Question

Two oxides of sulphur, A and B, were analysed, and the results obtained showed that in oxide A, 3.50g of sulphur combined with2.80g of oxygen; oxide B, 2.80g of sulphur combined with 2.80g of oxygen. Show that these results illustrate the law of multiple proportions. Write their formulae.

Answer 1

Before starting some information required :-

atm wt. of S = 32

atm wt. of O = 16

For the first oxide :-

moles of sulpher atom = mass of sulpher atom/atomic wt of S

=3.5g/32g=7/64=35/320

moles of oxygen atom = mass of oxygen atom/atomic wt of O

=2.8g/16=7/40=56/320

So we get its formulae as S(35/320)O(56/320)

and finally S5O8

For the second oxide :-

moles of sulpher atom = mass of sulpher atom/atomic wt of S

=2.8g/32g=7/80

moles of oxygen atom = mass of oxygen atom/atomic wt of O

=2.8g/16=7/40=14/80

So we get its formulae as S(7/80)O(14/80)

which simplifies to SO2

So in forst oxide sulpher and oxygen combined in 5:8 ratio whereas in second oxide they combined in 1:2 ratio ,which satisfys law of multiple proportion.

Hope it may help.