19 public questions tagged with this topic.
Octatriyne (CH≡C-C≡C-C≡C-CH₂-CH₃) has three triple bonds, giving three degrees of unsaturation, fitting C₈H₁₂.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Aromatic Hydrocarbons - Structure of Benzene and Resonance
Butane (C₄H₁₀), a saturated alkane, has 10 hydrogens, the maximum for 4 carbons, compared to unsaturated hydrocarbons like butene.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Aromatic Hydrocarbons - Electrophilic Substitution Reactions
Butene (CH₂=CH-CH₂-CH₃) is an alkene with the formula C₄H₈, fitting CnH₂n.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkenes - Preparation and Properties and Addition Reactions
Cyclopentane (C₅H₁₀) requires 15/2 O₂ (C₅H₁₀ + 15/2 O₂ -> 5CO₂ + 5H₂O), less than pentane’s 8 O₂.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkanes - Properties - Conformations and Reactions
For C₄H₁₀ (butane): C₄H₁₀ + 13/2 O₂ -> 4CO₂ + 5H₂O, producing 4 CO₂ and 5 H₂O.
Ref: NCERT Class 11 Chemistry > Chapter 9: Hydrocarbons > Topic: Alkanes - Properties - Conformations and Reactions
5 carbons (from 5 CO₂). CH₃CH₂CH₂CH₂CH₃ (pentane) has 9 σ (8 C-H, 1 C-C from single bonds), no π, and fits C₅H₁₂.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Fundamental Concepts - Reaction Mechanism - Fission and Reaction Intermediates
4 carbons (from 4 CO₂). Degree of unsaturation = 2 suggests two double bonds or one triple bond. CH₃C≡CCH₃ (but-2-yne, C₄H₆) fits with 2 π, 8 σ, and 4 CO₂.
Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Electronic Effects - Inductive Mesomeric Hyperconjugation and Resonance
Mass of C ≈ 0.84 g; mass of H = 0.14 g. Total = 0.98 g. Moles: C = 0.07, H = 0.14; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 98/14 = 7; molecular formula = C₇H₁₄.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Mass of C ≈ 0.48 g; mass of H = 0.04 g. Moles: C = 0.04, H = 0.04; ratio = 1 : 1; empirical formula = CH, mass = 13 g/mol. n ≈ 4; molecular formula = C₄H₄.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Mass of C ≈ 0.72 g; mass of H = 0.12 g. Total = 0.84 g. Moles: C = 0.06, H = 0.12; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 84/14 = 6; molecular formula = C₆H₁₂.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Mass of C ≈ 0.36 g; mass of H = 0.06 g. Total = 0.42 g. Moles: C = 0.03, H = 0.06; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 42/14 = 3; molecular formula = C₃H₆.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
Mass of C ≈ 0.36 g; mass of H = 0.06 g. Total = 0.42 g. Moles: C = 0.03, H = 0.06; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 42/14 = 3; molecular formula = C₃H₆.
Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition
