In an experiment, 1.288 g of barium sulphate was obtained from 1.03 g of barium. In another experiment, 3.672 g of barium sulphate gave, on reduction, 2.938 g of barium. Show that these figures verify the law of constant proportions.
In first experiment:
Mass of barium sulphate = 1.288g
Mass of barium left = 1.03g
Mass of Sulphur present = 1.288 – 1.03 = 0.25g
Percentage of sulphur present in 
⇒ Percenatage of sulphur is 20%
In second experiment:
Mass of barium sulphate = 3.672g
Mass of barium left = 2.938g
Mass of Sulphur present = 3.672 – 2.938 = 0.73g
Percentage of sulphur present 
⇒ Percenatage of sulphur is 20%
In both the experiments, the percentage of sulphur is same.
Hence, it is verified that these figures verify the law of constant proportions.
Note: Law of constant proportions states that “in a chemical
substance, the elements are always present in definite
proportions by mass.