Question

An organic compound on analysis gave $C = 54.2\%$, $H = 9.2\%$ by mass. Its empirical formula is.

• A. $CHO_{2}$
• B. $CH_{2}O$
• C. $C_{2}H_{8}O$
• D. $C_{2}H_{4}O$

Answer 1

Answer: (D)

$C_{2}H_{4}O$

Answer 2

(4) :

Element	%	No. of moles	Mole Ratio	Whole No.ratio
C	54.2	54.2/124.5	4.5/2.3 = 2	2
H	9.2	9.2/1 = 9.2	9.2/2.3 = 4	4
O	36.6 36.6	36.6/16 = 2.3	2.3/2.3 = 1	1

Empirical formula=$C_{2}H_{4}O$