A 1.534g sample of a component of the light petroleum distillate called naphtha

Question

A 1.534g sample of a component of the light petroleum distillate called naphtha is found to yield 4.702g CO2(g) and 2.245g 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 degree C, boiling point of 60.3 degree C, density of 0.6532 g/mL at 20 degree C, specific heat of 2.25 J/(g . degree C), and 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.

Explanation / Answer

4.702 g CO2 x (12.01g C/44.01g CO2) = 1.283 g C

1.283 g C / 12.01 = 0.107 mol C atoms

2.245g H2O x (2.02g H/18.01g H2O) = 0.252 g H

0.252 g H / 1.008 =.0.250 mol H atoms

for H: 0.250 /0.107 = 2.33 mole

for C: 0.107/0.107 = 1

C1H2.33

C3H7=empirical formula

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