# A student found that 1.19 g of chromium (Cr) formed 1.74 g f chromium oxide. The molar mass of Cr is 52.00 g/mol. What is the empirical formula of chromium oxide?

Nov 28, 2016

$C {r}_{2} {O}_{3}$

#### Explanation:

As with all these problems, we calculate the molar quantities of each substituent, and normalize them according to the masses of the atomic constituents:

$\text{Moles of metal}$ $=$ $\frac{1.19 \cdot g}{52.00 \cdot g \cdot m o {l}^{-} 1} = 0.0229 \cdot m o l$

$\text{Moles of oxygen}$ $=$ $\frac{1.74 \cdot g - 1.19 \cdot g}{16.00 \cdot g \cdot m o {l}^{-} 1} = 0.0343 \cdot m o l$

And we divide thru by the SMALLEST molar quantity, that of chromium, to give:

Cr, (0.0229*mol)/(0.0229*mol)=1; O, (0.0343*mol)/(0.0229*mol)=1.50.

But by definition, the empirical formula is the smallest WHOLE number ratio that defines constituent atoms in a species. To get a whole number ratio, clearly we mulitply the empirical ratio by $2$ to $C {r}_{2} {O}_{3}$ as required. Capisce?

How did I know there were $0.55 \cdot g$ of oxygen present?