Answer:
[tex]C_4H_5[/tex]
Explanation:
Hello!
In this case, since the combustion analysis provides us the yielded mass of both water and carbon dioxide, as the sources of hydrogen and carbon in the hydrocarbon, we are able to compute the moles of each via the following mole-mass relationships:
[tex]n_H=1.36gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.151molH\\\\n_C=5.22gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.119molC[/tex]
Now, we divide those moles by the fewest ones (those of carbon) in order to compute their subscripts in the empirical formula:
[tex]C: \frac{0.119}{0.119} = 1 \\\\H:\frac{0.151}{0.119} = 1.27[/tex]
However we need whole numbers, that is why we multiply each subscript by 4 to get 4 for carbon and 5.08 ≅ 5 for hydrogen:
[tex]C_4H_5[/tex]
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