# A sample of an unknown compound is vaporized at 150degree C . The gas produced has a volume of...

## Question:

A sample of an unknown compound is vaporized at 150degree C . The gas produced has a volume of 960.mL at a pressure of 1.00atm , and it weighs 0.941g . Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Round your answer to 3 significant digits.

## Ideal Gas Equation

The ideal gas equation applies to ideal gases which do have any intermolecular forces. This equation can also be used to calculate the density of gases under standard conditions.

Given Data:

• The mass of compound is 0.941 g.
• The pressure is 1 atm.
• The temperature is {eq}150\;^\circ {\rm{C}} {/eq}.
• The volume of solution is 960 mL(0.960 L).

The molar mass can be calculated using the ideal gas equation, which is shown below.

{eq}\begin{align*} PV &= nRT\\ PV &= \dfrac{{{\rm{Given}}\;{\rm{mass}}}}{{{\rm{Molecular}}\;{\rm{mass}}}} \times RT \end{align*} {/eq}

Where,

• {eq}V {/eq} is the volume of {eq}{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq}.
• {eq}n {/eq} is the moles of {eq}{\rm{C}}{{\rm{O}}_{\rm{2}}} {/eq}.
• {eq}P {/eq} is the pressure.
• {eq}T {/eq} is the temperature.
• {eq}R {/eq} is the gas constant (0.0821 L-atm/mol-K).

The temperature is in degrees

The conversion of degrees into Kelvin is shown below.

{eq}0^\circ {\rm{C}} = 273\;{\rm{K}} {/eq}

The conversion of {eq}150^\circ {\rm{C}} {/eq} into Kelvin is shown below

{eq}\begin{align*} 150^\circ {\rm{C}} &= \left( {150 + 273} \right)\;{\rm{K}}\\ &= {\rm{423}}\;{\rm{K}} \end{align*} {/eq}

Substitute the values in the above formula to calculate the molar mass.

{eq}\begin{align*} {\rm{1}}\;{\rm{atm}} \times {\rm{0}}{\rm{.960}}\;{\rm{L}} &= \dfrac{{0.941\;{\rm{g}} \times 0.0821\;\dfrac{{{\rm{L}} \cdot {\rm{atm}}}}{{{\rm{mol}} \cdot {\rm{K}}}} \times 423\;{\rm{K}}}}{{{\rm{Molecular}}\;{\rm{weight}}\left( {{\rm{g/mol}}} \right)}}\\ &= 34.04\;{\rm{g}} \end{align*} {/eq}

The molar mass of the unknown compound up to 3 significant figures is {eq}\underline {34.0\;{\rm{g}}} {/eq}.