4. Complete combustion of 1,00 g of the hydrocarbon pagodane gives 0.0768 moles carbondioxide, and therefore contained 0.0768 moles of carbon. What is the empirical formulaof pagodane? Recall the combustion of a generic hydrocarbon looks like this:CxHy(g) + O2(g) → CO2 (g) + H2O (8)a. CHb. CHa C₂H;4 CHA

4 Complete combustion of 100 g of the hydrocarbon pagodane gives 00768 moles carbondioxide and therefore contained 00768 moles of carbon What is the empirical f class=

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Answer:

CH. Option A is correct

Explanations:

Given the reaction for the complete combustion of pagodane expressed as:

[tex]C_xH_y(g)+O_2(g)\rightarrow CO_2(g)+H_2O[/tex]

In order to determine the empirical formula of the hydrocarbon, we must know the moles of carbon and hydrogen produced in the given reaction

Given the following parameter:

Moles of carbon = 0.0768 moles

Determine the moles of hydrogen produced

Determine the mass of carbon using moles * molar mass

[tex]\begin{gathered} mass\text{ of C=0.0768}\times12 \\ mass\text{ of C=0.9216grams} \end{gathered}[/tex]

Mass of Hygrogen = 1.0g - 0.9216g

Mass of Hydrogen = 0.0784grams

Determine the moles of hydrogen

Moles of hydrogen = 0.0784 /1 = 0.0784moles

Divide the moles by the least molar amount to normalize

Carbon = 0.0768/0.0768 = 1

Hydrogen = 0.0784/0.0768 = 1.021

Since the molar amount is in the ratio 1:1, hence the empirical formula of pagodane is CH

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