# Determining sulfate content in Mohr's salt

I am practicing problems for my chemistry test and I am stuck on the following problem:

The value of $$x$$ in $$\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}$$ can be found by determining the amount, in moles, of sulfate in the compound.

A $$\pu{0.982 g}$$ sample was dissolved in water and excess $$\ce{BaCl2(aq)}$$ was added. The precipitate of $$\ce{BaSO4}$$ was separated and dried and found to weigh $$\pu{1.17 g}.$$

Calculate the amount (in moles) of $$\ce{BaSO4}$$ in the $$\pu{1.17 g}$$ precipitate. And calculate the amount (in moles) of sulfate in the $$\pu{0.982 g}$$ sample of $$\ce{Fe(NH4)2(SO4)2 ⋅ xH2O}.$$

I was able to do the first part of the question:

$$n (\ce{BaSO4}) = \frac{\pu{1.17 g}}{\pu{233.38 g mol-1}} = \pu{0.005 mol}$$

But, I am unsure of the second part. Any help would be appreciated.

• Write a reaction equation for the precipitation (include it in the post), then track the sulfate from the product to the educt. You should then be able to find the amount of the iron salt. – Martin - マーチン Feb 12 '20 at 10:36
• And please choose a descriptive title for your question, it currently values no information at all and is not helpful. – Martin - マーチン Feb 12 '20 at 10:38
• Use the number of moles sulfate precipitated to calculate the number of moles of Mohr´s Salt in the original sample. Using the molecular weight of the anhydrous salt you can calculate how much that part weighs, and the difference up to 0,982g is due to the water. – FrankS Feb 12 '20 at 10:42
• @FrankS Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms". – Martin - マーチン Feb 12 '20 at 17:32

If there is $$\ce{0.00501 mol}$$ of $$\ce{BaSO4}$$ in the precipitate, this means that there was $$\pu{0.00501 mol}$$ of sulfate in the Mohr's salt solution and $$\pu{0.00501 mol}$$ of sulfate in $$\pu{0.982 g}$$ of Mohr's salt. And there was one half, that is $$\pu{0.0025 mol}$$ of $$\ce{Fe(NH4)2(SO4)2.xH2O}$$ in $$\pu{0.982 g}$$ of Mohr's salt. This allows you to calculate the molar mass of the salt, and to deduce $$x$$.