Calculating change in enthalpy of combustion of 1.12g of hexane

I will calculate the change in enthalpy of combustion of 1.12g of hexane.


**1)** Calculate the energy transferred to 200g of water using the equation: *J = mass of water in grams (**200g**) x specific heat capacity of water (**4.18 J/gK**) x temperature increase in Kelvin (24C = **297.15K**)*. This results in the amount of energy transferred to the water of a total of 248,417.4J or **248.4kJ**.


**2)** Calculate number of moles of hexane burnt using *n=m/M*. Weight of hexane before experiment = 222.07g. Weight of hexane after experiment = 220.92g. Therefore **1.15g** of hexane was burnt during the experiment. 1.15g of hexane = **0.013mol** of hexane.


**3)** Find enthalpy change of reaction. Combustion of 1mol of hexane produces (by my calculations) **-4194 kJ/mol.** Therefore, 0.013mol will produce (0.013/-4194) **-3.1x10-6 kJ/mol**.


So this is the procedure that I used to find out the answer! Was it correct?
Because, apparently, you can use the values of steps 1 and 2 to find out the enthalpy change?




I can't figure out exactly what you are trying to calculate, but if you want to know how much energy is created when burning 1.15 g of hexane by measuring an temperature increase of water, then this is the right way:


1.) 0.2 kg Water was icreased by 24 °C:
U = 0.2 kg \* 4187 J/(kgK) \* 24 K = 20,100 J = 20.100 kJ 


2.) n(Hexane) = 1.15 g / (86.18 g/mol) = 0.013 mol 


3.) 1.15 g caused the temperature increase of 24 °C:
20.100 kJ / 0.013 mol = 1,510 kJ/mol


So you will get 1.510 MJ out of burning 1 mol (86.18 g) hexane. 


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