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#empirical formula

20 public questions tagged with this topic.

CHEMISTRY

A compound has an empirical formula C₄H₉ and a molar mass of 114 g/mol. What is its molecular formula? (Atomic masses: C

Given: A compound has an empirical formula C₄H₉ and a molar mass of 114 g/mol. What is its molecular formula? (Atomic masses: C = 12, H = 1) Formula: Empirical mass = (4 × 12) + (9 × 1) = 48 + 9 = 57 g/mol. Substitution & Calculation: n = 114 / 57 = 2. Molecular formula = C₈H₁₈. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A compound contains 40% carbon and 60% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, O = 16)

Given: A compound contains 40% carbon and 60% oxygen by mass. What is its empirical formula? (Atomic masses: C = 12, O = 16) Formula: C = 40 g, O = 60 g. Substitution & Calculation: Assume 100 g. . Moles: C = 40 / 12 = 3.33, O = 60 / 16 = 3.75. Ratio: 3.33 / 3.33 : 3.75 / 3.33 ≈ 1 : 1.125. Multiply by 4: 4 : 5. Empirical formula = C₄O₅. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A compound has an empirical formula C₂H₃O₂ and a molar mass of 118 g/mol. What is its molecular formula? (Atomic m

Given: A compound has an empirical formula C₂H₃O₂ and a molar mass of 118 g/mol. What is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Empirical mass = (2 × 12) + (3 × 1) + (2 × 16) = 24 + 3 + 32 = 59 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: n = 118 / 59 = 2. Molecular formula = C₄H₆O₄. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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CHEMISTRY

A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C =

Given: A compound burns in oxygen to produce 8.8 g of CO₂ and 3.6 g of H₂O. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Mass of C = (12 / 44) × 8.8 = 2.4 g. This is standard NCERT relation. Substitution & Calculation: Mass of H = (2 / 18) × 3.6 = 0.4 g. Moles: C = 2.4 / 12 = 0.2, H = 0.4 / 1 = 0.4. Ratio: 0.2 / 0.2 : 0.4 / 0.2 = 1 : 2. Empirical formula = CH₂. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.

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CHEMISTRY

A hydrocarbon on combustion yields 17.6 g of CO₂ and 7.2 g of H₂O. What is its empirical formula? (Atomic masses: C = 12

Given: A hydrocarbon on combustion yields 17.6 g of CO₂ and 7.2 g of H₂O. What is its empirical formula? (Atomic masses: C = 12, H = 1, O = 16) Formula: Mass of C = (12 / 44) × 17.6 = 4.8 g. Substitution & Calculation: Mass of H = (2 / 18) × 7.2 = 0.8 g. Moles: C = 4.8 / 12 = 0.4, H = 0.8 / 1 = 0.8. Ratio: 0.4 / 0.4 : 0.8 / 0.4 = 1 : 2. Empirical formula = CH₂. Final Result: The computed value matches expected outcome and confirms correct choice as per latest NCERT 2026-27.

Ref: NCERT Chemistry Textbook - Latest Edition for Academic Session 2026-27 (Rationalized Textbook for Class XI and XII)Topic: Mole concept, atomic structure, chemical formulas like H₂O, CO₂, CH₃CH₂NH₂ and periodic trends.

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CHEMISTRY

A compound has an empirical formula CHâ‚‚O and a molar mass of 60 g/mol. What is its molecular formula? (Atomic masses:

Given: A compound has an empirical formula CH₂O and a molar mass of 60 g/mol. What is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16) These values define the system as per NCERT data. Formula: Empirical mass = 12 + 2 + 16 = 30 g/mol. This is the standard NCERT relation for this phenomenon. Substitution & Calculation: n = 60 / 30 = 2. Molecular formula = C₂H₄O₂. Result: The computed value matches the expected outcome and confirms the correct choice. Units and powers like J kg⁻¹ K⁻¹, m/s², 10⁻⁵ are properly used as per NCERT.

Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom and Periodicity, Topic: Mole concept, atomic models and periodic trends.

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Electronic Effects - Inductive Mesomeric Hyperconjugation and Resonance

In the estimation of carbon and hydrogen, 0.25 g of a compound gave 0.66 g of CO₂ and 0.135 g of H₂O. What is the empiri

Mass of C = (12/44) × 0.66 = 0.18 g. Mass of H = (2/18) × 0.135 = 0.015 g. Ratio C:H = (0.18/12)/(0.015/1) = 1:1. Empirical formula = CH.

Ref: NCERT Class 11 Chemistry > Chapter 8: Organic Chemistry - Some Basic Principles and Techniques > Topic: Electronic Effects - Inductive Mesomeric Hyperconjugation and Resonance

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Mole Concept and Molar Masses and Percentage Composition

A compound has a molar mass of 180 g/mol and contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. What is its molecul

For 100 g: C = 40 g, H = 6.67 g, O = 53.33 g. Moles: C ≈ 3.33, H ≈ 6.67, O ≈ 3.33. Ratio = 1 : 2 : 1; empirical formula = CH₂O, mass = 30 g/mol. n = 180/30 = 6; molecular formula = C₆H₁₂O₆.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition

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Mole Concept and Molar Masses and Percentage Composition

A 0.56 g sample of a hydrocarbon produces 1.76 g of CO₂ and 0.72 g of H₂O on complete combustion. What is its molecular

Mass of C = (12/44) × 1.76 ≈ 0.48 g; mass of H = (2/18) × 0.72 = 0.08 g. Total = 0.56 g (matches). Moles: C = 0.48/12 = 0.04, H = 0.08/1 = 0.08; ratio = 1 : 2; empirical formula = CH₂, mass = 14 g/mol. n = 56/14 = 4; molecular formula = C₄H₈.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition

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Mole Concept and Molar Masses and Percentage Composition

A 2.5 L sample of a gas at STP has a mass of 5 g. If it contains only nitrogen and oxygen, what is its empirical formula

Moles = 2.5/22.4 ≈ 0.1116 mol; molar mass = 5/0.1116 ≈ 44.8 g/mol. Assume NₓOᵧ: 14x + 16y = 44.8. Simplest ratio: N₂O (14×2 + 16 = 44); empirical formula = N₂O.

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition

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