# Stoichiometry - 4 | Chemical Calculations

## General Chemistry 2 - Stoichiometry - 4

A 19.33 g sample of Mo_{2}O_{3} 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:__

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__:

Mo_{2}O_{3} only reacts with O_{2} ⇒ 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 MoO_{3}