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⋅g44.01⋅g⋅mol−1 = 0.123⋅mol ≡ 1.47⋅g⋅C
Moles (i) and mass (ii) of hydrogen = 2×2.22⋅g18.01⋅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⋅g3.69⋅g×100%=40.00%
%H = 0.249⋅g3.69⋅g×100%=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⋅g12.011⋅g⋅mol−1 = 3.33⋅mol⋅C.
Moles of hydrogen = 6.75⋅g1.00794⋅g⋅mol−1 = 6.70⋅mol⋅H.
Moles of oxygen = 53.41⋅g15.999⋅g⋅mol−1 = 3.34⋅mol⋅H.
If we divide thru by the smallest molar quantity, we get, an empirical formula of CH2O. I am not terribly satisifed with this question. A molecular mass should have been quoted. This is a lot of work for simple sugar.
