Un compuesto formado por carbono, hidrógeno y oxígeno tiene una masa de 4,6 g. Se hace reaccionar con 9,6 g de oxígeno dando 8,8 g de CO2 y 5,4 g de agua. Si cogemos 9,2 g de un compuesto en un volumen 5,80l en P= 780 mmHg a una temperatura de 90ºC. Calcula la fórmula empírica y molecular.

Answers

Answer 1

Answer:

La fórmula empírica y molecular es: C₂H₆O.

Explanation:

Para calcular la formula empírica y molecular del compuesto debemos primero plantear la reacción:  

[tex] C_{x}H_{y}O_{z} + O_{2} \rightarrow CO_{2} + H_{2}O [/tex]

Necesitamos encontrar "x", "y" y "z". Para ello, tenemos que recordar que la masa de carbono e hidrógeno producida está relacionada con la cantidad de C y H inicial (del compuesto):

Para el H:

CHO → H₂O

y          5,4g

[tex] \frac{2*1 g}{18 g} = \frac{y}{5,4 g} \rightarrow y = 0,6 g [/tex]

Para C:  

CHO → CO₂

 x         8,8g

[tex] \frac{12 g}{44 g} = \frac{x}{8,8 g} \rightarrow x = 2,4 g [/tex]

Para el O:

[tex] z = 4,6 g - 2,4 g - 0,6 g = 1,6 g [/tex]

Ahora mediante el calculo de los moles del C, H y O podemos encontrar la fórmula empírica:

Para el H:

[tex] n_{y} = \frac{m}{Pm} = \frac{0,6 g}{1 g/mol} = 0,6 moles [/tex]

Para el C:

[tex] n_{x} = \frac{m}{Pm} = \frac{2,4 g}{12 g/mol} = 0,2 moles [/tex]

Para el O:

[tex] n_{z} = \frac{m}{Pm} = \frac{1,6 g}{16 g/mol} = 0,1 moles [/tex]

[tex] C_{\frac{n_{x}}{n_{z}}}H_{\frac{n_{y}}{n_{z}}}O_{\frac{n_{z}}{n_{z}}} = C_{\frac{0,2}{0,1}}H_{\frac{0,6}{0,1}}O_{\frac{0,1}{0,1}} = C_{2}H_{6}O_{1} [/tex]

Entonces, la fórmula empírica del commpuesto formado es C₂H₆O.

Ahora para determinar la fórmula molecular podemos usar la siguiente relación:

[tex] \frac{Pm}{Pm_{e}} = n [/tex]

[tex] F_{m} = n*F_{e} [/tex]

[tex] F_{m} = \frac{Pm}{Pm_{e}}*F_{e} [/tex]    

En donde Fm (fórmula molecular) y Fe (fórmula empírica) están relacionadas por n.

El valor de Pm lo obtenemos de la ecuación del gas ideal:

[tex]PV = nRT = \frac{m}{Pm}RT[/tex]

[tex] Pm = \frac{mRT}{PV} = \frac{9,2 g*0,082 L*atm/(K*mol)*(90 + 273 K)}{1.02 atm*5,80 L} = 46,3 g/mol [/tex]

[tex] F_{m} = \frac{46,3 g/mol}{(2*12 + 6*1 + 16)g/mol}*C_{2}H_{6}O_{1} = 1.00*C_{2}H_{6}O_{1} = C_{2}H_{6}O_{1} [/tex]    

Por lo tanto, la fórmula molecular es la misma que la fórmula empírica, a saber C₂H₆O.

Espero que te sea de utilidad!


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Answers

Answer:

The percent error is 0.023%.

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