Question #425a5

The first thing to do here is to figure out the empirical formula of the unknown carboxylic acid.

Start by converting the percent composition to masses by picking a #"100-g"# sample of this compound.

You will have

• #"57.83 g C"#
• #"3.64 g H"#
• #"38.52 g O"#

Next, use the molar masses of the three elements to convert the masses to moles

#"For C: " 57.83 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011color(red)(cancel(color(black)("g")))) = "4.815 moles C"#

#"For H: " 3.64 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.008color(red)(cancel(color(black)("g")))) = "3.611 moles H"#

#"For O: " 38.52 color(red)(cancel(color(black)("g"))) * "1 mole O"/(15.9994color(red)(cancel(color(black)("g")))) = "2.408 moles O"#

Next, divide all three values by the smallest one to get the mole ratio that exists between the three elements in the compound

#"For C: " (4.815 color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1.9996 ~~ 2#

#"For H: " (3.611 color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1.4996 ~~ 1.5#

#"For O: " (2.408color(red)(cancel(color(black)("moles"))))/(2.408color(red)(cancel(color(black)("moles")))) = 1#

Now, in order to get the empirical formula of the compound, you need the smallest whole number ratio that exists between its constituent elements.

In your case, you have

#"C : H : O = "2 : 1.5 : 1#

To get the smallest whole number ratio, multiply all three values by #2#. You will have

#"C : H : O = "4 : 3 : 2#

You can thus say that the empirical formula of the compound is

#"C"_4"H"_3"O"_2#

Now, the thing to keep in mind about the molecular formula is that it is always a multiple of the empirical formula.

#"molecular formula" = color(blue)(n) * "empirical formula"#

This means that the molar mass of the compound is equal to a multiple of the molar mass of the empirical formula.

The molar mass of the empirical formula is

#4 * "12.011 g mol"^(-1) + 3 * "1.008 g mol"^(-1) + 2 * "15.9994 g mol"^(-1)#

# = "83.07 g mol"^(-1)#

This means that you have

#"molar mass" = color(blue)(n) * "83.07 g mol"^(-1)#

Since you know that you have

#"molar mass" in ["158 g mol"^(-1), "167 g mol"^(-1)]#

you can say that #color(blue)(n)# will vary between

#color(blue)(n) = (158 color(red)(cancel(color(black)("g mol"^(-1)))))/(83.07color(red)(cancel(color(black)("g mol"^(-1))))) = 1.902#

and

#color(blue)(n) = (167 color(red)(cancel(color(black)("g mol"^(-1)))))/(83.07color(red)(cancel(color(black)("g mol"^(-1))))) = 2.01#

Since it's obvious that the only whole number present in this interval is

#color(blue)(n = 2)#

you can say that the molecular formula will be

#"molecular formula" = color(blue)(2) * "C"_4"H"_3"O"_2 = color(darkgreen)(ul(color(black)("C"_8"H"_6"O"_4)))#