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(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{mass of substance}}{	ext{molar mass}} 	imes 	ext{num

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$18 \, g$ of $CH_4$

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(B) The number of atoms is calculated using the formula: $	ext{Number of atoms} = \frac{	ext{mass of substance}}{	ext{molar mass}} 	imes 	ext{number of atoms in a molecule} 	imes N_A$.$(a)$ For $18 \, g$ of $H_2O$: $	ext{Molar mass} = 18 \, g/mol$. Atoms per molecule $= 3$. $	ext{Atoms} = \frac{18}{18} 	imes 3 	imes N_A = 3 N_A$.$(b)$ For $18 \, g$ of $CH_4$: $	ext{Molar mass} = 16 \, g/mol$. Atoms per molecule $= 5$. $	ext{Atoms} = \frac{18}{16} 	imes 5 	imes N_A = 5.625 N_A$.$(c)$ For $18 \, g$ of $CO_2$: $	ext{Molar mass} = 44 \, g/mol$. Atoms per molecule $= 3$. $	ext{Atoms} = \frac{18}{44} 	imes 3 	imes N_A \approx 1.23 N_A$.$(d)$ For $18 \, g$ of $O_2$: $	ext{Molar mass} = 32 \, g/mol$. Atoms per molecule $= 2$. $	ext{Atoms} = \frac{18}{32} 	imes 2 	imes N_A = 1.125 N_A$.Comparing the values,$5.625 N_A$ is the maximum. Thus,$18 \, g$ of $CH_4$ has the maximum number of atoms.

Which of the following has the maximum number of atoms?

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