Question

What is the density of gaseous ethane, ${{\text{C}}_{2}}{{\text{H}}_{6}}$ at STP?

Answer 1

Hint: Density is defined as the mass per unit volume occupied by one mole of a gas. Using the ideal gas law and assuming standard pressure and temperature (STP), the density of gaseous ethane can be calculated:
\[\text{Density (}\rho )=\dfrac{\text{Molar mass (M)}}{\text{Volume of gas at STP (V)}}\]

Complete answer:
Density is defined as the mass per unit volume of a substance. At STP, the volume of a gas is only dependent on the molar mass of that gas. By using the ideal gas law, we can calculate the density $\left( \rho \right)$ of a gas using only its molar mass. The ideal gas equation is an equation relating state variables of gas and it is written as given below:
\[\text{pV}=\text{nRT}\]
Where p is the pressure of the gas, V is volume, n is the number of moles, R is the universal gas constant and T is the absolute temperature.
Now, the volume of a gas is related to its density by the following equation:
\[\rho =\dfrac{\text{m}}{\text{V}}\]
On rearranging for volume, we get,
\[\text{V}=\dfrac{\text{m}}{\rho }\]
Then, on substituting this value of volume in the ideal gas equation and solving for density we get:
\[\begin{align}
  & \text{p}\dfrac{\text{m}}{\text{ }\!\!\rho\!\!\text{ }}=\text{nRT} \\
 & \Rightarrow \text{ }\!\!\rho\!\!\text{ }=\dfrac{\text{pm}}{\text{nRT}} \\
 & \because \text{Molar mass (M)}=\dfrac{\text{m}}{\text{n}}\Rightarrow \text{m}=\text{Mn} \\
 & \therefore \text{ }\!\!\rho\!\!\text{ }=\dfrac{\text{pM}}{\text{RT}} \\
\end{align}\]
Now at STP, we have
\[\begin{align}
  & \text{p}=1\text{ atm, R}=0.082\text{ L atm mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}}\text{, T}=273\text{K} \\
 & \therefore \text{ }\!\!\rho\!\!\text{ }=\dfrac{\left( \text{1 atm} \right)\text{M}}{\left( 0.082\text{ L atm mo}{{\text{l}}^{-1}}\text{ }{{\text{K}}^{-1}} \right)\left( 273\text{K} \right)} \\
 & \Rightarrow \text{ }\!\!\rho\!\!\text{ }=\dfrac{\text{M}}{22.4\text{ L mo}{{\text{l}}^{-1}}} \\
\end{align}\]
Using this information and assuming that ethane behaves as an ideal gas, we can find its density.
\[\begin{align}
  & \text{Molar mass of }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}\text{ (M)}=2\left( \text{mass of C} \right)+6\left( \text{mass of H} \right) \\
 & \Rightarrow \text{M}=2\left( 12 \right)+6\left( 1 \right) \\
 & \Rightarrow \text{M}=30\text{ g mo}{{\text{l}}^{-1}} \\
 & \therefore \text{ }\!\!\rho\!\!\text{ }=\dfrac{30\text{ g mo}{{\text{l}}^{-1}}}{22.4\text{ L mo}{{\text{l}}^{-1}}}=1.34\text{ g }{{\text{L}}^{-1}} \\
\end{align}\]
Hence, the density of gaseous ethane, ${{\text{C}}_{2}}{{\text{H}}_{6}}$ at STP is 1.34 grams per liter.

Note:
Since gases occupy the same volume at STP, the density of a particular gas is dependent on its molar mass. A gas with a small molar mass value will have a lower density than a gas with a large molar mass.