Question

3. Combustion of 0.5gm of Hydrocarbon produced 1.515 gm CO2 and 0.77gm of H20. If the molecular mass of the compound is 58 a.m.u., determine its Molecular formula. (2014)​

Answer 1

The molecular formula is C₄H₁₂

Given:

Mass of hydrocarbon = 0.5 g

Mass of CO2 produced on combustion = 1.515 g

Mass of H20 produced on combustion = 0.77 g

The molar mass of the compound = 58 amu

Find:

Molecular formula

Solution:

• The equation of combustion is as follows
• CₐBₓ→ H₂O + CO₂
• CO₂→C
• 44g CO2 = 12 g C
• 1.515 = X
• X = 0.413g
• Number of moles of carbon = $\frac{0.413}{12}$ = 0.034
• H₂O→ 2H
• 18g H20 = 2g H2
• 0.77g =X
• x = 0.0856g
• Number of moles of hydrogen = $\frac{0.856}{1}$ =0.0856
• Empirical formula = C (0.034/0.034) H (0.0856/0.034)
•                                =  CH3
• 58/15 = 3.866 ≈ 4
• So molecular formula is C4H12
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