Question

An organic compound has carbon 68.84%, oxygen 26.24% and hydrogen 4.92% (At. mass: C=12, O=16, H=1). The empirical formula is

Answer 1

Answer:

Mass percent of  oxygen atom = 26.24%  

Relative moles of atom of oxygen = 26.24/16 = 1.64

Mass percent of  hydrogen atom = 4.92%  .

Relative moles of atom of   hydrogen atom 4.92 / 1 = 4.92

To find the empirical formula one need to find the simple molar ratio of both atoms for which Relative moles of both atom is divided by the smallest value of Relative moles of atom of oxygen atom  

Simple molar ratio of oxygen atom = 1.64/1.64 = 1

Simple molar ratio of hydrogen atom =  4.92/1.64 = 3

Hence, empirical formula  for given compound according to the question  H₃O

Explanation:

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