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:

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