A 1738 g sample of a component of the light petroleum distil

A 1.738 g sample of a component of the light petroleum distillate called naphtha is found to yield 5.327 g CO2(g) and 2.544 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 .

Part C

Calculate the enthalpy of combustion, Hcomb, for C6H14. You\'ll first need to determine the balanced chemical equation for the combustion of C6H14.

Express your answer to four significant figures and include the appropriate units.

Solution

(a)complete combustion of C6H14 is given as

2C₆H_{14 (g)}+19O₂ -12CO₂ +14H₂O+energy

from the balanced chemical equation we can calculate standard heat of combustion for C₆H₁₄ is :

heat of formation of C₆H₁₄ =12(heat of formation of CO₂)+14(heat of formtion of water)+heat of combustion

standard heat of formation of CO₂=-94.0 kcal

standard heat of formation of water=-68 kcal

heat of formation of C6H14

C(g)-c(s)=-172kcal/mol

6C(g) - 6C(s) -1032kcal/mol

H(g) -1/2H(g) -52.09kcal/mol

14H(g) -7H2 - 729.26kcal/mol

adding above equation we get

6C+14H -C₆H₁₄    -1761.26

now heat of combustion is given as

-H =12(-94.0)+14(-68)+1761.26

=-2080+1761.26

=-318.74kcal/mol =-1333.6kj/m0l