A 0.3200 g sample of a carboxylic acid is burned in oxygen, producing 0.5706 g of CO2 and 0.2349 g of H2O. How would you determine the empirical formula of carboxylic acid?

1 Answer
Dec 18, 2015

Answer:

#"C"_3"H"_6"O"_2#

Explanation:

The idea here is that you need to figure out how much carbon and hydrogen were present in the initial sample, then use the sample's mass to figure out how much oxygen it contained.

Once you know how many grams of each element you had in the original sample, use the molar masses of the three elements to determine how many moles of each you have.

So, what you need to do is figure out the percent composition of hydrogen in water and of carbon in carbon dioxide. This will allow you to figure out how many grams of hydrogen and carbon, respectively, were produced by the combustion reaction.

As you know, one mole of water, #"H"_2"O"#, contains #2# moles of hydrogen. This means that the percent composition of hydrogen in water will be

#(2 xx 1.00794 color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = "11.19% H"#

What this means is that every #"100 g"# of water will contain #"11.19 g"# of hydrogen. In your case, the #"0.2349-g"# sample of water will contain

#0.2349 color(red)(cancel(color(black)("g water"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g water")))) = "0.026285 g H"#

Similarly, one mole of carbon dioxide, #"CO"_2#, contains one mole of carbon. The percent composition of carbon in carbon dioxide will be

#(1 xx 12.011 color(red)(cancel(color(black)("g/mol"))))/(44.01 color(red)(cancel(color(black)("g/mol")))) xx 100 = "27.29% C"#

Once again, this tells you that every #"100 g"# of carbon dioxide will contain #"27.29 g"# of carbon. In your case, you will have

#0.5706 color(red)(cancel(color(black)("g H"_2"O"))) * "27.29 g C"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "0.15572 g C"#

The mass of carbon + hydrogen originally present in the carboxylic acid sample will be

#m_"C + H" = "0.026285 g" + "0.15572 g" = "0.18201 g"#

The initial mass of the sample was equal to #"0.3200 g"#. This tells you that the sample also contained

#m_"oxygen" = "0.3200 g" - "0.18201 g" = "0.13799 g O"#

So, now that you know how many grams of each element were present in the initial sample, use their molar masses to convert these to moles

#"For C: " (0.15572 color(red)(cancel(color(black)("g"))))/(12.011color(red)(cancel(color(black)("g")))/"mol") = "0.012965 moles C"#

#"For H: " (0.026285 color(red)(cancel(color(black)("g"))))/(1.00794 color(red)(cancel(color(black)("g")))/"mol") = "0.026078 moles H"#

#"For O: " (0.13799 color(red)(cancel(color(black)("g"))))/(15.9994color(red)(cancel(color(black)("g")))/"mol") = "0.0086247 moles O"#

Now, to get the compound's empirical formula, you need to find the smallest whole number ratio that exists between the number of moles of each element in the compound.

To do that, divide these values by the smallest one to get

#"For C: " (0.012965 color(red)(cancel(color(black)("moles"))))/(0.0086247 color(red)(cancel(color(black)("moles")))) = 1.5032 ~~ 1.5#

#"For H: " (0.026078 color(red)(cancel(color(black)("moles"))))/(0.0086247 color(red)(cancel(color(black)("moles")))) = 3.0236 ~~ 3#

#"For O: " (0.0086247 color(red)(cancel(color(black)("moles"))))/(0.0086247 color(red)(cancel(color(black)("moles")))) = 1#

This means that you have

#"C"_1.5"H"_3"O"_1#

Since you're looking for the smallest whole number ratio that exists between the elements, multiply each subscript by #2# to get

#("C"_1.5"H"_3"O"_1)_2 implies color(green)("C"_3"H"_6"O"_2)#