# Chemistry - How to find the mass of the solute given a concentration as a mass fraction?

## Solution 1:

You have been given a mass fraction, $w_{\ce{CH2O}}=37\%~\mathrm{w/w}$, and the mass of the solute, $m_{\ce{CH2O}}=250~\mathrm{g}$. You calculated the mass of the solution wrong.

The mass fraction is defined as
$$w_i = \frac{m_i}{m}\tag1$$
with $m$ being the total mass and therefore
$$m = \sum_i m_i.$$
The latter is also the mass of your solution, the sum of the solute and the solvent.
$$m=m_{\ce{H2O}}+m_{\ce{CH2O}} = m_{\ce{H2O}} + 250~\mathrm{g}$$
Substitute what you know into $(1)$ and transform:
\begin{align}
w_{\ce{CH2O}} &= \frac{m_{\ce{CH2O}}}{m_{\ce{H2O}}+m_{\ce{CH2O}}}\\
0.37 &= \frac{250~\mathrm{g}}{m_{\ce{H2O}} + 250~\mathrm{g}}\\
m_{\ce{H2O}} &= \frac{250~\mathrm{g}}{0.37}-250~\mathrm{g} = 426~\mathrm{g}
\end{align}

## Solution 2:

You have a couple of problems. In your equation, you've written the mass of the solution is 250 g, but there is 250 g of methanal and the total mass of the solution is 250 g + $m_{\ce{H2O}}$.

Second, the solution is 37% methanal not 37% water so the you've written the equation, you're solving for the wrong thing.

These should help you get on track:

$$\begin{align}\\ f_\mathrm{meth} &= \frac{m_\mathrm{meth}}{m_\mathrm{solution}}\\ m_\mathrm{solution}&=m_\mathrm{meth}+m_\mathrm{\ce{H2O}}\\ \%_\mathrm{meth} &= f_\mathrm{meth}\times100\% \end{align}$$