The empirical formula and molecular mass of a compound are $CH_2O$ and 180 g respectively. What will be the molecular formula of the compound? (i) $C_9H_{18}O_9$ (ii) $CH_2O$ (iii) $C_6H_{12}O_6$ (iv) $C_2H_4O_2$

The empirical formula and molecular mass of a compound are CH₂O and 180 g respectively. What will be the molec

NCERT Exemplar · Q10

CH_2O

(i)

C_9H_{18}O_9

(ii)

CH_2O

(iii)

C_6H_{12}O_6

(iv)

C_2H_4O_2

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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 molecular formula is a whole-number multiple of the empirical formula. By calculating the empirical formula mass and comparing it to the given molecular mass, we find a scaling factor of 6, leading to the molecular formula $\boxed{C_6H_{12}O_6}$ .

In chemistry, formulas tell us about the composition of a compound. An empirical formula represents the simplest whole-number ratio of atoms in a compound. For example, if a compound has 2 carbon atoms and 4 hydrogen atoms, its empirical formula would be $CH_2$ (a 1:2 ratio).

The molecular formula, on the other hand, shows the actual number of atoms of each element in a molecule. For the example above, if the actual molecule has 2 carbon and 4 hydrogen atoms, its molecular formula would be $C_2H_4$ . Notice that $C_2H_4$ is $(CH_2)_2$ . This illustrates a key relationship: the molecular formula is always an integer multiple of the empirical formula.

\text{Molecular Formula} = (\text{Empirical Formula})_n

Here, $n$ is a whole number (1, 2, 3, ...). This means the molecular mass will also be $n$ times the empirical formula mass.

\text{Molecular Mass} = n \times (\text{Empirical Formula Mass})

Our goal is to find this scaling factor $n$ , and then use it to convert the empirical formula into the molecular formula.

Identify the given information.

We are given:
- Empirical formula: $CH_2O$
- Molecular mass: 180 g/mol
Calculate the empirical formula mass.

To do this, we need the atomic masses of the elements involved:
- Carbon (C): 12 g/mol
- Hydrogen (H): 1 g/mol
- Oxygen (O): 16 g/mol
The empirical formula $CH_2O$ contains 1 carbon atom, 2 hydrogen atoms, and 1 oxygen atom.

Empirical formula mass $= (1 \times \text{Atomic mass of C}) + (2 \times \text{Atomic mass of H}) + (1 \times \text{Atomic mass of O})$

Empirical formula mass $= (1 \times 12 \text{ g/mol}) + (2 \times 1 \text{ g/mol}) + (1 \times 16 \text{ g/mol})$

Empirical formula mass $= 12 \text{ g/mol} + 2 \text{ g/mol} + 16 \text{ g/mol}$

Empirical formula mass $= 30 \text{ g/mol}$
Determine the scaling factor 'n'.

The scaling factor $n$ is the ratio of the molecular mass to the empirical formula mass. …

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Q.The empirical formula and molecular mass of a compound are $CH_2O$ and 180 g respectively. What will be the molecular formula of the compound?

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: Molecular Formula from Empirical Formula

Method: Empirical Formula Multiplier Method

Steps

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting to calculate the empirical formula mass

Mistake 2: Using the wrong ratio formula

Mistake 3: Stopping at $n$ and not multiplying

Mistake 4: Confusing empirical and molecular formulas

Q.The empirical formula and molecular mass of a compound are CH2OCH_2OCH2​O and 180 g respectively. What will be the molecular formula of the compound?

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Q.The empirical formula and molecular mass of a compound are $CH_2O$ and 180 g respectively. What will be the molecular formula of the compound?

Unlock everything free for 14 days