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In the previous task, you calculated the values for the hydrate mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) , shown in the chart below. Using this information, answer the following questions. Type in the answers using one decimal place. M.M. mathrm(FeSO)_(4) & M.M. 7 mathrm(H)_(2) mathrm(O) & M.M. mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) 151.92 mathrm(~g) / mathrm(mol) & 126.14 mathrm(~g) / mathrm(mol) & 278.06 mathrm(~g) / mathrm(mol) What is the percent by mass of water in the hydrate? square % What is the percent by mass of the anhydrous salt, mathrm(FeSO)_(4) , in the hydrate? square %

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In the previous task, you calculated the values for the hydrate mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) , shown in the chart below. Using this information, answer the following questions. Type in the answers using one decimal place.

 M.M. mathrm(FeSO)_(4) & M.M. 7 mathrm(H)_(2) mathrm(O) & M.M. mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) 
 151.92 mathrm(~g) / mathrm(mol) & 126.14 mathrm(~g) / mathrm(mol) & 278.06 mathrm(~g) / mathrm(mol) 


What is the percent by mass of water in the hydrate? square % 
What is the percent by mass of the anhydrous salt, mathrm(FeSO)_(4) , in the hydrate? square %

In the previous task, you calculated the values for the hydrate mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) , shown in the chart below. Using this information, answer the following questions. Type in the answers using one decimal place. M.M. mathrm(FeSO)_(4) & M.M. 7 mathrm(H)_(2) mathrm(O) & M.M. mathrm(FeSO)_(4) cdot 7 mathrm(H)_(2) mathrm(O) 151.92 mathrm(~g) / mathrm(mol) & 126.14 mathrm(~g) / mathrm(mol) & 278.06 mathrm(~g) / mathrm(mol) What is the percent by mass of water in the hydrate? square % What is the percent by mass of the anhydrous salt, mathrm(FeSO)_(4) , in the hydrate? square %

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EloiseProfessional · Tutor for 6 years

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# Explanation<br /><br />To determine the percent by mass of water in the hydrate \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\), we use the following formula:<br /><br />\[\text{Percent by mass of water} = \left( \frac{\text{Mass of water in the hydrate}}{\text{Total mass of the hydrate}} \right) \times 100\]<br /><br />Given:<br />- M.M. of \(7 \mathrm{H}_{2} \mathrm{O}\) = \(126.14 \mathrm{~g} / \mathrm{mol}\)<br />- M.M. of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) = \(278.06 \mathrm{~g} / \mathrm{mol}\)<br /><br />Substituting these values into the formula:<br /><br />\[\text{Percent by mass of water} = \left( \frac{126.14 \mathrm{~g} / \mathrm{mol}}{278.06 \mathrm{~g} / \mathrm{mol}} \right) \times 100\]<br /><br />Calculating this:<br /><br />\[\text{Percent by mass of water} = \left( \frac{126.14}{278.06} \right) \times 100 \approx 45.4\%\]<br /><br />Next, to determine the percent by mass of the anhydrous salt \(\mathrm{FeSO}_{4}\) in the hydrate, we use a similar formula:<br /><br />\[\text{Percent by mass of } \mathrm{FeSO}_{4} = \left( \frac{\text{Mass of } \mathrm{FeSO}_{4} \text{ in the hydrate}}{\text{Total mass of the hydrate}} \right) \times 100\]<br /><br />Given:<br />- M.M. of \(\mathrm{FeSO}_{4}\) = \(151.92 \mathrm{~g} / \mathrm{mol}\)<br />- M.M. of \(\mathrm{FeSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) = \(278.06 \mathrm{~g} / \mathrm{mol}\)<br /><br />Substituting these values into the formula:<br /><br />\[\text{Percent by mass of } \mathrm{FeSO}_{4} = \left( \frac{151.92 \mathrm{~g} / \mathrm{mol}}{278.06 \mathrm{~g} / \mathrm{mol}} \right) \times 100\]<br /><br />Calculating this:<br /><br />\[\text{Percent by mass of } \mathrm{FeSO}_{4} = \left( \frac{151.92}{278.06} \right) \times 100 \approx 54.6\%\]<br /><br /># Answer<br /><br />The percent by mass of water in the hydrate is \(45.4 \%\).<br /><br />The percent by mass of the anhydrous salt, \(\mathrm{FeSO}_{4}\), in the hydrate is \(54.6 \%\).
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