As with all these problems, we assume that there are 100*g100⋅g unknown material:
First we calculate an empirical formula...........
"Moles of carbon:"Moles of carbon: == (52.2*g)/(12.011*g*mol^-1)=4.35*mol52.2⋅g12.011⋅g⋅mol−1=4.35⋅mol.
"Moles of hydrogen:"Moles of hydrogen: == (13.0*g)/(1.079*g*mol^-1)=12.04*mol13.0⋅g1.079⋅g⋅mol−1=12.04⋅mol.
"Moles of oxygen:"Moles of oxygen: == (34.8*g)/(15.999*g*mol^-1)=2.18*mol34.8⋅g15.999⋅g⋅mol−1=2.18⋅mol.
Note (i) that here we have simply divided the atomic masses thru by the "atomic mass"atomic mass, and (ii), normally "% oxygen content"% oxygen content would not be measured. You would be given "% carbon content"% carbon content, and "% hydrogen content"% hydrogen content, and "% nitrogen content"% nitrogen content, and "% oxygen content"% oxygen content would be assessed by the balance.
And now we normalize the formula by dividing thru by the lowest molar ratio, that of oxygen to get the empirical formula:
O=(2.18*mol)/(2.18*mol)O=2.18⋅mol2.18⋅mol == 11
C=(4.35*mol)/(2.18*mol)C=4.35⋅mol2.18⋅mol == 22
H=(12.04*mol)/(2.18*mol)H=12.04⋅mol2.18⋅mol == 66
And thus our empirical formula is C_2H_6OC2H6O.
But the molecular formula is always a mulitple of the empirical formula:
i.e. "(empirical formula)"xxn="(molecular formula)"(empirical formula)×n=(molecular formula).
We KNOW the molecular mass because it has been measured in a separate experiment, and provided for us:
So nxx(2xx12.011+6xx1.00794+15.999)*g*mol^-1=46.0*g*mol^-1n×(2×12.011+6×1.00794+15.999)⋅g⋅mol−1=46.0⋅g⋅mol−1,
and we solve for nn.
Clearly n=1n=1, and here
"molecular formula "-=" empirical formula"=C_2H_6Omolecular formula ≡ empirical formula=C2H6O.
