I wonder whether the following relation is true:

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}},$$

where $M$ is the molar mass of the component, $n$ is the amount of substance and $m$ is the mass. It was derived assuming $n = m/M,$ $n = n_\mathrm{solvent} + n_\mathrm{solute}$ and $m = m_\mathrm{solvent} + m_\mathrm{solute}.$

I don't think this is true, but I wanted to be sure before doing anything weird on a test.

  • 1
    $\begingroup$ No, I'm afraid it's completely wrong. Try re-doing it by substituting n_solvent=m_solvent/M_solvent, and n_solute=m_solute/M_solute. Also, include the units as a check. Note that your expression doesn't have the right dimensions. $\endgroup$
    – theorist
    Feb 13, 2021 at 23:47
  • $\begingroup$ Molar mass of solute and solvent is different, therefore, based on that alone, your relationship is incorrect. $\endgroup$ Feb 14, 2021 at 0:45
  • 4
    $\begingroup$ Even if M for both solvent and solute was accidentally equal, it would be still wrong. As molar amount is proportional to mass, not reciprocal to mass. $\endgroup$
    – Poutnik
    Feb 14, 2021 at 6:25

1 Answer 1


To sum up the comments, only the following relation for the total amount of solution $n_\mathrm{tot}$ is universally true:

$$n_\mathrm{tot} = n_\mathrm{solvent} + n_\mathrm{solute} = \frac{m_\mathrm{solvent}}{M_\mathrm{solvent}} + \frac{m_\mathrm{solute}}{M_\mathrm{solute}}\tag{1}$$

The best you can do is to assume that $n_\mathrm{tot}\approx n_\mathrm{solvent}$ for the diluted solutions of small molecules. Also, if the molar masses are similar $(M_\mathrm{solvent}\approx M_\mathrm{solute}\approx \bar{M}),$ the expression can be lead to a common denominator:

$$n_\mathrm{tot} \approx \frac{m_\mathrm{solvent} + m_\mathrm{solute}}{\bar{M}}\tag{2}$$

This can be the case, for example, for the solution of ammonium nitrate $(M(\ce{NH4NO3}) = \pu{80.043 g mol^-1})$ in dimethyl sulfoxide $(M(\text{DMSO}) = \pu{78.13 g mol^-1}).$

Algebra aside, your mistake was also neglecting $M_\mathrm{solvent}.$ Keep in mind you can always check dimensions. For your proposed formula

$$n_\mathrm{solvent} + n_\mathrm{solute} = \frac{M}{m_\mathrm{solvent} + m_\mathrm{solute}}$$

it doesn't work out well:

$$\mathrm{dim}~n_\mathrm{tot} = \mathsf{M}\cdot\mathsf{N}^{-1}\cdot\mathsf{M}^{-1} = \mathsf{N}^{-1} \neq \mathsf{N}\tag{3}$$

Illustrating with common units used in chemistry:

$$[n_\mathrm{tot}] = \frac{[M]}{[m_\mathrm{solvent}] + [m_\mathrm{solute}]} = \frac{\pu{g mol^-1}}{\pu{g}} = \pu{mol^-1} \neq \pu{mol}\tag{4}$$

  • $\begingroup$ I made a typo in the latex, I was referring to if this was true $n_{solvent}+n_{solute}=\frac{m_{solvent}+m_{solute}}{M}$ and not $n_{solvent}+n_{solute}=\frac{M}{m_{solvent}+m_{solute}}$ still not correct, right? $\endgroup$
    – Xetrez
    Feb 14, 2021 at 12:50
  • $\begingroup$ @Xetrez In general, no, the expression is still not correct with a minor exception I addressed in my edit. $\endgroup$
    – andselisk
    Feb 14, 2021 at 13:13
  • $\begingroup$ and what about it is wrong? is it that the $m_{total}\neq m_{solvent}+m_{solute}$? $\endgroup$
    – Xetrez
    Feb 15, 2021 at 11:14

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