Question

A compound on analysis gave the following percentage composition by weight: Hydrogen $=9.09$%, Oxygen $=36.36$%, Carbon $=54.55$%. It's vapour density is $44$. If the molecular formula of the compound is $C_x{H}_y{O}_z$, then find the value of $x,y$ and $z$.

• A. $x=4,y=8,z=2$
• B. $x=3,y=6,z=3$
• C. $x=5,y=10,z=4$
• D. None of the above

Answer 1

Answer: (A)

$x=4,y=8,z=2$

Element	Moles	Least ratio
$C$	$\frac{54.55}{12}=4.54$	$\frac{4.54}{2.27}=2$
$H$	$\frac{9.09}{1}=9.09$	$\frac{9.09}{2.27}=4$
$O$	$\frac{36.36}{16}=2.27$	$\frac{2.27}{2.27}=1$

From the above calculations, the empirical formula $=C_2{H}_{4}O$.
Empirical formula weight $=12\times 2+1\times 4+16=44$ g.
Vapour density ($VD$) $=44$ g.
Molecular weight $=2\times VD=2\times 44=88$ g.
$n=\frac{\text{Molecular weight}}{\text{Empirical formula weight}}=\frac{88}{44}=2$
Molecular Formula $=2\times$ Empirical formula$=2\times \left(C_2{H}_{4}O\right)=C_4{H}_8{O}_2$.