Molecular Comparison: $O_2$, $CO_2$, CO, $H_2$

The question asks to arrange four substances—16g of $O_2$, 16g of $CO_2$, 16g of CO, and 16g of $H_2$—in decreasing order based on the number of molecules they contain.

The number of molecules is directly proportional to the number of moles. For a fixed mass, the substance with the lowest molar mass will have the highest number of moles and thus the highest number of molecules.

Calculating Molar Masses

• $O_2$: Molar mass = $2 \times (\text{atomic mass of O}) = 2 \times 16.0 \text{ g/mol} = 32.0 \text{ g/mol}$
• $CO_2$: Molar mass = $(\text{atomic mass of C}) + 2 \times (\text{atomic mass of O}) = 12.0 + 2 \times 16.0 \text{ g/mol} = 44.0 \text{ g/mol}$
• CO: Molar mass = $(\text{atomic mass of C}) + (\text{atomic mass of O}) = 12.0 + 16.0 \text{ g/mol} = 28.0 \text{ g/mol}$
• $H_2$: Molar mass = $2 \times (\text{atomic mass of H}) = 2 \times 1.0 \text{ g/mol} = 2.0 \text{ g/mol}$

Calculating Number of Moles

Number of moles ($n$) is calculated using the formula: $n = \frac{\text{mass}}{\text{molar mass}}$

• (A) 16 g of $O_2$: $n_A = \frac{16 \text{ g}}{32.0 \text{ g/mol}} = 0.5 \text{ mol}$
• (B) 16 g of $CO_2$: $n_B = \frac{16 \text{ g}}{44.0 \text{ g/mol}} \approx 0.364 \text{ mol}$
• (C) 16 g of CO: $n_C = \frac{16 \text{ g}}{28.0 \text{ g/mol}} \approx 0.571 \text{ mol}$
• (D) 16 g of $H_2$: $n_D = \frac{16 \text{ g}}{2.0 \text{ g/mol}} = 8.0 \text{ mol}$

Determining Decreasing Order

Comparing the number of moles calculated:

• $n_D$ (8.0 mol) > $n_C$ (≈0.571 mol) > $n_A$ (0.5 mol) > $n_B$ (≈0.364 mol)

Therefore, the decreasing order of the number of molecules is (D), (C), (A), (B).