IChO 1984 · Theory problem 4

Formula of a gaseous hydrocarbon from its combustion

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15 cm³ of a gaseous hydrocarbon \(\ce{C_xH_y}\) are mixed with 120 cm³ oxygen and ignited. After the reaction the burned gases are shaken with concentrated aqueous \(\ce{KOH}\) solution. A part of the gases is completely absorbed while 67.5 cm³ gases remain. It has the same temperature and pressure as the original unburned mixture.

4.1 What is the composition of the remaining gas? Explain.

4.2 How large is the change in the amount of substance per mole of a hydrocarbon \(\ce{C_xH_y}\) when this is burned completely?

4.3 What is the chemical formula of the hydrocarbon used for the experiment?

Give the steps of the calculation.

Official English text of the problem.

Answers

  • 4.1: The remaining gas is oxygen, \(\ce{O2}\), SMILES O=O, with volume \(67.5\ \mathrm{cm^3}\), neglecting residual water vapour.
  • 4.2: Counting all combustion products, the change is \(\left(\frac y4-1\right)\ \mathrm{mol}\) per mole of hydrocarbon. After removal of water and carbon dioxide, the measured gas amount decreases by \(\left(1+x+\frac y4\right)\ \mathrm{mol}\) per mole of hydrocarbon.
  • 4.3: The intended hydrocarbon is ethane, \(\ce{C2H6}\), SMILES CC, under the usual assumption of an ordinary stable gaseous hydrocarbon.

Solution

4.1

Complete combustion produces carbon dioxide, \(\ce{CO2}\), SMILES O=C=O, and water, \(\ce{H2O}\), SMILES O:

\[ \ce{C_xH_y + (x + y/4) O2 -> x CO2 + (y/2) H2O}. \]

Water is treated as removed from the measured gas by cooling and contact with the concentrated aqueous solution. Carbon dioxide is absorbed by potassium hydroxide, \(\ce{KOH}\), SMILES [K+].[OH-]:

\[ \ce{CO2 + 2KOH -> K2CO3 + H2O}. \]

The salt formed is potassium carbonate, \(\ce{K2CO3}\), represented by the ionic SMILES [K+].[K+].[O-]C(=O)[O-].

Under this complete-combustion model, the remaining gas must therefore be excess oxygen or unreacted hydrocarbon. If oxygen had been exhausted, the volume of unreacted hydrocarbon could not exceed its initial volume, \(15\ \mathrm{cm^3}\), at the same temperature and pressure. Since

\[ 67.5\ \mathrm{cm^3}>15\ \mathrm{cm^3}, \]

oxygen was in excess. The hydrocarbon burned completely, leaving oxygen after the products were removed.

4.2

For one mole of hydrocarbon, the balanced combustion equation gives

\[ n_{\mathrm{reactants}} =\left(1+x+\frac y4\right)\mathrm{mol}, \]

and, counting both carbon dioxide and water,

\[ n_{\mathrm{products}} =\left(x+\frac y2\right)\mathrm{mol}. \]

Hence the change, defined as products minus reactants, is

\[ \boxed{\Delta n_{\mathrm{total}} =\left(\frac y4-1\right)\mathrm{mol} \quad\text{per mole of hydrocarbon}.} \]

For interpreting the gas-volume experiment, the stages must be distinguished:

Stage Change in gas amount per mole of hydrocarbon
Combustion, with all water remaining gaseous \(\left(\frac y4-1\right)\ \mathrm{mol}\)
After removal of water, before carbon dioxide absorption \(-\left(1+\frac y4\right)\ \mathrm{mol}\)
After removal of both water and carbon dioxide \(-\left(1+x+\frac y4\right)\ \mathrm{mol}\)

The last expression applies to the measured initial and final gas volumes.

4.3

At equal temperature and pressure, gas volumes are proportional to amounts of substance. The oxygen consumed is

\[ V_{\ce{O2},\,\mathrm{consumed}} =120-67.5 =52.5\ \mathrm{cm^3}. \]

Thus each mole of hydrocarbon consumes

\[ \frac{n_{\ce{O2},\,\mathrm{consumed}}}{n_{\mathrm{hydrocarbon}}} =\frac{52.5\ \mathrm{cm^3}}{15\ \mathrm{cm^3}} =3.5 \]

moles of oxygen. From the combustion equation,

\[ x+\frac y4=3.5, \]

so

\[ \boxed{4x+y=14}. \]

Equivalently, using the gas contraction after product removal,

\[ 1+x+\frac y4 =\frac{(15+120)-67.5}{15} =4.5. \]

For an ordinary neutral hydrocarbon, \(x,y\) are positive integers and the maximum hydrogen count is \(y=2x+2\). Combining this bound with \(y=14-4x\) gives

\[ 14-4x\le 2x+2 \quad\Longrightarrow\quad x\ge2, \]

while \(y>0\) requires \(x\le3\). The formal possibilities are therefore

\[ (x,y)=(2,6)\quad\text{or}\quad(3,2). \]

Of these, \(\ce{C2H6}\) is the ordinary stable gaseous hydrocarbon: ethane, SMILES CC. The formal alternative \(\ce{C3H2}\) requires unusual reactive or highly strained species, outside the intended scope of this elementary analysis.

Ethane burns according to

\[ \boxed{\ce{2C2H6 + 7O2 -> 4CO2 + 6H2O}}. \]

The volume check is

\[ V_{\ce{O2},\,\mathrm{consumed}} =\frac72(15\ \mathrm{cm^3}) =52.5\ \mathrm{cm^3}, \]

\[ V_{\ce{O2},\,\mathrm{remaining}} =120-52.5 =\boxed{67.5\ \mathrm{cm^3}}, \]

exactly reproducing the measurement.

The solution was checked against the official solution.