Question

Are the two gases he made the same?

A chemist reacts 50.0 g sulfur with 50.0 g oxygen to make 83.4 grams of a gas, along with some leftover sulfur. In a different experiment, he reacts 20.0 grams of sulfur with 30.0 grams of oxygen to make 49.9 grams of a gas, along with some leftover oxygen.

Answer 1

Yes....both scenarios yield `SO_3`, `"sulfur trioxide...."`

Explanation:

Initially `83.4*g` of gas was made from `33.4*g` sulfur, and `50.0*g` oxygen....

And thus `"moles of sulfur"=(33.4*g)/(32.06*g*mol^-1)=1.04*mol`

And `"moles of oxygen"=(50.0*g)/(16.00*g*mol^-1)=3.125*mol`

We divide thru by the SMALLER molar quantity to get an empirical formula of `S_((1.04*mol)/(1.04*mol))O_((3.125*mol)/(1.04*mol))=SO_3`

And in the second scenario, we use `20.0*g` sulfur and `29.9*g` oxygen to make `49.9*g` of compound....

And thus `"moles of sulfur"=(20.0*g)/(32.06*g*mol^-1)=0.624*mol`

And `"moles of oxygen"=(29.9*g)/(16.00*g*mol^-1)=1.87*mol`

We go thru the same rigmarole....to get an empirical formula of `S_((0.624*mol)/(0.624*mol))O_((1.87*mol)/(0.624*mol))=SO_3`.

What was the other possible sulfur oxide?