What is the empirical formula of methane given methane can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen?

1 Answer
Oct 30, 2016

Answer:

#"CH"_4#

Explanation:

Your goal when trying to figure out a compound's empirical formula is to find the smallest whole number ratio that exists between its constituent elements.

The problem tells you that a sample of methane underwent combustion and produced

  • #"4.5 g carbon, C"#
  • #"1.5 g hydrogen, H"#

The first thing to do here is to convert these masses to moles by using the molar masses of the two elements. You will have

#"For C: " 4.5 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.3747 moles C"#

#"For H: " 1.5 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "1.4881 moles H"#

Now, in order to find the mole ratio that existed between these two elements in the sample that underwent combustion, you must divide both values by the smallest one.

You will have

#"For C: " (0.3747 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 1#

#"For H: " (1.4881 color(red)(cancel(color(black)("moles"))))/(0.3747color(red)(cancel(color(black)("moles")))) = 3.97 ~~ 4#

Since #1:4# is already the smallest whole number ratio that can exist between these two elements, you can say that the empirical formula for methane is

#color(green)(bar(ul(|color(white)(a/a)color(black)("C"_1"H"_4 implies "CH"_4)color(white)(a/a)|)))#