Q.Calculate the amount of water

The key idea is to use the stoichiometric ratio from the balanced chemical equation for combustion.

Step 1: Write the balanced equation.

Step 2: Find moles of methane.

Molar mass of $\text{CH}_4 = 12 + 4 \times 1 = 16 \text{ g/mol}$.

Moles of $\text{CH}_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol}$.

Step 3: Use mole ratio to find moles of water.

From the equation, 1 mol $\text{CH}_4$ produces 2 mol $\text{H}_2\text{O}$.

So, moles of $\text{H}_2\text{O} = 2 \text{ mol}$.

Step 4: Convert moles of water to mass.

Molar mass of $\text{H}_2\text{O} = 2 \times 1 + 16 = 18 \text{ g/mol}$.

Mass of water $= 2 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g}$.

Concept and Intuition

This problem is a classic stoichiometry calculation. The core idea is that a balanced chemical equation tells us the exact mole-to-mole relationship between reactants and products. For every 1 mole of methane ($CH_4$) that burns completely, 2 moles of water ($H_2O$) are produced. So, if we know how many moles of methane we start with, we can directly find how many moles of water are formed. Then, we convert those moles of water into grams using its molar mass.

The "why" behind this approach: Chemistry happens in fixed ratios of particles (molecules, atoms, ions). We can't directly count particles, but we can measure mass. Molar mass is the bridge between the mass we measure in the lab and the number of moles (which represents a fixed number of particles, $6.022 \times 10^{23}$). Once we're in the world of moles, the balanced equation gives us the conversion factor.

Step-by-Step Solution

1. Write and balance the chemical equation for the combustion of methane.

Combustion means burning in the presence of oxygen ($O_2$). The products are always carbon dioxide ($CO_2$) and water ($H_2O$). The unbalanced equation is:

Balancing it: We have 1 carbon on each side, 4 hydrogens on the left require 2 water molecules on the right, and that gives 4 oxygens on the right, so we need 2 oxygen molecules on the left.

2. Calculate the number of moles of methane in the given 16 g sample.

We need the molar mass of methane ($CH_4$).

• Carbon (C): 1 atom × 12 g/mol = 12 g/mol
• Hydrogen (H): 4 atoms × 1 g/mol = 4 g/mol
• Molar mass of $CH_4$ = 12 + 4 = 16 g/mol

Now, use the formula: $\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$

3. Use the mole ratio from the balanced equation to find the moles of water produced.

From the balanced equation: 1 mole of $CH_4$ produces 2 moles of $H_2O$.

So, if we have 1 mole of $CH_4$, the moles of $H_2O$ produced are:

4. Convert the moles of water into grams.

We need the molar mass of water ($H_2O$).

• Hydrogen (H): 2 atoms × 1 g/mol = 2 g/mol
• Oxygen (O): 1 atom × 16 g/mol = 16 g/mol
• Molar mass of $H_2O$ = 2 + 16 = 18 g/mol

Now, use the formula: $\text{Mass} = \text{Moles} \times \text{Molar Mass}$

Concept: Molecular Mass Calculation & Stoichiometry

Step 1: Write the balanced chemical equation.

Combustion of methane:

Step 2: Find molar masses.

• Molar mass of $\text{CH}_4 = 12 + 4 \times 1 = 16 \text{ g/mol}$
• Molar mass of $\text{H}_2\text{O} = 2 \times 1 + 16 = 18 \text{ g/mol}$

Step 3: Use mole ratio from equation.

1 mole $\text{CH}_4$ (16 g) produces 2 moles $\text{H}_2\text{O}$ (2 × 18 = 36 g).

Given 16 g of $\text{CH}_4$ is exactly 1 mole, so water produced = 36 g.

Final Answer:

36 g of water is produced.

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 $180 \, \text{amu}$)