FeSO4 · 7H2O(s) \(\overset{\Delta}{\rightarrow}\) FeSO4(s) + 7H2O(g)
\[\begin{aligned} n_\mathrm{FeSO_4 \cdot 7H_2O} &= \frac {m_\mathrm{FeSO_4 \cdot 7H_2O}}{M_\mathrm{FeSO_4 \cdot 7H_2O}} = \\ &= \frac {1,26\mathrm{g}}{(55,85+32,07+16,00\cdot 4+7\cdot (1,008\cdot 2+16,00))\mathrm{g/mol}} = \\ &= 0,00453185\mathrm{mol}\end{aligned}\]
\(n_\mathrm{FeSO_4 \cdot 7H_2O}:n_\mathrm{FeSO_4} = 1:1\)
\(n_\mathrm{FeSO_4} = n_\mathrm{FeSO_4 \cdot 7H_2O} = 0,00453185\mathrm{mol}\)
\[\begin{aligned} m_\mathrm{FeSO_4} &= M_\mathrm{FeSO_4} \cdot n_\mathrm{FeSO_4} = \\ &= (55,85+32,07+16,00\cdot 4)\frac {\mathrm{g}}{\mathrm{mol}} \cdot 0,00453185\mathrm{mol} = \\ &= 0,68847902\mathrm{g} ≈ 0,688\mathrm{g}\end{aligned}\]
