IChO 1997 · Theory problem 5

Allotropes of iron and the density of martensite

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Iron metal melts at 1811 K. Between room temperature and its melting point, iron metal can exist in different allotropic or crystalline forms. From room temperature to 1185 K, the crystal structure of iron metal exists as a body-centred cubic (bcc) lattice known as α-iron. From 1185 K to 1667 K, the structure becomes face-centred cubic (fcc) and is called γ-iron. Above 1667 K, and up to its melting point, iron reverts to a bcc structure similar to that of α-iron. The latter phase is called δ-iron.

Given that the density of pure iron metal is \(7.874\ \mathrm{g\,cm^{-3}}\) at 293 K:

5.1 Calculate the atomic radius of iron (expressed in cm).

5.2 Calculate its density (expressed in \(\mathrm{g\,cm^{-3}}\)) at 1250 K.

Notes: Ignore the small effects due to the thermal expansion of the metal.

Clearly define any symbols which you use, e.g. \(r\) = atomic radius of \(\ce{Fe}\).

Steel is an alloy of iron and carbon in which some of the interstitial spaces (“holes”) of the crystal lattice (iron) are occupied by small atoms (carbon). Its carbon content typically ranges from 0.1 % to 4.0 %. In a blast-furnace, the melting of iron is facilitated when it contains 4.3 % of carbon by mass. If this mixture is cooled too rapidly the carbon atoms remain dispersed within the α-iron phase. This new solid, called martensite, is extremely hard and brittle. Although is it slightly distorted, the size of the unit cell of this solid (martensite) is the same as that of α-iron (bcc).

Assuming that the carbon atoms are evenly distributed in the iron structure:

5.3 Calculate the average number of carbon atoms per unit cell of α-iron in martensite containing 4.3 % \(\ce{C}\) by mass.

5.4 Calculate the density (expressed in \(\mathrm{g\,cm^{-3}}\)) of this material.

Molar masses and constants:

\(M(\ce{Fe}) = 55.847\ \mathrm{g\,mol^{-1}}\)

\(M(\ce{C}) = 12.011\ \mathrm{g\,mol^{-1}}\)

\(N_{\mathrm A} = 6.02214\times10^{23}\ \mathrm{mol^{-1}}\)

Official English text of the problem.

Answers

  • 5.1: \(r=\boxed{1.241\times10^{-8}\ \mathrm{cm}}\).
  • 5.2: \(\rho_\gamma=\boxed{8.572\ \mathrm{g\,cm^{-3}}}\) at 1250 K.
  • 5.3: \(\boxed{0.42\text{ carbon atoms per unit cell, on average}}\) (calculated value \(0.4178\)).
  • 5.4: \(\rho_{\mathrm{martensite}}=\boxed{8.23\ \mathrm{g\,cm^{-3}}}\).

Solution

5.1 Atomic radius of iron

Let \(r\) be the atomic radius of iron, \(a_\alpha\) the edge length of the cubic α-iron unit cell, and \(\rho_\alpha\) its density. \(M(\ce{Fe})\) is the molar mass of iron and \(N_{\mathrm A}\) is the Avogadro constant.

At 293 K, iron has a bcc structure. Its conventional unit cell contains eight corner atoms, each shared by eight cells, and one atom entirely inside the cell:

\[ \text{Number of iron atoms per cell} =8\left(\frac18\right)+1=2. \]

Consequently,

\[ \rho_\alpha=\frac{2M(\ce{Fe})}{N_{\mathrm A}a_\alpha^3}, \]

so

\[ a_\alpha =\left[\frac{2M(\ce{Fe})}{N_{\mathrm A}\rho_\alpha}\right]^{1/3} =\left[ \frac{2(55.847\ \mathrm{g\,mol^{-1}})} {(6.02214\times10^{23}\ \mathrm{mol^{-1}}) (7.874\ \mathrm{g\,cm^{-3}})} \right]^{1/3} =2.866561\times10^{-8}\ \mathrm{cm}. \]

In a bcc lattice, atoms touch along the body diagonal. This diagonal has length \(\sqrt3\,a_\alpha\) and spans four atomic radii:

\[ \sqrt3\,a_\alpha=4r. \]

Therefore,

\[ \boxed{ r=\frac{\sqrt3}{4}a_\alpha =1.241\times10^{-8}\ \mathrm{cm} }. \]

5.2 Density at 1250 K

Since \(1185<1250<1667\ \mathrm K\), the metal is γ-iron and has an fcc structure. Let \(a_\gamma\) be its unit-cell edge length and \(\rho_\gamma\) its density.

An fcc unit cell contains eight corner atoms and six face-centred atoms, each face-centred atom shared by two cells:

\[ \text{Number of iron atoms per cell} =8\left(\frac18\right)+6\left(\frac12\right)=4. \]

Atoms touch along a face diagonal, giving

\[ \sqrt2\,a_\gamma=4r, \qquad a_\gamma=2\sqrt2\,r. \]

Using the same atomic radius and neglecting thermal expansion,

\[ a_\gamma=3.510806\times10^{-8}\ \mathrm{cm}. \]

Thus,

\[ \rho_\gamma =\frac{4M(\ce{Fe})}{N_{\mathrm A}a_\gamma^3} =\frac{4(55.847\ \mathrm{g\,mol^{-1}})} {(6.02214\times10^{23}\ \mathrm{mol^{-1}}) (3.510806\times10^{-8}\ \mathrm{cm})^3}, \]

\[ \boxed{\rho_\gamma=8.572\ \mathrm{g\,cm^{-3}}}. \]

5.3 Average number of carbon atoms per unit cell

Let \(x\) be the average number of carbon atoms per martensite unit cell and \(w_{\mathrm C}=0.043\) the carbon mass fraction.

Carbon occupies interstitial sites, so the cell retains its two iron atoms. Its average iron and carbon masses are therefore

\[ m_{\mathrm{Fe}}=\frac{2M(\ce{Fe})}{N_{\mathrm A}}, \qquad m_{\mathrm C}=\frac{xM(\ce{C})}{N_{\mathrm A}}. \]

The carbon mass fraction is

\[ w_{\mathrm C} =\frac{m_{\mathrm C}}{m_{\mathrm{Fe}}+m_{\mathrm C}} =\frac{xM(\ce{C})}{2M(\ce{Fe})+xM(\ce{C})}. \]

Solving for \(x\),

\[ x=\frac{2M(\ce{Fe})w_{\mathrm C}} {M(\ce{C})(1-w_{\mathrm C})} =\frac{2(55.847)(0.043)} {(12.011)(0.957)} =0.417837. \]

Hence,

\[ \boxed{x\approx0.42\text{ carbon atoms per unit cell}}. \]

This fractional value is an average over many cells; individual cells contain integer numbers of carbon atoms.

5.4 Density of martensite

The problem specifies that the unit-cell volume remains \(a_\alpha^3\). The iron mass per cell also remains unchanged, while carbon adds mass.

Because iron accounts for the mass fraction \(1-w_{\mathrm C}=0.957\), the total cell mass is \(m_{\mathrm{Fe}}/(1-w_{\mathrm C})\). Therefore,

\[ \rho_{\mathrm{martensite}} =\frac{m_{\mathrm{Fe}}}{(1-w_{\mathrm C})a_\alpha^3} =\frac{\rho_\alpha}{1-w_{\mathrm C}} =\frac{7.874}{0.957}\ \mathrm{g\,cm^{-3}} =8.227795\ \mathrm{g\,cm^{-3}}. \]

Thus,

\[ \boxed{\rho_{\mathrm{martensite}}=8.23\ \mathrm{g\,cm^{-3}}}. \]

The solution was checked against the official solution.