Question

Water is added to $3.52$ g of $U{F}_6$. The products formed are $3.08$ g of a solid (containing only $U,O$ and $F$) and only $0.8$ g of gas. The gas (containing fluorine and hydrogen only) contains $95$ percent by mass fluorine. Assume that the empirical formula is the same as the molecular formula. ($U=238,F=19$).

The empirical formula of the gas is:

• A. $H{F}_2$
• B. $H_{2}F$
• C. $HF$
• D. $H{F}_3$

Answer 1

Answer: (C)

$HF$

Mass of one mole of H atoms is $1$ g.

Mass of one mole of fluorine atoms is $19$ g.

The gas contains $95$ percent by mass of fluorine.

Thus, $100$ g of gas contains $95$ g of $F$ and $5$ g of $H$.

This corresponds to $95=19\times 5$ or $5$ gram atoms of $F$ and $5=1\times 5$ or, $5$ gram atoms of $H$.

Thus, the empirical formula of the gas is $HF$.