Calculate the weight of CO having the same number of oxygen atoms as are present in 22g of $C{O}_2$

Molar Mass of $C{O}_2$=44g

1 mole of $C{O}_2$ [ 44g ] contains 6.022 $\times {10}^{23}$ molecules.

1 molecule of $C{O}_2$ = 2 atoms of Oxygen

6.022 x ${10}^{23}$molecules of $C{O}_2$ = 2 $\times$ 6.022x ${10}^{23}$ atoms of Oxygen
so, 44g of carbon dioxide = 2$\times$6.022$\times$${10}^{23}$ atoms of Oxygen
1 g of carbon dioxide = $\frac{2\times 6.022\times {10}^{23}}{44}$
22g of carbon dioxide = $\frac{2\times 6.022\times {10}^{23}}{44}$ $\times 22$
= $6.022\times {10}^{23}$ atoms of Oxygen.

1 mole of CO =$6.022\times {10}^{23}$atoms of Oxygen
1 mole of CO=28g
∴28g of CO contains $6.022\times {10}^{23}$ atoms of Oxygen
The weight of CO is 28 g having the same number of oxygen atoms as are present in 22g of$C{O}_2$