# How do you rewrite molarity to weight percentage or volume percentage?

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 $$vol\%$$

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

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}

Legend:

• $$c$$ - 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)
• Thank you so much! What would be the difference between $\varphi_1$ and $\varphi$? Jun 13, 2022 at 20:01
• Aside of general and particular context, it was just a typo of missing index in the equation. Jun 13, 2022 at 21:37
• 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? Jun 15, 2022 at 11:25
• 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. Jun 15, 2022 at 11:38