When 0.482g of gaseous hydrocarbon is injected into a 334.09 cm³ container, at 102°C, it exerts a pressure of 102.352 kNm⁻². Calculate the molar mass of the hydrocarbon. (R=8.314 Jmol⁻¹K)

The ideal gas law requires consistent units.

• Mass ($m$) = $0.482 \text{ g}$
• Volume ($V$) = $334.09 \text{ cm}^3 = 334.09 \times 10^{-6} \text{ m}^3 = 0.00033409 \text{ m}^3$
• Temperature ($T$) = $102 \text{ }^\circ\text{C} = 102 + 273.15 \text{ K} = 375.15 \text{ K}$
• Pressure ($P$) = $102.258 \text{ kN/m}^2 = 102.258 \times 10^3 \text{ N/m}^2 = 102258 \text{ Pa}$
• Gas constant ($R$) = $8.314 \text{ J mol}^{-1} \text{ K}^{-1}$

The ideal gas law is:

$$PV = nRT$$

Where $n$ is the number of moles. The number of moles can also be expressed as:

$$n = \frac{m}{M}$$

Where $m$ is the mass and $M$ is the molar mass.

Substituting $n = \frac{m}{M}$ into $PV = nRT$:

$$PV = \frac{m}{M}RT$$

Rearranging to solve for $M$:

$$M = \frac{mRT}{PV}$$

$$M = \frac{(0.482 \text{ g}) \times (8.314 \text{ J mol}^{-1} \text{ K}^{-1}) \times (375.15 \text{ K})}{(102258 \text{ Pa}) \times (0.00033409 \text{ m}^3)}$$ $$M = \frac{1503.266028 \text{ g} \cdot \text{J} \cdot \text{mol}^{-1}}{34.16100422 \text{ J}}$$ $$M \approx 44.006 \text{ g/mol}$$

Rounding to three significant figures, consistent with the mass given:

$$M \approx 44.0 \text{ g/mol}$$

The molar mass of the hydrocarbon is $\text{44.0 g/mol}$.