Question

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.

Answer 1

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

SO₂ → S + O₂    and  SO₃→ S + $\frac{3}{2}$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.