22g of a sample on complete combustion gives 44g $C{O}_2$ and 18g of $H_{2}O$. The empirical formula of the compound is $C_x{H}_y{O}_z$ then $\frac{x+y}{z}$  is : 

$C{O}_2=0.5mol of C{O}_2=2mole of C.$

$H_{2}O=1mol of H_{2}O=2mol of H$

Mass of C + Mass of  H + Mass of O =22
$12+2+x=22$
   $\therefore x=8$
So, molecular formula $=C_2{H}_{4}O$
$\frac{x+y}{z}=\frac{2+4}{1}=6$