Question #19944

1 Answer
Oct 2, 2015




The idea here is that you can use the produced carbon dioxide and the produced water to determine how much carbon and hydrogen, respectively, was present in the original #"1.00-g"# sample of aspirin.

Once you know that, you can determine the mass of oxygen by using the total mass of the sample.

So, if you assume that all the carbon that was present in the aspirin is now a part of the carbon dioxide, you can use carbon dioxide's percent composition of carbon to find how much carbon you had.

Use carbon and carbon dioxide's molar masses to get

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

This means that 2.20 g of #"CO"_2# will contain

#2.20color(red)(cancel(color(black)("g CO"""_2))) * "27.29 g C"/(100color(red)(cancel(color(black)("g CO"""_2)))) = "0.6004 g C"#

Do the same for water and hydrogen to get hydrogen's percent composition in water

#(2 xx 1.008color(red)(cancel(color(black)("g/mol"))))/(18.015color(red)(cancel(color(black)("g/mol")))) xx 100 = "11.19% H"#

This means that 0.400 g of water contain

#0.400color(red)(cancel(color(black)("g H"_2"O"))) * "11.19 g H"/(100color(red)(cancel(color(black)("g H"_2"O")))) = "0.04476 g H"#

The mass of oxygen found in the original aspirin sample was

#m_"total" = m_"C" + m_"H" + m_"O"#

<#m_"O" = 1.00 - 0.6004 - 0.04476 = "0.3548 g O"#

Now you need to focus on finding how many moles of each element your original sample contained.

To do that, use the element's respective molar masses

#"For C: " (0.6004color(red)(cancel(color(black)("g"))))/(12.011color(red)(cancel(color(black)("g")))/"mol") = "0.04999 moles C"#

#"For H: " (0.04476color(red)(cancel(color(black)("g"))))/(1.01color(red)(cancel(color(black)("g")))/"mol") = "0.04432 moles H"#

#"For O: " (0.3548color(red)(cancel(color(black)("g"))))/(16.0color(red)(cancel(color(black)("g")))/"mol") = "0.02218 moles O"#

Divide all the values by the smallest one to get

#"For C: " (0.04999color(red)(cancel(color(black)("moles"))))/(0.02218color(red)(cancel(color(black)("moles")))) = 2.25#

#"For H: " (0.04432color(red)(cancel(color(black)("moles"))))/(0.02218 color(red)(cancel(color(black)("moles")))) = 1.998 ~~ 2#

#"For O: " (0.02218color(red)(cancel(color(black)("moles"))))/(0.02218color(red)(cancel(color(black)("moles")))) = 1#

The mole ratio that exists between the elements in the sample of aspirin is


Now, a compound's empirical formula shows you the smallest integer ratio that exists between the elements that make up a compound.

This means that you will need to multiply what you got in order to get all-integes for the mole ratios.

Notice that if you multiply everything by #4#, you have

#"C"_(2.5 * 4)"H"_(2 * 4)"O"_(1 * 4) -> "C"_9"H"_8"O"_4#

Now, to get the molecular formula, you need to take a look at aspirin's molar mass, which is given to you as somewhere between #"170"# and #"190 g/mol"#.

Add the molar masses of all the atoms that you have in the empirical formula to get

#9 xx 12.011 + 8 xx 1.01 + 4 xx 16.0 = "180.179 g/mol"#

That's a perfect match, smack in the middle of the extreme values given to you for the molar mass of the compound.

This means that the molecular formula of aspirin is

#("C"_9"H"_8"O"_4)_1 = color(green)("C"_9"H"_8"O"_4)#