As with all with these problems, we assume a mass of 100*g100⋅g, and we address the elemental makeup by dividing thru by the "atomic mass"atomic mass of each component.
"Moles of carbon"=(64.65*g)/(12.011*g*mol^-1)=5.38*molMoles of carbon=64.65⋅g12.011⋅g⋅mol−1=5.38⋅mol.
"Moles of hydrogen"=(13.52*g)/(1.00794*g*mol^-1)=13.41*molMoles of hydrogen=13.52⋅g1.00794⋅g⋅mol−1=13.41⋅mol.
"Moles of oxygen"=(21.62*g)/(15.999*g*mol^-1)=1.351*molMoles of oxygen=21.62⋅g15.999⋅g⋅mol−1=1.351⋅mol.
They are spoon-feeding you a bit here, in that combustion analysis gives %C,H,N%C,H,N. %O%O would normally NOT be measured, and determined by the difference (100-%C-%H-%N)%(100−%C−%H−%N)%. And now we divide thru by the SMALLEST MOLAR QUANTITY, that of oxygen to give C,H,OC,H,O
C:(5.38*mol)/(1.351*mol)=3.98~=4C:5.38⋅mol1.351⋅mol=3.98≅4
H:(13.41*mol)/(1.351*mol)=9.93~=10H:13.41⋅mol1.351⋅mol=9.93≅10
O:(1.351*mol)/(1.351*mol)=1O:1.351⋅mol1.351⋅mol=1
And thus an "empirical formula"empirical formula, the simplest whole number defining constituent atoms in a species of C_4H_10OC4H10O.
