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One mole of a hydrocarbon is combusted. The products obtained are cooled down to STP and occupy a volume of $89.6\:\ell$. Oxygen required for conbustion was $145.6\:\ell$ at STP. Find the molecular formula of the hydrocarbon.

Let the hydrocarbon be $\ce{C}_x\ce{H}_y$. The combustion reaction can be written as: $$ \ce{C}_x\ce{H}_y+\ce{O_2} \rightarrow \ce{H_2O} + \ce{CO_2} $$ Keep in mind that this is not yet balanced.

We can now find out the amount of reactants and products. Since one mole of any gas occupies $22.4\:\ell$ of space at STP, we can say that we have: $$ \frac{89.6\:\ell}{22.4\:\ell/\mathrm{mol}} = 4\:\mathrm{mol} $$ of products. As for the oxygen reacted: $$ \frac{145.6\:\ell}{22.4\:\ell/\mathrm{mol}} = 6.5\:\mathrm{mol} $$. We can now deduce that the hydrocarbon will react with $13\ce{O}_2$. Therefore, the equation becomes. $$ \ce{C}_x\ce{H}_y + 13\ce{O2} \rightarrow \ce{H2O} + \ce{CO2} $$ But, There are $x$ $\ce{C}$s one the left hand side so the right hand side also has $x$ $\ce{C}$s: $$ \ce{C}_x\ce{H}_y + 13\ce{O2} \rightarrow \ce{H2O} + x\ce{CO2} $$ There are $y$ $\ce{H}$'s on the left hand side. Same for the right hand-side: $$ \ce{C}_x\ce{H}_y + 13\ce{O2} \rightarrow \frac{y}{2}\ce{H2O} + x\ce{CO2} $$ Since the amount of $\ce{O}$ has to balance as well, we have $26$ oxygens on the LHS and therefore $26$ on the RHS. Which means: $$ 26 = \frac{y}{2} + 2x \implies \frac{y}{2} = 26 - 2x $$ We have already figured out that $4\:\mathrm{mol}$ of products are obtained. i.e.: $$ 4 = \frac{y}{2} + x = 26 - 2x + x = 26 - x \implies x = 26 - 4 = 22 $$ But this would mean that $y$ is negative. Where did I go wrong?

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  • $\begingroup$ For the sake of completeness, keep in mind that the IUPAC has changed the reference pressure at STP from $\mathrm{atm}$ to $\mathrm{bar}$, and so the current accepted molar volume is $V_m=22.717\:\mathrm{L\:mol^{−1}}$. $\endgroup$ – alandella Sep 1 '13 at 12:02
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From your interpretation I have to say that giving the stochiometric coefficent of $13$ is wrong because this number is not related to the molecules of the species (therefore the suspect that you multiplied the moles obtained by two is where lies the main error).

Since we are talking about a combustion, we can safely assume that every reagent is completely oxidised, so we can write the complete reaction with the generic coefficients: $$ \ce{C}_{\color{blue}{x}}\ce{H}_{\color{blue}{y}(g)}+\left[\color{blue}{\frac{y}{4}}+\color{blue}{x}\right]\ce{O2_{($g$)}}\to\color{blue}{\frac{y}{2}}\ce{H2O}_{(g)}+\color{blue}{x}\ce{CO2_{($g$)}} $$ Note that the rule of given saturated hydrocarbons implies that $y=2x+2$, but this is left intentionally generalized because the presence of specific double/triple bonds is unknown.

The purpose of writing this equation is to specify the relation between the reactants, so now we can assume the molar relation between the oxygen needed to the first hydrocarbon to obtain a simple equation in terms of $x$ and $y$: $$ n_{\ce{O2}}=\left[\frac{y}{4}+x\right]n_{\ce{C}_{x}\ce{H}_{y}} $$ Since, as you calculated, there are needed $6.5\:\mathrm{mol}$ of oxygen to oxidise $1\:\mathrm{mol}$ of the unknown hydrocarbon; we can substitute those values inside the above equation, giving: $$ 6.5=\frac{y}{4}+x\tag{1} $$ After the complete combustion, the total amount of moles is indicated as you derived correctly, and there can be expressed in relation to the unknown reagent, and then added together to obtain the second equation as follows: $$ n_{\ce{H2O}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}\quad n_{\ce{CO2}}=x\:n_{\ce{C}_{x}\ce{H}_{y}} $$ Since the total volume is additive, then (because $\ce{C}_{x}\ce{H}_{y}=1:\mathrm{mol}$): $$ 4\:\mathrm{mol}=n_{\ce{H2O}}+n_{\ce{CO2}}=\frac{y}{2}n_{\ce{C}_{x}\ce{H}_{y}}+x\:n_{\ce{C}_{x}\ce{H}_{y}}=\frac{y}{2}+x $$ It now produces the second needed equation: $$ 4=\frac{y}{2}+x\tag{2} $$ Solving the equations $(1)$ and $(2)$ gives the values of $x$ and $y$ needed to derive the brute formula for the unknown hydrocarbon: $$ y=-10,\quad x=9 $$ Since the result is negative for one term, we can assume that the water produced is not in the vapor phase, hence we eliminate water term inside the equation $(2)$, $y/2\:n_{\ce{C}_{x}\ce{H}_{y}}$, giving the immediate value of $x$ because there is no effort given by the water: $$ x=4,\quad y=10\to \ce{C}_{4}\ce{H}_{10}\quad\mathrm{butane} $$ This result is describing a saturated hydrocarbon, because it follows the rule $\ce{C}_{n}\ce{H}_{2n+2}$.

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