Question

Calculate the weight of CO having same number of oxygen atoms as present in 88g of $C{O}_2$ ?

Answer 1

Molar mass of CO: - 44 g/mol
Therefore number of moles of $C{O}_2$ present in $88g=\frac{mass}{molarmassofC{O}_2}=\frac{88}{44}=2moles$
As there are $6.023\times {10}^{23}$ molecules present in one mole of a compound therefore 2 moles of $C{O}_2$ will contain $2\times 6.023\times {10}^{23}$ molecules $12.046\times {10}^{23}$ molecules of $C{O}_2$
Now 1 molecule of $C{O}_2$ contains 2 atoms of oxygen. Therefore $12.046\times {10}^{23}$ molecules of $C{O}_2$ will contain $2\times 12.044\times {10}^{23}$ atoms of oxygen.
Thus, $12.046\times {10}^{23}$ molecules of $C{O}_2.4092\times {10}^{24}$ atoms of oxygen.
So, $2.4092\times {10}^{24}$ atoms of oxygen are present in 889 of carbon dioxide
Now molar mass of CO = 28 g/mole
There are $6.023\times {10}^{23}$ molecules of CO present in 28 g of CO. As one molecule of CO contains 1 atom of oxygen, therefore there are $6.023\times {10}^{23}$ atoms of oxygen present in $6.023\times {10}^{23}$ molecules of CO.
So $6.023\times {10}^{23}$ atoms of Oxygen $\to$ 28 g of Oxygen
1 atom of oxygen $\to \frac{28}{[6023\times {10}^{23}]}g$ therefore $2.4092\times {10}^{24}$ atoms of oxygen $\to 28\times \frac{[2.4092\times {10}^{24}}{[6.023\times {10}^{23}]g}=112g$of CO
Thus 112 g of CO will have the same number of oxygen atoms as present in 88g of $C{O}_2$