To start, we know our reaction is going to look something like this:
#C_xH_yO_z# + #O_2# #-># #CO_2# + #H_2O#
We are given the following information:
Beginning mass of #C_xH_yO_z#: .047 g
Ending mass of #CO_2#: .0989 g
Ending mass of #H_2O#: .03599 g
Let's begin by converting our #CO_2# and #H_2O# into moles.
#.0989 g CO_2 xx (1 mol CO_2)/(44 g CO_2) = .00225# mol #CO_2#
#.03599 g H_2O xx (1 mol H_2O)/(18 g H_2O) = .002# mol #H_2O#
There is one atom of #C# for every one molecule of #CO_2#, so we know that our sample of #C_xH_yO_z# contains #.00225# mol #C#.
There are two atoms of #H# for every one molecule of #H_2O#, so we know our sample of #C_xH_yO_z# contains #.002 xx 2 = .004# mol #H#.
To find the amount of #O# in #C_xH_yO_z#, we must subtract the masses of #C# and #H# from our original sample mass.
#"Mass of O" = "Original mass" - .00225 mol C xx (12 g C)/(1 mol C) - .004 mol H xx (1 g H)/(1 mol H)#
Giving #"Mass of O" = .016 "g"#, and now we must convert the mass of #O# to moles of #O#:
#.016 g O xx (1 mol O)/(16 g O) = .001# mol O.
We now know the number of moles of C,H, and O, and can rewrite #C_xH_yO_z# as #C_.00225H_.004O_.001#.
To get this into the empirical form, we divide all numbers by the smallest one, in this case .001, giving #C_2.25H_4O#.
Finally, we eliminate the .25 in #C_2.25# by multiplying all numbers by 4, giving us the final answer of #C_9H_16O_4#.
