AS with all these problems, it is useful to ASSUME that there are 100*g100⋅g of unknown compound. And we divide thru by the ATOMIC masses of the constituent elements to find the "empirical formula"empirical formula.
There are thus (62.07*g)/(12.011*g*mol^-1)=5.17*mol*"carbon"62.07⋅g12.011⋅g⋅mol−1=5.17⋅mol⋅carbon
And (10.34*g)/(1.00794*g*mol^-1)=10.26*mol*"hydrogen"10.34⋅g1.00794⋅g⋅mol−1=10.26⋅mol⋅hydrogen
Now, we still have a missing percentage, and this is assumed to be oxygen (note that this is a standard assumption; there are few ways to analyse oxygen gas by combustion in that we have to burn the organic compound in oxygen to oxidize it, and so typically the oxygen percentage is calculated by difference, i.e. it is the missing percentage).
And thus(100*g-62.07*g-10.34*g)/(15.999*g*mol^-1)=1.72*mol*"oxygen"100⋅g−62.07⋅g−10.34⋅g15.999⋅g⋅mol−1=1.72⋅mol⋅oxygen
So to get the empirical formula, we divide thru by the SMALLEST molar quantity (that of oxygen) to get:
C:(5.17*mol)/(1.72*mol)=3.0C:5.17⋅mol1.72⋅mol=3.0
H:(10.26*mol)/(1.72*mol)=5.96H:10.26⋅mol1.72⋅mol=5.96
O:(1.72*mol)/(1.72*mol)=1.0O:1.72⋅mol1.72⋅mol=1.0
And thus we get the "empirical formula"empirical formula, the simplest whole number ratio defining constituent atoms in a species as:
C_3H_6OC3H6O
Now we further know that the "molecular formula"molecular formula is a WHOLE number mulitple of the "empirical formula"empirical formula.
And so 116*g*mol^-1=nxx(3xx12.011+6xx1.00794+15.999)*g*mol^-1=nxx58*g*mol^-1116⋅g⋅mol−1=n×(3×12.011+6×1.00794+15.999)⋅g⋅mol−1=n×58⋅g⋅mol−1.
Clearly, n=2n=2, and the "molecular formula"-=C_6H_12O_2molecular formula≡C6H12O2.
