# A chloride of rhenium contains 63.6% rhenium. What is the name of this compound?

Nov 10, 2015

Empirical formula is $R e C {l}_{3}$.

#### Explanation:

In $100$ $g$ of the metal species, there are $\frac{63.6 \cdot g}{186.21 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.342$ $m o l$ $R e$, and $\frac{36.4 \cdot g}{35.45 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.03$ $m o l$ $C l$.

If we divide through by the lowest molar quantity ($0.342$ $m o l$), we get an empirical formula of $R e C {l}_{3}$, rhenium trichloride or rhenium (III) chloride. Do I win £5-00?

Nov 10, 2015

#### Explanation:

We will be determining the empirical formula for this compound containing the elements rhenium (Re) and chlorine (Cl).

Since the only other element is chlorine, we can subtract the percentage of rhenium from 100% to calculate the percent of chlorine in the compound.

100%.0-63.6%=36.4% "Cl"

Since the percentages add to 100, we can assume a 100-g sample of the compound, and the percentages become mass in grams.

Moles of Each Element
Determine the moles of each element by dividing the mass by its molar mass (atomic weight on the periodic table in g/mol).

$63.6 \cancel{\text{g Re"xx(1"mol Re")/(186.207cancel"g Re")="0.342 mol Re}}$

$36.4 \cancel{\text{g Cl"xx(1"mol Cl")/(35.453cancel"g Cl")="1.027 mol Cl}}$

Mole Ratios and Empirical Formula
Determine the mole ratio of each element by dividing the moles of each element by the least number of moles.

$\text{Re} :$(0.342cancel"mol")/(0.342cancel"mol")="1.00"

$\text{Cl} :$(1.027cancel"mol")/(0.342cancel"mol")="3.00"

The empirical formula is $\text{ReCl"_3}$.

This compound has an empirical formula mass of 292.6 g/mol. According to Sigma-Aldrich, a chemical supplier, this compound is rhenium(III) chloride.

http://www.sigmaaldrich.com/catalog/product/aldrich/309184?lang=en&region=US