Question #05543

1 Answer
Nov 26, 2015

Answer:

#"CH"_2"O#

Explanation:

In order to determine the empirical formula of the compound, you need to know how many moles of each element are present in your sample.

You can get the number of moles of each element present in that sample by using the elements' respective molar masses.

Now, judging from the values given to you, I will assume that you're supposed to use these molar masses

  • carbon #-># #"12.0 g/mol"#
  • hydrogen #-># #"1.00 g/mol"#
  • oxygen #-> # #"16.0 g/mol"#

So, the molar mass of an element tells you what the mass of one mole of that element is. For example, carbon has a molar mass of #"12.0 g/mol"#.

This means that every mole of carbon has a mass of #"12.0 g"#.

This means that you will have

#"For C: " 24 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.0color(red)(cancel(color(black)("g")))) = "2 moles C"#

#"For H: " 4 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00color(red)(cancel(color(black)("g")))) = "4 moles H"#

#"For O: " 32 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16.0color(red)(cancel(color(black)("g")))) = "2 moles O"#

So, your sample contains #2# moles of carbon, #4# moles of hydrogen, and #2# moles of oxygen.

The important thing to keep in mind is that the empirical formula expresses the smallest whole number ratio that exists between the elements that make up a certain compound.

In your case, the smallest whole number ratio between the three elements will not be #2:4:2#, since you can divide all values by #2# to get #1:2:1#.

This means that the compound's empirical formula will be

#"C"_1"H"_2"O"_1 implies "CH"_2"O"#