Question

Metallocene are metals sandwiched by two cyclopentadienyl ions, (C5H5)–. A metallocene sample of unknown metal, MC5H5 (all 5 are subscripts) was burned in air. Products of combustion found to contain 46.08488% carbon dioxide, 9.42644% water vapour and the rest are metal (I) oxide. All products are in mass percent. The mass ratio of stoichiometric amount of oxygen to the burned metallocene is 1:1.295. What is the name of the metal in the metallocene sample? Prove by showing your complete solution.

Answer 1

The reaction of combustion of the given metallocene (MeCp) is following:

$2 \ MeC_5H_5 + 13O_2=10CO_2+5H_2O+Me_2O$

The stoichiometric ratio between MeCp and O₂ is:

$\frac{n(MeCp)}{n(O_2)}=\frac{2}{13}$

As $n(MeCp)=\frac{m(MeCp)}{M(MeCp)}$ and $n(O_2)=\frac{m(O_2)}{M(O_2)}$ ,the above equation transforms to the following:

$\frac{m(MeCp)}{m(O_2)}\cdot \frac{M(O_2)}{M(MeCp)}=\frac{2}{13}$

From the given conditions the mass ratio of stoichiometric amount of oxygen to the burned metallocene is 1:1.295, i.e. $\frac{m(O_2)}{m(MeCp)}=\frac{1}{1.295}$

Thus two equations above transform to the following:

$\frac{1.295}{1}\cdot \frac{M(O_2)}{M(MeCp)}=\frac{2}{13}$

from which we can find molar mass of metallocene:

$M(MeCp)=\frac{1.295\cdot 13 \cdot 32\frac{g}{mol}}{2}=269.36\frac{g}{mol}$

Taking into account that $M(MeCp)=M(Me)+M(Cp)=M(Me)+M(C_5H_5)=M(Me)+65 \ (\frac{g}{mol})$ it is easy to find the molar mass of metal:

$M(Me)+65=269.36$

$M(Me)=204.36 \frac{g}{mol}$

This molar mass of metal corresponds to thallium, that is known to form a metallocene that adopts a sandwich-like polymeric structure: