Sulfur and oxygen form both sulfur dioxide and sulfur trioxide. when samples of these are decomposed, the sulfur dioxide produces 3.46 g oxygen and 3.47 g sulfur, while the sulfur trioxide produces 8.25 g oxygen and 5.50 g sulfur. a) calculate the mass of oxygen per gram of sulfur for sulfur dioxide.
b) calculate the mass of oxygen per gram of sulfur for sulfur trioxide.

For A: The mass of oxygen per gram of sulfur for sulfur dioxide is 0.997 g

For B: The mass of oxygen per gram of sulfur for sulfur trioxide is 1.5 g

Explanation:

The chemical equation for decomposition of sulfur dioxide follows:

We are given:

Mass of sulfur = 3.47 g

Mass of oxygen = 3.46 g

To calculate the mass of oxygen per gram of sulfur, we apply unitary method:

For every 3.47 g of sulfur, 3.46 g of oxygen will form.

So, for 1 gram of sulfur, of oxygen will form.

Hence, the mass of oxygen per gram of sulfur for sulfur dioxide is 0.997 g

The chemical equation for decomposition of sulfur trioxide follows:

We are given:

Mass of sulfur = 5.50 g

Mass of oxygen = 8.25 g

To calculate the mass of oxygen per gram of sulfur, we apply unitary method:

For every 5.50 g of sulfur, 8.25 g of oxygen will form.

So, for 1 gram of sulfur, of oxygen will form.

Hence, the mass of oxygen per gram of sulfur for sulfur trioxide is 1.5 g

imma come back to you hold on

1. for the first question, we must find the mass of the anhydrous salt, mgso₄. the molar mass for mgso₄·7h₂o is 246.47 g/mol, while that of mgso₄ is 120.37 g/mol.7.834 g mgso₄·7h₂o * 1 mol mgso₄·7h₂o/246.47 g * 1 mol mgso₄/1 mol mgso₄·7h₂o * 20.37 g mgso₄/mol = 0.647 g mgso₄

2. mass of water = mass of sample - mass of anhydrous salemass of water = 7.834 - 0.647 = 7.19 gpercent water in hydrate = 7.19/7.834 * 100 = 91.74%