Question

What is the empirical formula for a compound if a sample contains 1.0 g of S and 1.5 g of O

Answer 1

To answer this question, you need to follow these steps:

1.) Convert each of the 'mass' values associated with each of the elements (S and O) into moles. To do this, we divide each individually by their respective atomic masses (For S, it's 32 g/mol, whilst for O it's 16 g/mol)
2.) We now divide each of these molar values by whichever is the smallest of the two. So, for example, if we have less number of moles of sulfur than oxygen, then we divide the number of moles of sulfur and oxygen by the number of moles of oxygen respectively.

\[\frac{ Moles Of Sulfur }{ Smallest Molar Value }: \frac{ Moles of Oxygen }{ Smallest Molar Value}\]

3.) Clearly, this will leave us with a value of '1' for one of the elements, and some value greater than 1 for the other. We round this value, if required, to the nearest whole number so that each element is now represented by a whole number value.

These values are the number of atoms present in the empirical formula for the compound, which is the simplest whole number ratio of atoms of different elements present in it. So, if we have a whole number ratio for S:O in our compound of x:y, then the empirical formula would be:
\[S _{x}O _{y}\]
Hope that helps! :)

Answer 2

Sorry, I think I made a little mistake in Point 2.) above....I meant to say:

"So, for example, if we have less number of moles of sulfur than oxygen, then we divide the number of moles of sulfur and oxygen each by the number of moles of SULFUR respectively."

My apologies! :)