Assertion (A): The empirical mass of ethene is half of its molecular mass. Reason (R): The empirical formula r

NCERT Exemplar · Q38

Q. Assertion (A): The empirical mass of ethene is half of its molecular mass. Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound. Choose the correct option out of the choices given below.

(i) Both A and R are true and R is the correct explanation of A.

(ii) A is true but R is false.

(iii) A is false but R is true.

(iv) Both A and R are false.

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Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

Imagine you're at a market buying apples. You don't weigh each apple individually — you put a dozen on a scale. The total weight tells you something about the apples, but it also depends on how many apples you have.

Atoms and molecules are unimaginably tiny. A single water molecule ( $H_2O$ ) weighs about $3 \times 10^{-23}$ grams. That number is useless for practical chemistry. So instead of working with individual molecules, chemists count them in huge fixed numbers — specifically, $6.022 \times 10^{23}$ molecules, which is called one mole.

Molecular mass is simply the mass of one mole of a substance, expressed in grams per mole (g/mol). It answers the question: "If I have $6.022 \times 10^{23}$ molecules of this compound, how much would they weigh on a lab balance?"

Note

The number $6.022 \times 10^{23}$ is Avogadro's constant. It's chosen so that the mass of one mole of carbon-12 atoms is exactly 12 grams — matching the atomic mass unit scale perfectly.

The Precise Definition

Molecular mass (also called molar mass) is the mass of one mole of a molecular substance. It is numerically equal to the sum of the atomic masses of all atoms in the molecule, expressed in g/mol.

For example:

Water ( $H_2O$ ): 2 hydrogen atoms + 1 oxygen atom
Atomic mass of H = 1.008 g/mol
Atomic mass of O = 16.00 g/mol
Molecular mass of $H_2O$ = $2(1.008) + 16.00 = 18.016$ g/mol

\text{Molecular mass} = \sum (\text{number of atoms of each element} \times \text{atomic mass of that element})

How to Calculate It — Step by Step

Let's take glucose, $C_6H_{12}O_6$ , as a worked example.

Step 1: Identify each element and its count

Carbon (C): 6 atoms
Hydrogen (H): 12 atoms
Oxygen (O): 6 atoms

Step 2: Look up atomic masses (from the periodic table)

C: 12.01 g/mol
H: 1.008 g/mol
O: 16.00 g/mol

Step 3: Multiply and add

\text{Molecular mass} = 6(12.01) + 12(1.008) + 6(16.00)

= 72.06 + 12.096 + 96.00

= 180.156 \text{ g/mol}

Tip

Always keep at least 2 decimal places from the periodic table. For exam problems, they usually give you atomic masses — use exactly what's provided.

Why This Matters

Molecular mass is the bridge between the microscopic world (atoms and molecules) and the macroscopic world (grams you can weigh). Once you know the molecular mass, you can:

Convert grams to moles: $\text{moles} = \frac{\text{mass in grams}}{\text{molecular mass}}$
Convert moles to grams: $\text{mass} = \text{moles} \times \text{molecular mass}$
Determine the number of molecules: $\text{molecules} = \text{moles} \times 6.022 \times 10^{23}$

Watch out

Do not confuse molecular mass with atomic mass. Atomic mass refers to a single element (like oxygen = 16.00 g/mol). Molecular mass refers to a compound (like $CO_2$ = 44.01 g/mol). Also, for ionic compounds like NaCl, we use formula mass (same calculation, but the substance isn't molecular).

Common Exam Pitfalls

Forgetting to multiply by the subscript. In $H_2SO_4$ , there are 2 hydrogens, not 1. …

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

Let's build this from the ground up — not as a list of formulas to memorise, but as a logical chain of reasoning.

1. The Core Question: What is a Mole?

A mole is simply a counting unit, like a dozen (12) or a gross (144). But instead of 12, a mole contains $6.022 \times 10^{23}$ particles (Avogadro's number, $N_A$ ).

Why this number?

It was chosen so that 1 mole of any substance has a mass in grams equal to its atomic/molecular mass in amu.

Example: 1 atom of carbon-12 has mass 12 amu.
1 mole of carbon-12 has mass 12 grams.

This is the bridge between the microscopic (atoms/molecules) and the macroscopic (grams we can weigh).

2. The Fundamental Relationship

The key formula is:

n = \frac{m}{M}

Where:

$n$ = number of moles
$m$ = mass of substance (in grams)
$M$ = molar mass (in g/mol)

Why does this work?

Think of it as a conversion factor:

If 1 mole of a substance weighs $M$ grams, then $m$ grams contains $\frac{m}{M}$ moles.

Derivation logic:

Molar mass $M$ tells you: "1 mol = $M$ g"
So the conversion factor is $\frac{1 \text{ mol}}{M \text{ g}}$
Multiply mass $m$ by this factor: $m \times \frac{1}{M} = \frac{m}{M}$ moles

3. Connecting to Number of Particles

n = \frac{N}{N_A}

Where:

$N$ = number of particles (atoms, molecules, ions)
$N_A = 6.022 \times 10^{23}$ particles/mol

Why?

1 mole = $N_A$ particles
So $N$ particles = $\frac{N}{N_A}$ moles

Combine both formulas:

\frac{m}{M} = \frac{N}{N_A}

This single equation ties mass, molar mass, number of particles, and Avogadro's number together.

4. The Gas Volume Connection (for gases at STP)

For gases only:

n = \frac{V}{22.4 \text{ L/mol}}

Why 22.4 L?

From the ideal gas law: $PV = nRT$

At STP (Standard Temperature and Pressure: 0°C, 1 atm):

$P = 1$ atm
$T = 273.15$ K
$R = 0.0821$ L·atm/(mol·K)

For $n = 1$ mole:

V = \frac{nRT}{P} = \frac{1 \times 0.0821 \times 273.15}{1} \approx 22.4 \text{ L}

So 1 mole of any ideal gas occupies 22.4 L at STP. This is a consequence of the gas laws, not a definition.

5. The Stoichiometry Chain: From One Substance to Another

In a balanced chemical equation like:

aA + bB \rightarrow cC + dD

The coefficients tell you the mole ratio:

\frac{\text{moles of A}}{\text{moles of B}} = \frac{a}{b}

Why this works: …

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Ethene $C_2H_4$ has empirical formula $CH_2$ (empirical mass $14$ u), which is exactly half the molecular mass ( $28$ u). So the Assertion is true, the Reason is true, and the Reason correctly explains the Assertion — option (i).

Empirical formula of ethene

Ethene is $C_2H_4$ . Reducing the subscripts to the simplest whole-number ratio, $C:H = 2:4 = 1:2$ , so the empirical formula is $CH_2$ .

Compare the masses

\text{Empirical mass}=12+2(1)=14\ \text{u},\qquad \text{Molecular mass of }C_2H_4=2(12)+4(1)=28\ \text{u}

\frac{\text{Molecular mass}}{\text{Empirical mass}}=\frac{28}{14}=2

So the empirical mass is exactly half the molecular mass — Assertion (A) is true.

The Reason …

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Here’s a breakdown of the common mistakes students make on this specific Assertion-Reason question, along with how to avoid each one.

The Correct Answer First

Assertion (A): True. Ethene is $\text{C}_2\text{H}_4$ . Its empirical formula is $\text{CH}_2$ . The empirical mass ( $12 + 2 = 14 \text{ g/mol}$ ) is exactly half of the molecular mass ( $24 + 4 = 28 \text{ g/mol}$ ).
Reason (R): True. This is the standard definition of an empirical formula.
Conclusion: Both A and R are true, and R is the correct explanation of A.

Common Mistake #1: Thinking the Assertion is False

The error: Students calculate the molecular mass of ethene incorrectly (e.g., forgetting that ethene is $\text{C}_2\text{H}_4$ , not $\text{CH}_2$ ) or assume the empirical mass must always be a simple fraction like one-third or one-fourth. They then mark the Assertion as false.

Why it happens: Rote memorization without checking the actual numbers. Many students remember that "empirical mass is less than molecular mass" but don't verify the ratio.

How to avoid: Always calculate explicitly.

Write the molecular formula: $\text{C}_2\text{H}_4$ .
Find the empirical formula: Divide subscripts by the HCF (2) → $\text{CH}_2$ .
Compute both masses:
- Empirical mass = $12 + 2(1) = 14$
- Molecular mass = $2(12) + 4(1) = 28$
Ratio = $\frac{14}{28} = \frac{1}{2}$ . The assertion is correct.

Common Mistake #2: Confusing "Empirical Mass" with "Molecular Mass"

The error: Students think the empirical mass of ethene is 28 (the molecular mass) because they confuse the two terms. They then mark the Assertion as false.

Why it happens: Weak vocabulary. "Empirical mass" and "molecular mass" sound similar, and students often use them interchangeably.

How to avoid: Define each term before answering.

Empirical mass: Sum of atomic masses from the empirical formula ( $\text{CH}_2$ ).
Molecular mass: Sum of atomic masses from the molecular formula ( $\text{C}_2\text{H}_4$ ).
Write both formulas side by side on rough paper before comparing.

Common Mistake #3: Saying the Reason is False

The error: Students claim the Reason is false because they think the empirical formula represents the "actual number of atoms" or the "molecular formula."

Why it happens: Misreading the definition. The Reason says "simplest whole number ratio" — some students skip the word "simplest" and think it means the actual formula.

How to avoid: Memorize the exact definition.

Empirical formula = simplest whole number ratio of atoms.
Molecular formula = actual number of atoms.
For ethene: empirical = $\text{CH}_2$ , molecular = $\text{C}_2\text{H}_4$ .
The Reason is a textbook definition — it is always true.

Common Mistake #4: Saying the Reason Does Not Explain the Assertion

The error: Students correctly identify both as true, but then choose "A is true, R is true, but R is not the correct explanation of A." …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
$3\times10^{22}$
›Reveal solutionSolution
Converting each quantity to grams shows 0.5 mol of ozone (24 g) has the highest mass among the four options.

Concept and Intuition

This tests careful unit conversion between grams, moles, molecules, and STP volumes into a common basis (mass in grams) so they can be directly compared.

Step-by-Step Solution

(A) 0.5 g-atom of oxygen = 0.5 mol O atoms $\times$ 16 g/mol = 8 g.

(B) 0.5 mol of ozone (O₃, molar mass 48 g/mol) = $0.5\times48=24$ g.

(C) $3\times10^{22}$ molecules of N₂: moles $=\dfrac{3\times10^{22}}{6.022\times10^{23}}\approx0.0498$ mol $\times28\ \text{g/mol}\approx1.39$ g.

(D) 5.6 L CO₂ at STP: moles $=\dfrac{5.6}{22.4}=0.25$ mol $\times44\ \text{g/mol}=11$ g. …
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AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
$^{54}Fe$
›Reveal solutionSolution
Atomic mass is the abundance-weighted average of isotopic masses; for Fe here it works out to $55.85$ .

Concept and Intuition

The atomic mass listed on the periodic table is not any single isotope's mass — it is the average of all naturally occurring isotopes, weighted by how abundant each one is.

Step-by-Step Solution

Multiply each isotope's mass by its fractional abundance: $^{54}Fe: 0.10\times54=5.4$ ; $^{56}Fe: 0.85\times56=47.6$ ; $^{57}Fe: 0.05\times57=2.85$ .

Sum: $5.4+47.6+2.85=55.85$ . …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$NO_2$
›Reveal solutionSolution
Using vapour density to get the average molar mass of the $NO_2$ / $N_2O_4$ mixture and a mole-fraction balance gives $0.437$ mol of $NO_2$ in $100$ g of mixture.

Concept and Intuition

Vapour density relates to molar mass by $M=2\times VD$ ; for a mixture of two related gases, the observed (average) molar mass is a mole-fraction-weighted average of the pure components' molar masses.

Step-by-Step Solution

Average molar mass of the mixture: $M_{avg}=2\times38.3=76.6\ g/mol$ .

Let $x$ = mole fraction of $NO_2$ ( $M=46$ ), so $(1-x)$ = mole fraction of $N_2O_4$ ( $M=92$ ): $46x+92(1-x)=76.6$ .

Solve: $92-46x=76.6\Rightarrow46x=15.4\Rightarrow x=0.3348$ .

Total moles of mixture in 100 g: $n=\dfrac{100}{76.6}=1.305\ mol$ . …
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AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
$6.643 \times 10^{-23}\ g$
›Reveal solutionSolution
Tests converting single-atom mass to molar mass via Avogadro's number, then finding moles in a bulk sample. Answer: 1250 moles.

Concept and Intuition

The mass of a single atom, multiplied by Avogadro's number ( $6.022\times10^{23}$ , the number of atoms in one mole), gives the molar mass of the element. Once we know the molar mass, converting a bulk mass into moles is a straightforward division.

Step-by-Step Solution

Molar mass $M = (\text{mass of one atom}) \times N_A = 6.643\times10^{-23}\ g \times 6.022\times10^{23}\ mol^{-1}$ .

$M \approx 6.643 \times 6.022 \approx 40.01\ g/mol$ (this is calcium, atomic mass 40). …
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AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
$Fe$
›Reveal solutionSolution
Equivalent weight of an element in a compound is its atomic mass divided by its valence (oxidation number) in that compound; for Fe(III) in $Fe_2O_3$ , this gives $56/3 \approx 18.6$ .

Concept and Intuition

Equivalent weight expresses how much mass of an element corresponds to a single 'unit of combining power' (one unit of charge/valence). It is defined as $\text{Equivalent weight} = \dfrac{\text{Atomic mass}}{\text{Valence}}$ . The valence to use is the oxidation state the element actually has in the given compound.

Step-by-Step Solution

In $Fe_2O_3$ , oxygen is $-2$ ; overall neutral, so $2\times(\text{Fe oxidation state}) + 3\times(-2) = 0 \Rightarrow$ Fe oxidation state $=+3$ .

Equivalent weight of Fe $= \dfrac{\text{Atomic mass of Fe}}{\text{valence}} = \dfrac{56}{3}$ . …
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AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
$3.011 \times 10^{22}$
›Reveal solutionSolution
Converting the given number of atoms to moles (via Avogadro's number) and dividing the given mass by that mole count gives an atomic mass of $23\ amu$ — consistent with sodium.

Concept and Intuition

The mole concept links a countable number of atoms to a measurable mass via Avogadro's number ( $6.022\times10^{23}$ per mole) and molar mass (mass per mole). Given both the atom count and the corresponding mass, we can directly compute the molar (atomic) mass.

Step-by-Step Solution

Convert atom count to moles:

$n = \frac{3.011\times10^{22}}{6.022\times10^{23}} = 0.05\ mol$

Atomic mass = mass per mole:

$M = \frac{\text{given mass}}{n} = \frac{1.15\ g}{0.05\ mol} = 23\ g/mol$

So the atomic mass is $23\ amu$ .

Common Mistakes …
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←Previous questionQ37 — Match the following physical quantities with units: Physical quantity:…→Next questionQ39 — Assertion (A): One atomic mass unit is defined as one twelfth of the m…

Q.Assertion (A): The empirical mass of ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
Choose the correct option out of the choices given below.

Concept understanding — Molecular Mass Calculation

What is Molecular Mass? The Intuition

The Precise Definition

How to Calculate It — Step by Step

Why This Matters

Common Exam Pitfalls

Why this formula?

Stoichiometry & Mole Calculation: The "Why" Behind the Formula

1. The Core Question: What is a Mole?

2. The Fundamental Relationship

Why does this work?

3. Connecting to Number of Particles

Why?

4. The Gas Volume Connection (for gases at STP)

Why 22.4 L?

5. The Stoichiometry Chain: From One Substance to Another

Why this works: …

Concept: Empirical Formula vs Molecular Formula

Method: Empirical Formula Determination

The Correct Answer First

Common Mistake #1: Thinking the Assertion is False

Common Mistake #2: Confusing "Empirical Mass" with "Molecular Mass"

Common Mistake #3: Saying the Reason is False

Common Mistake #4: Saying the Reason Does Not Explain the Assertion

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Q.Assertion (A): The empirical mass of ethene is half of its molecular mass. Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound. Choose the correct option out of the choices given below.

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Q.Assertion (A): The empirical mass of ethene is half of its molecular mass.
Reason (R): The empirical formula represents the simplest whole number ratio of various atoms present in a compound.
Choose the correct option out of the choices given below.

Unlock everything free for 14 days