A 1.678 g sample of a component of the light petroleum distillate called naphtha is found to yield 5.143 g CO2 (g) and 2.456 g H2O (l) on complete combustion. This particular compound is also found to be an alkane with one methyl group attached to a longer carbon chain and to have a molecular formula twice its empirical formula. The compound also has the following properties: melting point of -154 C , boiling point of 60.3 C , density of 0.6532 g/mL at 20 C , specific heat of 2.25 J/(g*C), and -204.6 kJ/mol Use the masses of carbon dioxide, CO2, and water, H2O , to determine the empirical formula of the alkane component. Express your answer as a chemical formula?

At first, based on the given information, one proceeds to compute the moles of carbon that the alkane has by considering the yielded amount of carbon dioxide as follows:

[tex]5.143gCO_2*\frac{1molCO_2}{44gCO_2}*\frac{1molC}{1molCO_2}=0.1169molC[/tex]

Next, the moles of hydrogen coming from the yielded water as:

[tex]2.456gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{2molH}{1molH_2O} =0.2729molH[/tex]

Now, dividing by the carbon's moles, one computes the empirical formula as follows:

[tex]C=\frac{0.1169}{0.1169}=1\\H=\frac{0.2729}{0.1169}=2.33[/tex]

The closest whole-numbered factor, by multiplying by 3 is:

[tex]C_3H_7[/tex]

So it is its empirical formula which allows us to determine the molecular formula if needed which is:

[tex]C_6H_{14}[/tex]

As it is twice the empirical formula.

Best regards.