Calculation of VOCs concentration (ppm) in the headspace of a bottle

What is the right way to calculate the concentration of volatile compounds in the air in a closed container?

Suppose we have a $$\pu{50 mL}$$ bottle with $$\pu{25 mL}$$ of $$100\%$$ ethanol $$(\ce{EtOH})$$ in it.

In this case, will it be okay to use $$PV = nRT?$$

Doing so, it will be:

\begin{align} p(\ce{EtOH}) &= \pu{0.05872 atm}\\ V &= \pu{0.025 L}\\ R &= \pu{0.08206 L atm mol-1 K-1}\\ T &= \pu{298 K}\\ \end{align}

giving us $$n = \pu{6.00314E-5 mol}.$$

The ppm: $$\pu{1 ppm} = \pu{1 mg l-1}$$

$$M(\ce{EtOH}) = \pu{46.07 g mol-1}$$

$$\pu{6.00314E-5 mol} × \pu{46.07 g mol-1} × 1000 = \pu{2.766 mg}$$

This lead to $$[\pu{ppm}] = \frac{\pu{2.766 mg}}{\pu{0.025 l}} = \pu{110.6 ppm}$$

• You have calculated the number of moles of ethanol. Do you need something else to find the concentration? Sep 23, 2019 at 16:30
• I just want to verify that it is the right way to calculate concentration of volatile liquids in the air? Thank you! Sep 23, 2019 at 16:32
• Just a couple more steps and you're done. Sep 23, 2019 at 16:43
• Thank you, made them, is it right? As this calculator giving me completely different answer madur.com/index.php?page=/partial_pressure Sep 23, 2019 at 16:54
• Don't you think approx.0.06 atm should lead to approx.60000 ppm v/v, even without any calculation ? Sep 24, 2019 at 7:33

Usage of ppm/ppb etc units is discouraged, as they are ambiguous without the explicit context. As generally, 1 ppm (w/w) <> 1 ppm (V/V) <> 1 ppm (n/n) <> 1 ppm (w/V).

It is recommended to use explicit units, like e.g mg/L.

Expecting ideal gas behaviour $$pV = nRT$$ is not reasonable. Ideal gases are not in equilibrium with their liquid phase.

There should be rather used a real gas equation.The easiest to use is the van Der Waals equation, but even that has significant error near condensation state.

Additionally, the sample(!) temperature has to be strictly controlled, otherwise $$n$$ would be variable.

Why not rather to use small volume of ethanol to be known to fully evaporate in given volume ? Or perhaps other solvent, unless ethanol is waterless. Note that ethanol has significant deviation from the Raoult's law.

I prefer $$R = \pu{8.31446 J K-1 mol-1}$$

Note that by May 2019 redefinition, It's exact value is $$\pu{8.31446261815324 J K-1 mol-1}$$

• Engineer here. Using the vapour pressure and the ideal gas law (IGL) will usually give a sufficiently accurate solution for gases for which $\dfrac{R\cdot T}{P} > 20 \frac{L}{g\ mol}$. That said, if you want to correlate vapour-liquid equilibrium, using the IGL is not a good idea. Sep 26, 2019 at 9:44