Question

In a gas S and O are 50%by mass .What is their mol ratio?

Answer 1

Well, we got `SO_2` as the empirical formula....`S:O-=1:2`

Explanation:

We ASSUME an `100*g` mass of the gas....

And we interrogate the molar quantities of each element:

`"Moles of S"=(50.0*g)/(32.06*g*mol^-1)=1.56*mol`

`"Moles of O"=(50.0*g)/(16.00*g*mol^-1)=3.125*mol`

To get the empirical formula, we divide thru by the LOWEST molar quantity, i.e. that of sulfur...

`S_((1.56*mol)/(1.56*mol))O_((3.125*mol)/(1.56*mol))-=SO_2.`