In a compound C=40 % and H=6.66 %, what is the empirical formula of the compound?

Question

In a compound C=40 % and H=6.66 %, what is the empirical formula of the compound? (A) C 4 HO 2 (B) C 4 HO 4 (C) CHO (D) CH 2 O

Tardigrade Admin · Accepted Answer

Symbol % Atomic mass Relative no. of atoms Simple Reaction C 40 12 (40/12)=3.33 (3.33/3.33)=1 H 6.66 1 (6.66/1)=6.66 (6.66/3.33)=2 O 53.34 16 (53.34/16) 3.33 (3.33/3.33)=1

Symbol	$%$	Atomic mass	Relative no. of atoms	Simple Reaction
$C$	$40$	$12$	$\frac{40}{12} = 3.33$	$\frac{3.33}{3.33} = 1$
$H$	$6.66$	$1$	$\frac{6.66}{1} = 6.66$	$\frac{6.66}{3.33} = 2$
$O$	$53.34$	$16$	$\frac{53.34}{16} 3.33$	$\frac{3.33}{3.33} = 1$

Q. In a compound $C = 40%$ and $H = 6.66%,$ what is the empirical formula of the compound?

$C_{4} H O_{2}$

$C_{4} H O_{4}$

$C H O$

$C H_{2} O$

Symbol $%$ Atomic mass Relative no. of atoms Simple Reaction

$C$ $40$ $12$ $\frac{40}{12} = 3.33$ $\frac{3.33}{3.33} = 1$

$H$ $6.66$ $1$ $\frac{6.66}{1} = 6.66$ $\frac{6.66}{3.33} = 2$

$O$ $53.34$ $16$ $\frac{53.34}{16} 3.33$ $\frac{3.33}{3.33} = 1$

Q. In a compound C=40% and H=6.66%, what is the empirical formula of the compound?

C4​HO2​

C4​HO4​

CHO

CH2​O