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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.

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?

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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}

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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:

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

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