Question

A gaseous compound Y contains carbon and hydrogen only and has a density
of 1.696 g dm-3 at 95.3 kPa pressure and 298K. Assume that Y behaves ideally,
calculate its molar mass and deduce its molecular formula. Given 1kPa =1 x 103Nm-2

.

Answer 1

Answer:

$M=44.06\frac{g}{mol}$

Explanation:

Hello.

In this case, knowing the temperature, density and temperature of Y which behaves ideally, we can write the ideal gas equation:

$PV=nRT$

Whereas the moles are equal to the mass over the molar mass of Y:

$PV=\frac{m}{M} RT$

Thus, solving for the molar mass we write:

$M=\frac{mRT}{PV}$

Yet, since density is mass of over volume, we then write:

$M=\frac{\rho RT}{P}$

Considering the pressure in atm and the density in g/L:

$P=95.3kPa*\frac{1atm}{101.325kPa}=0.941atm\\\\\rho=1.696\frac{g}{dm^3} *\frac{1dm^3}{1L}=1.696\frac{g}{L}$

Therefore, by plugging the values in, we obtain:

$M=\frac{1.696\frac{g}{L} *0.082\frac{atm*L}{mol*K}*298K}{0.941atm}\\\\M=44.06\frac{g}{mol}$

Thus, the gas may be propane ($C_3H_8$) since it molar mass is 44.11 g/mol.

Best regards!