Understanding the Concept
To solve this problem, we rely on the fundamental relationship between gas volume, mass, and molar mass. At NTP (Normal Temperature and Pressure), which is defined as a temperature of 273.15 K ($0^\circ C$) and a pressure of 1 atm, one mole of any ideal gas occupies a standard volume of 22.4 Liters.
This is known as the Molar Volume of a gas at STP/NTP.
The Formula
We can relate the mass of a substance to its molar mass using the following relationship:
$$\text{Number of moles (n)} = \frac{\text{Mass (m)}}{\text{Molecular Weight (M)}} = \frac{\text{Volume at NTP (V)}}{22.4 \text{ L}}$$
By rearranging this formula, we can solve for the Molecular Weight (M):
$$M = \frac{\text{Mass} \times 22.4 \text{ L}}{\text{Volume at NTP}}$$
Step-by-Step Solution
Given the data from the question:
- Mass (m) = $14 \text{ g}$
- Volume (V) = $11.2 \text{ L}$
Step 1: Calculate the number of moles Since 22.4 L corresponds to 1 mole, we calculate how many moles are in 11.2 L:
$$\text{n} = \frac{11.2 \text{ L}}{22.4 \text{ L/mol}} = 0.5 \text{ moles}$$
Step 2: Calculate the molecular weight Now, use the formula $\text{n} = \frac{\text{mass}}{\text{molecular weight}}$ to solve for M:
$$0.5 \text{ mol} = \frac{14 \text{ g}}{\text{M}}$$
$$M = \frac{14 \text{ g}}{0.5 \text{ mol}}$$
$$M = 28 \text{ g/mol}$$
Conclusion
The molecular weight of the gas is 28 g/mol. This value corresponds to common diatomic gases like Nitrogen ($N_2$) or Carbon Monoxide ($CO$).