A sample of sulfur having a mass of #1.28# g combines with oxygen to form a compound with a mass #3.20# g. What is the empirical formula of the compound?

Your goal when dealing with empirical formulas is to determine the smallest whole number ratio that exists between the elements that make up the compound.

In order to do that, you first need to know exactly how many moles of each element are present in your sample.

You are told that a #"1.28-g"# sample of sulfur combines with oxygen gas to form a compound that has a mass of #"3.20 g"#.

Right from the start, with the Law of conservation of mass in mind, you can say that the mass of oxygen must be equal to the difference between the final mass of the compound and the mass of the sulfur.

#m_"compound" = m_"sulfur" + m_"oxygen"#

In this case, you can say that

#m_"oxygen" = "3.20 g" - "1.28 g" = "1.92 g"#

So, your compound contains #"1.28 g"# of sulfur and #"1.92 g"# of oxygen. Next, use the molar masses of the two elements to determine how many moles of each you have

#1.28 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.066color(red)(cancel(color(black)("g")))) = "0.03992 moles S"#

#1.92 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "0.1200 moles O"#

To get the mole ratio that exists between the two elements, divided both values by the smallest one

#"For S: " (0.03992 color(red)(cancel(color(black)("moles"))))/(0.03992color(red)(cancel(color(black)("moles")))) = 1#

#"For O: " (0.1200color(red)(cancel(color(black)("moles"))))/(0.03992color(red)(cancel(color(black)("moles")))) = 3.01 ~~3#

Since you can't have a smallest whole number ratio between the two elements, it follows that the empirical formula of the compound is

#"S"_1"O"_3 implies color(green)("SO"_3)#