Question

Certain non-metal X forms two oxides I and II. The mass percentage of oxygen in $I\left(X_4{O}_6\right)$ is $43.7$ which is same as that X in the $2nd$ oxide. Find the formula of $2nd$ oxide.

Answer 1

Mass percentage of X in oxide 2 $=43.7$

Mass percentage of oxygen in oxide 2 $=56.3$

Hence atoms of oxygen $=\frac{6}{43.7}\times 56.3=7.73$

Atoms of X$=\frac{4}{56.3}\times 43.7=3.1$

Therefore, ratio of oxygen to X $=\frac{7.73}{3.1}=\frac{5}{2}$

Formula$=X_2{O}_5$