Question

The fluoride of a particular element contains $32.4$% fluorine. An oxide of the same element contains $11.8$% oxygen. Suggest what would be the atomic weight of metal. Molecular weight of metal oxides and metal fluorides are $270$ and $352u$ respectively. Also suggest the molecular formulas of metal fluoride and metal oxide.

Answer 1

Let the metal be $M$.Its fluoride is $M{F}_x$ and oxide is $M{O}_y$.

% fluorine= $\frac{19\times Numberoffluorineatoms}{Molecularweight}\times 100$

$\Rightarrow 32.4=\frac{19x}{352}\times 100$

$\therefore x=6$

% oxygen= $11.8=\frac{16y}{270}\times 100$

$\Rightarrow y\approx 2$

Its fluoride is $M{F}_6$ and oxide is $M{O}_2$

Let A be its atomic weight.

$A+19\times 6=352$

$\therefore A=238$