Exercise: Convert mole fraction to mass fraction

Select a basis of \(\SI{100}{gmol}\) of exhaled gas. As explained in Introduction to Chemical Engineering: Tools for Today and Tomorrow (5th Edition), we select that number because it is convenient and because it makes it easy to calculate how many \(\si{gmol}\) of each substance are present in that quantity of material.

\[ \begin{align}\begin{aligned}n_{O_{2}} = y_{O_{2}} \, n = \left( \frac{ \SI{18}{gmol \, O_{2}} } { \SI{100}{gmol \, total} } \right) (\SI{100}{gmol \, total}) = \SI{18}{gmol} \ce{O2}\\n_{N_{2}} = y_{N_{2}} \, n = \left( \frac{ \SI{78}{gmol \, N_{2}} } { \SI{100}{gmol \, total} } \right) (\SI{100}{gmol \, total}) = \SI{78}{gmol} \ce{N2}\\n_{CO_{2}} = y_{CO_{2}} \, n = \left( \frac{ \SI{4}{gmol \, O_{2}} } { \SI{100}{gmol \, total} } \right) (\SI{100}{gmol \, total}) = \SI{4}{gmol} \ce{CO2}\end{aligned}\end{align} \]

Now, now calculate the number of grams of each substance:

\[ \begin{align}\begin{aligned}m_{O_{2}} = MW_{O_{2}} \, n_{O_{2}} = (\SI{32}{g \, O_{2}/gmol})(\SI{18}{gmol \, O_{2}}) = \SI{576}{g}\\m_{N_{2}} = MW_{N_{2}} \, n_{N_{2}} = (\SI{28}{g \, N_{2}/gmol})(\SI{78}{gmol \, N_{2}}) = \SI{2184}{g}\\m_{CO_{2}} = MW_{CO_{2}} \, n_{CO_{2}} = (\SI{44}{g \, CO_{2}/gmol})(\SI{4}{gmol \, CO_{2}}) = \SI{176}{g}\end{aligned}\end{align} \]

Summing these values, we get a total mass of \(\SI{2936}{g}\). Therefore, the mass fractions are as follows:

\[ \begin{align}\begin{aligned}x_{O_{2}} = \frac{\SI{576}{g \, O_{2}}} {\SI{2936}{g}} = 0.196\\x_{N_{2}} = \frac{\SI{2184}{g \, N_{2}}} {\SI{2936}{g}} = 0.744\\x_{CO_{2}} = \frac{\SI{176}{g \, CO_{2}}} {\SI{2936}{g}} = 0.060\end{aligned}\end{align} \]

These fractions sum to \(1\) (one of our important checks).