Question

Wjat is the percent of mass of magnesium sulfate in mgso4×7h2o
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Answer 1

`\large \boxed{\text{48.83 \% by mass}}`

Step-by-step explanation:

`\text{Percent by mass} = \frac{\text{Mass of component}}{\text{Total mass}} * 100 \%\\\\= \frac{\text{Mass of MgSO}_(4)}{\text{Mass of MgSO$_(4)\cdot$ 7H$_(2)$O}} * 100 \%`

Let's rewrite the formula of MgSO₄·7H₂O as MgSO₄·H¬₁₄O₇ and calculate the masses.

`\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}`

`\text{Percent by mass} = \frac{\text{120.36}}{\text{246.47}} * 100 \% = 48.83 \%\\\\\text{The relative amount of MgSO$_(4)$ is }\large \boxed{\textbf{48.83 \% by mass}}`