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Grade 11IIT JEE Entrance Exam

Complete combustion of a sample of hydrocarbon gives a 0.66 g of CO2 and 0.36 g of H2O the empirical formula of the compound is

Profile image of Sudhanva G V
7 Years agoGrade 11
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Profile image of Arun
ApprovedApproved Tutor Answer7 Years ago
Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol 
Molar mass of H₂O = (1.0×2 + 16.0) g/mol = 18.0 g/mol 

n(C in X) = n(C in CO₂) = n(CO₂) = (0.66 g) / (44 g/mol) = 0.015 mol 
n(H in X) = n(H in H₂O) = 2 × n(H₂O) = 2 × (0.36 g) / (18.0 g/mol) = 0.04 mol 

Mole ratio C : H = 0.015 : 0.04 = 3 : 8 
Empirical formula = C₃H₈

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