Butyric acid empirical formula


Asked by maham237 @ in Chemistry viewed by 264 People


The foul odor of rancid butter is due largely to butyric acid, a compound containing carbon, hydrogen, and oxygen. combustion analysis of a 4.30-g sample of butyric acid produced 8.59 g co2 and 3.52 g h2o. part a find the empirical formula for butyric acid. express your answer as a chemical formula.

Answered by maham237 @



Mass of CO₂ = 8.59 g

Molar mass of CO₂ = 44 g/mol

Moles of CO₂ = Mass of CO₂ / Molar mass of CO₂ = 8.59 g / 44 g/mol

= 0.1952 moles

Now there is 1 mole of C in 1 mole of CO₂

Therefore, moles of carbon = moles of CO₂ = 0.1952 moles

Mass of carbon = moles of C x the atomic weight of C

= 0.1952 moles x 12 g/mol

= 2.3427 g

Mass of H₂O = 3.52g

Molar mass of H₂O = 18g/mol

Moles of H₂O = 3.52g /18g/mol = 0.1956 moles

Now there is 2 moles of H in 1 mole of H₂O

Moles of H = 0.1956 moles X 2 = 0.3912 moles

Mass of H = 0.3912 moles X atomic mass of H

= 0.3912 moles X 1 g/mol = 0.3912 g

Mass of butyric acid = 4.30g

Mass of O = mass of butyric acid - (mass of C + mass of H)


= 4.30 g - 2.3427 g - 0.3912 g = 1.5661 g

Moles of O = mass of O / atomic mass of O = 1.5661 g / 16 g/mol

= 0.0978 moles

Divide by smallest mole to get the molar ratio

C = 0.1952 moles/0.0978 moles = 2

H = 0.3912 moles/0.0978 moles = 4

O = 0.0978 moles / 0.0978 moles = 1

Empirical formula of butric acid is C₂H₄O


Molecular formula of butric acid is C₄H₈O₂



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