Calculate the weight of carbon monoxide having the same number of oxygen atoms as are present in 22g of carbon dioxide

Explanation

Step I. No. of the oxygen atoms in 22 g of carbon dioxide (`CO_(2)`) :
Molar mass of `CO_(2) = 12 + 2 xx 16 = 44 g`
44 g of `CO_(2)` represent = 1 mol
22 g of `CO_(2)` represent = 1 (mol) `xx (22 g)/(44 g) = 0.5` mol
Now, 1 mole of `CO_(2)` contains oxygen atoms `=2 xx N_(0) = 2 xx 6.022 xx 10^(23)`
`therefore` 0.5 mole of `CO_(2)` will contain oxygen atoms = `2 xx 6.022 xx 10^(23) xx 0.5 = 6.022 xx 10^(23)` atoms
Step II. Weight of carbon monoxide `6.022 xx 10^(23)` atoms of oxygen are present in `6.022 xx 10^(23)` molecules of carbon monoxide (CO) By definition, `6.022 xx 10^(23)` molecules of CO have mass = Molar mass of CO = 12 + 16 = 28 g