Suppose I have a molar concentration $[C]$ in $mol/m^3$ in a continuous reactor, from which the differential equation is as follows:

$ \frac{\partial [C]}{\partial t} = k[A][B] + \phi C_{,in} - \phi C_{,out}$

Where the first term is the rate law with rate constant $k$, and with concentrations of reactants $A$ and $B$, as $A+B \longrightarrow C$.

The second and third terms are inflows and outflows of substance $C$ in a continuous reactor.

The outflow of $C$ is measured, and therefore I need to have each term correlate with each other by having the correct dimensions. I have the following question:

Suppose $C$ is a gaseous substance, and the measured output is in $vol\%$ (volume percentage), how would I convert $\phi C_{,out}$ or $\phi C_{,in}$, which is in $[Nm^3/s]$ (normal cubic meter per second), to $vol\%C$? I found the following:

$\hspace{30pt} \phi_{C_{,out}} = \dfrac{p\dfrac{vol\% C}{100\%}}{RT}\phi_{out}\hspace{10pt}$ but this leads to a wrong dimension $[m^3/mol]$ for the $vol\%$

  • $\begingroup$ Try to derive both percentages from molar concentration from the scratch, using the respective definitions and known ideal gas laws. $\endgroup$
    – Poutnik
    Jun 3, 2022 at 6:55
  • $\begingroup$ Could you maybe give a hint? I'm a bit stuck $\endgroup$
    – user313866
    Jun 8, 2022 at 9:29
  • $\begingroup$ Describe being stuck. // Write down for yourself definitions of considered quantities (molarity, weight percentage, volume percentage) and relations between them. Then you are done. $\endgroup$
    – Poutnik
    Jun 8, 2022 at 9:33

1 Answer 1


Relation of molarity and molar fraction ( = volume fraction for ideal gas approximation), using ideal gas state equation:

\begin{align} c &= \frac{n_1 }{ V} \tag{1}\\ pV &= nRT\tag{2}\\ c = \frac{n_1 }{ \frac{(n1+n2)RT}{p}}&=\frac{p}{RT}\frac{n_1}{n_1+n_2}\tag{3} \end{align}

Computing volume fraction and percentage from molar concentration:

\begin{align} \varphi&=c \cdot \frac{RT}{p}\tag{4}\\ vol\%&=\varphi \cdot 100\tag{5} \end{align}

Dimension of $\frac{cRT}{p}$ is $$\frac{\pu{mol m-3}\pu{J K-1 mol-1}\pu{K}}{\pu{N m-2}}= \frac{\pu{ m-3 J }}{\pu{(N m)(m-1 m-2)}}= \frac{\pu{ m-3 J }}{\pu{J m-3}}=1\tag{6}$$

Computing mass fraction and percentage from volume fraction:

\begin{align} w_1&=\frac{\varphi_1 M_1}{\varphi_1 M_1 + (1-\varphi_1)M_2}\tag{7}\\ mass\%&=w \cdot 100\tag{8} \end{align}


  • $c$ - molar concentration $\pu{mol m-3}$
  • $n_1$ - analyte molar amount [$\pu{mol}$]
  • $n_2$ - molar amount of the mixture but the analyte[$\pu{mol}$]
  • $V$ - volume of mixture [$\pu{m3}$]
  • $R$ - gas constant [$\pu{8.314 J K-1 mol-1}$]
  • $p$ - gas pressure [$\ce{Pa}$]
  • $\varphi$ - volume fraction
  • $\varphi_1$ - the volume fraction of the analyte
  • $w$ - mass fraction
  • $w_1$ - the mass fraction of the analyte
  • $M_1$ - molar mass of the analyte [$\pu{g/mol}$]
  • $M_2$ - molar mass of the major gas [$\pu{g/mol}$]
    ( or the effective mean molar mass of used gas mixture, e.g. $\pu{28.8 g mol-1}$ for air)
  • $\begingroup$ Thank you so much! What would be the difference between $\varphi_1$ and $\varphi$? $\endgroup$
    – user313866
    Jun 13, 2022 at 20:01
  • $\begingroup$ Aside of general and particular context, it was just a typo of missing index in the equation. $\endgroup$
    – Poutnik
    Jun 13, 2022 at 21:37
  • $\begingroup$ That seems logical, thanks. If I understand correctly, this is only useful for gaseous analyte $A$ in a mixture or major gas $B$ right? What would be different if it would be a solid mass for the mass fraction? $\endgroup$
    – user313866
    Jun 15, 2022 at 11:25
  • $\begingroup$ For volume/mass fraction/percentage, you would need density of the components and the mixture at given conditions. Rest is trivial high school knowledge. // Do not expect receiving answers to your questions on StackExchange network without explicit a priori effort to answer it yourself. $\endgroup$
    – Poutnik
    Jun 15, 2022 at 11:38

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