Which of the following has the maximum number of atoms?

To determine which substance has the maximum number of atoms among the options given, we need to calculate the number of moles of each substance and then use Avogadro's number to find the number of atoms.

18 g of CO₂:

Calculate the molar mass of CO₂:
Carbon (C): 1 atom x 12.01 g/mol = 12.01 g/mol
Oxygen (O): 2 atoms x 16.00 g/mol = 32.00 g/mol
Total molar mass = 12.01 g/mol + 32.00 g/mol = 44.01 g/mol
Number of moles =$\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}}$  = $\frac{18 g}{44.01}$≈ 0.409 moles
 

Number of atoms = Number of moles x Avogadro's number 

= 0.409 moles x 6.022 x 10^23 atoms/mol ≈ $2.46 \times {10}^{23}$ atoms.
 

18 g of H₂O₂:

Calculate the molar mass of H₂O₂:

Hydrogen (H): 2 atoms x 1.008 g/mol = 2.016 g/mol
Oxygen (O): 2 atoms x 16.00 g/mol = 32.00 g/mol
Total molar mass = 2.016 g/mol + 32.00 g/mol = 34.016 g/mol
Number of moles =$\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}}$ = $\frac{18 g}{34.01}$ ≈ 0.529 moles
 

Number of atoms = Number of moles x Avogadro's number 

= 0.529 moles x $6.023 \times {10}^{23}$ atoms/mol ≈ $3.18 \times {10}^{23}$ atoms
 

18 g of O₂:

Calculate the molar mass of O₂:

Oxygen (O): 2 atoms x 16.00 g/mol = 32.00 g/mol
Number of moles =$\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}}$ = $\frac{18 g}{32 g}$ ≈ 0.563 moles
Number of atoms = Number of moles x Avogadro's number 

= 0.563 moles x $6.023 \times {10}^{23}$atoms/mol ≈ $\mathbf{3}\mathbf{.}\mathbf{39}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$ atoms
 

18 g of CH₄:

Calculate the molar mass of CH₄:

Carbon (C): 1 atom x 12.01 g/mol = 12.01 g/mol
Hydrogen (H): 4 atoms x 1.008 g/mol = 4.032 g/mol
Total molar mass = 12.01 g/mol + 4.032 g/mol = 16.042 g/mol
Number of moles =$\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{mass}}$ = $\frac{18 g}{16 g}$ ≈ 1.123 moles
Number of atoms = Number of moles x Avogadro's number 

= 1.123 moles x $6.023 \times {10}^{23}$ atoms/mol ≈ $\mathbf{6}\mathbf{.}\mathbf{75}\mathbf{ }\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}$ atoms