Sulphur and oxygen are known to form two compounds. The sulphur content in one of these is 51 prcnt while in the other is 41 prcnt. Show that this data is in agreement with the law of multiple proportions.
Hint:ratio of oxygen in oxides of sulphur is 2:3
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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!
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!
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