What percent of FeSO_(4)cdot 6H_(2)O is Fe? 55.68% 12.68% 21.48% 29.89%

To determine the percentage of iron (Fe) in $FeSO_{4}\cdot 6H_{2}O$, we need to calculate the molar mass of the compound and then find the proportion of iron in it.<br /><br />First, let's calculate the molar mass of $FeSO_{4}\cdot 6H_{2}O$:<br /><br />- Molar mass of Fe: 55.85 g/mol<br />- Molar mass of S: 32.07 g/mol<br />- Molar mass of O: 16.00 g/mol<br />- Molar mass of H: 1.01 g/mol<br /><br />The molar mass of $FeSO_{4}\cdot 6H_{2}O$ is calculated as follows:<br /><br />\[ \text{Molar mass of } FeSO_{4} = 55.85 + 32.07 + (4 \times 16.00) = 151.92 \, \text{g/mol} \]<br /><br />Next, we calculate the molar mass of $6H_{2}O$:<br /><br />\[ \text{Molar mass of } 6H_{2}O = 6 \times (2 \times 1.01 + 16.00) = 108.12 \, \text{g/mol} \]<br /><br />Now, we add the molar masses of $FeSO_{4}$ and $6H_{2}O$ to get the molar mass of $FeSO_{4}\cdot 6H_{2}O$:<br /><br />\[ \text{Molar mass of } FeSO_{4}\cdot 6H_{2}O = 151.92 + 108.12 = 260.04 \, \text{g/mol} \]<br /><br />Finally, we calculate the percentage of iron in $FeSO_{4}\cdot 6H_{2}O$:<br /><br />\[ \text{Percentage of Fe} = \left( \frac{55.85}{260.04} \right) \times 100\% \approx 21.48\% \]<br /><br />Therefore, the correct answer is:<br /><br />\[ \boxed{21.48\%} \]