Question

Which of the following has maximum number of atoms ?

• A. \(18\text{g of H}_2\text{O}\)
• B. \(18\text{g of O}_2\)
• C. \(18\text{g of CO}_2\)
• D. \(18\text{g of CH}_4\)

Hint:

For a fixed mass of different substances, the substance with the: 1. Lowest molar mass will have the highest number of moles. 2. The highest number of atoms per molecule will contribute more atoms per mole. Combine these: Calculate (atoms per molecule / molar mass). The substance with the largest value of this ratio for a given mass will have the most atoms. For \(CH_4\): Molar mass = 16 g/mol, Atoms per molecule = 5. Ratio = 5/16 = 0.3125. For \(H_2O\): Molar mass = 18 g/mol, Atoms per molecule = 3. Ratio = 3/18 \(\approx\) 0.1667. \(CH_4\) has more moles of molecules (1.125 vs 1 for \(H_2O\)) AND more atoms per molecule (5 vs 3 for \(H_2O\)).

Answer 1

The Correct Option is D

Concept: To find the total number of atoms, we first need to calculate the number of moles of each substance, then the number of molecules, and finally the total number of atoms by considering the number of atoms per molecule. Number of moles (\(n\)) = Mass (\(m\)) / Molar mass (\(M\)). Number of molecules = Number of moles \(\times\) Avogadro's number (\(N_A \approx 6.022 \times 10^{23} \text{ mol}^{-1}\)). Total number of atoms = Number of molecules \(\times\) Number of atoms per molecule. Since the mass (18g) and Avogadro's number are constant for all options, the number of atoms will be proportional to (Number of atoms per molecule) / (Molar mass). A simpler approach for comparison is to calculate moles, then total moles of atoms. Step 1: Calculate Molar Masses (M)
\(H_2O\): \(2 \times 1 \text{ (for H)} + 16 \text{ (for O)} = 18 \text{ g/mol}\)
\(O_2\): \(2 \times 16 \text{ (for O)} = 32 \text{ g/mol}\)
\(CO_2\): \(12 \text{ (for C)} + 2 \times 16 \text{ (for O)} = 12 + 32 = 44 \text{ g/mol}\)
\(CH_4\): \(12 \text{ (for C)} + 4 \times 1 \text{ (for H)} = 12 + 4 = 16 \text{ g/mol}\) (Atomic masses: H=1, C=12, O=16 g/mol approximately) Step 2: Calculate number of moles (\(n\)) for 18g of each substance
\(H_2O\): \(n = 18\text{g} / 18\text{ g/mol} = 1 \text{ mol}\)
\(O_2\): \(n = 18\text{g} / 32\text{ g/mol} = 0.5625 \text{ mol}\)
\(CO_2\): \(n = 18\text{g} / 44\text{ g/mol} \approx 0.409 \text{ mol}\)
\(CH_4\): \(n = 18\text{g} / 16\text{ g/mol} = 1.125 \text{ mol}\) Step 3: Calculate total number of moles of atoms (Total moles of atoms = moles of substance \(\times\) number of atoms in one molecule)
\(H_2O\): 1 molecule has \(2 (H) + 1 (O) = 3\) atoms. Total moles of atoms = \(1 \text{ mol} \times 3 = 3 \text{ moles of atoms}\).
\(O_2\): 1 molecule has \(2 (O)\) atoms. Total moles of atoms = \(0.5625 \text{ mol} \times 2 = 1.125 \text{ moles of atoms}\).
\(CO_2\): 1 molecule has \(1 (C) + 2 (O) = 3\) atoms. Total moles of atoms = \(0.409 \text{ mol} \times 3 \approx 1.227 \text{ moles of atoms}\).
\(CH_4\): 1 molecule has \(1 (C) + 4 (H) = 5\) atoms. Total moles of atoms = \(1.125 \text{ mol} \times 5 = 5.625 \text{ moles of atoms}\). Step 4: Compare the total number of moles of atoms Comparing the values:
\(H_2O\): 3 moles of atoms
\(O_2\): 1.125 moles of atoms
\(CO_2\): \(\approx 1.227\) moles of atoms
\(CH_4\): 5.625 moles of atoms \(18\text{g of CH}_4\) has the maximum number of moles of atoms, and thus the maximum number of atoms.