# 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?

• When you convert °C to K then only add 273.15 when you have absolute values but not a difference (like in this case), because they have the same scale. Sep 5, 2015 at 15:52

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.

• Ah so you find the amount of energy transferred to the water (20.1 kJ) and divide this by the number of moles of fuel burnt (0.013 mol) to obtain the change in enthalpy of the exothermic reaction (1,510 kJ)! Thanks a lot! One thing that is throwing me, however, is that the paper tells me that the published enthalpy changes of combustion for hexane is -4163 kJ/mol, so how could we get -1,510 kJ from burning 0.013 mol? @AstronAUT Sep 6, 2015 at 13:11
• That is indeed a good question. If this was a real experiment i'd say it could be because of some energy being lost into the environment. That would mean your efficiency factor is around: 1,510 kJ / 4163 kJ = 36 %. Other factors that could have lead to an error are incomplete burning of the haxane or the fact that you are extrapolating your results from a rather small test amount of hexane (1.15 g), therefore also small errors have an bigger impact. Sep 6, 2015 at 13:55
• Ahh okay, I reckon it was simply down to not measuring the mass of the hexane immediately after burning and the small amount burnt. Thanks a lot for your help! Sep 6, 2015 at 14:08