A 1.24 g sample of a hydrocarbon, when completely burned in...

A $1.24 \mathrm{g}$ sample of a hydrocarbon, when completely burned in an excess of $\mathrm{O}_{2}(\mathrm{g}),$ yields $4.04 \mathrm{g} \mathrm{CO}_{2}$ and $1.24 \mathrm{g} \mathrm{H}_{2} \mathrm{O} .$ Draw a plausible structural formula for the hydrocarbon molecule. [Hint: There is more than one possible arrangement of the C and H atoms.]

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Combustion Analysis

This is the method used to determine the elemental composition of a compound by burning it in oxygen. The masses of the combustion products (CO2 and H2O) directly relate to the amounts of carbon and hydrogen in the original compound, allowing the calculation of the empirical formula.

Empirical Formula Determination

The process involves converting the masses of CO2 and H2O to moles of carbon and hydrogen respectively, then determining the simplest whole-number ratio of the elements. This ratio forms the empirical formula of the hydrocarbon.

Stoichiometry

Stoichiometry applies the law of conservation of mass to relate the mass and mole quantities of reactants and products in a reaction. In combustion analysis, it is used to convert measured product masses into moles of constituent elements, serving as the basis for empirical formula calculations.

Hydrocarbon Structural Formulas

With the empirical formula determined, one can propose various structures by arranging the carbon and hydrogen atoms in different ways while obeying the rules of chemical bonding. This illustrates that the molecular structure can have isomeric forms, each representing a plausible arrangement of atoms in a hydrocarbon.

Transcript

Please give Ace some feedback