MCQ
An organic compound on analysis gave the following results : $C = 54.5\%, \,O = 36.4\%, \,H = 9.1\%$. The Empirical formula of the compound is
- A$C{H_3}O$
- ✓${C_2}{H_4}O$
- C${C_3}{H_4}O$
- D${C_4}{H_8}O$
✓
Answer
Correct option: B.
${C_2}{H_4}O$
b
(b)
(b)
Element No. of Moles Simple Ratio
|
$C = 54.5$ |
$54.5/12 = 4.54$ |
$2$ |
|
$H = 9.1$ |
$9.1/1 = 9.1$ |
$4$ |
|
$O = 36.4$ |
$36.4/16 = 2.27$ |
$1$ |
Hence, $C_2H_4O$.
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