# Stoichiometry - 4 | Chemical Calculations

A 19.33 g sample of Mo2O3 is converted completely to another molybdenum oxide by reaction with dioxygen. The new oxide has a mass of 23.20 g. What is the empirical formula of this new oxide?

Sample 1:

nMo2O3 = $\frac{{\mathrm{m}}_{\mathrm{Mo}2\mathrm{O}3}}{{\mathrm{M}}_{\mathrm{Mo}2\mathrm{O}3}}$  with MMo2O3 = 2 MMo + 3 MO = 245.9 g.mol-1
⇒ nMo2O3 = $\frac{19.33}{245.9}$ = 7.861 x 10-2 mol

nMo = 2 nMo2O3 = 1.572 x 10-1 mol

nO = 3 nMo2O3 = 2.358 x 10-1 mol

Sample 2:

Mo2O3 only reacts with O2 ⇒ no Mo atom is added

nO2 reacting = $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ =
nO2 =  = 2.419 x 10-1 mol

nMo (sample 2) = nMo (sample 1) = 1.572 x 10-1 mol

nO (sample 2) = nO (sample 1) + nO2 = 2.358 x 10-1 + 2.419 x 10-1 = 4.777 x 10-1 mol

= 3

The empirical formula of the new oxide is MoO3