Stoichiometry - 4 | Chemical Calculations

Sample 1:

n_Mo2O3 = $\frac{{\mathrm{m}}_{\mathrm{Mo}2\mathrm{O}3}}{{\mathrm{M}}_{\mathrm{Mo}2\mathrm{O}3}}$  with M_Mo2O3 = 2 M_Mo + 3 M_O = 245.9 g.mol^-1
⇒ n_Mo2O3 = $\frac{19.33}{245.9}$ = 7.861 x 10^-2 mol

n_Mo = 2 n_Mo2O3 = 1.572 x 10^-1 mol

n_O = 3 n_Mo2O3 = 2.358 x 10^-1 mol

Sample 2:

Mo₂O₃ only reacts with O₂ ⇒ no Mo atom is added

n_O2 reacting = $\frac{{\mathrm{m}}_{\mathrm{O}2}}{{\mathrm{M}}_{\mathrm{O}2}}$ = $\frac{{\mathrm{m}}_{\mathrm{new} \mathrm{oxide}} (\mathrm{sample} 2) - {\mathrm{m}}_{\mathrm{Mo}2\mathrm{O}3} (\mathrm{sample} 1)}{{\mathrm{M}}_{\mathrm{O}2}}$
n_O2 = $\frac{23.20 - 19.33}{16.00}$ = 2.419 x 10^-1 mol

n_Mo (sample 2) = n_Mo (sample 1) = 1.572 x 10^-1 mol

n_O (sample 2) = n_O (sample 1) + n_O2 = 2.358 x 10^-1 + 2.419 x 10^-1 = 4.777 x 10^-1 mol

$\frac{{\mathrm{n}}_{\mathrm{O}} (\mathrm{sample} 2)}{{\mathrm{n}}_{\mathrm{Mo}} (\mathrm{sample} 2)}$ = 3

The empirical formula of the new oxide is MoO₃