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kow [346]
3 years ago
15

Se someten a combustion 0,452g de un compuesto de C,H y N de masa molecular 80. Al recoger el CO2 y el H2O producidas obtenemos

0,994 g y 0,203 respectivamente. Determine formula empirica y formula molecular
Chemistry
1 answer:
NARA [144]3 years ago
3 0

Answer:

Fórmula empírica: C₂H₂N

Fórmula molecular: C₄H₄N₂

Explanation:

Un compuesto que contiene carbono hidrógeno y nitrógeno con fórmula CₐHₓNₙ es sometido a combustion produciendo:

CₐHₓNₙ + O₂ → aCO₂ + x/2 H₂O + nNO₂

Con la masa de dióxido de carbono y agua podemos encontrar las moles de carbono e hidrógeno y su aporte a los 0.452g de muestra que fueron puestos en combustión, así:

<em>Moles C:</em>

Moles C = Moles CO₂ = 0.994g CO₂ ₓ (1mol / 44g) = 0.0226 moles C

Masa C: 0.0226 moles C ₓ (12.01g / mol) = 0.271g Carbono hay en la muestra

<em>Moles H:</em>

Moles H = 2 Moles H₂O = 0.203g H₂O ₓ (1mol / 18g) = 0.0113 moles H₂O = 0.0226 moles H

Masa H: 0.0226 moles H ₓ (1.01g / mol) = 0.023g Hidrógeno hay en la muestra

Así, la masa de nitrógeno en la muestra y sus moles son:

Masa N = 0.452g - 0.271g C - 0.023g H

Masa N = 0.158g Nitrógeno

Y su moles son:

0.158g ₓ (1 mol / 14.01g) = 0.0113 moles N

Con las moles de C, H y N podemos determinar la formula empírica que se define como: "La relación de números enteros más simple entre la cantidad de átomos presentes en una mólecula. Si usamos como base las moles de nitrógeno (Valor menor):

Relación H/N: 0.0226 mol / 0.0113 mol = 2

Relación C/N: 0.0226 mol / 0.0113 mol = 2

Relación N/N: 0.0113 mol / 0.0113 mol = 1

Así, la <em>fórmula empírica es:</em>

<h3>C₂H₂N</h3>

Esta fórmula empírica tiene una masa molar de:

2C = 2*12 g/mol = 24g/mol

2H = 2*1g/mol = 2g/mol

N = 14g/mol

24+14+2 = 40g/mol

Como la masa molecular del compuesto es 80g/mol (Dos veces la de la fórmula empírica, la <em>fórmula molecular es 2 veces la fórmula empírica:</em>

<h3>C₄H₄N₂</h3>
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The vapor pressure at 60.21°C is 327 mmHg.

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We need to find vapor pressure at 60.21°C.

The Clausius-Clapeyron equation is often used to find the vapor pressure of pure liquid.

ln(\frac{P_2}{P_1})=\frac{\Delta_{vap}H}{R}(\frac{1}{T_1}-\frac{1}{T_2})

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P_1=102\ mmHg

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And R is the Universal Gas Constant.

R=0.008 314 kJ/Kmol

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Taking inverse log both side we get,

\frac{P_2}{102}=e^{1.165}\\\\P_2=102\times 3.20\ mmHg\\P_2=327\ mmHg

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