## alphabeta one year ago I am asked to calculate the enthalpy change for the combustion of 1 mole of sucrose. What I know from before is this; Mass of sample of sucrose, m = 0.4385 g Heat capacity of the system, c = 10.114 kJ K^-1 I have also already calculated the amount of sucrose in moles to be 1.281*10^-3 mol (and I know from the mark scheme that this is correct). So, how to calculate the enthalpy change of combustion of one mole of sucrose?

1. alphabeta

Another thing; I thought I could use this formula; Delta(H) = mass * capacity * delta(temperature), and then divide this by the number of moles (1.281*10^-3). However, the mark scheme doesn't include the mass in its calculations. Why?

2. Zale101

$$−q=\Delta H∗n$$ <-- this is the enthalpy change of burned sucrose, hence the negative q. n is the moles of sucrose (you were given the mass of sucrose, now you can easily calculate the mols by molar mass to mols conversion factor) But, we are solving for the enthalpy change ΔH. $$\Delta H=\Large\frac{−q}{n}$$ We know that $$q=m∗c∗\Delta H$$. So, we plug what equals to q into the enthalpy change formula $$ΔH=\large \frac{−(m∗c∗\Delta T)}{n}$$

3. alphabeta

Thank you for your response Zale101! However, the mark scheme says that I should do it this way to calculate the enthalpy change of combustion; ΔHc = (c*ΔT)/n When I first calculated it, I did it in the same way as you describe. However, for some reason, mass is not included in the formula (as the mark scheme says), and I don't understand why :-o

4. cuanchi

In this case you have already the ΔH of a certain amount of glucose. It is correct calculate the moles in that grams. then is a simple conversion 1 mol of glucose x ( 10.114 kJ K^-1/1.281*10^-3 mol of glucose) = ????

5. Photon336

Q = mc(delta t) Delta H = -q * n So mass would be in your formula if you substitute mCDelta T into q you get Delta H = -(m*C*delta T)/(n)

6. alphabeta

As mentioned before, I have already used the formula "DeltaH = m*c*deltaT" and then divided it by the number of moles, but this doesn't give the correct answer, and I struggle to understand why :-/ (Btw, deltaT = 0.67) The answer should be -5.3*10^3 kJ mol^-1

7. cuanchi

That is the wrong formula to use to solve that problem!!!!

8. cuanchi

In this case you have already the ΔH of a certain amount of sucrose. you already correctly calculate the moles in that grams. then is a simple conversion 1 mol of sucrose x ( 10.114 kJ K^-1/1.281*10^-3 mol of sucrose) = ????