Which of the following reactions is not correct according to the law of conservation of mass. (i) 2Mg(s) + O₂(

NCERT Exemplar · Q14

Q. Which of the following reactions is not correct according to the law of conservation of mass.

(i)

2Mg(s) + O_2(g) \rightarrow 2MgO(s)

(ii)

C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)

(iii)

P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)

(iv)

CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)

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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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The law of conservation of mass requires equal numbers of each type of atom on both sides of a chemical equation. Option (ii) is unbalanced: $C_3H_8 + O_2 \rightarrow CO_2 + H_2O$ has 3 carbons and 8 hydrogens on the left but only 1 carbon and 2 hydrogens on the right.

The law of conservation of mass states that matter cannot be created or destroyed in a chemical reaction. In practical terms, this means every atom present before the reaction must be accounted for after the reaction. A correctly balanced chemical equation reflects this principle: the number of atoms of each element on the reactant side must equal the number on the product side.

When an equation violates this law, it's not that nature is breaking rules—it's that the equation is incorrectly written and doesn't represent what actually happens.

Let me check each reaction by counting atoms of each element on both sides.

Checking Each Reaction

1. Option (i): $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Left side: 2 magnesium atoms, 2 oxygen atoms

Right side: 2 magnesium atoms (in $2MgO$ ), 2 oxygen atoms (in $2MgO$ )

Every atom is accounted for. This equation is balanced.

2. Option (ii): $C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$

Left side: 3 carbon atoms, 8 hydrogen atoms, 2 oxygen atoms

Right side: 1 carbon atom (in $CO_2$ ), 2 hydrogen atoms (in $H_2O$ ), 3 oxygen atoms (2 in $CO_2$ + 1 in $H_2O$ )

The carbon and hydrogen counts don't match. We started with 3 carbons but ended with only 1; we started with 8 hydrogens but ended with only 2. This equation violates conservation of mass.

Watch out

A common mistake is to assume that if an equation is written down, it must be correct. Always verify by counting atoms—especially in combustion reactions where multiple products form.

3. Option (iii): $P_4(s) + 5O_2(g) \rightarrow P_4O_{10}(s)$

Left side: 4 phosphorus atoms, 10 oxygen atoms …

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Concept: Law of Conservation of Mass

The law states that mass is neither created nor destroyed in a chemical reaction. This means every balanced chemical equation must have the same number of atoms of each element on both sides.

Method: Atom-Counting Method

Steps

List all elements appearing in the reaction.
Count atoms of each element on the reactant side (left).
Count atoms of each element on the product side (right).
Compare the counts. If any element has unequal numbers, the equation is not balanced — and therefore violates the law of conservation of mass.

Applying the method

(i) $2Mg(s) + O_2(g) \rightarrow 2MgO(s)$

Element	Reactants	Products
Mg	2	2
O	2	2

✓ Balanced

(ii) $C_3H_8(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)$

Element	Reactants	Products
C	3	1
H	8	2
O	2	3

✗ Not balanced — carbon, hydrogen, and oxygen all differ.

--- …

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Common Mistakes & How to Avoid Them

Mistake 1: Not balancing the equation before checking conservation of mass

The error: Students look at the given equation and assume it is already balanced, then check if atoms match. But in option (ii), the equation is unbalanced — and that is exactly the trap.

How to avoid: Always count atoms on both sides systematically before concluding anything.

Left side: C = 3, H = 8, O = 2
Right side: C = 1, H = 2, O = 3

Clearly, the numbers don't match. This violates the law of conservation of mass.

✓ Correct approach: Write a quick tally for each element. If any element has unequal counts, the equation is incorrect.

Mistake 2: Confusing "not correct" with "not possible"

The error: Some students think the question asks which reaction cannot happen chemically. They may rule out (iii) thinking $P_4O_{10}$ is unusual, or rule out (i) thinking Mg doesn't react with $O_2$ that way.

How to avoid: The law of conservation of mass is about atom balance, not reaction feasibility. Focus only on whether the number of atoms of each element is the same on both sides.

(i) ✓ Balanced: 2 Mg, 2 O on both sides
(iii) ✓ Balanced: 4 P, 10 O on both sides
(iv) ✓ Balanced: 1 C, 4 H, 4 O on both sides
(ii) ✗ Unbalanced — this is the only one that violates the law.

Mistake 3: Miscounting atoms in polyatomic molecules

The error: In option (ii), students miscount oxygen atoms in $CO_2$ and $H_2O$ , or forget that $O_2$ has 2 oxygen atoms per molecule.

How to avoid: Write the count clearly:

Element	Left side	Right side
C	3	1
H	8	2
O	2	2 + 1 = 3

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Q.Which of the following reactions is not correct according to the law of conservation of mass.

Checking Each Reaction

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