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A compound consists of 65.45% c, 5.492% h and 29.06%o. it has a molar mass of approximately 110 amu. determine the empirical and molecular formula of the compound

User Timitry
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1 Answer

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The empirical and molecular formula of the compound is calculated as follows

the empirical formula calculation step

find the mole of each element= % composition of an element/ molar mass of an element
carbon(C) = 65.45/12=5.454 moles
Hydrogen(H) = 5.492/1= 5.492 moles
oxygen(O)= 29.06/ 16= 1.816 moles

find the mole ratio by diving each mole by the smallest number of mole(1.816 moles) and round off to the nearest integer

that is Carbon(C) = 5.454/1.816=3
Hydrogen(H)= 5.492/1.816= 3
oxygen(O) =1.816/1.816=1

the empirical formula is therefore= C3H3O


molecular formula calculation
110 amu is the total molar mass of C3H3O

therefore (C3H3O)n = 110
find for n

={(12 x3) + (1 x3)+ ( 16 x1)}n= 110

55n= 110
divide both side by 55
n=2
the molecular formula = (c3H3O)2 = C6H6O2



User Hetman
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