 18 amu.

Question 19:
Give an example of a diatomic molecule.
Answer:

An example of a diatomic molecule is Oxygen (O2), which consists of two oxygen atoms chemically bonded together.

Question 20:
What is the symbol for the element Gold?
Answer:

The symbol for the element Gold is Au, derived from its Latin name Aurum.

Question 21:
How many atoms are present in a molecule of sulfur (S8)?
Answer:

A molecule of sulfur (S8) consists of 8 atoms of sulfur bonded together in a ring structure.

Question 22:
What is the valency of an element?
Answer:

The valency of an element is the combining capacity of its atoms, determined by the number of electrons it can lose, gain, or share to achieve a stable electron configuration.

Question 23:
Calculate the formula unit mass of NaCl (Atomic masses: Na = 23 u, Cl = 35.5 u).
Answer:

The formula unit mass of NaCl is calculated as:
Mass of Na = 23 u
Mass of Cl = 35.5 u
Total mass = 23 + 35.5 = 58.5 u.

Question 24:
What is the difference between an atom and a molecule?
Answer:

An atom is the smallest unit of an element that retains its properties, while a molecule is a group of two or more atoms chemically bonded together, which can be of the same or different elements.

Question 25:
Name the polyatomic ion present in sulfuric acid (H2SO4).
Answer:

The polyatomic ion present in sulfuric acid (H2SO4) is the sulfate ion (SO42-).

Question 26:
What is the Avogadro number?
Answer:

The Avogadro number is 6.022 × 1023, which represents the number of atoms, molecules, or ions in one mole of a substance.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Calculate the molecular mass of H2SO4.
Answer:

Molecular mass of H2SO4 is calculated as:
2 × (Atomic mass of H) + 1 × (Atomic mass of S) + 4 × (Atomic mass of O)
= 2 × 1 + 1 × 32 + 4 × 16
= 2 + 32 + 64
= 98 amu.

Question 2:
What is the symbol of the element Gold?
Answer:

The symbol of Gold is Au, derived from its Latin name Aurum.

Question 3:
How many atoms are present in a molecule of Ozone (O3)?
Answer:

A molecule of Ozone (O3) contains 3 oxygen atoms chemically bonded together.

Question 4:
What is the valency of Nitrogen in NH3?
Answer:

The valency of Nitrogen in NH3 is 3, as it forms three bonds with hydrogen atoms.

Question 5:
What is the chemical formula of Magnesium Chloride?
Answer:

The chemical formula of Magnesium Chloride is MgCl2, as magnesium has a valency of 2 and chlorine has a valency of 1.

Question 6:
How many moles are there in 36 grams of water (H2O)?
Answer:

Molar mass of H2O = 2 × 1 + 16 = 18 g/mol.
Number of moles = Given mass / Molar mass
= 36 g / 18 g/mol
= 2 moles.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Define atomic mass unit (u) and explain its significance in chemistry.
Answer:

The atomic mass unit (u) is a standard unit used to express the masses of atoms and molecules.
1 u is defined as one-twelfth the mass of a carbon-12 atom.
Its significance lies in:

  • Providing a unified scale to compare atomic masses.
  • Simplifying calculations in chemical reactions.
  • Helping determine molecular formulas and stoichiometry.

Question 2:
Differentiate between molecule of an element and molecule of a compound with examples.
Answer:

Molecule of an element: Contains two or more identical atoms bonded together.
Example: O2 (oxygen), N2 (nitrogen).
Molecule of a compound: Contains two or more different atoms chemically combined.
Example: H2O (water), CO2 (carbon dioxide).
Key difference: Elements have same atoms; compounds have different atoms.

Question 3:
Calculate the molecular mass of sulphuric acid (H2SO4).
Answer:

Molecular mass = Sum of atomic masses of all atoms.

Step 1: Atomic mass of H = 1 u (2 atoms → 2 × 1 = 2 u).
Step 2: Atomic mass of S = 32 u (1 atom → 32 u).
Step 3: Atomic mass of O = 16 u (4 atoms → 4 × 16 = 64 u).
Step 4: Total = 2 + 32 + 64 = 98 u.

Question 4:
Explain Law of Conservation of Mass with an example.
Answer:

The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction.
Example: Burning magnesium ribbon in air.

  • Mass of magnesium + oxygen = mass of magnesium oxide formed.
  • Total mass remains unchanged before and after the reaction.

This law is foundational for balancing chemical equations.

Question 5:
What is the chemical formula of a compound? Write the steps to derive the formula of calcium chloride.
Answer:

A chemical formula represents the types and numbers of atoms in a compound.

Steps for calcium chloride:
Step 1: Identify valencies: Calcium (Ca2+), Chlorine (Cl-).
Step 2: Cross-multiply valencies: Ca1Cl2.
Step 3: Simplify if needed (not required here).
Final formula: CaCl2.

Question 6:
Why do atoms form molecules? Explain with reference to noble gases.
Answer:

Atoms form molecules to achieve stability by completing their outermost electron shell.

  • Most atoms bond to attain octet (8 electrons) or duplet (2 electrons) configuration.
  • Noble gases (e.g., He, Ne) already have stable electron configurations, so they rarely form molecules.

Example: Hydrogen (H2) shares electrons to achieve stability like helium.

Question 7:
What is the mole concept? How is it useful in chemistry?
Answer:

The mole concept is a method to count particles (atoms, molecules, ions) using Avogadro's number (6.022 × 1023).

1 mole = 6.022 × 1023 particles = Gram molecular mass.

Usefulness:

  • Simplifies measurement of substances in chemical reactions.
  • Helps relate macroscopic quantities to microscopic particles.
  • Essential for stoichiometric calculations in labs and industries.

Question 8:
Describe the postulates of Dalton's atomic theory relevant to modern chemistry.
Answer:

Key postulates:

  • All matter is made of tiny, indivisible particles called atoms.
  • Atoms of the same element are identical in mass and properties.
  • Atoms combine in simple whole-number ratios to form compounds.
  • Atoms cannot be created or destroyed in chemical reactions (modified by modern discoveries).

Modern relevance:
  • Foundation for understanding chemical bonding.
  • Basis for the periodic table and stoichiometry.
  • Though some postulates are revised (e.g., atoms are divisible), the core ideas remain vital.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Define atomic mass and molecular mass. How are they calculated? Give one NCERT example and a real-world application.
Answer:
Concept Overview

Atomic mass is the mass of an atom, while molecular mass is the sum of atomic masses in a molecule. We studied that atomic mass is measured in atomic mass units (u).

Process Explanation
  • For atomic mass: Carbon-12 is taken as a reference (12 u).
  • For molecular mass: Add atomic masses of all atoms. Example: H2O = 2(1) + 16 = 18 u.
Real-world Application

Molecular mass helps in calculating medicine doses, like paracetamol (C8H9NO2 = 151 u).

Question 2:
Explain law of conservation of mass with an experiment. How does it apply to burning magnesium?
Answer:
Concept Overview

The law states mass remains constant in a chemical reaction. Our textbook shows this with barium chloride and sodium sulphate.

Process Explanation
  • Experiment: React BaCl2 + Na2SO4 → BaSO4 + 2NaCl. Total mass stays same.
  • For magnesium: 2Mg + O2 → 2MgO. Mass of Mg + O2 = Mass of MgO.
Real-world Application

Used in industries to balance chemical equations for manufacturing, like cement production.

Question 3:
What is a mole? Calculate the number of moles in 90g of water (H2O).
Answer:
Concept Overview

A mole is a unit representing 6.022 × 1023 particles (atoms/molecules). We studied it connects mass to number of particles.

Process Explanation
  • Step 1: Find molecular mass of H2O = 18 u.
  • Step 2: Moles = Given mass/Molar mass = 90/18 = 5 moles.
Real-world Application

Chemists use moles to prepare solutions, like saline (0.9% NaCl) in hospitals.

Question 4:
Differentiate between empirical and molecular formula using glucose (C6H12O6) as an example.
Answer:
Concept Overview

Empirical formula shows simplest ratio, while molecular formula gives actual atom counts. Our textbook uses glucose.

Process Explanation
  • Glucose molecular formula: C6H12O6.
  • Empirical formula: Divide by 6 → CH2O.
Real-world Application

Used in analyzing fuels. Example: Ethane (C2H6) has empirical formula CH3.

Question 5:
Describe Dalton’s atomic theory and its limitations with modern examples.
Answer:
Concept Overview

Dalton proposed atoms are indivisible and combine in fixed ratios. We studied this in Chapter 3.

Process Explanation
  • Key points: Atoms of same element are identical, form compounds.
  • Limitations: Isotopes exist (e.g., Cl-35, Cl-37), atoms can be divided (nuclear reactions).
Real-world Application

Modern atomic models explain radioactivity (e.g., uranium decay), which Dalton’s theory couldn’t.

Question 6:
Define atomic mass unit (u) and explain how it is used to express the mass of atoms and molecules. Provide an example from NCERT.
Answer:
Concept Overview

An atomic mass unit (u) is a standard unit for measuring atomic and molecular masses. It is defined as one-twelfth the mass of a carbon-12 atom.

Process Explanation

We studied that 1 u equals 1.66 × 10⁻²⁷ kg. It simplifies comparing masses of atoms, like oxygen (16 u) or hydrogen (1 u).

Real-world Application

Our textbook shows water (H₂O) has a molecular mass of 18 u (2 × 1 u + 16 u). This helps in chemical calculations.

Question 7:
Describe Dalton’s atomic theory and list its postulates. How does it explain the law of conservation of mass?
Answer:
Concept Overview

Dalton’s atomic theory states that matter is made of indivisible atoms.

Process Explanation
  • Atoms of an element are identical.
  • Atoms combine in fixed ratios.

Our textbook explains it supports the law of conservation of mass as atoms rearrange but aren’t destroyed.

Real-world Application

In burning wood, mass remains constant as atoms form new compounds like CO₂.

Question 8:
What is a mole? Calculate the number of moles in 36 g of water (H₂O). Use NCERT concepts.
Answer:
Concept Overview

A mole is a unit representing 6.022 × 10²³ particles (atoms/molecules).

Process Explanation

We studied that 1 mole of water (H₂O) has a mass of 18 g (2 + 16).

Real-world Application

For 36 g of water, moles = mass/molar mass = 36/18 = 2 moles. This helps in lab measurements.

Question 9:
Explain chemical formula with an example. How is it derived for magnesium chloride (MgCl₂)?
Answer:
Concept Overview

A chemical formula represents elements and their ratios in a compound.

Process Explanation

Our textbook shows MgCl₂ forms as Mg²⁺ and Cl⁻ ions combine in a 1:2 ratio for neutrality.

Real-world Application

Common salt (NaCl) has a 1:1 ratio. Formulas help predict reactions.

Question 10:
Differentiate between atoms and molecules. Give one NCERT example for each.
Answer:
Concept Overview

Atoms are the smallest units of elements, while molecules are groups of bonded atoms.

Process Explanation
  • Example of atom: Helium (He).
  • Example of molecule: Oxygen (O₂).
Real-world Application

Our textbook shows water (H₂O) as a molecule, while gold (Au) is atomic.

Question 11:
Explain the Law of Conservation of Mass with an example. How does this law support the concept of atoms and molecules?
Answer:

The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants is always equal to the total mass of the products.


Example: When calcium carbonate (CaCO3) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO2). If 100g of CaCO3 is heated, it produces 56g of CaO and 44g of CO2. The total mass remains conserved (100g = 56g + 44g).


This law supports the concept of atoms and molecules because it shows that atoms are simply rearranged during a reaction, not created or destroyed. The same number and types of atoms exist before and after the reaction, forming new molecules while conserving mass.


Additional Insight: This principle is fundamental in balancing chemical equations, ensuring the same number of atoms of each element on both sides.

Question 12:
Define molecular mass and formula unit mass. Calculate the molecular mass of water (H2O) and the formula unit mass of sodium chloride (NaCl).
Answer:

Molecular mass is the sum of the atomic masses of all atoms in a molecule, while formula unit mass is the sum of atomic masses of all atoms in a formula unit of an ionic compound.


Calculation for H2O:
Atomic mass of Hydrogen (H) = 1 u
Atomic mass of Oxygen (O) = 16 u
Molecular mass of H2O = (2 × 1) + 16 = 18 u


Calculation for NaCl:
Atomic mass of Sodium (Na) = 23 u
Atomic mass of Chlorine (Cl) = 35.5 u
Formula unit mass of NaCl = 23 + 35.5 = 58.5 u

Question 13:
Differentiate between atoms and molecules with suitable examples. Why are molecules of noble gases considered monoatomic?
Answer:

Difference between atoms and molecules:

  • Atoms are the smallest indivisible particles of an element (e.g., Oxygen atom (O)).
  • Molecules are formed when two or more atoms combine chemically (e.g., Oxygen molecule (O2)).

Noble gases like Helium (He) and Neon (Ne) exist as single atoms because they have a stable electronic configuration. They do not form bonds with other atoms, so their molecules are monoatomic.

Question 14:
Define atomic mass unit (u) and explain how it is used to express the mass of atoms and molecules. Provide an example to illustrate your answer.
Answer:

The atomic mass unit (u) is a standard unit of mass used to express the masses of atoms and molecules. It is defined as one-twelfth the mass of a carbon-12 atom, which is approximately 1.66 × 10-24 grams.


This unit is convenient because atoms are extremely tiny, and their masses in grams are very small numbers. By using atomic mass units, we can compare the masses of different atoms and molecules easily. For example:


  • The mass of a hydrogen atom is about 1 u.
  • The mass of an oxygen atom is about 16 u.

When calculating the mass of molecules, we add up the atomic masses of all the atoms in the molecule. For instance, a water molecule (H2O) has a mass of 2 u (from hydrogen) + 16 u (from oxygen) = 18 u.

Question 15:
Describe the law of constant proportions with a suitable example. How does this law support the concept of chemical compounds?
Answer:

The law of constant proportions, also known as the law of definite proportions, states that a chemical compound always contains the same elements combined together in the same proportion by mass, regardless of its source or method of preparation.


For example, water (H2O) always consists of hydrogen and oxygen in a fixed mass ratio of 1:8. This means:


  • 1 gram of hydrogen combines with 8 grams of oxygen to form 9 grams of water.
  • No matter where the water comes from (rain, river, or lab), this ratio remains constant.

This law supports the concept of chemical compounds because it proves that compounds have a fixed composition, distinguishing them from mixtures. It also helps in determining the purity of a substance. If the mass ratio differs, the substance is either impure or not the expected compound.

Question 16:
Explain the Law of Conservation of Mass with an example. How does this law support the concept of balanced chemical equations?
Answer:

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 before the reaction is equal to the total mass of the products after the reaction.


Example: When calcium carbonate (CaCO3) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO2). If 100g of CaCO3 decomposes, the total mass of CaO (56g) and CO2 (44g) will still be 100g, proving the law.


This law supports balanced chemical equations because the number of atoms of each element must be equal on both sides of the equation. For example, in the reaction:
2H2 + O2 → 2H2O,
the total mass of hydrogen (4g) and oxygen (32g) equals the mass of water (36g) formed, ensuring the equation is balanced.


Value-added insight: This law is fundamental in stoichiometry, helping scientists predict the quantities of substances involved in reactions without any loss of mass.

Question 17:
Define mole and explain its significance in chemistry. Calculate the number of moles in 22g of carbon dioxide (CO2). (Given: Atomic masses - C = 12u, O = 16u)
Answer:

The mole is the SI unit for measuring the amount of a substance. One mole contains exactly 6.022 × 1023 particles (atoms, molecules, or ions), known as Avogadro's number.

Significance:

  • It helps relate microscopic particles to measurable quantities.
  • Enables calculation of reactants and products in chemical reactions.

To find moles in 22g of CO2:
Step 1: Calculate molar mass of CO2
C = 12u, O = 16u → CO2 = 12 + (16 × 2) = 44g/mol
Step 2: Use formula: Moles = Given mass / Molar mass
Moles = 22g / 44g/mol = 0.5 moles

Question 18:
Explain the Law of Conservation of Mass with an example. How does this law apply to a chemical reaction?
Answer:

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 before the reaction is equal to the total mass of the products after the reaction.


Example: When calcium carbonate (CaCO3) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO2).
Reactants: CaCO3 (100 g)
Products: CaO (56 g) + CO2 (44 g)
Total mass before (100 g) = Total mass after (56 g + 44 g = 100 g).


Application in Chemical Reactions:

  • Atoms are rearranged, not created or destroyed.
  • The number and type of atoms remain the same on both sides.
  • This law is the foundation for balancing chemical equations.

Additional Insight: This law was experimentally verified by Antoine Lavoisier, who used precise measurements to show mass conservation in reactions.

Question 19:
Explain the Law of Conservation of Mass with an example. How does this law support the concept of balanced chemical equations?
Answer:

The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. The total mass of the reactants is always equal to the total mass of the products.


Example: When calcium carbonate (CaCO3) is heated, it decomposes into calcium oxide (CaO) and carbon dioxide (CO2). If 100g of CaCO3 decomposes, the total mass of CaO and CO2 produced will also be 100g.


This law supports the concept of balanced chemical equations because the number of atoms of each element must be the same on both sides of the equation, ensuring mass conservation. For example, in the reaction:
2H2 + O2 → 2H2O, the total mass of hydrogen and oxygen before and after the reaction remains the same.

Question 20:
Define molecular mass and formula unit mass. Calculate the molecular mass of sulphuric acid (H2SO4) and the formula unit mass of sodium chloride (NaCl).
Answer:

Molecular mass is the sum of the atomic masses of all atoms in a molecule, while formula unit mass is the sum of atomic masses of all atoms in a formula unit of an ionic compound.


Calculation for H2SO4:
Atomic mass of H = 1u
Atomic mass of S = 32u
Atomic mass of O = 16u
Molecular mass = (2 × 1) + 32 + (4 × 16) = 2 + 32 + 64 = 98u


Calculation for NaCl:
Atomic mass of Na = 23u
Atomic mass of Cl = 35.5u
Formula unit mass = 23 + 35.5 = 58.5u

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
In a lab, students observed that 2g of hydrogen combines with 16g of oxygen to form water. Using Dalton’s atomic theory, explain why the ratio is fixed and calculate the mass of water formed.
Answer:
Case Summary

Hydrogen and oxygen combine in a fixed ratio to form water.

Scientific Principle
  • Dalton’s theory states compounds form in fixed mass ratios.
  • Our textbook shows H2O has a 1:8 mass ratio (H:O).
Solution Approach

2g H + 16g O = 18g water. The ratio (2:16) simplifies to 1:8, matching the formula.

Question 2:
A molecule of sulfur (S8) has 8 atoms. If the atomic mass of sulfur is 32u, find its molecular mass. How does this relate to mole concept?
Answer:
Case Summary

Sulfur exists as S8 with atomic mass 32u.

Scientific Principle
  • Molecular mass = atomic mass × number of atoms.
  • 1 mole = 6.022×1023 particles (NCERT example).
Solution Approach

Molecular mass = 32u × 8 = 256u. 1 mole of S8 weighs 256g, linking mass to particles.

Question 3:
Rust forms when iron reacts with oxygen. Write the balanced equation and calculate the mass of oxygen needed to react with 112g of iron (Fe: 56u, O: 16u).
Answer:
Case Summary

Iron reacts with oxygen to form rust (Fe2O3).

Scientific Principle
  • Balanced equation: 4Fe + 3O2 → 2Fe2O3.
  • Mass calculations use atomic masses (NCERT).
Solution Approach

112g Fe = 2 moles. For 4Fe, 3O2 is needed. Thus, 1.5 moles O2 (48g) reacts.

Question 4:
A compound has the formula X2O3 and contains 68.4% oxygen by mass. Find the atomic mass of X using the given data (O: 16u).
Answer:
Case Summary

Compound X2O3 has 68.4% oxygen.

Scientific Principle
  • % composition = (mass of element/total mass) × 100.
  • Similar to NCERT’s urea example.
Solution Approach

Let atomic mass of X = A. Total mass = 2A + 48. (48/(2A+48)) × 100 = 68.4%. Solving, A ≈ 27u.

Question 5:
In a lab, students observed that 2g of hydrogen combines with 16g of oxygen to form water. Using Dalton’s atomic theory, explain why the ratio is fixed.
Answer:
Case Summary

Hydrogen and oxygen combine in a fixed mass ratio (1:8) to form water.

Scientific Principle
  • Dalton’s theory states atoms combine in simple whole-number ratios.
  • Our textbook shows H2O has 2:1 atom ratio (H:O).
Solution Approach

Since 1 H atom combines with 1 O atom, 2g H (1 mole) needs 16g O (1 mole). The fixed ratio confirms atomic combination rules.

Question 6:
A molecule of sulfur (S8) has 8 atoms. Calculate its molecular mass and compare it with oxygen (O2).
Answer:
Case Summary

Sulfur (S8) and oxygen (O2) are polyatomic molecules.

Scientific Principle
  • Molecular mass = sum of atomic masses.
  • NCERT example: O2 = 16u × 2 = 32u.
Solution Approach

S8 = 32u × 8 = 256u. O2 is lighter (32u). This shows how atomic arrangement affects mass.

Question 7:
Rust forms when iron (Fe) reacts with oxygen (O2). Write the balanced equation and identify the law of conservation of mass.
Answer:
Case Summary

Iron reacts with oxygen to form rust (Fe2O3).

Scientific Principle
  • Atoms rearrange but total mass remains constant.
  • NCERT example: 4Fe + 3O2 → 2Fe2O3.
Solution Approach

4Fe (224u) + 3O2 (96u) = 2Fe2O3 (320u). Mass is conserved as per the law.

Question 8:
A compound has 40% carbon and 60% oxygen by mass. Determine its empirical formula using atomic masses (C=12u, O=16u).
Answer:
Case Summary

A compound’s mass composition is given to find its simplest formula.

Scientific Principle
  • Empirical formula shows simplest whole-number ratio.
  • NCERT example: CO2 has 1:2 ratio.
Solution Approach

Moles of C = 40/12 ≈ 3.33, O = 60/16 ≈ 3.75. Ratio ≈ 1:1.125 ≈ 1:1 (CO).

Question 9:
In a lab, students observed that 2g of hydrogen combines with 16g of oxygen to form water. Using atomic mass units (u), explain the ratio of their masses. How does this relate to the law of constant proportions?
Answer:
Case Summary

Hydrogen and oxygen combine in a fixed mass ratio (1:8) to form water.

Scientific Principle
  • Hydrogen (H2) has atomic mass 1u, oxygen (O2) 16u.
  • Our textbook shows the ratio of masses follows their atomic masses.
Solution Approach

The 1:8 ratio matches the law of constant proportions, proving compounds have fixed composition. Real-world example: H2O always has 11% H by mass.

Question 10:
A molecule of sulfur (S8) contains 8 atoms. Calculate its molecular mass. How does this differ from oxygen (O2) in terms of atomicity?
Answer:
Case Summary

Sulfur (S8) is polyatomic, while oxygen (O2) is diatomic.

Scientific Principle
  • Atomic mass of S = 32u, O = 16u.
  • Molecular mass of S8 = 8 × 32u = 256u.
Solution Approach

O2 has molecular mass 32u. Atomicity of S8 is 8 (like our NCERT example), while O2 is 2. Real-world: Ozone (O3) has atomicity 3.

Question 11:
A student writes the chemical formula of magnesium oxide as MgO2. Identify the error and correct it using valency rules. Compare with NCERT’s example of sodium chloride.
Answer:
Case Summary

MgO2 is incorrect as it violates valency rules.

Scientific Principle
  • Mg has valency +2, O has −2.
  • Our textbook shows formulas must balance charges (e.g., NaCl).
Solution Approach

Correct formula is MgO (2+ and 2− cancel). Like NaCl (Na+ + Cl−), valency determines ratios. Real-world: CaO (lime) follows the same rule.

Question 12:
In an experiment, 5g of calcium carbonate (CaCO3) decomposes to 2.8g calcium oxide (CaO). Calculate the mass of CO2 released, applying the law of conservation of mass. Relate to NCERT’s burning magnesium example.
Answer:
Case Summary

CaCO3 decomposes to CaO + CO2, conserving mass.

Scientific Principle
  • Total mass before = after (5g = 2.8g + CO2).
  • CO2 mass = 5g − 2.8g = 2.2g.
Solution Approach

Like NCERT’s Mg + O2 → MgO example, mass is conserved. Real-world: Rusting iron also follows this law.

Question 13:

In a chemistry lab, Riya observed that when 2.4g of magnesium (Mg) is burnt in 1.6g of oxygen (O2), it forms 4.0g of magnesium oxide (MgO). Using this data:

  • Show that this reaction follows the Law of Conservation of Mass.
  • Calculate the mass of magnesium oxide formed if 4.8g of Mg is burnt in 3.2g of O2.
Answer:

Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. Here:


Mass of reactants (Mg + O2) = 2.4g + 1.6g = 4.0g
Mass of product (MgO) = 4.0g
Since both are equal, the law is verified.

For the second part:
2.4g Mg reacts with 1.6g O2 → 4.0g MgO
Thus, 4.8g Mg (double the initial amount) will react with 3.2g O2 (double the initial amount) to form 8.0g MgO.
Question 14:

A student writes the chemical formula of water as H2O and carbon dioxide as CO2.

  • Explain how the valency of elements helps in writing these formulas.
  • Why is the formula of sodium chloride written as NaCl and not as Na2Cl2?
Answer:

Valency is the combining capacity of an element. In H2O:


Hydrogen (H) has a valency of 1+, Oxygen (O) has a valency of 2-.
To balance charges, 2 H atoms (2×1+) combine with 1 O atom (2-).

Similarly, in CO2:
Carbon (C) has a valency of 4+, Oxygen (O) has 2-.
To balance, 1 C atom (4+) combines with 2 O atoms (2×2-).

For NaCl:
Sodium (Na) has valency 1+, Chlorine (Cl) has 1-.
One Na+ and one Cl- cancel each other’s charge, so the simplest ratio is 1:1. Writing Na2Cl2 is unnecessary as it represents the same ratio.
Question 15:
In a chemistry lab, Riya observed that when 12g of carbon is burned in air, it reacts with 32g of oxygen to form 44g of carbon dioxide. Using this data, answer the following:
(a) Show that this reaction follows the Law of Conservation of Mass.
(b) Calculate the mass of oxygen required to react completely with 24g of carbon.
Answer:

(a) According to the Law of Conservation of Mass, the total mass of reactants must equal the total mass of products.
Mass of carbon (reactant) = 12g
Mass of oxygen (reactant) = 32g
Total mass of reactants = 12g + 32g = 44g
Mass of carbon dioxide (product) = 44g
Since the mass of reactants (44g) equals the mass of products (44g), the reaction obeys the Law of Conservation of Mass.

(b) From the given data, 12g of carbon reacts with 32g of oxygen.
Thus, 1g of carbon reacts with 32/12 = 2.67g of oxygen.
For 24g of carbon, the mass of oxygen required = 24 × 2.67 = 64g.
Alternatively, since the ratio is 12:32, doubling the carbon (24g) means doubling the oxygen (64g).

Question 16:
A molecule of water (H2O) consists of 2 atoms of hydrogen and 1 atom of oxygen. Answer the following:
(a) Calculate the molecular mass of water.
(b) If 18g of water contains 6.022 × 1023 molecules, find the mass of one molecule of water in grams.
Answer:

(a) The molecular mass of water is calculated as follows:
Mass of 2 hydrogen atoms = 2 × 1u = 2u
Mass of 1 oxygen atom = 1 × 16u = 16u
Total molecular mass of water = 2u + 16u = 18u.

(b) Given that 18g of water contains 6.022 × 1023 molecules, the mass of one molecule is:
Mass of one molecule = Total mass / Number of molecules
= 18g / 6.022 × 1023
= 2.99 × 10-23g (approximately).
This shows how tiny the mass of a single molecule is!

Question 17:
Rahul conducted an experiment to determine the number of molecules in 18 grams of water (H2O). He knew the molar mass of water is 18 g/mol. Help him calculate the number of molecules present using Avogadro's number (6.022 × 1023). Explain each step clearly.
Answer:

To calculate the number of molecules in 18 grams of water, follow these steps:


Step 1: Determine the number of moles of water.
Given mass = 18 g
Molar mass of water (H2O) = 18 g/mol
Number of moles = Given mass / Molar mass = 18 g / 18 g/mol = 1 mole

Step 2: Use Avogadro's number to find the number of molecules.
1 mole of any substance contains 6.022 × 1023 molecules.
Number of molecules = Number of moles × Avogadro's number = 1 × 6.022 × 1023 = 6.022 × 1023 molecules

Thus, 18 grams of water contains 6.022 × 1023 molecules.

Question 18:
Priya observed that 12 grams of carbon (C) combines with 32 grams of oxygen (O2) to form carbon dioxide (CO2). Verify whether this reaction follows the Law of Conservation of Mass. Show calculations and justify your answer.
Answer:

To verify the Law of Conservation of Mass, we compare the total mass of reactants and products:


Step 1: Calculate the mass of reactants.
Mass of carbon (C) = 12 g
Mass of oxygen (O2) = 32 g
Total mass of reactants = 12 g + 32 g = 44 g

Step 2: Determine the mass of the product (CO2).
1 molecule of CO2 contains 1 carbon atom (12 u) and 2 oxygen atoms (2 × 16 u = 32 u).
Molar mass of CO2 = 12 + 32 = 44 g/mol.
Since 12 g of C and 32 g of O2 form 44 g of CO2, the mass remains conserved.

Thus, the reaction follows the Law of Conservation of Mass as the total mass of reactants (44 g) equals the mass of the product (44 g).

Question 19:
In a chemistry lab, Riya observed that 12 grams of carbon combines with 32 grams of oxygen to form carbon dioxide (CO2). Using this data, answer the following:

(a) Calculate the mass of carbon dioxide formed.
(b) If 24 grams of carbon is used instead, how much oxygen will be required for complete reaction? Justify your answer using the Law of Conservation of Mass.
Answer:

(a) According to the Law of Conservation of Mass, the total mass of reactants equals the total mass of products.
Mass of carbon = 12 g
Mass of oxygen = 32 g
Mass of carbon dioxide formed = Mass of carbon + Mass of oxygen = 12 g + 32 g = 44 g.

(b) The ratio of carbon to oxygen in CO2 is fixed (12:32 or 3:8 by mass).
If 24 g of carbon is used (double the original amount), the required oxygen will also double.
Mass of oxygen needed = 32 g × 2 = 64 g.
Justification: The Law of Conservation of Mass states that mass is neither created nor destroyed in a chemical reaction. Thus, the ratio of masses remains constant.

Question 20:
A molecule of water (H2O) contains 2 atoms of hydrogen and 1 atom of oxygen. Analyze the following scenarios:

(a) How many moles of water molecules are present in 36 grams of water?
(b) Calculate the number of hydrogen atoms in this sample. (Given: Molar mass of water = 18 g/mol, Avogadro's number = 6.022 × 1023)
Answer:

(a) Molar mass of water (H2O) = 18 g/mol.
Number of moles = Given mass / Molar mass = 36 g / 18 g/mol = 2 moles.

(b) Each water molecule (H2O) contains 2 hydrogen atoms.
Number of water molecules in 2 moles = 2 × 6.022 × 1023 = 12.044 × 1023.
Number of hydrogen atoms = 2 × Number of water molecules = 2 × 12.044 × 1023 = 2.4088 × 1024 atoms.

Note: Avogadro's number helps relate macroscopic measurements (grams) to microscopic entities (atoms/molecules).

Question 21:
A molecule of water (H2O) contains 2 atoms of hydrogen and 1 atom of oxygen. Based on this, answer the following:

(a) Calculate the molecular mass of water.
(b) If a sample contains 18 grams of water, how many moles of water molecules are present? (Given: Atomic masses - H = 1 u, O = 16 u)
Answer:

(a) The molecular mass of water (H2O) is calculated as:
Mass of 2 hydrogen atoms = 2 × 1 u = 2 u
Mass of 1 oxygen atom = 1 × 16 u = 16 u
Total molecular mass = 2 u + 16 u = 18 u.

(b) Number of moles = Given mass / Molar mass
Given mass of water = 18 g
Molar mass of water = 18 g/mol
Number of moles = 18 g / 18 g/mol = 1 mole.
Thus, 18 grams of water contains 6.022 × 1023 molecules (Avogadro's number).

Question 22:
A student performed an experiment to determine the molecular mass of water (H2O). Given the atomic masses: H = 1 u, O = 16 u, explain the steps to calculate the molecular mass of water. Also, state why the molecular mass is considered a relative mass.
Answer:

To calculate the molecular mass of water (H2O), follow these steps:

1. Identify the number of atoms of each element in the molecule: 2 Hydrogen (H) atoms and 1 Oxygen (O) atom.
2. Multiply the number of atoms by their respective atomic masses:
- For Hydrogen: 2 × 1 u = 2 u
- For Oxygen: 1 × 16 u = 16 u
3. Add the masses of all atoms: 2 u + 16 u = 18 u.

The molecular mass of water is 18 u.

Molecular mass is called a relative mass because it is measured relative to the mass of a carbon-12 atom (which is taken as 12 u). It compares the mass of a molecule to 1/12th the mass of a carbon-12 atom, making it a unitless ratio.

Question 23:
In a chemistry lab, two samples were labeled as 1 mole of oxygen gas (O2) and 1 mole of ozone gas (O3). Compare the number of molecules and atoms present in each sample. Justify your answer using Avogadro's number (6.022 × 1023).
Answer:

Both samples contain 1 mole of their respective gases, so the number of molecules in each is the same:

- O2: 1 mole = 6.022 × 1023 molecules of oxygen.
- O3: 1 mole = 6.022 × 1023 molecules of ozone.

However, the number of atoms differs due to their molecular structures:

- O2: Each molecule has 2 atoms of oxygen.
Total atoms = 6.022 × 1023 × 2 = 1.2044 × 1024 atoms.
- O3: Each molecule has 3 atoms of oxygen.
Total atoms = 6.022 × 1023 × 3 = 1.8066 × 1024 atoms.

This difference arises because Avogadro's number counts molecules, but the atomic composition varies based on the molecular formula.

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