Question

1. Obtengan las fórmulas empíricas de las sustancias que tienen las siguientes composiciones porcentuales: (a) Fe = 63,53 %, S = 36,47 %; (b) Fe = 46,55 %, S = 53,45 %; (c) Fe = 53,73 %, S = 46,27 %.

Answer 1

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Resolución

$a)~Fe=63,53\%, S=36,47\% \\ \\ Fe=masa..molar=56g/mol=\ \textgreater \ \frac{63,53}{56} =1,13 \\ S=masa..molar=32g/mol=\ \textgreater \ \frac{36,47}{32} =1,14 \\ Ahora~dividiremos~por~el~menor~valor: \\ Fe= \frac{1,13}{1,13} =1 \\ S= \frac{1,14}{1,13}= 1 \\ Entonces~la~formula~sera: \\ Fe _{1} S _{1} ==\ \textgreater \ \boxed{FeS}$
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$b)~Fe=46,55\%~, S=53,45\% \\ \\ Fe=masa..molar=56g/mol=\ \textgreater \ \frac{46,55}{56} =0,83 \\ \\ ~S=masa..molar=32g/mol=\ \textgreater \ \frac{53,45}{32} =1,67 \\ Dividiremos~por~el~menor~valor \\ Fe= \frac{0,83}{0,83} =1 \\ \\ S= \frac{1,67}{0,83} =2 \\ Entonces~la~formula~sera: \\ Fe _{1} S_{2} ==\ \textgreater \ \boxed{Fe S_{2} }$

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$c)~Fe=53,73\%~S=46,27\% \\ \\ Fe=masa ~molar=56g/mol=\ \textgreater \ \frac{53,73}{56} =0,96 \\ \\ S=masa~molar=32g/mol=\ \textgreater \ \frac{46,27}{32} =1,45 \\ dividiremos~por~el~menor~valor. \\ Fe= \frac{0,96}{0,96} =1 \\ \\ S= \frac{1,45}{0,96} =1,5 \\ La~formula~es: \\ Fe _{1} S _{1,5} =\ \textgreater \ multiplicamos~por~(2)~los~atomos~tenemos. \\ \boxed{Fe _{2} S_{3} }$

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