Textbook problem with molar mass of carbon dioxide

Question

$\ce{K2CO3}$ and $\ce{HCl}$ react to produce $\ce{CO2}$. In an empty flask ( m = $\pu{85.431 g}$ ) we put in the produced gas and the mass of the flask is now $\pu{85.510 g}$. After that we fill the flask with water. Volume of the flask filled with water is $\pu{122 ml}$. Pressure is $\pu{101325 Pa}$, temperature of air is $\pu{300 K}$. Density of water is $\pu{0.99893 g cm-3}$, and the molar mass of air is $\pu{28.8 g mol-1}$. Find the molar mass of carbon dioxide.

I have calculated the mass of $\ce{CO2}$, using the mass of the flask filled with gas and subtracting the mass of the empty flask. I have figured out that somehow through $pV = nRT$ I am supposed to get to $n$ and take into consideration that $n$ is not only the number of moles of $\ce{CO2}$ but also of $\ce{H2O}$ and air.
What I don't understand is the volume of water used to fill the flask. In that volume do we only have $\ce{H2O}$ or is that volume of 122 mL also the volume of air + $\ce{CO2}$ + $\ce{H2O}$?

Any hint is appreciated.

Answer 1

The problem is that, when the flask is empty, it weighs $85.431\ \mathrm g$. But this weight is the weight of the empty flask plus the air included. So the first thing to do is to calculate the weight of the air, and the corresponding number of moles of air in the flask before adding $\ce{CO2}$. The volume of the air is the same as the volume of water when it is filled with water, namely $122\ \mathrm{mL}$.

The amount of air is $$n = pV/RT = \frac{122\times10^{-6}\ \mathrm{m^3}\times101325\ \mathrm{Pa}}{8.316\times300\ \mathrm{J/mol}} = 4.956\times10^{-3}\ \ \mathrm{mol}$$ The mass of this air is : $$m = 4.956\times10^{-3}\ \mathrm{mol}\times28.8\ \mathrm{g/mol} = 0.1427\ \mathrm g$$ So the mass $m_0$ of the empty flask without air is : $$m_0 = 85.431\ \mathrm g - 0.1427\ \mathrm g = 85.2883\ \mathrm g$$

A s consequence, the mass of the $\ce{CO2}$ in the flask is : $$m(\ce{CO2}) = 85.510\ \mathrm g - 85.2883\ \mathrm g = 0.2227\ \mathrm g$$

The molar mass of this $\ce{CO2}$ is: $$M = \frac{m}{n} = \frac{0.2227\ \mathrm g}{4.956\times10^{-3}\ \mathrm{mol}}= 44.8\ \mathrm{g/mol}$$

This is nearly the expected value for the molar mass of $\ce{CO2}$. If you take into account the fact that the volume $122\ \mathrm{mL}$ is known with a precision $\pm1\ \mathrm{mL}$, or $\pm1\ \%$, the final result will be defined with an uncertainty of $\pm1\ \mathrm{g/mol}$. It is: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$