Part A: Determine the empirical formula when the elemental composition by mass shows: 40.0% C, 6.71% H, 53.3% 0. For that I got CH2010 simplified to CHO5 Part B: Determine the molecular formula when the mole mass of the compound is 90.08 g/mol. (Help?)

1 Answer
Sep 29, 2015

Answer:

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

Explanation:

To get the molecular formula of the compound, you need to use its empirical formula and its molar mass.

The empirical formula tells you what the smallest integer ratio of atoms is for the elements that make up your compound.

Now, from what I can tell, you didn't get the correct empirical formula. When you are given the percent composition by mass of the compound, you can take a 100-g sample to make the calculations easier.

If you sample has a mass of 100 g, then you know that it contains

  • 40.0 grams of carbon
  • 6.71 g of hydrogen
  • 53.3 grams of oxygen

Use the molar masses of the three elements to find how many moles of each you get in that sample

#"For C: " (40.0color(red)(cancel(color(black)("g"))))/(12.011color(red)(cancel(color(black)("g")))/"mol") = "3.33 moles C"#

#"For H: " (6.71color(red)(cancel(color(black)("g"))))/(1.01color(red)(cancel(color(black)("g")))/"mol") = "6.64 moles H"#

#"For O: " (53.3color(red)(cancel(color(black)("g"))))/(16.0color(red)(cancel(color(black)("g")))/"mol") = "3.33 moles O"#

Divide all the values by the smallest one to get

#"For C: " (3.33color(red)(cancel(color(black)("moles"))))/(3.33color(red)(cancel(color(black)("moles")))) = 1#

#"For H: " (6.64color(red)(cancel(color(black)("moles"))))/(3.33 color(red)(cancel(color(black)("moles")))) = 1.994 ~~ 2#

#"For O: " (3.33color(red)(cancel(color(black)("moles"))))/(3.33color(red)(cancel(color(black)("moles")))) = 1#

The empirical formula will thus be

#"C"_1"H"_2"O"_1" "#, or #" " "CH"_2"O"#

So, you know that the smallest integer ratio of atoms in your compound is #1:2:1#, but you don't know exactly how many atoms of each element are needed to form one molecule.

This is where the molar mass of the compound comes into play. You know that one mole of molecules has a mass of #"90.08 g"#, and that you ahve a minimum of one mole of carbon, two moles of hydrogen, and one mole of oxygen for every mole of the compound.

This means that you can write

#(1 xx M_"M carbon" + 2 xx M_"M hydrogen" + 1 xx M_"M oxygen") * color(blue)(n) = "90.08 g"#

#(1 xx "12.011 g" + 2 xx "1.01 g" + 1 xx "16.0 g") * color(blue)(n) = "90.08 g"#

This will get you

#color(blue)(n) = (90.08color(red)(cancel(color(black)("g"))))/(30.03color(red)(cancel(color(black)("g")))) = 2.9997 ~~ 3#

The molecular formula will thus be

#("CH"_2"O")_3 = color(green)("C"_3"H"_6"O"_3#