Calculate the molar mass of the following: (i) H₂O (ii) CO₂ (iii) CH₄

Exercises · 1.1

Q. Calculate the molar mass of the following:

(i)

H_2O

(ii)

CO_2

(iii)

CH_4

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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.
Using atomic number instead of atomic mass. Atomic number (protons) is not mass.
Rounding too early. Keep 2-3 decimal places until the final answer.
Confusing molecular mass with molecular weight. They mean the same thing — both are in g/mol.

Quick Reference Table

Substance	Formula	Calculation	Molecular Mass (g/mol)
Oxygen gas	$O_2$	$2(16.00)$	32.00
Carbon dioxide	$CO_2$	$12.01 + 2(16.00)$	44.01
Methane	$CH_4$	$12.01 + 4(1.008)$	16.042
Sodium chloride	$NaCl$	$22.99 + 35.45$	58.44

The last one is a formula mass (ionic compound), but the calculation is identical.

The Big Picture

Molecular mass is not a property you measure directly — it's a calculated value from the periodic table. Every molecule of a given compound has the same molecular mass. When you weigh out that many grams, you know exactly how many moles (and therefore how many molecules) you have. That's the foundation of all stoichiometry.

Searches like "molecular mass calculation formula chemistry" and "mole concept class 11 chemistry" are extremely common, since this is one of the very first skills taught in the Some Basic Concepts of Chemistry chapter of the NCERT/CBSE Class 11 curriculum. Molecular mass calculations underpin virtually every stoichiometry question in board exams, JEE Main, and NEET.

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:

Coefficients represent relative numbers of molecules (or moles of molecules)
If $a$ molecules of A react with $b$ molecules of B, then $a$ moles of A react with $b$ moles of B
The ratio is fixed by the balanced equation

The complete problem-solving chain:

\text{Mass of A} \xrightarrow{\div M_A} \text{Moles of A} \xrightarrow{\times \frac{c}{a}} \text{Moles of C} \xrightarrow{\times M_C} \text{Mass of C}

Each step uses one of the relationships above.

Summary: The Logical Flow

What you know	Formula	Why it works
Mass of substance	$n = m/M$	Molar mass is the conversion factor between grams and moles
Number of particles	$n = N/N_A$	Avogadro's number is the conversion factor between particles and moles
Volume of gas (STP)	$n = V/22.4$	Derived from ideal gas law at standard conditions
Moles of one reactant	$n_C = n_A \times (c/a)$	Balanced equation gives fixed mole ratios

The mole is the universal translator — it converts between mass, particle count, and gas volume, allowing you to move seamlessly through a chemical reaction.

Molar mass is the mass of one mole of a substance, found by summing the atomic masses of all atoms in its formula. For $H_2O$ it is $18.016\ \text{g/mol}$ , for $CO_2$ it is $44.01\ \text{g/mol}$ , and for $CH_4$ it is $16.043\ \text{g/mol}$ .

The idea behind molar mass is simple: a mole is just a counting unit — $6.022 \times 10^{23}$ particles. If you know the mass of a single atom (its atomic mass in atomic mass units, u), then one mole of those atoms has the same numerical mass in grams. So to find the molar mass of a molecule, you add up the atomic masses of all the atoms in it, and the result is in grams per mole.

Let’s walk through each compound.

Water ( $H_2O$ )

Water has two hydrogen atoms and one oxygen atom.
- Atomic mass of hydrogen: $1.008\ \text{u}$
- Atomic mass of oxygen: $16.00\ \text{u}$ So the molecular mass = $2 \times 1.008 + 16.00 = 2.016 + 16.00 = 18.016\ \text{u}$ . Therefore, the molar mass of water is $18.016\ \text{g/mol}$ .
Carbon dioxide ( $CO_2$ )

One carbon atom and two oxygen atoms.
- Atomic mass of carbon: $12.01\ \text{u}$
- Atomic mass of oxygen: $16.00\ \text{u}$ Molecular mass = $12.01 + 2 \times 16.00 = 12.01 + 32.00 = 44.01\ \text{u}$ . So molar mass of $CO_2$ is $44.01\ \text{g/mol}$ .
Methane ( $CH_4$ )

One carbon atom and four hydrogen atoms.
- Carbon: $12.01\ \text{u}$
- Hydrogen: $1.008\ \text{u}$ Molecular mass = $12.01 + 4 \times 1.008 = 12.01 + 4.032 = 16.042\ \text{u}$ . (Using more precise values: $12.011 + 4 \times 1.008 = 12.011 + 4.032 = 16.043\ \text{u}$ ) So molar mass of $CH_4$ is $16.043\ \text{g/mol}$ .

Watch out

A common mistake is to forget that the atomic mass of hydrogen is about $1.008$ , not exactly $1$ . That small difference adds up, especially in molecules with many hydrogens. Also, always use the same precision for all atoms — don’t mix $12.01$ with $1.008$ inconsistently.

Tip

You don’t need to memorise every atomic mass. For exams, the periodic table is usually provided. But the common ones — H, C, O, N, Cl — are worth remembering to save time.

✓Final answer

\boxed{18.016\ \text{g/mol}}

Common Mistakes in Molecular Mass Calculation (Glucose)

Here are the most frequent errors students make when calculating the molecular mass of glucose ( $C_6H_{12}O_6$ ), along with how to avoid each.

1. Using Wrong Atomic Mass Values

The Mistake:

Students often use approximate values like $C = 12$ , $H = 1$ , $O = 16$ without checking the problem's requirement. Some exam questions expect precise values (e.g., $C = 12.01$ , $H = 1.008$ , $O = 16.00$ ).

How to Avoid:

Always read the question carefully — if it says "atomic masses" or gives a table, use those exact values.
For standard CBSE/ICSE exams, the accepted values are:
- $C = 12.0 \, \text{u}$
- $H = 1.0 \, \text{u}$
- $O = 16.0 \, \text{u}$
If no values are given, use the standard ones above.

2. Miscounting the Number of Atoms

The Mistake:

Misreading the subscript — for example, thinking glucose has 6 oxygen atoms (correct) but accidentally using 12 for hydrogen (correct) or 6 for carbon (correct). The error often comes from rushing.

How to Avoid:

Write the formula clearly: $C_6H_{12}O_6$
Count each element separately:
- Carbon: 6 atoms
- Hydrogen: 12 atoms
- Oxygen: 6 atoms
Double-check by adding subscripts: $6 + 12 + 6 = 24$ atoms total.

3. Forgetting to Multiply Atomic Mass by Number of Atoms

The Mistake:

Adding atomic masses directly without multiplying by the subscript. For example:

$12 + 1 + 16 = 29 \, \text{u}$ (wrong!)

How to Avoid:

Use the formula: $\text{Molecular mass} = (\text{atomic mass of C} \times 6) + (\text{atomic mass of H} \times 12) + (\text{atomic mass of O} \times 6)$
Write each term separately:
- Carbon: $12 \times 6 = 72$
- Hydrogen: $1 \times 12 = 12$
- Oxygen: $16 \times 6 = 96$
Then add: $72 + 12 + 96 = 180 \, \text{u}$

4. Arithmetic Errors in Addition

The Mistake:

Simple addition mistakes, e.g., $72 + 12 = 84$ (correct), then $84 + 96 = 180$ (correct), but sometimes students write $180$ as $190$ or $170$ .

How to Avoid:

Do the addition step-by-step:
- $72 + 12 = 84$
- $84 + 96 = 180$
Verify by adding in a different order: $96 + 72 = 168$ , then $168 + 12 = 180$ .
Use a calculator if allowed, but always recheck mentally.

5. Confusing Molecular Mass with Molar Mass

The Mistake:

Writing the answer as $180 \, \text{g}$ instead of $180 \, \text{u}$ (atomic mass units). Molecular mass is dimensionless in u, while molar mass is in g/mol.

How to Avoid:

Molecular mass = sum of atomic masses in u (e.g., $180 \, \text{u}$ )
Molar mass = same numerical value but in g/mol (e.g., $180 \, \text{g/mol}$ )
In the question "Calculate the molecular mass," the answer must be in u (or amu).

6. Not Showing Steps (Losing Method Marks)

The Mistake:

Writing only the final answer $180$ without showing the multiplication and addition steps.

How to Avoid:

Always write the full calculation:
- Step 1: $C: 12 \times 6 = 72$
- Step 2: $H: 1 \times 12 = 12$
- Step 3: $O: 16 \times 6 = 96$
- Step 4: Total $= 72 + 12 + 96 = 180 \, \text{u}$
This ensures partial credit even if the final answer is wrong.

Quick Summary Checklist

Mistake	How to Avoid
Wrong atomic masses	Use standard values or given data
Miscount atoms	Write formula and count carefully
Forget multiplication	Multiply each atomic mass by subscript
Addition errors	Add step-by-step and verify
Wrong units	Answer in u for molecular mass
No steps shown	Show all multiplication and addition

Final Correct Answer:

$\boxed{180 \, \text{u}}$ (or $180 \, \text{amu}$ )

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

Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2023Set ANNUAL2 marks
$Q. Pressure of 1g of an ideal gas 'A' at 27°C is found to be 2 bar. When 2g of another ideal gas 'B' is introduced in the same flask at the same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.$
›Reveal solutionSolution

Setting up PV = nRT for each gas separately and dividing the equations gives M_B = 4 x M_A.

Given: 1 g of ideal gas A at 27 deg C (300 K) gives a pressure of 2 bar in a flask of volume V. Then 2 g of ideal gas B is added at the SAME temperature and volume, and the total pressure becomes 3 bar.

Step 1 -- Partial pressure of B (Dalton's Law of partial pressures):

P(total) = P(A) + P(B)

3 = 2 + P(B) => P(B) = 1 bar

Step 2 -- Apply the ideal gas equation to gas A:

P(A) V = n(A) R T = (1 g / M_A) R T

2V = (1/M_A) R T ... (i)

Step 3 -- Apply the ideal gas equation to gas B:

P(B) V = n(B) R T = (2 g / M_B) R T

1.V = (2/M_B) R T ... (ii)

Step 4 -- Divide equation (i) by equation (ii):

(2V) / (1V) = [(1/M_A) R T] / [(2/M_B) R T]

2 = (1/M_A) x (M_B/2)

2 = M_B / (2 M_A)

M_B = 4 M_A

✓Final answer
$M_B = 4 x M_A -- the molar mass of gas B is four times that of gas A.$
Council of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2020Set ANNUAL2 marks
$Q. At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?$
›Reveal solutionSolution

Using density = PM/RT at the same temperature for both gases, equal densities mean P1·M1 = P2·M2; solving with the known values gives a molar mass of 70 g/mol for the oxide.

From the ideal gas equation, PV = nRT = (mass/M)RT, so density (d = mass/V) is:

d = PM/RT

Given both gases are at the same temperature (0°C = 273 K):

For the oxide: d_oxide = P_oxide × M_oxide / RT, with P_oxide = 2 bar

For N2: d_N2 = P_N2 × M_N2 / RT, with P_N2 = 5 bar, M_N2 = 28 g/mol

Since d_oxide = d_N2 and RT is the same for both (same temperature):

P_oxide × M_oxide = P_N2 × M_N2

2 × M_oxide = 5 × 28

2 × M_oxide = 140

M_oxide = 70 g/mol

✓Final answer
$The molecular mass of the oxide is 70 g/mol.$

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Q.Calculate the molar mass of the following:

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

Quick Reference Table

The Big Picture

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:

Summary: The Logical Flow

Concept: Molecular Mass Calculation

Common Mistakes in Molecular Mass Calculation (Glucose)

1. Using Wrong Atomic Mass Values

2. Miscounting the Number of Atoms

3. Forgetting to Multiply Atomic Mass by Number of Atoms

4. Arithmetic Errors in Addition

5. Confusing Molecular Mass with Molar Mass

6. Not Showing Steps (Losing Method Marks)

Quick Summary Checklist

Q.Calculate the molar mass of the following:

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