Copper will react with sulfur to form a copper sulfide, if 1.500g copper reacts with sulfur to form 1.880g of copper sulfide, what is the empirical formula of the copper sulfide?

The idea here is that you need to use the mass of copper and the mass of the copper sulfide to determine how much sulfur the produced compound contains, then convert these masses to a mole ratio.

So, assuming that sulfur is in excess, you know that all the number of moles of copper that reacted will now be a part of the copper sulfide.

This mens that the difference between the recorded mass of the sulfide and the mass of the copper will be the mass of the sulfur.

#m_"copper sulfide" = m_"sulfur" + m_"copper"#

#m_"sulfur" = "1.880 g" - "1.500 g" = "0.380 g S"#

So, the mass of copper sulfide contains

• #"1.500 g"# of copper
• #"0.380 g"# of sulfur

Use the molar masses of these two elements to find how many moles of each you'd get in this sample

#1.500color(red)(cancel(color(black)("g"))) * "1 mole Cu"/(63.546color(red)(cancel(color(black)("g")))) = "0.023605 moles Cu"#

and

#0.380color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.065color(red)(cancel(color(black)("g")))) = "0.011851 moles S"#

To get the mole ratio that exists between these two elements in the copper sulfide, divide both values by the smallest one

#"For Cu: " (0.023605color(red)(cancel(color(black)("moles"))))/(0.011851color(red)(cancel(color(black)("moles")))) = 1.9918 ~~ 2#

#"For S: " (0.011851color(red)(cancel(color(black)("moles"))))/(0.011851color(red)(cancel(color(black)("moles")))) = 1#

This means that the empirical formula of this copepr sulfide will be

#"Cu"_2"S " -># copper(I) sulfide