Question

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Answer 1

Here is your answer—

In the first compound % of S = 51% , hence % of O will be = 49%
therefore if % of S = 1% , the % of O will be = 49 /51= 0.96.

In the second compound
the % of S = 41% , hence % of O will be = 59%
if % of S = 1% , % of O = 59/ 41 = 1.439

therefore the simple ratio between O: O in these oxide = 1.439/0.96 : 0.96/ 0.96 = 1.5 : 1 = 3:2 hence the law
Ratio of oxygen in oxides of sulphur is 3:2