What is the empirical chemical formula of a compound that has has mass percent 62.1% carbon, 10.5% hydrogen, and 27.4% oxygen?

1 Answer
Mar 17, 2016

Answer:

#"C"_3"H"_6"O"_1#

Explanation:

In order to get a compound's empirical formula, you basically need to determine the smallest whole number ratio that exists between the number of moles of its constituent elements.

This means that your goal here will be to convert the given percent composition to grams, then convert the grams to moles by using the molar masses of the elements.

In order to make the calculations easier, pick a #"100-g"# sample of this unknown compound. According to the given percent composition, this sample will contain

  • #"62.1 g " -># carbon
  • #"10.5 g " -># hydrogen
  • #"27.4 g " -># oxygen

Now, use the molar masses of the three elements to determine how many moles of each you have in this sample.

#"For C: " 62.1 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "5.1703 moles C"#

#"For H: " 10.5color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "10.417 moles H"#

#"For O: " 27.4color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "1.7126 moles O"#

To get the mole ratio that exists between the elements in the compound, divide all values by the smallest one. This will get you

#"For C: " (5.1703color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 3.019 ~~ 3#

#"For H: " (10.417color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 6.083 ~~ 6#

#"For O: " (1.7126color(red)(cancel(color(black)("moles"))))/(1.7126color(red)(cancel(color(black)("moles")))) = 1#

Since #3:6:1# represents the smallest whole number ratio that can exist between the three elements, the empirical formula of the unknown compound will be

#color(green)(|bar(ul(color(white)(a/a)"C"_3"H"_6"O"_1color(white)(a/a)|)))#