Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. when samples of these were decomposed the sulfur dioxide produced 3.49 g oxygen and 3.50 g sulfur, while the sulfur trioxide produced 6.75 g oxygen and 4.50 g sulfur. calculate the mass of oxygen per gram of sulfur for each sample and show that these results are consistent with the law of multiple proportions.

The reaction for decomposition of sulfur dioxide (SO₂) and sulfur trioxide (SO₃) are given below:

SO₂ → S + O₂    and  SO₃→ S + O₂

In SO₂, for 3.50 g sulfur,  oxygen produced is equal to 3.49 g, so, for 1 g sulfur 3.49/3.50 g= 0.997 g of O₂

In SO₃, for 4.50 g sulfur, oxygen produced is 6.75 g, so for 1 g sulfur, oxygen p;roduced will be equal to  6.75/4.50g=1.5 g of O₂.

With 1 g of sulfur, the ratio of the amount of oxygen present in SO₂ and SO₃ is 0.997:1.50= 2:3, this is a simple ratio and small whole number. These results are consistent with the Law of Multiple Proportion.

19.52 ; 2.80 ; 1.28 is the correct group of like decimals.

Hope this helps:)