Question

A molecular compound is composed of 60.4% Xe, 21.1% O and 17.5% F, by mass. If the molecular weight is 217.3 u, what is the molecular formula?

• A. $Xe{O}_3{F}_2$
• B. $Xe{O}_{3}F$
• C. $Xe{O}_2{F}_3$
• D. None of these

Answer 1

The correct option is A $Xe{O}_3{F}_2$

Atomic weights of $Xe=131,O=16,F=19$
So, first calculate to find the number of atoms:

$Xe=\frac{60.4}{131}=0.461$

$O=\frac{21.1}{16}=1.318$

$F=\frac{17.5}{19}=0.921$

Divide the three numbers through by the smallest, 0.461 and you get 1, 3, and 2.

So, the empirical formula is $Xe{O}_3{F}_2$.

The formula weight of $Xe{O}_3{F}_2$ is $217$, so the molecular formula is identical with the empirical formula.