IChO 1983 · Theory problem 2

Composition of a CO-CO2 mixture

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In a gaseous mixture of \(\ce{CO}\) and \(\ce{CO2}\), a mass ratio of carbon : oxygen = 1 : 2 was determined.

2.1 Calculate the mass percent composition.

2.2 Calculate the volume percent composition.

2.3 Indicate values of the carbon: oxygen ratios for which both gases cannot be present simultaneously.

Official English text of the problem.

Answers

  • 2.1 Carbon monoxide, \(\ce{CO}\) ([C-]#[O+]): 38.89 mass %; carbon dioxide, \(\ce{CO2}\) (O=C=O): 61.11 mass %.
  • 2.2 50 volume % \(\ce{CO}\) and 50 volume % \(\ce{CO2}\).
  • 2.3 Defining \(r=m_{\ce{C}}/m_{\ce{O}}\), both gases cannot be present simultaneously when \[ \boxed{r\leq \frac38\quad\text{or}\quad r\geq\frac34.} \] Equality corresponds to a pure gas; values outside these bounds are impossible for a mixture containing only these two gases.

Solution

2.1 Mass percent composition

Using the conventional atomic molar masses \(M(\ce{C})=12\,\mathrm{g\,mol^{-1}}\) and \(M(\ce{O})=16\,\mathrm{g\,mol^{-1}}\), \[ M(\ce{CO})=28\,\mathrm{g\,mol^{-1}},\qquad M(\ce{CO2})=44\,\mathrm{g\,mol^{-1}}. \]

Let \(n_1\) and \(n_2\) be the amounts of \(\ce{CO}\) and \(\ce{CO2}\), respectively. Each molecule contains one carbon atom, while their oxygen atom counts are one and two. Therefore, \[ \frac{m_{\ce{C}}}{m_{\ce{O}}} = \frac{(12\,\mathrm{g\,mol^{-1}})(n_1+n_2)} {(16\,\mathrm{g\,mol^{-1}})(n_1+2n_2)} =\frac12. \] Consequently, \[ 24(n_1+n_2)=16(n_1+2n_2) \quad\Longrightarrow\quad n_1=n_2. \]

On a basis of \(1\,\mathrm{mol}\) of each gas, their masses are \(28\,\mathrm{g}\) and \(44\,\mathrm{g}\), giving a total mass of \(72\,\mathrm{g}\). Thus, \[ w(\ce{CO})=\frac{28\,\mathrm{g}}{72\,\mathrm{g}}\times100\% =\boxed{38.89\%}, \] \[ w(\ce{CO2})=\frac{44\,\mathrm{g}}{72\,\mathrm{g}}\times100\% =\boxed{61.11\%}. \]

2.2 Volume percent composition

For ideal gases compared at the same temperature and pressure, \(V=nRT/P\), so volume fractions equal mole fractions. Since \(n_1=n_2\), \[ \varphi(\ce{CO})=\frac{n_1}{n_1+n_2}=\frac12, \qquad \varphi(\ce{CO2})=\frac{n_2}{n_1+n_2}=\frac12. \]

The composition is therefore \[ \boxed{50\text{ volume \% }\ce{CO} \quad\text{and}\quad 50\text{ volume \% }\ce{CO2}.} \]

2.3 Ratios excluding simultaneous presence

Let \(x\) be the mole fraction of \(\ce{CO2}\). The carbon-to-oxygen mass ratio is \[ r=\frac{12}{16(1-x)+32x} =\frac{3}{4(1+x)}. \]

Both gases are present only if \(0<x<1\), which gives \[ \boxed{\frac38<r<\frac34.} \]

The endpoints correspond to:

  • Pure \(\ce{CO2}\): \(r=12/32=3/8\), or carbon : oxygen \(=3:8\).
  • Pure \(\ce{CO}\): \(r=12/16=3/4\), or carbon : oxygen \(=3:4\).

Hence simultaneous presence is excluded for \(r\leq3/8\) or \(r\geq3/4\). At either endpoint one component is absent; outside the interval even a pure gas or mixture of these two gases cannot reproduce the specified elemental ratio.

Equivalently, if carbon : oxygen is written \(1:a\), simultaneous presence requires \[ \frac43<a<\frac83, \] and is excluded for \(a\leq4/3\) or \(a\geq8/3\).

The solution was checked against the official solution.