Assuming an ideal gas, calculate the density in $\mathrm{g/cm^3}$ of propane, $\ce{C3H8}$ at $0.000~^\circ\mathrm{C}$ and $1.000~\mathrm{atm}$. (3 significant figures)

I have figured out that I need this formula: $$\frac{n}{V} = \frac{p}{RT}$$

I have a total of $44.09562~\mathrm{g/mol}$

I am stuck at this point.


As you have already figured out that we have to use the ideal gas law $PV= nRT$, which is you have modified as $$\frac{n}{V}= \frac{P}{RT}$$

To find the density, we can further modify the equation as, \begin{align} \frac{m}{MV} &= \frac{P}{RT}\\ \frac{\rho}{M}&= \frac{P}{RT}\\ \rho &= \frac{PM}{RT}\\ \end{align}

Now we just have to replace the parameters with their respective values.

The value of temperature should be converted to Kelvin as $(0 + 273)~\mathrm{K} = 273~\mathrm{K}$.

Likewise, the value of gas constant $R$ should be chosen as $0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}}$. \begin{align} \rho &= \frac{1~\mathrm{atm} \cdot 44.09562~\mathrm{g~mol^{-1}}}{{0.8206~\mathrm{L~atm~K^{-1}{mol}^{-1}} \cdot 273~\mathrm{K}}}\\ &=1.9683~\mathrm{g/L}\\ &=0.0019683~\mathrm{g/cm^3}\\ \end{align}

  • $\begingroup$ You can use other methods for getting the answer since you are already given 44.09562 g/mol . $\endgroup$ – CCR Dec 19 '14 at 4:56
  • $\begingroup$ R above is out by 10, Use 0.08206 L atm K-1 mol-1 $\endgroup$ – Michel Mar 9 '20 at 6:05

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