Number of atoms present in 64 grams of oxygen ${}_8{}^{16}\text{O}$ is

We know that 16 g of oxygen atom is equal to 1 mol of O . The weight of one mol of oxygen atom is equal to $6.023\times {10}^{23}$ atoms. The number of moles present in 64g of oxygen is

$\frac{1\text{\hspace{0.17em}mol}}{16\text{\hspace{0.17em}g\hspace{0.17em}of\hspace{0.17em}O}}\times 64\text{\hspace{0.17em}g\hspace{0.17em}of O = 4 mol}$

So, there are  $4\times 6.023\times {10}^{23}$ oxygen atoms are present in 64 g of oxygen.