# How to calculate a molar mass based on density?

I have been asked to calculate the molar mass and mole fraction of a 35 w/w% HClO4 solution with a density of 1.3g per ml.

This is the equation I used to try to solve it:

$\frac{1}{M} = \frac{0.35}{100.46} + \frac{0.65}{18.02}$

Is this correct?

• Welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. This appears to be a homework question, please share your thoughts and attempts towards the solution. This question has some information that can help you. – Martin - マーチン Jan 13 '16 at 7:45

What we are talking here are 3 measures of concentration:

What we know that the mass fraction is $0.35w/w$ and the density is $1.3g/mL$. You can count the molar mass of water and HClO4 I assume, let's call them $M_{solvent} (M_{H_2O})$ and $M_{solute} (M_{HClO_4})$.

I will use the following equations by the calculations:

$$m_{solvent} = m_{solution} - m_{solute}$$ $$n_{solute} = \frac{m_{solute}}{M_{solute}}$$ $$n_{solvent} = \frac{m_{solvent}}{M_{solvent}}$$ $$n_{solution} = n_{solute} + n_{solvent} = \frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}$$

From the mass fraction you can get the mole fraction this way:

$$\frac{n_{solute}}{n_{solution}} = \frac{\frac{m_{solute}}{M_{solute}}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{1}{1+\frac{\frac{m_{solvent}}{M_{solvent}}}{\frac{m_{solute}}{M_{solute}}}} = \frac{1}{1+\frac{m_{solvent} \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(m_{solution} - m_{solute}) \cdot M_{solute}}{M_{solvent} \cdot m_{solute}}} = \frac{1}{1+\frac{(1 - \frac{m_{solute}}{m_{solution}}) \cdot M_{solute}}{M_{solvent} \cdot \frac{m_{solute}}{m_{solution}}}}$$

You can count the molar mass the following way:

$$\frac{m_{solution}}{n_{solution}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solvent}}{M_{solvent}}} = \frac{m_{solution}}{\frac{m_{solute}}{M_{solute}} + \frac{m_{solution} - m_{solute}}{M_{solvent}}} = \frac{1}{\frac{\frac{m_{solute}}{m_{solution}}}{M_{solute}} + \frac{1 - \frac{m_{solute}}{m_{solution}}}{M_{solvent}}}$$

You don't need the density for these calculations.

$$M_{HClO_4} = 100.46g/mol$$ $$M_{H_2O} = 18.02g/mol$$ $$m_{HClO_4}/m_{solution} = 0.35$$ $$\frac{n_{HClO_4}}{n_{solution}} = \frac{1}{1+\frac{(1 - \frac{m_{HClO_4}}{m_{solution}}) \cdot M_{HClO_4}}{M_{H_2O} \cdot \frac{m_{HClO_4}}{m_{solution}}}} = \frac{1}{1+\frac{(1 - 0.35) \cdot 100.46g/mol}{18.02g/mol \cdot 0.35}} = 0.088$$ $$M_{solution} = \frac{m_{solution}}{n_{solution}} = \frac{1}{\frac{\frac{m_{HClO_4}}{m_{solution}}}{M_{HClO_4}} + \frac{1 - \frac{m_{HClO_4}}{m_{solution}}}{M_{H_2O}}} = \frac{1}{\frac{0.35}{100.46g/mol} + \frac{1 - 0.35}{18.02g/mol}} = 25.28g/mol$$

• First of all thank you for your answer. Could you please confirm the correctness of this equation I used to solve my problem: $\frac{1}{M} = \frac{0.35}{100.46} + \frac{0.65}{18.02}$ ? – ohiliouh Jan 13 '16 at 12:40
• @JaqcuesMartin It's correct. – inf3rno Jan 14 '16 at 6:53