Complete Stoichiometry Dashboard – Class 9

Complete Stoichiometry Dashboard

Class 9 Chemistry – Full Exercise Solutions

Master stoichiometry with comprehensive notes and complete exercise solutions

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Comprehensive Key Notes

Stoichiometry Fundamentals

  • Stoichiometry is the calculation of reactants and products in chemical reactions using balanced equations.
    Example:

    In the reaction: 2H₂ + O₂ → 2H₂O

    Stoichiometry tells us that 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of water.

  • Based on the law of conservation of mass – matter cannot be created or destroyed in a chemical reaction.
    Law of Conservation of Mass

    Total mass of reactants = Total mass of products

    This fundamental law was established by Antoine Lavoisier in the late 18th century.

  • Stoichiometric coefficients in balanced equations provide the mole ratio between reactants and products.
    Example:

    In: N₂ + 3H₂ → 2NH₃

    The coefficients tell us: 1 mole N₂ : 3 moles H₂ : 2 moles NH₃

Chemical Formulas

  • Elements exist in different forms:
    • Atomic elements: Na, Ca, C, Fe (exist as individual atoms)
    • Molecular elements: O₂, H₂, N₂, Cl₂ (exist as diatomic molecules)
    • Polyatomic elements: O₃ (ozone), P₄ (phosphorus), S₈ (sulfur)
  • Ionic compounds are represented by formula units showing the simplest ratio between ions:
    • NaCl (1:1 ratio of Na⁺ and Cl⁻ ions)
    • CaCl₂ (1:2 ratio of Ca²⁺ and Cl⁻ ions)
    • Al₂O₃ (2:3 ratio of Al³⁺ and O²⁻ ions)
  • Covalent compounds exist as discrete molecules:
    • H₂O (water) – 2 hydrogen atoms bonded to 1 oxygen atom
    • NH₃ (ammonia) – 1 nitrogen atom bonded to 3 hydrogen atoms
    • CH₄ (methane) – 1 carbon atom bonded to 4 hydrogen atoms

Empirical vs Molecular Formulas

  • Empirical formula shows the simplest whole number ratio of atoms in a compound.
    Example:

    Benzene: Molecular formula = C₆H₆, Empirical formula = CH

    Hydrogen peroxide: Molecular formula = H₂O₂, Empirical formula = HO

  • Molecular formula shows the actual number of atoms of each element in a molecule.
    Example:

    Glucose: Molecular formula = C₆H₁₂O₆, Empirical formula = CH₂O

  • Relationship: Molecular formula = n × (Empirical formula)
    n = Molar mass / Empirical formula mass

Problem Solving Strategies

Balancing Chemical Equations

Use the criss-cross method for ionic compounds:

  1. Write symbols of cation and anion with their charges
  2. Cross the numerical value of charges (ignore signs)
  3. Reduce to simplest ratio if necessary
Example: Aluminum Oxide

Al³⁺ and O²⁻ → Cross 3 and 2 → Al₂O₃

Molecular Formula from Empirical Formula

Use the formula: Molecular formula = n × (Empirical formula)

Where: n = Molar mass / Empirical formula mass

Example:

Empirical formula = CH₂O (mass = 30 g/mol)

Molar mass = 180 g/mol → n = 180/30 = 6

Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ (Glucose)

Mole Concept Visualization

H₂
+
O₂
→
H₂O

2 molecules H₂ + 1 molecule O₂ → 2 molecules H₂O

2 moles H₂ + 1 mole O₂ → 2 moles H₂O

4g H₂ + 32g O₂ → 36g H₂O

Limiting Reactant

The reactant that is completely consumed in a reaction, determining the maximum amount of product that can be formed.

Percentage Yield

Percentage Yield = (Actual Yield / Theoretical Yield) × 100%

Measures the efficiency of a chemical reaction.

Complete Exercise Solutions

Multiple Choice
Short Answer
Constructed Response
Descriptive Questions

Multiple Choice Questions (Solved)

1(i) How many atoms are present in one gram of H₂O?

(a) 1002 × 10¹³ atoms
(b) 6.022 × 10¹³ atoms
(c) 0.334 × 10²³ atoms
(d) 2.004 × 10²³ atoms

Answer: (c) 0.334 × 10²³ atoms

Step-by-step solution:

Molar mass of H₂O = (2×1) + 16 = 18 g/mol
Number of moles in 1g = 1/18 = 0.0556 mol
Number of molecules = 0.0556 × 6.022 × 10²³ = 0.334 × 10²³ molecules
Each H₂O molecule has 3 atoms (2H + 1O)
Total atoms = 3 × 0.334 × 10²³ = 1.002 × 10²³ atoms

Note: The question asks for atoms, but option (c) gives the number of molecules. The correct number of atoms would be approximately 1.002 × 10²³.

1(ii) Which is the correct formula of calcium phosphate?

(a) CaP
(b) Ca₂P₃
(c) Ca₃P₂
(d) Ca₃(PO₄)₂

Answer: (d) Ca₃(PO₄)₂

Step-by-step solution:

Calcium ion: Ca²⁺
Phosphate ion: PO₄³⁻
Using criss-cross method: Ca₃(PO₄)₂
Check: 3×(+2) + 2×(-3) = 6 – 6 = 0 (neutral compound)

1(iv) Structural formula of 2-hexene is CH₃-CH=CH-(CH₂)₂-CH₃. What will be its empirical formula?

(a) C₆H₁₂
(b) CH
(c) C₃H₆
(d) CH₂

Answer: (d) CH₂

Step-by-step solution:

Molecular formula of 2-hexene: C₆H₁₂
Simplest ratio C:H = 6:12 = 1:2
Empirical formula = CH₂

1(v) How many moles are there in 25 g of H₂SO₄?

(a) 0.765 moles
(b) 0.51 moles
(c) 0.255 moles
(d) 0.4 moles

Answer: (c) 0.255 moles

Step-by-step solution:

Molar mass of H₂SO₄ = (2×1) + 32 + (4×16) = 98 g/mol
Number of moles = mass / molar mass = 25 / 98
Number of moles = 0.255 moles

1(vi) A necklace has 6g of diamonds in it. What are the number of carbon atoms in it?

(a) 6.02 × 10²²
(b) 12.04 × 10²²
(c) 1.003 × 10²³
(d) 3.01 × 10²³

Answer: (d) 3.01 × 10²³

Step-by-step solution:

Molar mass of carbon = 12 g/mol
Moles in 6g = 6/12 = 0.5 moles
Number of atoms = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ atoms

1(vii) What is the mass of Al in 204g of aluminium oxide, Al₂O₃?

(a) 26g
(b) 27g
(c) 54g
(d) 108g

Answer: (d) 108g

Step-by-step solution:

Molar mass of Al₂O₃ = (2×27) + (3×16) = 102 g/mol
Mass of Al in 102g of Al₂O₃ = 54g
204g is exactly 2 × 102g
Mass of Al in 204g = 2 × 54g = 108g

1(viii) Which one of the following compounds will have the highest percentage of the mass of nitrogen?

(a) CO(NH₂)₂ (Urea)
(b) N₂H₄ (Hydrazine)
(c) NH₃ (Ammonia)
(d) NH₂OH (Hydroxylamine)

Answer: (c) NH₃ (Ammonia)

Step-by-step solution:

Calculate nitrogen percentage in each compound:
Urea (CO(NH₂)₂): Molar mass = 60 g/mol, N% = (28/60)×100 = 46.67%
Hydrazine (N₂H₄): Molar mass = 32 g/mol, N% = (28/32)×100 = 87.5%
Ammonia (NH₃): Molar mass = 17 g/mol, N% = (14/17)×100 = 82.35%
Hydroxylamine (NH₂OH): Molar mass = 33 g/mol, N% = (14/33)×100 = 42.42%
Highest percentage is NH₃ with 82.35%

Short Answer Questions (Solved)

2(i) Write down the chemical formula of barium nitride.

Answer: Ba₃N₂

Step-by-step solution:

Barium ion: Ba²⁺
Nitride ion: N³⁻
Using criss-cross method: Ba₃N₂
Check: 3×(+2) + 2×(-3) = 6 – 6 = 0 (neutral compound)

2(iii) How many molecules are present in 1.5 g H₂O?

Answer: 5.02 × 10²² molecules

Step-by-step solution:

Molar mass of H₂O = 18 g/mol
Moles in 1.5g = 1.5/18 = 0.0833 mol
Number of molecules = 0.0833 × 6.022 × 10²³
Number of molecules = 5.02 × 10²² molecules

2(iv) What is the difference between a mole and Avogadro’s number?

Answer:

Key Differences:

Avogadro’s number (6.022 × 10²³) is a constant – the number of particles in one mole of a substance
A mole is a unit of measurement for amount of substance
Avogadro’s number is the bridge between the microscopic and macroscopic worlds
One mole of any substance contains exactly Avogadro’s number of particles

2(v) Write down the chemical equation of the following reaction: Copper + Sulphuric acid → Copper sulphate + Sulphur dioxide + Water

Answer: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Balancing the equation:

Left side: Cu, 2H, 2S, 😯
Right side: Cu, S, 4O (from CuSO₄) + S, 2O (from SO₂) + 2H, 1O (from H₂O)
Total right: Cu, 2H, 2S, 7O → Need to balance oxygen
Add coefficient 2 to H₂O: CuSO₄ + SO₂ + 2H₂O
Now right side: Cu, 4H, 2S, 😯
Left side needs 2H₂SO₄ to match: Cu + 2H₂SO₄
Balanced equation: Cu + 2H₂SO₄ → CuSO₄ + SO₂ + 2H₂O

Constructed Response Questions (Solved)

3(i) Different compounds will never have the same molecular formula but they can have the same empirical formula. Explain

Answer:

Explanation:

Molecular formula represents the actual number of atoms of each element in a molecule
Each unique compound has a unique molecular formula that defines its chemical identity
Empirical formula shows only the simplest whole number ratio of atoms
Different compounds can have the same ratio of atoms but different actual numbers
Examples:
– Benzene (C₆H₆) and acetylene (C₂H₂) both have empirical formula CH
– Formaldehyde (CH₂O) and acetic acid (C₂H₄O₂) both have empirical formula CH₂O
These compounds have different properties despite same empirical formula

3(ii) Write down the chemical formulas of the following compounds: Calcium phosphate, Aluminium nitride, Sodium acetate, Ammonium carbonate and Bismuth sulphate.

Answer:

Formulas with explanations:

Calcium phosphate: Ca₃(PO₄)₂ (Ca²⁺ and PO₄³⁻)
Aluminium nitride: AlN (Al³⁺ and N³⁻)
Sodium acetate: CH₃COONa or NaC₂H₃O₂ (Na⁺ and CH₃COO⁻)
Ammonium carbonate: (NH₄)₂CO₃ (NH₄⁺ and CO₃²⁻)
Bismuth sulphate: Bi₂(SO₄)₃ (Bi³⁺ and SO₄²⁻)

3(iii) Why does Avogadro’s number have an immense importance in chemistry?

Answer:

Importance of Avogadro’s number:

Connects microscopic world of atoms/molecules to macroscopic measurements
Enables counting particles by weighing measurable amounts
Foundation for stoichiometric calculations in chemical reactions
Allows determination of molecular and formula masses
Essential for understanding gas laws and behavior of ideal gases
Fundamental to concept of molar volume of gases
Critical for quantitative analysis in chemistry

3(iv) When 8.657g of a compound were converted into elements, it gave 5.217g of carbon, 0.962g of hydrogen and 2.478g of oxygen. Calculate the percentage of each element present in this compound.

Answer:

Step-by-step solution:

Total mass of compound = 8.657g
Percentage of carbon = (5.217/8.657) × 100 = 60.24%
Percentage of hydrogen = (0.962/8.657) × 100 = 11.11%
Percentage of oxygen = (2.478/8.657) × 100 = 28.62%
Check: 60.24% + 11.11% + 28.62% = 99.97% ≈ 100%

3(v) How can you calculate the masses of the products formed in a reversible reaction?

Answer:

Calculating product masses in reversible reactions:

Reversible reactions reach equilibrium where both forward and reverse reactions occur
At equilibrium, the reaction does not go to completion
To calculate product masses, we need the equilibrium constant (K)
Set up an ICE table (Initial, Change, Equilibrium)
Use the equilibrium constant expression
Solve for the equilibrium concentrations
Convert concentrations to masses using molar masses
Without K, we can only calculate theoretical maximum yield assuming complete reaction

Descriptive Questions (Solved)

4(i) Which conditions must be fulfilled before writing a chemical equation for a reaction?

Answer:

Conditions for writing a chemical equation:

Experimental verification that the reaction occurs
Identification of all reactants and products
Correct chemical formulas for all substances
Knowledge of physical states (s, l, g, aq)
The equation must be balanced (law of conservation of mass)
Indication of reaction direction (→ or ⇌)
Inclusion of important reaction conditions if needed
For redox reactions, verification of electron transfer

4(ii) Explain the concepts of Avogadro’s number and mole.

Answer:

Avogadro’s Number:

Defined as 6.022 × 10²³ particles per mole
Based on number of atoms in exactly 12g of carbon-12
Universal constant that applies to all substances

The Mole:

SI unit for amount of substance
One mole contains exactly Avogadro’s number of particles
Mass of one mole equals molar mass in grams
Allows conversion between microscopic and macroscopic scales

Examples:

1 mole of carbon atoms = 6.022 × 10²³ C atoms = 12.01g
1 mole of water molecules = 6.022 × 10²³ H₂O molecules = 18.02g
1 mole of NaCl formula units = 6.022 × 10²³ NaCl units = 58.44g

4(iv) How many moles of coal are needed to produce 10 moles of CO according to the following equation? 3C(s) + O₂(g) + H₂O(g) → H₂(g) + 3CO(g)

Answer: 10 moles of coal (carbon)

Step-by-step solution:

Balanced equation: 3C + O₂ + H₂O → H₂ + 3CO
From the equation: 3 moles C produce 3 moles CO
Mole ratio C:CO = 3:3 = 1:1
To produce 10 moles CO, we need 10 moles C

4(v) How much SO₂ is needed in grams to produce 10 moles of sulphur? 2H₂S(g) + SO₂(g) → 2H₂O(l) + 3S(s)

Answer: 213.3 grams of SO₂

Step-by-step solution:

Balanced equation: 2H₂S + SO₂ → 2H₂O + 3S
From the equation: 1 mole SO₂ produces 3 moles S
To produce 10 moles S, we need (10/3) moles SO₂
Molar mass of SO₂ = 32 + 32 = 64 g/mol
Mass of SO₂ needed = (10/3) × 64 = 213.3 g

4(vi) How much ammonia is needed in grams to produce 1 kg of urea fertilizer? 2NH₃(aq) + CO₂(aq) → CO(NH₂)₂(aq) + H₂O(l)

Answer: 566.7 grams of ammonia

Step-by-step solution:

Balanced equation: 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O
From the equation: 2 moles NH₃ produce 1 mole urea
Molar mass of urea (CO(NH₂)₂) = 12 + 16 + (2×14) + (4×1) = 60 g/mol
Moles in 1 kg (1000g) of urea = 1000/60 = 16.67 moles
Moles of NH₃ needed = 2 × 16.67 = 33.33 moles
Molar mass of NH₃ = 14 + 3 = 17 g/mol
Mass of NH₃ needed = 33.33 × 17 = 566.7 g

4(vii) Calculate the number of atoms in the following: (a) 3 g of H₂ (b) 3.4 moles of N₂ (c) 10 g of C₆H₁₂O₆

Answer:

(a) 3 g of H₂:

Molar mass of H₂ = 2 g/mol
Moles = 3/2 = 1.5 moles
Molecules = 1.5 × 6.022 × 10²³ = 9.033 × 10²³ molecules
Each H₂ molecule has 2 atoms
Total atoms = 2 × 9.033 × 10²³ = 1.8066 × 10²⁴ atoms

(b) 3.4 moles of N₂:

Molecules = 3.4 × 6.022 × 10²³ = 2.047 × 10²⁴ molecules
Each N₂ molecule has 2 atoms
Total atoms = 2 × 2.047 × 10²⁴ = 4.094 × 10²⁴ atoms

(c) 10 g of C₆H₁₂O₆:

Molar mass of C₆H₁₂O₆ = (6×12) + (12×1) + (6×16) = 180 g/mol
Moles = 10/180 = 0.0556 moles
Molecules = 0.0556 × 6.022 × 10²³ = 3.347 × 10²² molecules
Each C₆H₁₂O₆ molecule has 6+12+6 = 24 atoms
Total atoms = 24 × 3.347 × 10²² = 8.033 × 10²³ atoms