How to calculate the molecular weight for a volatile substance introduced into a Dumas bulb?

Question

In a Dumas bulb a volatile substance is introduced. After a few minutes when the liquid has evaporated, the bulb is sealed. It is known that the initial weight of the bulb with air is $12.0468~\mathrm{g}$, $12.4528~\mathrm{g}$ with the volatile substance and $350.6264~\mathrm{g}$ with water. Calculate the molar mass of the substance if all these measurements were done at $25~^\circ\mathrm{C}$ and $1~\mathrm{atm}$ of pressure.

The correct answer is $29~\mathrm{g/mol}$.

What I've done is the following.

I've constructed a system of two equations to know the mass of the bulb and the volume it can contain.

$m_b+\rho_\mathrm{air}V = 12.0468~\mathrm{g}$
$m_b+\rho_\mathrm{water}V = 350.6264~\mathrm{g}$

The solutions are $m_b=11.6144~\mathrm{g}$ and $V=0.339012~\mathrm{m^3}$

With this I can find the molar mass of the volatile substance by knowing that its true mass was $m=0.8384~\mathrm{g}$ and the volume it occupied was $0.339~\mathrm{m^3}$ and using the ideal gas equation.

The ideal gas law, can be used to say

$M=\frac{m}{pV}RT=0.060~\mathrm{g/mol}$

Which is wrong. I've done all sorts of other things and can't get $29~\mathrm{g/mol}$. Maybe this is the craziest thing I've done until now.

Answer 1

There are a couple of things wrong.

First, the mass of the volatile gas is not $0.8384\ \mathrm g$. It's the difference between $12.4528\ \mathrm g$ and $12.0468\ \mathrm g$.

Second, the volume is incorrect. $11.6144\ \mathrm g + 999.97\ \mathrm{kg/m^3} \times 0.339012\ \mathrm{m^3} \ne 350.6264\ \mathrm g$