Question

An organic compound has the empirical formula CH$_2$O. Its vapour density is 45. The molecular formula of compound is

• A. CH$_2$O
• B. C$_2$H$_5$O
• C. C$_2$H$_4$O$_2$
• D. C$_3$H$_6$O$_3$

Hint:

In competitive exams, if you've already calculated the molecular mass ($90$), you can quickly check the options to see which one sums to that mass. For (D): $(3 \times 12) + (6 \times 1) + (3 \times 16) = 36 + 6 + 48 = 90$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound.
The molecular formula gives the actual number of atoms of each element.
The molecular formula is always a whole number multiple (\(n\)) of the empirical formula: \(\text{Molecular Formula} = n \times (\text{Empirical Formula})\).
Step 2: Key Formula or Approach:
1. \(\text{Molecular Mass} = 2 \times \text{Vapour Density}\)
2. \(n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}\)
Step 3: Detailed Explanation:
1. Calculate Empirical Formula Mass:
Empirical Formula = \(CH_2O\).
Masses: \(C = 12, H = 1, O = 16\).
Empirical Mass \(= 12 + (2 \times 1) + 16 = 30\).
2. Calculate Molecular Mass:
Given Vapour Density = 45.
Molecular Mass \(= 2 \times 45 = 90\).
3. Calculate the multiplier \(n\):
\[ n = \frac{90}{30} = 3 \]
4. Determine Molecular Formula:
\[ \text{Molecular Formula} = 3 \times (CH_2O) = C_3H_6O_3 \]
Step 4: Final Answer:
The molecular formula of the compound is \(C_3H_6O_3\).