Question

Find empirical formula c 54.55%, H 9.09% O 36.6%

Answer 1

Hey !

Solution :

Given : Carbon = 54.55 %, Hydrogen = 9.09 %, Oxygen = 36.6 %

So For 100 g of the compound, 54.55 g are of Carbon, 9.09 g are of Hydrogen, 36.6 g are of Oxygen.

Therefore, Number of Moles = Given Mass / Molar Mass

Carbon :

Given Mass - 54.55 g
Molar Mass - 12 g

=> Number of moles = 54.55 g / 12 g = 4.54 moles

Hydrogen :

Given Mass = 9.09 g
Molar Mass = 1 g

=> Number of moles = 9.09 g / 1 g = 9.09 moles

Oxygen :

Given Mass = 36.6 g
Molar Mass = 16 g

=> Number of moles = 36.6 g / 16 g = 2.28 moles.

So we must divide the number of moles by the least number of moles obtained. On obtaining we get,

Least Number of moles = 2.28 moles

Hence on dividing we get,

Carbon = 4.54 mol / 2.28 mol = 1.99 which is equal to 2 molecules.

Hydrogen = 9.09 mol / 2.28 mol = 3.98 which is equal to 4 molecules 

Oxygen = 2.28 mol / 2.28 mol = 1 molecule.

Hence the Empirical Formula = C₂H₄O.

Hope the answer helped :-)