$\ce{K2CO3}$ 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.

  • $\begingroup$ Hint: Find the density of carbon dioxide. You know mass of the CO2 added and the volume of the flask. Using pV=nRT find out the molar volume of carbon dioxide. What is the relation between density and molar volume. Can we get molar mass of CO2 from that? $\endgroup$ – M. Farooq Sep 13 '20 at 5:32
  • $\begingroup$ Molar volume of CO2 implies that n=1 in PV=nRT. $\endgroup$ – M. Farooq Sep 13 '20 at 5:37
  • $\begingroup$ Ok. Thanks. And what am I supposed to do with the volume of water added? That's what's bugging me. $\endgroup$ – user33683 Sep 13 '20 at 7:01
  • $\begingroup$ Please refrain from using the homework tag, it was removed a long time ago. $\endgroup$ – Safdar Sep 13 '20 at 7:04
  • $\begingroup$ For formatting, See here and here. For a more detailed MathJax guide, look here, minor other details $\endgroup$ – Safdar Sep 13 '20 at 7:04

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: $$M(\ce{CO2}) = 44.8\ \mathrm{g/mol} \pm 1\ \mathrm{g/mol}$$

  • 1
    $\begingroup$ I would rather say 122 mL means +- 0.5 mL, as the more different value would be estimated rather as 121 mL or 123 mL. $\endgroup$ – Poutnik Sep 13 '20 at 11:50
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    $\begingroup$ The question should have explicitly stated the meaning of "Empty" flask. Empty as in vacuum or empty in a non-scientific way. $\endgroup$ – M. Farooq Sep 13 '20 at 13:51
  • $\begingroup$ @M. Farooq Empty as in a lab way, filled by air at lab conditions. :-) $\endgroup$ – Poutnik Sep 14 '20 at 8:15
  • $\begingroup$ @Poutnik, vacuum techniques? And how could they ensure carbon dioxide filled it completely? $\endgroup$ – M. Farooq Sep 14 '20 at 13:26
  • $\begingroup$ @M. Farooq sure, vacuum techniques are possible, but not necessery. If filled from the bottom, flushed by a multiple of the volume, it will be. $\endgroup$ – Poutnik Sep 14 '20 at 13:28

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