$Phần$ $trăm$ $của$ $đồng$ $vị$ $thứ$ $3$ $là$
$100$ $-$ $99,757$ $-$ $0,039$ $=$ $0,204$
$A_{O_{2}}$ $=$ $\frac{16.99,757 + 17.0,039+18.0,204}{100}$
$A_{O_{2}}$ $=$ $16,00447$ ≈ $16$
$n_{O_{2}}$ $=$ $\frac{8,96}{22,4}$ $=$ $0,4$ mol
$m_{O_{2}}$ $=$ $0,4$ $.$ $16$ $=$ $6,4$ g