Answer:
[tex]\large \boxed{\text{48.83 \% by mass}}[/tex]
Explanation:
[tex]\text{Percent by mass} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times 100 \%\\\\= \dfrac{\text{Mass of MgSO}_{4}}{\text{Mass of MgSO$_{4}\cdot$ 7H$_{2}$O}} \times 100 \%[/tex]
Let's rewrite the formula of MgSO₄·7H₂O as MgSO₄·H¬₁₄O₇ and calculate the masses.
[tex]\begin{array}{rrr}\textbf{Atom} & \textbf{Mass/u} & \textbf{Contribution/u}\\\text{1Mg} & 24.30 & 24.30\\\text{1S} & 32.06 & 32.06\\\text{4O} & 16.00 & 64.00\\\text{Subtotal}&=& \mathbf{120.36}\\\text{14H} & 1.01 & 14.11\\\text{7O} & 16.00 & 112.00\\\text{Subtotal} &=& \mathbf{126.11}\\\text{TOTAL} & = & \mathbf{246.47}\\\end{array}[/tex]
[tex]\text{Percent by mass} = \dfrac{\text{120.36}}{\text{246.47}} \times 100 \% = 48.83 \%\\\\\text{The relative amount of MgSO$_{4}$ is }\large \boxed{\textbf{48.83 \% by mass}}[/tex]
