Question

Which one of the following gas contains the same number of molecules as 16 g of oxygen? (1) \(16 \mathrm{~g}\) of \(\mathrm{O}_{3}\) (2) \(16 \mathrm{~g}\) of \(\mathrm{SO}_{2}\) (3) \(32 \mathrm{~g}\) of \(\mathrm{SO}_{2}\) (4) \(\Lambda\) ll the above

Short answer

Option (3): 32 g of \text{SO}_2

Step-by-step solution

Determine the number of moles in 16 g of oxygen gas (\text{O}_2)

Oxygen gas (\text{O}_2) has a molar mass of approximately 32 g/mol. Calculate the number of moles in 16 g of \text{O}_2 using the formula: \ \[ n = \frac{m}{M} \] where \( n \) is the number of moles, \( m \) is the mass, and \( M \) is the molar mass. \ \[ n = \frac{16 \, \text{g}}{32 \, \text{g/mol}} = 0.5 \, \text{moles} \]

Determine the number of molecules in 0.5 moles of \text{O}_2

Use Avogadro's number to calculate the number of molecules: \ \[ N = n \times N_A \] where \( N \) is the number of molecules and \( N_A \) is Avogadro's number (\(6.022 \times 10^{23} \) molecules/mol). \ \[ N = 0.5 \, \text{moles} \times 6.022 \times 10^{23} \, \text{molecules/mol} = 3.011 \times 10^{23} \, \text{molecules} \]

Compare the number of molecules in the given options

Now compare the number of molecules in each option using their molar masses: \ (1) \text{O}_3: Molar mass = 48 g/mol \ \[n = \frac{16 \, \text{g}}{48 \, \text{g/mol}} = 0.333 \, \text{moles} \] \ \[ N = 0.333 \times 6.022 \times 10^{23} = 2.007 \times 10^{23} \, \text{molecules}\] \ (2) \text{SO}_2: Molar mass = 64 g/mol \ \[ n = \frac{16 \, \text{g}}{64 \, \text{g/mol}} = 0.25 \, \text{moles} \] \ \[ N = 0.25 \times 6.022 \times 10^{23} = 1.505 \times 10^{23} \, \text{molecules} \] \ (3) \text{SO}_2: 32 g \ \[ n = \frac{32 \, \text{g}}{64 \, \text{g/mol}} = 0.5 \, \text{moles} \] \ \[ N = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \, \text{molecules} \] \ Since 32 g of \text{SO}_2 contains the same number of moles (and thus the same number of molecules) as 16 g of \text{O}_2, option (3) is correct.

Key concepts

Molar Mass

Molar mass is a core concept in chemistry, representing the mass of one mole of any substance. It is expressed in units of grams per mole (g/mol). To determine the molar mass, you sum the atomic masses of all atoms in a molecule.
For instance, the molar mass of oxygen gas (\text{O}_2) is calculated as follows: Each oxygen atom has an atomic mass of 16 g/mol. Since there are two oxygen atoms in \text{O}_2, the molar mass is:
\[ M_{\text{O}_2} = 16 \text{ g} / \text{mol} \times 2 = 32 \text{ g} / \text{mol} \]
Likewise, for sulfur dioxide (\text{SO}_2), where the atomic masses are 32 g/mol for sulfur and 16 g/mol for each oxygen atom:
\[ M_{\text{SO}_2} = 32 + 16 \times 2 = 64 \text{ g} / \text{mol} \]
Understanding molar mass is essential for converting between the mass of a substance and the number of moles, aiding in various chemical calculations.

Avogadro's Number

Avogadro's number (\text{N}_A) is an incredibly useful constant in chemistry, defining the number of entities (atoms, molecules, etc.) in one mole of a substance. The value of Avogadro's number is approximately:
\[ N_A = 6.022 \times 10^{23} \text{ per mole} \]
This means that one mole of any substance contains exactly 6.022 x 10^23 particles of that substance.
When you know the number of moles of a substance, you can easily find the number of molecules by multiplying by Avogadro's number:
\[ N = n \times N_A \]
For example, if you have 0.5 moles of oxygen gas (\text{O}_2), the number of molecules would be:
\[ N = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \text{ molecules} \]
This constant is pivotal for linking the macroscopic world of grams and liters to the microscopic world of atoms and molecules.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It helps predict the amounts of substances consumed and produced in a given reaction.
To solve stoichiometric problems, you often start by converting masses of substances to moles using their molar masses, and then use ratios from the balanced chemical equation. For example, solving the given exercise involves comparing the number of molecules in different gas samples:

• First, calculate the number of moles in 16g of oxygen gas (\text{O}_2): \[ n = \frac{m}{M} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \]
  
• Then, use Avogadro's number to find the number of molecules: \[ N = 0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23} \]
  
• For other gases, follow the same process: \text{SO}_2 with a molar mass of 64 g/mol and 32 g of mass gives: \[ n = \frac{32 \text{ g}}{64 \text{ g/mol}} = 0.5 \text{ moles} \] \br This calculation ensures you can compare the number of molecules between different gases, leading to the conclusion that 32 g of \text{SO}_2 contains the same number of molecules as 16 g of \text{O}_2.