Problem: 70 grams of pure, hydrated $\ce{MgSO_4}$ is dissolved in 90 grams of $\ce{H_2O}$. If the mole fraction of the solute is 0.039, find the molecular formula of the solute.

My flawed attempt at a solution:

Let the solute be $\ce{MgSO_4.xH_2O}$. In $70~\mathrm{g}$ of solute, the amount of solute is $$\frac{70}{120+18x}~\mathrm{mol}.$$

Likewise, in $\pu{90 g}$ of solvent, we have $$\frac{90}{18} = 5~\mathrm{mol}.$$

Plugging these values into the formula for mole fraction and equating it to 0.039, I thought I would get the value of $x$ as 7. Instead, I seem to be getting approximately 12.

Could somebody please help me solve this question? In general, when we have a hydrated solute in an aqueous solution, do we include the water of crystallisation in the total mass of available water while calculating the number of moles of water?

  • 1
    $\begingroup$ I think the idea here is that once the solid $\ce{MgSO_4.xH_2O}$ dissolves, its water content becomes part of the solvent. Does that help? $\endgroup$ – Nicolau Saker Neto May 22 '15 at 14:47
  • $\begingroup$ @NicolauSakerNeto I couldn't understand Sir. Could you please show me through calculations? $\endgroup$ – New Change May 22 '15 at 14:53
  • $\begingroup$ You've done a good job. You just need to take into account the following equation: $\ce{MgSO_4.xH_2O(s) \xrightarrow{water} MgSO_4(aq) + xH_2O(l)}$. That is, the total amount of solvent is not only the water initially added, but also has to take into account the water that was freed from the solid when dissolving. Thus, $n_{solvent}=5+\mathrm{something}$. If you can figure out what that something is, you'll be able to solve your problem. Try it out a bit more, if you don't manage I can provide some additional guidance. $\endgroup$ – Nicolau Saker Neto May 22 '15 at 15:11
  • $\begingroup$ @NicolauSakerNeto Sir, but then shouldn't the mass of the solute reduce from 70 g? Originally, $MgSO_4.xH_2O$ had a mass of 70 g and as part of it (the water of crystallistion) is lost to the solvent hence the mass of the solute should reduce. Therefore, the number of moles of the solute should also reduce. $\endgroup$ – New Change May 22 '15 at 16:01
  • $\begingroup$ The "solute", really, is only the $\ce{MgSO4}$, as the crystallized water in the hydrate becomes indistinguishable from the solvent once dissolved. The mass of $\ce{MgSO4}$ is indeed smaller than the initial amount of $\ce{MgSO_4.xH_2O}$, but by the stoichiometry of the dissolution in my previous comment, the number of moles of $\ce{MgSO4}$ is the same as $\ce{MgSO_4.xH_2O}$. $\endgroup$ – Nicolau Saker Neto May 22 '15 at 16:38

Alright, let’s take this one step at a time.

First of all, you want to determine a molecular formula, and among your known data you have a mole fraction. This suggests it’s a good idea to work with moles to solve the problem. As such, the first step is to convert the amount of water added and magnesium sulfate hydrate used. Calculating the number of moles of water added is easy:

$$90\ \mathrm{g}\ \text{water added} \times \frac{1\ \mathrm{mol}\ \ce{H2O}}{18\ \mathrm{g} \ \ce{H2O}} = 5\ \mathrm{mol}\ \ce{H2O}$$

The number of moles of $\ce{MgSO4.$x\,$H2O}$ is unknown because even though you know how much mass was dissolved, you don’t know its formula completely. However, $\ce{MgSO4}$ contributes $120\ \mathrm{g\ mol^{-1}}$ to the molar mass, and each additional molecule of water in the hydrate stoichiometry adds a further $18\ \mathrm{g\ mol^{-1}}$, so that the molar mass of $\ce{MgSO4.$x\,$H2O}$ is $120\ \mathrm{g\ mol^{-1}} + x \times 18\ \mathrm{g\ mol^{-1}} = (120+18x)\ \mathrm{g\ mol^{-1}}$. With this molar mass, the number of moles of $\ce{MgSO4.$x\,$H2O}$ is:

$$70\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O} \times \frac{1\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}}{(120+18x)\ \mathrm{g}\ \ce{MgSO4.$x\,$H2O}} = \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4.$x\,$H2O}$$

From the definition of mole fraction:


Now, one is tempted to write the following and solve for $x$:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}\ }{\frac{70}{120+18x}+5 \ }=0.039$$

However, this equation is incorrect. The problem is that the following process was not taken into account:

$$\ce{MgSO4.$x\,$H2O(s) \xrightarrow{water} MgSO4(aq) + $x$\ H2O(l)}$$

In other words, when the magnesium sulfate hydrate dissolves in water, the water molecules contained in the solid are released and become an indistinguishable part of the solvent; the final mixture is a simply a solution of the solute $\ce{MgSO4}$ in water. Therefore, the number of moles of solvent is not simply $5\ \mathrm{mol}$ as calculated at the start, but rather a larger value, which should include the number of moles of water released when the solid dissolves.

From the dissolution equation, we see that $1\ \mathrm{mol}$ of $\ce{MgSO4.$x\,$H2O}$ generates $1\ \mathrm{mol}$ of $\ce{MgSO4}$ and $x\ \mathrm{mol}$ of $\ce{H2O}$ from crystallized water. However, we know the amount of moles of $\ce{MgSO4.$x\,$H2O}$. This means the following:

$$n_\text{solute}=n_{\ce{MgSO4}}=1 \times n_{\ce{MgSO4.$x\,$H2O}}= \frac{70}{120+18x}\ \mathrm{mol}\ \ce{MgSO4}$$

$$n_\text{solvent} = n_\text{added water} + n_\text{crystallized water} = n_\text{added water} + x \times n_{\ce{MgSO4.$x\,$H2O}} = \bigg(5 + \frac{70x}{120+18x}\bigg)\ \mathrm{mol}\ \ce{H2O}$$

Now we can finally apply these values to the mole fraction equation:

$$X_\text{solute}=\frac{n_\text{solute}}{n_\text{solute}+n_\text{solvent}}=\frac{\frac{70}{120+18x}}{\frac{70}{120+18x} + \bigg(5 + \frac{70x}{120+18x}\bigg)}=0.039$$

Solving for $x$ provides the answer $x=7.03$, which means there are seven moles of water per mole of $\ce{MgSO4}$, and thus the formula for the solid is $\ce{MgSO4.7H2O}$, representing magnesium sulfate heptahydrate.

| improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.