Q.Calculate the molar mass of the following:

Concept: Molecular Mass Calculation – the sum of the atomic masses of all atoms in a molecule, using atomic masses from the periodic table (H = 1 u, C = 12 u, O = 16 u).

Step 1: Identify the number of each atom in the molecule.

Step 2: Multiply each atomic mass by its count.

Step 3: Add the contributions.

• (i) $H_2O$: $2 \times 1 + 1 \times 16 = 2 + 16 = 18$ u
• (ii) $CO_2$: $1 \times 12 + 2 \times 16 = 12 + 32 = 44$ u
• (iii) $CH_4$: $1 \times 12 + 4 \times 1 = 12 + 4 = 16$ u

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.

1. 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 $18.016\ \text{g/mol}$.

2. 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 $44.01\ \text{g/mol}$.

3. 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 $16.043\ \text{g/mol}$.

Concept: Molecular Mass Calculation

The molar mass is the sum of the atomic masses of all atoms in a molecule, expressed in g/mol.

Step 1: Identify atomic masses (from periodic table)

• H = 1 u, O = 16 u, C = 12 u

Step 2: Compute for each molecule

• (i) $H_2O$: $2 \times 1 + 16 = 18$ g/mol
• (ii) $CO_2$: $12 + 2 \times 16 = 44$ g/mol
• (iii) $CH_4$: $12 + 4 \times 1 = 16$ g/mol

Final Answer:

• (i) 18 g/mol
• (ii) 44 g/mol
• (iii) 16 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

Final Correct Answer:

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