What is the empirical formula of a compound of uranium and fluorine that is composed of 67.6% uranium and 32.4% fluorine?

Dec 16, 2015

The empirical formula for the compound containing uranium and fluorine is $\text{UF"_6}$.

Explanation:

The empirical formula of a compound represents the lowest whole number ratio of elements in the compound. This ratio is represented by subscripts in the formula. In order to determine the empirical formula, we first must determine the number of moles of each element and then determine the mole ratio of each element. The mole ratio will give us the subscripts for the empirical formula.

Since the percentages of each element add up to 100, we can assume a 100 g sample of the compound and convert the percentages to grams.

Masses
$\text{U} :$67.6%rArr"67.6 g"
$\text{F} :$32.4%rArr"32.4 g"

Molar Masses
$\text{U} :$$\text{238.02891 g/mol}$
$\text{F} :$$\text{18.99840316 g/mol}$

Moles
Divide the mass of each element by its molar mass (atomic weight on the periodic table in g/mol).

$\text{U} :$$67.6 \cancel{\text{g U"xx(1"mol U")/(238.02891cancel"g U")="0.284 mol U}}$
$\text{F} :$$32.4 \cancel{\text{g F"xx(1"mol F")/(18.99840316cancel"g F")="1.71 mol F}}$

Mole Ratios
Divide the number of moles of each element by the lowest number of moles.

$\text{U} :$$\left(0.284 \cancel{\text{mol")/(0.284cancel"mol}}\right) = 1.00$

$\text{F} :$$\left(1.71 \cancel{\text{mol")/(0.284cancel"mol}}\right) = 6.02 \approx 6$

The empirical formula for this compound is $\text{UF"_6}$.