Question

The elemental analysis of an organic compound gave:
C: 38.71%, H: 9.67%, O: 51.67%. What is the empirical formula of the compound?

• A. \( \text{CH}_2\text{O} \)
• B. \( \text{CH}_3\text{O} \)
• C. \( \text{CH}_4\text{O} \)
• D. \( \text{CHO} \)
• E. \( \text{CH}_5\text{O} \)

Hint:

To determine the empirical formula from elemental analysis data, first calculate the moles of each element, then find the smallest mole ratio and simplify it to get the empirical formula.

Answer 1

The Correct Option is B

Step 1: Calculate moles of each element.
We are given the mass percentages of carbon (C), hydrogen (H), and oxygen (O). To determine the empirical formula, we need to convert the mass percentages to moles. The molecular weights of carbon, hydrogen, and oxygen are as follows:
- C: 12 g/mol
- H: 1 g/mol
- O: 16 g/mol
Now, calculate the moles for each element: \[ \text{Moles of C} = \frac{38.71}{12} = 3.2258 \, \text{mol} \] \[ \text{Moles of H} = \frac{9.67}{1} = 9.67 \, \text{mol} \] \[ \text{Moles of O} = \frac{51.67}{16} = 3.2294 \, \text{mol} \]

Step 2: Determine the mole ratio.
Now, divide each number of moles by the smallest number of moles (which is 3.2258) to find the simplest ratio: \[ \frac{\text{Moles of C}}{3.2258} = \frac{3.2258}{3.2258} = 1 \] \[ \frac{\text{Moles of H}}{3.2258} = \frac{9.67}{3.2258} = 3 \] \[ \frac{\text{Moles of O}}{3.2258} = \frac{3.2294}{3.2258} = 1 \]

Step 3: Write the empirical formula.
The mole ratio is approximately 1:3:1 for C:H:O. Therefore, the empirical formula is \( \text{CH}_3\text{O} \).

Final Answer: (B) \( \text{CH}_3\text{O} \)