Chemistry 3/4 2013 Thread

[09Ti08]

The complete combustion of 24.0g of an organic compound produces 1.2mol of CO2 and 1.6mol of H2O. The compound was:
Propane
1-Propanol
Propanoic Acid
Methyl Ethanoate

Firstly, you should work out the ratio n(C):n(H) for each compound. They are: 3:8, 3:8, 3:6, and 3:6 respectively.

Then, you do some calculations:

n(C) in 24.0 g of the compound = n(CO2) produced
= 1.2 mol
n(H) in 24.0 g of the compound = 2 x n(H2O) produced
= 2 x 1.6
= 3.2 mol
Ratio n(C) : n(H)
= 1.2 : 3.2
= 1 : 2.67
= 3 : 8

I don't think this is too counterintuitive. At this stage, you only need to choose between propane and propan-1-ol. But how? Well, you need to plug in the molar mass of each compound to see if it matches your value for n(C) or n(H):

n(C3H8) in 24.0 g = m(C3H8) / M(C3H8)
= 24.0 g / 44.0 g mol-1
= 0.545 mol
n(C3H8O) in 24.0 g = m(C3H8O) / M(C3H8O)
= 24.0 g / 60.0 g mol-1
= 0.400 mol

I hope you can see that for propan-1-ol, we do have n(C)=0.4*3=0.12 mol or n(H)=0.4*8=3.2 mol. They are consistent with what we had earlier.

Volume of water: 440 ml
Initial mass of ethyne: 2.88g
Final mass of ethyne: 2.17g
Initial temperature of water: 23.5 C
Final temperature of water 42.6 C

During combustion of ethyne (C2H2), Use the change in temperature of the water to determine the amount of energy in KJ added during the heating.

What I did was E=4.184 x 440 x 19.1 = 35162 = 35.1KJ

but then the solution goes on to find the energy for one mole by dividing it by (0.71/26) giving 1.3 x 10^3 KJ. However isnt that for a mole of ethyne? and the question asks for energy in KJ added during the heating. ?
So shouldn't it be 35.1 KJ?

Yes, I agree with you. If the suggested answer was correct, its unit for energy (which is what the question asks) wouldn't have made sense.

[clıppy]

Thanks ti that clears some things up, but I'm still stuck here. What actually happens to get this ratio?

Ratio n(C) : n(H)
= 1.2 : 3.2
= 1 : 2.67
= 3 : 8

[KevinooBz]

Thanks ti that clears some things up, but I'm still stuck here. What actually happens to get this ratio?

You divide by 1.2 to get a 1:x ratio. Since you want a whole number for x, you then multiply by 3 and you get a 3:8 ratio. The aim is to get a ratio with whole numbers.

[clıppy]

You divide by 1.2 to get a 1:x ratio. Since you want a whole number for x, you then multiply by 3 and you get a 3:8 ratio. The aim is to get a ratio with whole numbers.

Thanks
I feel so silly know for not knowing something so simple 

[Jaswinder]

Yes, I agree with you. If the suggested answer was correct, its unit for energy (which is what the question asks) wouldn't have made sense.

huh? aren't both answers the same unit?

[09Ti08]

huh? aren't both answers the same unit?

I don't think that kJ and kJ/mol are the same thing...

[EspoirTron]

kJ is the amount of energy expresses in kilojoules; however, kJ/mol is the amount of energy per mol expressed in kJ and mole, this has a common association with the change in enthalpy (heat of reaction), delta H.

[Scooby]

Is it only the ions in the salt bridge that migrate to each electrode or does ion exchange occur between the two half cells as well (thatchemguy on Youtube seems to think so...)?

[Edward21]

Is it only the ions in the salt bridge that migrate to each electrode or does ion exchange occur between the two half cells as well (thatchemguy on Youtube seems to think so...)?

I remember in a practice exam I did, it mentioned that the existing ions in the solutions (spectator ions) of the respective half cells can migrate too. Ie. The sulfate ion would migrate towards the half-cell with the anode to restore electronegativity, together with the other anion in the salt bridge like nitrate if you're using (KNO3) 

[Holmes]

I remember in a practice exam I did, it mentioned that the existing ions in the solutions (spectator ions) of the respective half cells can migrate too. Ie. The sulfate ion would migrate towards the half-cell with the anode to restore electronegativity, together with the other anion in the salt bridge like nitrate if you're using (KNO3) 

So can a spectator ion travel from one half cell, through the salt bridge, to the other half cell?

[EspoirTron]

So can a spectator ion travel from one half cell, through the salt bridge, to the other half cell?

Interesting question, I like it. However, to my knowledge, no this cannot happen unless the polarity (not electrolysis) of the cell is reversed; that is, unless the reduction and oxidation reactions are reversed. The purpose of the salt bride is to distribute charges to maintain neutrality which is essential for the correct function of a galvanic cell. Each ion, for example, the positive Potassium ion will go towards the cathode to counteract the increasing negative charge and maintain neutrality. If, for example, the nitrate ion migrated towards the positive cathode, we would have an influx of negative charge (remember cathode is reduction so it is accumulating a negative charge), and this would disrupt out required neutrality, which in key would de a detriment to the effective operation of our galvanic cell.

[SocialRhubarb]

Yes, it will happen if you leave the cell on for long enough. But obviously it would take a while for ions to migrate up and across the salt bridge.

[Alwin]

Good one VCAA. *slow clapping*




congrats if you see it too

[clıppy]

I know right? Good one VCAA

Oh god i don't see it. What is it?

[Jeggz]

I know right? Good one VCAA

Oh god i don't see it. What is it?

LOL! Yeah I don't see it either :\
What is it, Oh Genius Alwin?