A 5.325 g sample of methyl benzoate, a compound in perfumes, was found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula? If its molar mass is about 136 g/mol, what is its molecular formula?

1 Answer
Mar 7, 2016


Here's what I got.


I propose an interesting alternative to the classic approach of finding the empirical formula first, then using the molar mass to find the molecular formula.

More specifically, we will find the molecular formula first, then backtrack to find the empirical formula.

Mind you, this approach may seem a little complicated compared with the classic one, but it can be good practice nonetheless.

So, you know that methyl benzoate contains carbon, hydrogen, and oxygen.

Your #"5.325 g"# of methyl benzoate is said to contain

  • #"3.758 g " -># carbon
  • #"0.316 g " -># hydrogen
  • #"1.251 g " -># oxygen

Use the mass of the sample and the mass of each element to determine the percent composition of the compound.

#"For C: " (3.758 color(red)(cancel(color(black)("g"))))/(5.325color(red)(cancel(color(black)("g")))) xx 100 = "70.573% C"#

#"For H: " (0.316color(red)(cancel(color(black)("g"))))/(5.325color(red)(cancel(color(black)("g")))) xx 100 = "5.934% H"#

#"For O: " (1.251color(red)(cancel(color(black)("g"))))/(5.325color(red)(cancel(color(black)("g")))) xx 100 = "23.493% O"#

Now, methyl benzoate has a molar mass of #"136 g mol"^(-1)#. This means that one mole of this compound will have a mass of #"136 g"#.

Use the compound's percent composition to determine how many moles of each element you get in one mole of the compound - this will give you its molecular formula.

You will have

#"For C: " 136color(red)(cancel(color(black)("g compound"))) * overbrace("70.573 g C"/(100color(red)(cancel(color(black)("g compound")))))^(color(blue)("70.573% C")) = "95.979 g C"#

#"For H: " 136color(red)(cancel(color(black)("g compound"))) * overbrace("5.934 g H"/(100color(red)(cancel(color(black)("g compound")))))^(color(blue)("5.934% H")) = "8.070 g H"#

#"For O: " 136color(red)(cancel(color(black)("g compound"))) * overbrace("23.493 g O"/(100color(red)(cancel(color(black)("g compound")))))^(color(blue)("23.493% O")) = "31.950 g O"#

Now use the molar masses of the three elements to find how many moles of each you get per mole of methyl benzoate

#"For C: " 95.979 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = 7.991 ~~ "8 moles C"#

#"For H: " 8.070 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = 8.006 ~~ "8 moles H"#

#"For O: " 31.950color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = 1.997 ~~ "2 moles O"#

Since these represent the number of moles of each element in one mole of methyl benzoate, the molecular formula of the compound will be


Now, a compound's empirical formula tells you the smallest whole number ratio that exists between the moles of each element that makes up the compound.

In your case, a ratio of #8:8:2# can be simplified to one of #4:4:1#. This means that methyl benzoate's empirical formula will be

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