Which of the following statements about a compound is incorrect? (i) A molecule of a compound has atoms of dif

NCERT Exemplar · Q12

Q. Which of the following statements about a compound is incorrect?

(i) A molecule of a compound has atoms of different elements.

(ii) A compound cannot be separated into its constituent elements by physical methods of separation.

(iii) A compound retains the physical properties of its constituent elements.

(iv) The ratio of atoms of different elements in a compound is fixed.

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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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A compound is a pure substance formed from atoms of different elements in a fixed ratio, with properties entirely different from its elements — so the statement that it retains the physical properties of its constituents is false.

The question tests your understanding of what a compound is at the most fundamental level. In chemistry, a compound is not just a mixture thrown together; it's a new substance with its own identity. The key idea is that when elements combine chemically, they lose their individual properties and give rise to something entirely new.

Let's examine each statement carefully.

Statement (i): "A molecule of a compound has atoms of different elements."

This is the very definition of a compound. Water ( $H_2O$ ) has hydrogen and oxygen atoms; carbon dioxide ( $CO_2$ ) has carbon and oxygen. If a molecule had only one kind of atom, it would be an element, not a compound. So (i) is correct.
Statement (ii): "A compound cannot be separated into its constituent elements by physical methods of separation."

Physical methods like filtration, distillation, or magnetism work on mixtures because the components keep their individual properties. In a compound, atoms are bonded chemically — you'd need a chemical reaction (like electrolysis or heating) to break those bonds. For example, you can't just filter salt out of salt water; you have to evaporate the water (physical) to get salt, but to get sodium and chlorine gas from salt, you need electrolysis (chemical). So (ii) is correct.
Statement (iii): "A compound retains the physical properties of its constituent elements." …

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Here are the common mistakes students make on this question, along with how to avoid each.

Mistake 1: Confusing "mixture" properties with "compound" properties

The error: Students pick statement (iii) as correct because they think a compound does retain the properties of its elements. This is the classic trap — it sounds intuitive, but it’s wrong.
Why it happens: In everyday life, mixing things (like sand and salt) keeps each component’s properties. Students incorrectly extend this idea to compounds.
How to avoid: Memorise the key distinction:
- Mixture: retains properties of its components.
- Compound: has entirely new properties, different from its elements.
- Example: Sodium (explosive metal) + Chlorine (poisonous gas) → Sodium chloride (table salt, harmless). The compound does not retain the properties of Na or Cl.

Mistake 2: Misreading "incorrect" as "correct"

The error: The question asks for the incorrect statement. Students sometimes rush and mark the first correct-looking option.
Why it happens: Exam pressure or skimming the question too quickly.
How to avoid:
- Underline the word "incorrect" in the question before reading the options.
- After choosing an answer, double-check: “Is this the one that is false?”

Mistake 3: Thinking "fixed ratio" means "always 1:1"

The error: Students reject statement (iv) because they think a compound can have variable ratios (e.g., in some oxides of nitrogen).
Why it happens: They confuse compounds with mixtures or non-stoichiometric compounds (rare exceptions not in standard syllabus).
How to avoid:
- For all standard exam purposes, a compound has a fixed ratio by mass and by atoms (Law of Constant Proportion).
- Example: Water is always $H_2O$ (2:1 ratio of H to O). If the ratio changes, it becomes a different compound (e.g., $H_2O_2$ ).

Mistake 4: Overthinking statement (ii) — "physical methods of separation"

The error: Some students think distillation or crystallisation can separate compounds (e.g., salt from water). …

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Q.Which of the following statements about a compound is incorrect?

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: Properties of Compounds (vs. Mixtures)

Method: Elimination by Definition

Mistake 1: Confusing "mixture" properties with "compound" properties

Mistake 2: Misreading "incorrect" as "correct"

Mistake 3: Thinking "fixed ratio" means "always 1:1"

Mistake 4: Overthinking statement (ii) — "physical methods of separation"

Q.Which of the following statements about a compound is incorrect?

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