Q.Determine the molecular formula of an oxide of iron, in which the mass per cent of iron and oxygen are 69.9 and 30.1, respectively.

The key idea is Empirical Formula Determination.

1. Convert mass percentages to moles: Assume a $100 \text{ g}$ sample.

   Moles of Fe $= \frac{69.9 \text{ g}}{55.85 \text{ g/mol}} = 1.2516 \text{ mol}$

   Moles of O $= \frac{30.1 \text{ g}}{16.00 \text{ g/mol}} = 1.8813 \text{ mol}$

2. Determine the simplest mole ratio: Divide by the smallest number of moles ($1.2516$).

   Fe: $\frac{1.2516}{1.2516} = 1$

   O: $\frac{1.8813}{1.2516} \approx 1.503 \approx 1.5$

3. Convert to whole number ratio: Multiply the ratio $(1 : 1.5)$ by $2$ to obtain whole numbers.

   Fe: $1 \times 2 = 2$

   O: $1.5 \times 2 = 3$ …

When we talk about the formula of a chemical compound, we are essentially describing the ratio of atoms of each element present in it. The mass percentages tell us how much mass each element contributes to the total mass of the compound. However, atoms combine in simple whole-number ratios of moles, not masses. Therefore, the core idea is to convert these given mass percentages into a mole ratio, which then directly gives us the subscripts in the chemical formula.

For ionic compounds like iron oxides, the term "molecular formula" is often used interchangeably with "empirical formula" because they do not exist as discrete molecules but rather as extended crystal lattices. The empirical formula represents the simplest whole-number ratio of atoms in the compound. Without additional information (like the molar mass of the compound), we can only determine the empirical formula from mass percentages.

Here's how we determine the formula:

1. Assume a basis for calculation.

   To make calculations straightforward, we assume we have a $100 \text{ g}$ sample of the compound. This allows us to directly convert the given mass percentages into actual masses.

   • Mass of Iron (Fe) = $69.9\%$ of $100 \text{ g} = 69.9 \text{ g}$
   • Mass of Oxygen (O) = $30.1\%$ of $100 \text{ g} = 30.1 \text{ g}$

2. Convert the mass of each element to moles.

   We use the atomic masses of iron and oxygen to convert their respective masses into moles. The atomic mass of Iron (Fe) is approximately $55.85 \text{ g/mol}$, and that of Oxygen (O) is approximately $16.00 \text{ g/mol}$.

   • Moles of Iron (Fe):

*   Moles of Oxygen (O):

3. Determine the simplest whole-number mole ratio.

   To find the simplest ratio, we divide the number of moles of each element by the smallest number of moles calculated. This normalizes one of the elements to '1'.

   • Smallest number of moles = $1.2516 \text{ mol}$ (for Fe)

   • Ratio for Fe:

*   Ratio for O:

So, the preliminary mole ratio of Fe : O is $1 : 1.5$.

4. Convert the ratio to the smallest whole numbers. …

Concept: Molecular Mass Calculation (Empirical Formula from Percent Composition)

Step 1: Assume 100 g of the compound

• Mass of iron (Fe) = 69.9 g
• Mass of oxygen (O) = 30.1 g

Step 2: Convert masses to moles

• Moles of Fe = $\frac{69.9}{55.85} \approx 1.25$
• Moles of O = $\frac{30.1}{16.00} \approx 1.88$

Step 3: Find simplest whole-number ratio

Divide by the smallest number of moles (1.25): …

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., $180 \, \text{u}$) …