What is the empirical formula of a compound containing #C, H#, and #O# if combustion of 3.69 g of the compound yields 5.40 g of #CO_2# and 2.22 g of #H_2O#?

1 Answer
Sep 9, 2016

Answer:

We get finally an empirical formula of #CH_2O#; I think the question is suspect.

Explanation:

ONLY the #C# and the #H# of the combustion can be presumed to derive from the unknown. (Why? Because the analysis is performed in air and typically an oxidant is added to the combustion.)

#"Moles (i) and mass (ii) of carbon"# #=# #(5.40*g)/(44.01*g*mol^-1)# #=# #0.123*mol# #-=# #1.47*g*C#

#"Moles (i) and mass (ii) of hydrogen"# #=# #2xx(2.22*g)/(18.01*g*mol^-1)# #=# #0.247*mol# #-=# #0.249*g*H#. Note that the hydrogen in the compound was combusted to water; this is why we multiply the molar quantity by 2.

And now, finally, we work out the percentage composition of #C,H,O# with respect to the original sample:

#%C# #=# #(1.47*g)/(3.69*g)xx100%=40.00%#

#%H# #=# #(0.249*g)/(3.69*g)xx100%=6.75%#

#%O# #=# #(100-39.84-6.75)%=53.25%#

Note that we cannot (usually) measure the percentage of oxygen in a microanalysis as extra oxidant is typically added. Thus #O%# is the percentage balance.

After all this wrok we start again. We assume that there were #100*g# of compound. And from this we work out the empirical formula.

#"Moles of carbon"# #=# #(40.0*g)/(12.011*g*mol^-1)# #=# #3.33*mol*C#.

#"Moles of hydrogen"# #=# #(6.75*g)/(1.00794*g*mol^-1)# #=# #6.70*mol*H#.

#"Moles of oxygen"# #=# #(53.41*g)/(15.999*g*mol^-1)# #=# #3.34*mol*H#.

If we divide thru by the smallest molar quantity, we get, an empirical formula of #CH_2O#. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar.