Q.95 Chemistry · Section B ⭐ 360° Option Elimination Active

NEET 2024 Chemistry Question 95 Solution

Topic: Some Basic Concepts of Chemistry · Official NTA Booklet Code T1

Question

A compound X contains 32% of A, 20% of B and remaining percentage of C. Then, the empirical formula of X is: (Given atomic masses of A = 64; B = 40; C = 32 u)

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Verified Against NTA Final Answer Key

Option (4) : ABC3

📚 Some Basic Concepts of Chemistry Easy

📋 Given Parameters & Problem Setup

Percentage of A 32%

Atomic mass of A 64 u

Percentage of B 20%

Atomic mass of B 40 u

Atomic mass of C 32 u

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The 30-Second Exam Hall Shortcut: ⚡ Relative Mole Ratio Mental Arithmetic

⏱️ Save ~35 seconds

% C = 100 - (32 + 20) = 48%! Moles: A = 32/64 = 0.5; B = 20/40 = 0.5; C = 48/32 = 1.5! Divide by 0.5 -> 1 : 1 : 3 -> ABC3!

1 Calculate percentage of element C: % C = 100 - (32 + 20) = 48%.

2 Compute relative number of moles of each element in 100 g:

3 - Moles of A: 32 / 64 = 0.5 mol.

4 - Moles of B: 20 / 40 = 0.5 mol.

5 - Moles of C: 48 / 32 = 1.5 mol.

6 Divide each mole value by the smallest value (0.5):

7 - A: 0.5 / 0.5 = 1.

8 - B: 0.5 / 0.5 = 1.

9 - C: 1.5 / 0.5 = 3.

10 Simplest whole number molar ratio is A : B : C = 1 : 1 : 3.

11 Empirical formula is ABC3 -> Option (4) in 10 seconds!

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Why You Can 100% Trust This Shortcut: NCERT Chemistry · Some Basic Concepts of Chemistry (Determination of empirical formula from mass percent composition).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining the Mass Percentage of Element C

The sum of percentages of all constituent elements in compound X must equal 100%:

\%C = 100\% - (\%A + \%B) = 100\% - (32\% + 20\%) = 48\%

Step 2

Calculating Moles of Each Element in 100 g of Compound

Dividing the mass percentage of each element by its respective atomic mass: - Element A: n_A = 32 g / 64 g/mol = 0.50 mol - Element B: n_B = 20 g / 40 g/mol = 0.50 mol - Element C: n_C = 48 g / 32 g/mol = 1.50 mol

n_A = \frac{32}{64} = 0.5, \quad n_B = \frac{20}{40} = 0.5, \quad n_C = \frac{48}{32} = 1.5

Step 3

Simplifying to the Lowest Whole-Number Atomic Ratio

Dividing all mole numbers by the smallest mole value (0.50):

A : B : C = \frac{0.5}{0.5} : \frac{0.5}{0.5} : \frac{1.5}{0.5} = 1 : 1 : 3

Step 4

Writing the Empirical Formula

The atomic ratio 1 : 1 : 3 yields the empirical formula ABC3:

\text{Empirical Formula} = \text{ABC}_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:

Incorrect atomic ratio.

⚠️

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

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Overestimates oxygen/C count.

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Student Fallacy (Why candidates pick this trap): Dividing 48 by 12 instead of atomic mass 32.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Inverts stoichiometry of A and C.

⚠️

Student Fallacy (Why candidates pick this trap): Multiplying moles instead of dividing by smallest.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Exact whole-number molar ratio: 0.5 : 0.5 : 1.5 = 1 : 1 : 3.

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Why this is airtight: None.

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NTA Trap Alert: Forgetting to Calculate % of C First

The question gives percentages for A and B only, leaving C as "remaining percentage". Always compute %C = 100 - (32 + 20) = 48% first.

🛡️ How to avoid the -1 mark: Check that sum of all percentages equals 100% before dividing by atomic weights.

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Core NEET Syllabus Standard

Chemistry Core Physical Chemistry · Some Basic Concepts of Chemistry Standard Benchmark

Core Scientific Principle Tested:

"Empirical formula represents the simplest whole number ratio of atoms of various elements present in one molecule of the compound."

Syllabus Focus: Empirical and Molecular Formulas

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Half, half, and one-and-a-half in line: Double each piece and ABC3 will shine!

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NEET 2024 Chemistry Question 95 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Relative Mole Ratio Mental Arithmetic

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining the Mass Percentage of Element C

Calculating Moles of Each Element in 100 g of Compound

Simplifying to the Lowest Whole-Number Atomic Ratio

Writing the Empirical Formula

360° Option Elimination Engine Key Feature

NTA Trap Alert: Forgetting to Calculate % of C First

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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