Question

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?

Answer 1

Answer:
`C_3H_7`

Step-by-step explanation:

Hello,

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:

`5.143gCO_2*(1molCO_2)/(44gCO_2)*(1molC)/(1molCO_2)=0.1169molC`

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

`2.456gH_2O*(1molH_2O)/(18gH_2O)*(2molH)/(1molH_2O) =0.2729molH`

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

`C=(0.1169)/(0.1169)=1\\H=(0.2729)/(0.1169)=2.33`

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

`C_3H_7`

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

`C_6H_(14)`

As it is twice the empirical formula.

Best regards.