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

Concept: The relationship between molecular formula and empirical formula is given by: \[ \text{Molecular Formula} = n \times (\text{Empirical Formula}) \] To find the multiplier $n$, we use the formula: \[ n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} \] Additionally, the molecular mass of a compound is twice its vapour density: \[ \text{Molecular Mass} = 2 \times \text{Vapour Density} \]

Step 1: Calculate the Empirical Formula Mass.
The empirical formula is CH$_2$O.
• Mass of C = $1 \times 12 = 12$
• Mass of H = $2 \times 1 = 2$
• Mass of O = $1 \times 16 = 16$
• Empirical Formula Mass = $12 + 2 + 16 = 30$

Step 2: Calculate the Molecular Mass.
Given vapour density is $45$. \[ \text{Molecular Mass} = 2 \times 45 = 90 \]

Step 3: Determine the value of $n$ and the Molecular Formula.
\[ n = \frac{90}{30} = 3 \] Now, multiply the empirical formula by $n$: \[ \text{Molecular Formula} = 3 \times (\text{CH}_2\text{O}) = \text{C}_3\text{H}_6\text{O}_3 \]