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

Question

An organic compound on analysis gave C = 54.2 %, H = 9.2 %A by mass. Its empirical formula is (A) C H O2 (B) CH2O (C) C2H8O (D) C2H4O

Tardigrade Admin · Accepted Answer

Element % No. of Moles Mole ratio Whole no. ratio C 54.2 54.2/12=4.5 4.5/2.3=2 2 H 9.2 9.2/1=9.2 9.2/2.3=4 4 O 36.6 36.6/16=2.3 2.3/2.1=1 1 Empirical Formula = C2H4O

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

$C H O_{2}$

$C H_{2} O$

$C_{2} H_{8} O$

$C_{2} H_{4} O$

Element % No. of Moles Mole ratio Whole no. ratio

C 54.2 54.2/12=4.5 4.5/2.3=2 2

H 9.2 9.2/1=9.2 9.2/2.3=4 4

O 36.6 36.6/16=2.3 2.3/2.1=1 1

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

Element	%	No. of Moles	Mole ratio	Whole no. ratio
C	54.2	54.2/12=4.5	4.5/2.3=2	2
H	9.2	9.2/1=9.2	9.2/2.3=4	4
O	36.6	36.6/16=2.3	2.3/2.1=1	1

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

CHO2​

CH2​O

C2​H8​O

C2​H4​O