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babymother [125]
3 years ago
8

A compound has an empirical formula of CH2 and a molar mass of 70.15. What is its molecular formula?

Chemistry
1 answer:
lina2011 [118]3 years ago
7 0
<h3>Answer:</h3>

Cyclopentane (C5H10)

<h3>Explanation:</h3>

The molar mass of CH2 is about 14.3, Since empirical is a ratio just divide the given molar mass by the molar mass of the empirical. 70.15/14.3=5 So now multiple everything by 5 in the empirical equation.

So, a compound that has an empirical formula of CH2 and a molar mass of 70.15 will have molecular formula of" C5H10".

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Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

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