A compound on analysis was found to contain `C = 34.6 %, H = 3.85 % and O = 61.55%` .Calculate its empirical formula.

Explanation

Step I : Calculation of simplest whole number ratios of the elements
`{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("C",34.6,12,(34.6)/(12)=2.88,(2.88)/(2.88)=1,3),("H",3.85,1,(3.85)/(1)=3.85,(3.85)/(2.88)=1.337or 4//3,4),("O",61.55,16,(61.55)/(16)=3.85,(3.85)/(2.88)=1.337or 4//3,4):}`
The simplest whole number ratios of the different elements are : `C : H : O : : 3 : 4: 4`
Step II. Writing the empirical formula of the compound.
The empirical formula of the compound `= C_(3)H_(4)O_(4)`.