Heat of combustion in CF calculations

[Asian_Superstar]

hey,
I am just a bit confused, when working out CF or other stuff which uses molar enthalphy
for example, 1-propanol is used to calibrate a bomb calorimeter. 1.86g of 1-propanol is reacted with excess oxygen in the bomb calorimeter, causing the temperature of the surrounding water to increase from 22.4 degrees to 63.5 degrees.
Calculate the CF of the bomb calorimeter in KJ/C

now if you write the equation you get two moles of 1-propanol so i multipled wthe heat of combustion (2016) by 2 but thats wrong?
can somebody please explain because i remember somewhere, you had to multiply by the number of moles
thanks!

[jasoN-]

firstly find the number of moles of propanol you have
n = 1.86 / 60 = 0.031 mol
Given the delta H value (-2016) this means that 2016 kJ is released when 1 mol of propanol reacts with x mol of oxygen (doesn't matter)
so 0.031 mol of propanol releases x kJ
ratios: x / 2016 = 0.031 / 1
x = 0.031 x 2016 = 62.496 kJ

delta(T) = 63.5 - 22.4 = 40.1
CF = E / delta(T) = 62.496 / 40.1 = 1.56 kJ / °C (convert to J if you wish)

If you multiplied it by 2, it would turn out to be the same.
ie. 2016 x 2 kJ released when 2 mol (given the equation 2CH3CH2CH2OH + ....)
as they're both multiplied by 2 it will cancel out

i hope this is right

[Asian_Superstar]

yep you're spot on
i didnt use ratios i used the formulas like deltaH = E/n(1-propanol)
i think ratios is the best method to use
thanks!