Combustion of a #20*g# mass of compound gives #44*g# #CO_2#, and #7.99*g# of water. Given that the molecular mass is #180*g*mol^-1# what are the empirical and molecular formulae?

1 Answer
Mar 27, 2017

Answer:

This question operates under an unspoken assumption. We get (eventually!) a molecular formula of #C_9H_8O_4#.

Explanation:

We can burn organic compounds in a furnace to produce (i) carbon dioxide; and (ii) water. The carbon and hydrogen content of these products COME directly from the hydrocarbon. Sometimes these are expressed as a percentage by mass and SOMETIMES, as here, the measured percentages do not add up to 100%. The missing percentage is presumed to be oxygen, which is NOT measured, but obtained by difference, i.e. #O%=100%-%C-%H#.

Here, the carbon content is measured by #CO_2#, and the hydrogen content is measured by WATER (I think you made a mistake when you posed the question!).

So let's see if we can get something useful.

In #20*g# compound, there are #(44*g)/(44*g*mol^-1)C=1*mol*C#.

In #20*g# compound, there are #(7.99*g)/(18.01*g*mol^-1)xx2H=0.89*mol*H#.

In #20*g# compound, there are presumed to be the following molar quantity of oxygen,

#(20.0*g-12.011*g-0.89*g)/(16*g*mol^-1)O=0.44*mol*O#. (That is the mass balance was due to oxygen!).

We divide thru by the SMALLEST molar quantity, that of oxygen to get a trial formula of #C_(2.27)H_2O#, which we must multiply by #4 # to get whole numbers, i.e. #C_9H_8O_4#, which is the #"empirical formula"#.

But we know that the #"molecular formula"# is a simple whole number multiple of the #"empirical formula"#

And thus............................,

#nxx(9xx12.011+8xx1.00794+4xx15.999)*g*mol^-1#

#=180*g*mol^-1#

Clearly, #n=1#, so here #"molecular formula"# #=# #"empirical formula"# #=# #C_9H_8O_4#

I am sorry for making such a meal of this answer, but if you are an undergrad, you should be aware of the background to this question. If you are at A level, your teacher has no business asking you these sorts of questions.