What is mol % and how do you calculate the mass of a compound needed to be a certain mol % relative to another compound at a particular volume? [closed]








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I know that there is some stoichiometry involved, but I am lost here. I know you need the molecular weight of both compounds and that's about it?




Assuming we have a binary mixture, or equivalently, a mixture that only involves two different components $A$ and $B$, the total mass of the mixture is simply the sum of the individual masses:


$$m=m\_A+m\_B$$


The same logic applies to the total amount of the mixture:


$$n=n\_A+n\_B$$


When you divide the mass of $A$ and $B$ (separately) by the total mass of the mixture, you obtain the mass fraction of $A$ and $B$ in the mixture:


$$X\_A=\frac{m\_A}{m}\quad\quad X\_B=\frac{m\_B}{m}$$


When you divide the amount of $A$ and $B$ (separately) by the total amount of the mixture, you obtain the mole fraction of $A$ and $B$ in the mixture:


$$Y\_A=\frac{n\_A}{n}\quad\quad Y\_B=\frac{n\_B}{n}$$


In a binary mixture, the sum of fractions of $A$ and $B$ is equal to 1:


$$X\_A+X\_B=1$$


$$Y\_A+Y\_B=1$$


The molar mass of a binary mixture can be calculated with either the mass fractions or mole fractions, and the molar masses of each component:


$$M=\left(\frac{X\_A}{M\_A}+\frac{X\_B}{M\_B}\right)^{-1}$$


$$M=Y\_A\;M\_A+Y\_B\;M\_B$$


Finally, if you want to convert a mass fraction into a mole fraction or vice-versa, you can use these formulas:


$$X\_A=Y\_A\;\frac{M\_A}{M}$$


$$X\_B=Y\_B\;\frac{M\_B}{M}$$


$$Y\_A=X\_A\;\frac{M}{M\_A}$$


$$Y\_B=X\_B\;\frac{M}{M\_B}$$


Note that the "mol percent" you're referring to is the mole fraction multiplied by 100, and neither fraction (mass or mole) depends on volume, assuming no reaction is taking place.


You should be able to solve your problem by using these formulas.


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