Question

Which of the following contains greatest number of oxygen atoms?

• A. 1g of O
• B. 1g of O₂
• C. 1g of O₃
• D. All have the same number of atoms

Answer 1

Answer: (D)

All have the same number of atoms

Answer 2

To determine which option contains the greatest number of oxygen atoms, we calculate the number of oxygen atoms present in 1 gram of each substance. Let $N_A$ be Avogadro's number. The atomic mass of oxygen (O) is approximately 16 g/mol.

(A) 1g of O (atomic oxygen): Molar mass of O = 16 g/mol. Number of moles of O = $\frac{1 \text{ g}}{16 \text{ g/mol}} = \frac{1}{16}$ mol. Number of oxygen atoms = $\frac{1}{16} \times N_A$.

(B) 1g of O₂ (molecular oxygen): Molar mass of O₂ = $2 \times 16 \text{ g/mol} = 32$ g/mol. Number of moles of O₂ = $\frac{1 \text{ g}}{32 \text{ g/mol}} = \frac{1}{32}$ mol. Each O₂ molecule contains 2 oxygen atoms. Number of oxygen atoms = $\frac{1}{32} \times N_A \times 2 = \frac{2}{32} \times N_A = \frac{1}{16} \times N_A$.

(C) 1g of O₃ (ozone): Molar mass of O₃ = $3 \times 16 \text{ g/mol} = 48$ g/mol. Number of moles of O₃ = $\frac{1 \text{ g}}{48 \text{ g/mol}} = \frac{1}{48}$ mol. Each O₃ molecule contains 3 oxygen atoms. Number of oxygen atoms = $\frac{1}{48} \times N_A \times 3 = \frac{3}{48} \times N_A = \frac{1}{16} \times N_A$.

Comparing the number of oxygen atoms in each case, we find that all three options contain the same number of oxygen atoms: $\frac{1}{16} N_A$.