Question

Three oxides of lead contain 92.83% ,90.65% and 86.10% lead. Prove that the law of multiple proportions is verified from these data.

Answer 1

According to Law of Multiple Proportions

When two elements combine to form more than one compound, the mass of the element, which combines with a fixed mass of the other element, will always be ratios of the whole number.

Explanation:
When two elements combine to form more than one compound [for example two elements be Carbon(C) and Oxygen(O) to form more than one compound like Carbon monoxide(CO), Carbon dioxide(CO2)], the mass of element (here, mass of oxygen) which combines with a fixed mass of the other element (here, fixed mass of carbon as no of carbon in both compound is same), will always be ratios of whole number [wt of O : wt of O2 i.e. 16:32 or 1:2, (the whole numbers)].

Solution
Three oxides of lead:
1) Lead(II) oxide, PbO, litharge
2) Lead(II, IV) oxide, Pb3O4, red lead
3) Lead dioxide (lead(IV) oxide), PbO2

Let the total mass be 100g. Then,

In the first compound,
the mass of Lead is 92.83 g,
and the mass of Oxygen will be 100 - 92.83 = 7.17g.
Therefore the mass of Oxygen that combines with 1 g of lead = 7.17/92.83 = 0.08

In the second compound,
the mass of Lead is 90.65 g,
and the mass of Oxygen will be 100 - 90.65 = 9.35g
Therefore the mass of Oxygen that combines with 1 g of lead = 9.35/90.65 = 0.10

In the third compound,
the mass of Lead is 86.10 g,
and the mass of Oxygen will be 100 - 86.10 = 13.90g
Therefore the mass of Oxygen that combines with 1 g of lead = 13.9/86.1 = 0.16

Thus the ratio of Oxygen combining with a fixed mass of lead is
=> 0.08: 0.10: 0.16
or, 8 : 10 : 16
or, 4 : 5 : 8 which proves the law of multiple proportions.