Which of the following has maximum number of atoms?

To find the total number of atoms, we use the formula:
$$\text{Total Atoms}=\frac{\text{Given Mass}}{\text{Molar Mass}}\times N_A\times \text{Atomicity}$$
Where $N_A$ is Avogadro's number and atomicity is the number of atoms present in one molecule.Let's compute this for each option given:Option A: $16{\text{g of N}}_2$ (Molar Mass = $28\text{g/mol}$, Atomicity = $2$)
$$\text{Atoms}=\frac{16}{28}\times 2\times N_A=\frac{32}{28}{N}_A\approx 1.14N_A$$
Option B: $16{\text{g of CO}}_2$ (Molar Mass = $44\text{g/mol}$, Atomicity = $1+2=3$)
$$\text{Atoms}=\frac{16}{44}\times 3\times N_A=\frac{48}{44}{N}_A\approx 1.09N_A$$
Option C: $16{\text{g of CH}}_4$ (Molar Mass = $16\text{g/mol}$, Atomicity = $1+4=5$)
$$\text{Atoms}=\frac{16}{16}\times 5\times N_A=5N_A$$
Option D: $16{\text{g of O}}_2$ (Molar Mass = $32\text{g/mol}$, Atomicity = $2$)
$$\text{Atoms}=\frac{16}{32}\times 2\times N_A=1N_A$$
Comparing the calculated coefficients, $16{\text{g of CH}}_4$ contains the maximum number of atoms ($5N_A$).