The mass of a sample of a magnesium sulfate hydrate was 14.081g.After heating with a bunsen burner to remove the water of hydration, the sample weighed 6.878g. what is the correct chemical formula for magnesium sulfate hydrate? show all calculations.ccc

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rass of \mathrm{MgSO}_{4} hydrate =14.081 \mathrm{~g}mass of antydrous \mathrm{MgSO}_{4}=6.878 \mathrm{~g}When the sample was heated, allwater contents got evapourated.mass of water = mass of hydrate -mass andydrate\begin{array}{l}=14.081 \mathrm{~g}-6.878 \mathrm{~g} \\=7.203 \mathrm{~g}\end{array}Now, find moles of \mathrm{H}_{2} \mathrm{O} and MgSOy in the sample.\text { moles of } \begin{aligned}\mathrm{H}_{2} \mathrm{O} & =\frac{\text { mass }}{\text { molarmass }}=\frac{7.203 \mathrm{~g}}{18.015 \frac{9}{\mathrm{~mol}}} \\& =0.39983 \mathrm{~mol}\end{aligned}Moles of \mathrm{MgSO}_{4}=\frac{6.8789}{120.3669 / \mathrm{mol}}=0.05714 \mathrm{~mol}Find ratio of moles of \mathrm{H}_{2} \mathrm{O} to \mathrm{MgSO}_{4} in the hydrate sample.\begin{array}{l}\text { moles of } \mathrm{H}_{2} \mathrm{O} \\\text { moles of } \mathrm{MgSO}_{4}=\frac{0.39983 \mathrm{~mol}}{0.05714 \mathrm{~mol}} \\=7.00=\frac{7}{1} \\\end{array}ie, For each \mathrm{MgSO}_{4}, there are 7 \mathrm{H}_{2} \mathrm{O}.Formula of hydrate is \mathrm{MgSO}_{4} \mathrm{MH}_{2} \mathrm{O} ...