IChO 1978 · Theory problem 1

Analysis of a chromium ore

a) A chromium ore which does not contain water, consists of: \(\ce{Fe(CrO2)2}\), \(\ce{Mg(CrO2)2}\), \(\ce{MgCO3}\), and \(\ce{CaSiO3}\).

b) It was found by analysis the ore contains 45.6 % of \(\ce{Cr2O3}\), 7.98 % of \(\ce{Fe2O3}\), and 16.12 % of \(\ce{MgO}\).

c) When the ore was treated with a concentrated hydrochloric acid, chromium compounds being present in the ore did not react with the acid.

d) When the reaction was finished, the ore was thoroughly washed with water (till the reaction with \(\ce{Cl-}\) was negative) and the solid residue was dried to a constant mass.

Problems:

1.1 Write stoichiometric and ionic equations for the reactions taking place when the ore is treated with the hydrochloric acid as given in c).

1.2 Calculate:

– the content of the compounds (in mass %) present in the ore,
– amounts of substances of the compounds present in the ore.

1.3 Calculate the content of \(\ce{Cr2O3}\) (in mass %) in the dried residue obtained according to d).

1.4 A glass tube was filled with a sufficient amount of granulated \(\ce{CaO}\), the total mass of the filled tube having been 412.02 g. A gas formed by the reaction as given in c), was dried and then transmitted through the glass tube. Calculate the mass of the glass tube with its filling after the reaction was finished.

Relative atomic masses: \(A_{\mathrm r}(\ce{Cr})=52.01;\quad A_{\mathrm r}(\ce{Fe})=55.85;\quad A_{\mathrm r}(\ce{Mg})=24.32;\quad A_{\mathrm r}(\ce{Ca})=40.08;\)

\(A_{\mathrm r}(\ce{Si})=28.09;\quad A_{\mathrm r}(\ce{C})=12.01;\quad A_{\mathrm r}(\ce{O})=16.00.\)

Official English text of the problem.

Answers

  • 1.1 The molecular equations are \[ \ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + CO2(g) + H2O(l)} \] \[ \ce{CaSiO3(s) + 2HCl(aq) -> CaCl2(aq) + H2SiO3(s)} \] and the net ionic equations are \[ \ce{MgCO3(s) + 2H+(aq) -> Mg^2+(aq) + CO2(g) + H2O(l)} \] \[ \ce{CaSiO3(s) + 2H+(aq) -> Ca^2+(aq) + H2SiO3(s)}. \] Here \(\ce{H2SiO3}\) conventionally represents precipitated hydrated silica. Both chromites remain unchanged.

  • 1.2 On a 100 g ore basis:

Compound Mass percentage Amount per 100 g ore
Iron(II) chromite, \(\ce{Fe(CrO2)2}\) \(22.4\%\) \(0.0999\ \mathrm{mol}\)
Magnesium chromite, \(\ce{Mg(CrO2)2}\) \(38.5\%\) \(0.200\ \mathrm{mol}\)
Magnesium carbonate, \(\ce{MgCO3}\) \(16.8\%\) \(0.200\ \mathrm{mol}\)
Calcium silicate, \(\ce{CaSiO3}\) \(22.3\%\) \(0.192\ \mathrm{mol}\)

For an ore sample of mass \(m\), multiply these amounts by \(m/(100\ \mathrm g)\).

  • 1.3 Assuming the hydrated silica is dried to anhydrous silicon dioxide, \(\ce{SiO2}\), \[ \boxed{w(\ce{Cr2O3})=63.0\%}. \]

  • 1.4 Calcium oxide absorbs the carbon dioxide: \[ \ce{CaO(s) + CO2(g) -> CaCO3(s)}. \] For an ore sample of mass \(m\), \[ \boxed{m_{\mathrm{tube,final}}=412.02\ \mathrm g+0.08792\,m}. \] If \(m=100\ \mathrm g\), the final mass is \(420.81\ \mathrm g\). The supplied statement does not specify the ore mass, so an unconditional numerical answer is unavailable.

Solution

1.1 — Reactions with hydrochloric acid

The two chromites do not react, as explicitly stated in the problem.

Magnesium carbonate reacts with acid because protonation of carbonate produces carbon dioxide and water. Magnesium chloride remains in solution:

\[ \ce{MgCO3(s) + 2HCl(aq) -> MgCl2(aq) + CO2(g) + H2O(l)}. \]

Removing the spectator chloride ions gives

\[ \ce{MgCO3(s) + 2H+(aq) -> Mg^2+(aq) + CO2(g) + H2O(l)}. \]

Calcium silicate reacts with acid to give soluble calcium chloride and insoluble hydrated silica, conventionally written as metasilicic acid:

\[ \ce{CaSiO3(s) + 2HCl(aq) -> CaCl2(aq) + H2SiO3(s)}. \]

Its net ionic equation is

\[ \ce{CaSiO3(s) + 2H+(aq) -> Ca^2+(aq) + H2SiO3(s)}. \]

The molecular and ionic species used here have the following names and SMILES:

Species Name SMILES
\(\ce{HCl}\) Hydrogen chloride; its aqueous solution is hydrochloric acid Cl
\(\ce{H+}\) Hydrogen ion [H+]
\(\ce{Cl-}\) Chloride ion [Cl-]
\(\ce{Mg^2+}\) Magnesium(II) ion [Mg+2]
\(\ce{Ca^2+}\) Calcium(II) ion [Ca+2]
\(\ce{CO3^2-}\) Carbonate ion, present in the carbonate salts [O-]C(=O)[O-]
\(\ce{CO2}\) Carbon dioxide O=C=O
\(\ce{H2O}\) Water O

Formal monomeric metasilicic acid, \(\ce{H2SiO3}\), can be represented by O[Si](=O)O. The actual precipitate is a hydrated silica network; this molecular SMILES does not describe its polymeric structure. Likewise, the crystalline salts, chromites, and solid oxides are extended solids whose crystal structures cannot be specified by a single molecular SMILES. Aqueous hydration is also not shown by the ionic SMILES.

1.2 — Composition and amounts of substances

Choose 100 g of ore as the calculation basis. This is a convenient normalization, since no actual sample mass is given.

The reported oxide percentages are analytical equivalents, not additional compounds in the ore. Thus the analysis corresponds to:

\[ m(\ce{Cr2O3})=45.6\ \mathrm g,\qquad m(\ce{Fe2O3})=7.98\ \mathrm g,\qquad m(\ce{MgO})=16.12\ \mathrm g. \]

Use the shorter formulas \(\ce{FeCr2O4}\) and \(\ce{MgCr2O4}\) for the two chromites. Let their amounts be \(x\) and \(y\), and let the amounts of magnesium carbonate and calcium silicate be \(z\) and \(t\), respectively.

The required molar masses, calculated from the supplied atomic masses, are:

Compound Molar mass / \(\mathrm{g\,mol^{-1}}\)
\(\ce{Cr2O3}\) 152.02
\(\ce{Fe2O3}\) 159.70
\(\ce{MgO}\) 40.32
\(\ce{FeCr2O4}\) 223.87
\(\ce{MgCr2O4}\) 192.34
\(\ce{MgCO3}\) 84.33
\(\ce{CaSiO3}\) 116.17

All iron occurs in iron(II) chromite. One mole of iron(III) oxide equivalent represents two moles of iron, so

\[ x=2\frac{7.98\ \mathrm g}{159.70\ \mathrm{g\,mol^{-1}}} =0.0999374\ \mathrm{mol}. \]

Each mole of either chromite contains two chromium atoms and therefore corresponds to one mole of chromium(III) oxide equivalent:

\[ x+y=\frac{45.6\ \mathrm g}{152.02\ \mathrm{g\,mol^{-1}}} =0.2999605\ \mathrm{mol}. \]

Hence

\[ y=0.2999605-0.0999374 =0.2000231\ \mathrm{mol}. \]

Magnesium occurs in magnesium chromite and magnesium carbonate. Each contains one magnesium atom per formula unit, giving

\[ y+z=\frac{16.12\ \mathrm g}{40.32\ \mathrm{g\,mol^{-1}}} =0.3998016\ \mathrm{mol}, \]

and therefore

\[ z=0.3998016-0.2000231 =0.1997784\ \mathrm{mol}. \]

Keeping guard digits, the masses are

\[ \begin{aligned} m(\ce{FeCr2O4}) &=(0.0999374)(223.87)\ \mathrm g =22.37298\ \mathrm g,\\ m(\ce{MgCr2O4}) &=(0.2000231)(192.34)\ \mathrm g =38.47245\ \mathrm g,\\ m(\ce{MgCO3}) &=(0.1997784)(84.33)\ \mathrm g =16.84732\ \mathrm g. \end{aligned} \]

Since these and calcium silicate are the only constituents,

\[ m(\ce{CaSiO3}) =100-22.37298-38.47245-16.84732 =22.30725\ \mathrm g, \]

so

\[ t=\frac{22.30725\ \mathrm g}{116.17\ \mathrm{g\,mol^{-1}}} =0.1920225\ \mathrm{mol}. \]

On the 100 g basis, the numerical masses in grams equal the mass percentages. Rounding gives the results listed in Answers.

1.3 — Chromium content of the dried residue

Washing removes dissolved magnesium chloride, calcium chloride, and excess hydrochloric acid. The residue contains the two unchanged chromites and the silica precipitate.

Using the conventional assumption that drying dehydrates the silica,

\[ \ce{H2SiO3(s) ->[\text{drying}] SiO2(s) + H2O(g)}. \]

Each mole of calcium silicate therefore leaves one mole of silicon dioxide. Its molar mass is

\[ M(\ce{SiO2})=28.09+2(16.00)=60.09\ \mathrm{g\,mol^{-1}}, \]

and its mass is

\[ m(\ce{SiO2}) =(0.1920225\ \mathrm{mol})(60.09\ \mathrm{g\,mol^{-1}}) =11.53863\ \mathrm g. \]

Thus

\[ m_{\mathrm{residue}} =22.37298+38.47245+11.53863 =72.38406\ \mathrm g. \]

All chromium remains in the residue, so its chromium(III) oxide equivalent is still 45.6 g:

\[ w(\ce{Cr2O3}) =\frac{45.6}{72.38406}\times100\% =62.9973\% \approx\boxed{63.0\%}. \]

This percentage expresses the chromium content as oxide equivalent; it does not imply that free chromium(III) oxide was produced.

1.4 — Final mass of the absorption tube

The evolved gas is carbon dioxide, produced solely by magnesium carbonate:

\[ n(\ce{CO2})=n(\ce{MgCO3}) =0.1997784\ \mathrm{mol} \]

per 100 g of ore.

The gas is dried, then absorbed by calcium oxide to form calcium carbonate:

\[ \ce{CaO(s) + CO2(g) -> CaCO3(s)}. \]

Because sufficient calcium oxide is present, the tube gains the entire mass of the carbon dioxide. Using

\[ M(\ce{CO2})=12.01+2(16.00)=44.01\ \mathrm{g\,mol^{-1}}, \]

the gain per 100 g of ore is

\[ \Delta m =(0.1997784\ \mathrm{mol})(44.01\ \mathrm{g\,mol^{-1}}) =8.79225\ \mathrm g. \]

Consequently, for an actual ore mass \(m\),

\[ m_{\mathrm{tube,final}} =412.02\ \mathrm g +(8.79225\ \mathrm g)\frac{m}{100\ \mathrm g}. \]

For a 100 g sample, this becomes

\[ m_{\mathrm{tube,final}} =412.02+8.79225 \approx\boxed{420.81\ \mathrm g}. \]

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.