# A 5.5*g mass of a chromium chloride compound contains 1.82*g of metal. What is its "empirical formula"?

Mar 13, 2017

$\text{Chromic chloride,}$ $C r C {l}_{3}$.

#### Explanation:

We simply find the empirical formula of the chromium halide, using the given masses. Because it is a binary compound, i.e. ONLY 2 components, we can easily work out the masses of metal and chloride in the sample, even tho we were only given the mass of metal. In each case we divide each contributing mass by its atomic mass:

$\text{Moles of metal}$ $=$ $\frac{1.82 \cdot g}{52.00 \cdot g \cdot m o {l}^{-} 1} = 0.0350 \cdot m o l$.

$\text{Moles of chlorine}$ $=$ $\frac{5.55 \cdot g - 1.82 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1} = 0.105 \cdot m o l$.

We divide thru by THE SMALLEST MOLAR QUANTITY, which is that of the metal:

$C r : \frac{0.0350 \cdot m o l}{0.0350 \cdot m o l} = 1$

$C l : \frac{0.105 \cdot m o l}{0.0350 \cdot m o l} = 3$

And thus we arrive at an empirical formula of $C r C {l}_{3}$.

Note that when use oxidation numbers to distinguish the compounds, we use ROMAN instead of ARABIC numerals......

$\text{Cr(III) chloride}$ $=$ $\text{chromic chloride}$.

$\text{Cr(II) chloride}$ $=$ $\text{chromous chloride}$.