Question

A Sample of M2O3 weighing 11.205g is converted to 15.380g of MCL3, where M is the unknown metal, then what is the identity of M? (answer is Au, but how?)

Answer 1

The balanced reaction to convert M2O3 into MCl3

2 M2O3 + 6 Cl2 = 4 MCl3 + 3 O2

Let the Molecular weight of M = x g/mol

Molecular weight of M2O3 = (2x + 48) g/mol

Molecular weight of MCl3 = (x + 106.5) g/mol

From the stoichiometry of the reaction

2*(2x + 48) gM2O3 produces = 4*(x + 106.5) g MCl3

From the problem given

11.205 gM2O3 produces

= [4*(x + 106.5)*11.205]/[2*(2x + 48)] g MCl3

[4*(x + 106.5)*11.205]/[2*(2x + 48)] = 15.380

[4*(x + 106.5)*11.205] = [15.380*2*(2x + 48)]

44.82x + 4773.33 = 61.52x + 1476.48

3296.85 = 61.52x - 44.82x

x = 197.42 g/mol

This is the molecular weight of Au

M = Au = gold