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


> 
> Formalin is a $37\%~\mathrm{W/W}$ solution of methanal in water. What mass of water must be added to $250~\mathrm{g}$ of methanal to make formalin.
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Here's what I've done so far. It's wrong though. According to my book answer is $430~\mathrm{g}$ $\ce{H2O}$.


$$\frac{Cw}{w} = \frac{\mathrm{m\ solute}}{\mathrm{m\ solution}} \cdot 100\%$$


$$37 \% = \mathrm{\frac{m\ solute}{250} \cdot 100\ \%}$$


$$\mathrm{\frac{37}{100} = \frac{m\ solute}{250}}$$


$$\mathrm{0.37 \cdot 250 = m\ solute}$$


$$\mathrm{m\ solute = 92.5\ g}$$


Where have I went wrong? Am I completely off or at least close? 




You have been given a [mass fraction](http://en.wikipedia.org/wiki/Mass_fraction_(chemistry)), $w\_{\ce{CH2O}}=37\%~\mathrm{w/w}$, 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}


 5

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:


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


 2