AS with all these problems, we assume a 100*g100⋅g mass of unknown compound and work out the molar quantities of each element....
"Moles of carbon"=(63.15*g)/(12.011*g*mol^-1)=5.26*molMoles of carbon=63.15⋅g12.011⋅g⋅mol−1=5.26⋅mol.
"Moles of hydrogen"=(5.30*g)/(1.00794*g*mol^-1)=5.26*molMoles of hydrogen=5.30⋅g1.00794⋅g⋅mol−1=5.26⋅mol.
"Moles of oxygen"=(31.55*g)/(15.999*g*mol^-1)=1.97*molMoles of oxygen=31.55⋅g15.999⋅g⋅mol−1=1.97⋅mol.
We divide thru by the smallest molar quantity to give a trial empirical formula of........
C_(2.67)H_(2.67)O_()C2.67H2.67O, but if we multiply this by 3 we get.........."whole numbers as required. "whole numbers as required.
......................C_8H_8O_3
