Question

Combustion analysis is performed on 0.50 g of a hydrocarbon, and 1.47 g of CO2 and 0.902 g of H2O are produced. What is the empirical formula of the hydrocarbon? If the molar mass is 30.00 g/mol , what is the molecular formula?

Answer 1

1. The empirical formula of the hydrocarbon is CH₃

2. The molecular formula of the hydrocarbon is C₂H₆

How to determine the mass of Carbon

• Mass of CO₂ = 1.47 g
• Molar mass of CO₂ = 44 g/mol
• Molar of C = 12 g/mol
• Mass of C =?

Mass of C = (12 / 44) × 1.47

Mass of C = 0.4 g

How to determine the mass of H

• Mass of compound = 0.5 g
• Mass of C = 0.4 g
• Mass of H = ?

Mass of H = (mass of compound) – (mass of C)

Mass of H = 0.5 – 0.4

Mass of H =0.1 g

1. How to determine the empirical formula

• C = 0.4 g
• H = 0.1 g
• Empirical formula =?

Divide by their molar mass

C = 0.4 / 12 = 0.03

H = 0.1 / 1 = 0.1

Divide by the smallest

C = 0.03 / 0.03 = 1

H = 0.1 / 0.03 = 3

Thus, the empirical formula of the compound is CH₃

2. How to determine the molecular formula

• Empirical formula = CH₃
• Molar mass = 30 g/mol
• Molecular formula =?

Molecular formula = empirical × n = mass number

[CH₃]n = 30

[12 + (3×1)]n = 30

15n = 30

Divide both side by 15

n = 30 / 15

n = 2

Molecular formula = [CH₃]n

Molecular formula = [CH₃]₂

Molecular formula = C₂H₆

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