Any suggestions how to solve this? $m(\ce{FeSO4. $x$~H2O})=20.0\:\mathrm{g}$

$N(\ce{H2O})=3.03 \cdot10^{23}$



$x=?$

What I've done:

(1.)


  \begin{aligned}
   m (\ce{H2O})&amp;= \frac{N(\ce{H2O}) \cdot M(\ce{H2O})}{N_{A}}\\
   &amp;= \frac{3.03 \cdot10^{23} \cdot 18.02$\:\mathrm{g/mol}}{6.02 \cdot 10^{23}/\:\mathrm{mol}}\\
   &amp;= 9.06987\:\mathrm{g}
  \end{aligned}


(2.) 


  \begin{aligned}
   m(\ce{FeSO4})&amp;= m(\ce{FeSO4} \cdot x \ce{H2O}) - m(\ce{H2O})\\
   &amp;= 20.00\:\mathrm{g} - 9.06987\:\mathrm{g}\\
   &amp;= 10.9301\:\mathrm{g}
  \end{aligned} 


Any suggestions what to do next?
Calculate $n(\ce{FeSO4})$ with $M(\ce{FeSO4})\approx 136\:\mathrm{g/mol}$


   $n(\ce{FeSO4})\approx0.08\:\mathrm{mol}$


Calculate $n(\ce{H2O})$


   $n(\ce{H2O})\approx0.5\:\mathrm{mol}$


Set them into relation $n(\ce{FeSO4}):n(\ce{H2O})= 1:x$


   $x\approx 6$