A 1.515 g sample of a component of the light petroleum distillate called naphtha

Question

A 1.515 g sample of a component of the light petroleum distillate called naphtha is found to yield 4.644 g CO2(g) and 2.218 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/(gC) , and

Hf=204.6kJ/mol .

QUESTION: Use the masses of carbon dioxide, CO2, and water, H2O, to determine the empirical formula of the alkane component.

No clue how to solve this problem.

Explanation / Answer

we know that

moles = mass / molar mass

so

moles of C02 = 4.644 / 44 = 0.105545

now

moles of C = moles of C02 = 0.105545

now consider H20

moles of H20 = 2.218 / 18 = 0.123222

now

moles of H = 2 x moles of H20

so

moles of H = 2 x 0.123222 = 0.246444

now

consider the ratio

C:H = 0.105545 : 0.246444

C :H = 1 : 2.33

C : H = 3 : 7

so

the empirical formula of the alkane component is C3H7

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