Como fazer formula empirica e molecular

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\[{{\text{C}}_{\text{x}}}{{\text{H}}_{\text{y}}}{{\text{O}}_{\text{z}}}{\text{Cl}} + {{\text{O}}_{\text{2}}} \to {\text{C}}{{\text{O}}_{\text{2}}} + {{\text{H}}_{\text{2}}}{\text{O}}\]

Daí, através de regras de três, determina-se a massa de cada elemento:

\[\eqalign{ & a = \dfrac{{\left( {12{\text{g de C}}} \right) \cdot \left( {0,0682{\text{ g de C}}{{\text{O}}_2}} \right)}}{{44{\text{ g de C}}{{\text{O}}_2}}} \cr & = 0,0186{\text{ g de C}} \cr & \cr & b = \dfrac{{\left( {{\text{2g de H}}} \right) \cdot \left( {0,0140{\text{ g de }}{{\text{H}}_2}{\text{O}}} \right)}}{{{\text{18 g de }}{{\text{H}}_2}{\text{O}}}} \cr & = 0,00156{\text{ g de H}} }\]

Ademais, \(55\%\) da massa é cloro, isto é, temos \(0,055\text{ g de Cl}\) e o restante, \(c\), é hidrogênio:

\[\eqalign{ & c = 0,10{\text{g}} - 0,055{\text{g}} - 0,00156{\text{g}} - 0,0186{\text{g}} \cr & = 0,{\text{02484 de }}{{\text{O}}_2} }\]

Por fim, basta dividir as massas encontradas pelas respectivas massas molares de cada elemento:

\[\eqalign{ & {\text{C}} = \dfrac{{0,0186{\text{g}}}}{{\dfrac{{{\text{12g}}}}{{{\text{mol}}}}}} \cong \boxed{0,0015\;{\text{mol}}} \cr & \cr & {\text{H}} = \dfrac{{0,00156{\text{g}}}}{{\dfrac{{1{\text{g}}}}{{{\text{mol}}}}}} \cong \boxed{0,0015{\text{ mol}}} \cr & \cr & {\text{Cl}} = \dfrac{{0,055{\text{g}}}}{{\dfrac{{35,45{\text{g}}}}{{{\text{mol}}}}}} \cong \boxed{0,0015{\text{ mol}}} \cr & \cr & {\text{O}} = \dfrac{{0,02484{\text{g}}}}{{\dfrac{{16{\text{g}}}}{{{\text{mol}}}}}} \cong \boxed{0,0015{\text{ mol}}} }\]