Q.78 Chemistry ⭐ 360° Option Elimination Active

NEET 2021 Chemistry Question 78 Solution

Topic: Some Basic Concepts of Chemistry · Official NTA Booklet Code M1

Question

An organic compound contains 78% (by wt.) carbon and remaining percentage of hydrogen. The right option for the empirical formula of this compound is: [Atomic wt. of C is 12, H is 1]

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Verified Against NTA Final Answer Key

Option (3) : CH3

📚 Some Basic Concepts of Chemistry Easy

📋 Given Parameters & Problem Setup

Carbon % by Weight 78%

Hydrogen % by Weight 100 - 78 = 22%

Atomic Masses C = 12, H = 1

⚡

The 30-Second Exam Hall Shortcut: ⚡ Mole Ratio: 78/12 = 6.5 and 22/1 = 22 $\implies$ 22/6.5 ≈ 3.38 $\implies$ CH3

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Moles of C $= 78 / 12 = 6.5$ . Moles of H $= 22 / 1 = 22$ . Simplest molar ratio $= \frac{22}{6.5} = 3.38 \approx 3$ ! Direct verification: in $CH_3$ , mass % of C is $\frac{12}{15} \times 100 = 80\%$ (extremely close to 78%). In $CH_4$ , C is $75\%$ ; in $CH_2$ , C is $85.7\%$ . Closest empirical formula is $CH_3$ ! Match Option (3)!

1 Assume 100 g sample: Mass of C

= 78\text{ g}

, Mass of H

= 22\text{ g}

2 Calculate moles:

n_{\text{C}} = 78 / 12 = 6.5\text{ mol}

;

n_{\text{H}} = 22 / 1 = 22\text{ mol}

3 Divide by smallest mole value (6.5): - C:

6.5 / 6.5 = 1.0

- H:

22 / 6.5 = 3.38 \approx 3

4 The closest integer empirical formula in options is

CH_3

5 Match Option (3).

🛡️

Why You Can 100% Trust This Shortcut: Canonical stoichiometry empirical formula algorithm and official NTA final key.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining Elemental Mass Percentages

For a 100 g sample of the hydrocarbon: - Mass of Carbon: $w_{\text{C}} = 78\text{ g}$ - Mass of Hydrogen: $w_{\text{H}} = 100 - 78 = 22\text{ g}$

Step 2

Calculating Relative Number of Moles

Divide the mass of each element by its respective atomic mass ( $C = 12\text{ g/mol}, H = 1\text{ g/mol}$ ):

n_{\text{C}} = \frac{78}{12} = 6.5\text{ mol}, \quad n_{\text{H}} = \frac{22}{1} = 22.0\text{ mol}

Step 3

Determining the Simplest Atomic Ratio

Divide by the smaller number of moles ( $6.5$ ): - Atomic ratio of C: $\frac{6.5}{6.5} = 1$ - Atomic ratio of H: $\frac{22.0}{6.5} = 3.38 \approx 3$ Comparing theoretical percentages of options: - $CH$ : $\%C = 12/13 \times 100 = 92.3\%$ - $CH_2$ : $\%C = 12/14 \times 100 = 85.7\%$ - $CH_3$ : $\%C = 12/15 \times 100 = 80.0\%$ (closest to $78\%$ ) - $CH_4$ : $\%C = 12/16 \times 100 = 75.0\%$

\mathbf{\text{Empirical Formula: } CH_3}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to $92.3\% C$ and $7.7\% H$ (ethyne/benzene), far from $78\%$ .

⚠️

Student Fallacy (Why candidates pick this trap): Assuming 1:1 ratio.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to $85.7\% C$ and $14.3\% H$ (alkenes).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming 1:2 ratio.

3 3

✓ 100% Correct

Verdict & Scientific Reason:

Matches the experimental ratio $C:H \approx 1:3$ , with theoretical $80\% C$ (closest to $78\%$ ).

💡

Why this is airtight: None; official NTA designated answer.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Corresponds to $75\% C$ and $25\% H$ (methane).

⚠️

Student Fallacy (Why candidates pick this trap): Assuming 1:4 ratio.

⚠️

NTA Trap Alert: Non-Integer Ratio Approximation Trap

The ratio $22 / 6.5 = 3.38$ is non-integer. Some students multiply by 3 to get $C_3H_{10}$ , but the prompt provides simple multiple choices ( $CH, CH_2, CH_3, CH_4$ ). Checking the mass percentage confirms $CH_3$ ( $80\% C$ ) is the closest empirical formula.

🛡️ How to avoid the -1 mark: Check percentages of choices:

CH_3 = 80\% C, 20\% H

, closest to

78\% C, 22\% H

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Core NEET Syllabus Standard

NCERT Chemistry · Some Basic Concepts of Chemistry Standard Benchmark

Core Scientific Principle Tested:

"An empirical formula represents the simplest whole-number ratio of various atoms present in a compound based on percent composition."

Syllabus Focus: Empirical and Molecular Formulae

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Chemistry Question 78 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Mole Ratio: 78/12 = 6.5 and 22/1 = 22 ⟹ \implies⟹ 22/6.5 ≈ 3.38 ⟹ \implies⟹ CH3

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining Elemental Mass Percentages

Calculating Relative Number of Moles

Determining the Simplest Atomic Ratio

360° Option Elimination Engine Key Feature

NTA Trap Alert: Non-Integer Ratio Approximation Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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The 30-Second Exam Hall Shortcut: ⚡ Mole Ratio: 78/12 = 6.5 and 22/1 = 22 $\implies$ 22/6.5 ≈ 3.38 $\implies$ CH3