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A container of volume $\pu{15 L}$ contains a hydrocarbon at $\pu{27 °C}$ and $\pu{2 atm}$ pressure. For the analysis, we take $\pu{10 g}$ of hydrocarbon and the pressure becomes $\pu{1.718 atm}$ (at the same temperature). Combustion with a stochiometrical quantity of air it results an gaseous mixture of $\ce{CO2}$ and $\ce{N2}$ in the mass ratio $\ce{CO2 : N2 = 0.241}$. Find the molecular formula for the hydrocarbon.

My solution:

Solution

The correct answer is $\ce{C4H10}$. It means that $m=10$ and $n=4$, but these values do not satisfy my equations.

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closed as off-topic by Mithoron, aventurin, user55119, Jon Custer, Tyberius Jan 9 at 22:48

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