Question

A 13.53 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 16.24 g.

Add subscripts below to correctly identify the empirical formula of the new oxide:

MoO

Answer 1

molarmass of Mo2O3 = 239.92 g/mol

atomicmass of Mo = 95.96 g/mol

so that ,

mass of Mo present in the given sample = 95.96*2*13.53/239.92 = 10.82 g

mass of Oxygen in the given sample = 16*3*13.53/239.92 = 2.707 g

mass of Oxygen added = 16.24 - 13.53 = 2.71 g

no of mole of Mo in new sample = 10.82/95.96 = 0.113 mole

no of mole of O-atom in new sample = (2.707+2.71)/16 = 0.34 mol

simplest ratio

Mo = 0.113/0.34 = 0.332 mole

O = 0.34/0.34 = 1

simplest whole number ratio

Mo = 0.33*3 = 0.99   = 1

O = 3

so that , empirical formula = MoO3