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.
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 $x$.