Question

Two oxide of sulphur,A and B were analyzed and the results obtained showed that in oxide A,3.50g of sulphur combined with 6.00g of oxygen and in oxide B,2.80g of sulphur combined with 9.55g.Show that this result illustrate the law of multiple proportion

Answer 1

Answer:

See Explanaton

Explanation:

The law of multiple proportions states that when two same elements form more than a compound, the different masses of one element that combine with the same mass of the other element are in the ratio of small whole numbers.

Oxide A

3.50g of sulphur combined with 6.00g of oxygen

Oxygen:Sulphur = 6 : 3.5

Oxide B

2.80g of sulphur combined with 9.55g

Oxygen : Sulphur = 9.55 : 2.8

Therefore:

The ratio of Oxygen to Sulphur in Oxides A and B is:

$\dfrac{\text{Oxygen}}{\text{Sulphur}} \implies \dfrac{6}{3.5}:\dfrac{9.55}{2.8} = \dfrac{1.7}{1}:\dfrac{3.4}{1}=1:2$

Ratio of Oxygen=1:2

There is exactly twice in Oxide B as in Oxide A.

This result illustrates the law of multiple proportions.