What is the empirical formula of a compound that contains 53.73% Fe and 46.27% of S?

1 Answer
Sep 5, 2016

Answer:

#"Fe"_2"S"_3#

Explanation:

Your strategy here will be to

  • pick a sample of this ionic compound
  • use the molar masses of iron and of sulfur to find how many moles of each it contains
  • find the smallest whole number ratio that exists between iron and sulfur in the compound

To make the calculations easier, pick a #"100-g"# sample of this unknown compound. This sample with contain, as given by the known percent composition

  • #"53.73 g Fe"#
  • #"46.27 g S"#

Convert these masses to moles

#"For Fe: " 53.73 color(red)(cancel(color(black)("g"))) * "1 mole Fe"/(55.845color(red)(cancel(color(black)("g")))) = 0.962#

#"For S: " 46.27 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32.066color(red)(cancel(color(black)("g")))) = 1.44#

Next, divide both values by the smallest one

#"For Fe: " (0.95 color(red)(cancel(color(black)("moles"))))/(0.95color(red)(cancel(color(black)("moles")))) = 1#

#"For S: " (1.44 color(red)(cancel(color(black)("moles"))))/(0.95color(red)(cancel(color(black)("moles")))) = 1.52#

Now, to find the smallest whole number ratio that exists between the two elements in the compound, multiply both values by #2#. You will have

#"For Fe: " 2 xx 1 = 2#

#"For S: " 2 xx 1.52 = 3.04 ~~ 3#

The empirical formula of the unknown compound will thus be

#color(green)(bar(ul(|color(white)(a/a)color(black)("Fe"_2"S"_3)color(white)(a/a)|)))#