When barium metal is burned in air, an oxide is formed with
81.1% by mass of barium. Calculate the formula of the compound. [Mr
of barium=137, oxygen=16]
The answer is simple (BaO2), but I'm looking for a proper
working instead of guess & check.
I tried doing this but I failed:
Let compound be Ba(x)O(y)
137x / (137x + 16y) = 0.811
...