Study the following table 

	Compound (Molecular weight)	Weight of compound (in g) taken
I)	${\mathrm{CO}}_2\left(44\right)$	4.4
II)	${\mathrm{NO}}_2\left(46\right)$	2.3
III)	${\mathrm{H}}_2{\mathrm{O}}_2\left(46\right)$	6.8
IV)	${\mathrm{SO}}_2\left(64\right)$	1.6

Which two compounds have least weight of oxygen ? (molecular weights of compounds are given in brackets)

$\text{ In }{\mathrm{CO}}_2\text{, }$ 

$\because 44{\mathrm{gCO}}_2\text{ contains }{\mathrm{O}}_2=32\mathrm{g}$

$\therefore 4.4{\mathrm{gCO}}_2{\text{ contains }}_2=\frac{32}{44}\times 4.4=3.2\mathrm{g}$

$\text{ In }{\mathrm{NO}}_2\text{, }$$\because 46\mathrm{g}\text{ of }{\mathrm{NO}}_2\text{ contains }{\mathrm{O}}_2=32\mathrm{g}$

$\therefore 2.3{\mathrm{gNO}}_2\text{ contains }{\mathrm{O}}_2=\frac{32\times 2.3}{46}\times 4.4=1.6\mathrm{g}$

$\text{ In }{\mathrm{H}}_2{\mathrm{O}}_2\text{, }$

$\because 34{\mathrm{gH}}_2{\mathrm{O}}_2\text{ contains }{\mathrm{O}}_2=32\mathrm{g}$

$\therefore 6.8{\mathrm{gH}}_2{\mathrm{O}}_2\text{ contains }{\mathrm{O}}_2=\frac{32\times 6.8}{34}\times 4.4=6.4\mathrm{g}$

$\text{ In }{\mathrm{SO}}_2\text{, }$

$\because 64\text{ g of }{\mathrm{SO}}_2\text{ contains }{\mathrm{O}}_2=32\mathrm{g}$

$\therefore 1.6{\mathrm{gSO}}_2\text{ contains }{\mathrm{O}}_2=\frac{32\times 1.6}{64}=0.8\mathrm{g}$

Hence, molecule II and IV $\left({\mathrm{NO}}_2\text{ and }{\mathrm{SO}}_2\right)$ have least weight of oxygen