brightsky's Chem Thread

[Aurelian]

I'm going to be VERY picky here...
Your relation with delta G and electrode potentials is slightly inconsistent. Either the Gibbs energy change is the standard change, or the electrode potential is not the standard electrode potential. I don't think we can have a mix

Haha actually I fixed that as soon as I posted, if you'll notice =)

[brightsky]

i think i'm going delirious. question 10 vcaa 2013 sample exam. reagent A is obviously something with OH-(aq), like NaOH(aq). but how does compound 1, the di-alcohol, become the molecule they've given after oxidation in the presence of acidified dichromate solution? where did the extra carbon atoms come from?

[lzxnl]

You have fallen victim to a disorder known as vcaaism, in which you start suspecting everything around you of being wrong after being subjected to VCAA exams.

No seriously, you're right, the question is stupid, and VCAA deserves to have their own examiners sit chemistry exams and fail. It's a reflection of the declining of the once proud chemistry curriculum...

[brightsky]

1. when asked to give the semistructural formula for ethene, do we write CH2CH2 or CH2=CH2? is the double bond symbol '=' necessary? if so, must we write H2C=CH2, since technically it is a C=C double bond?
2. will we be expected to know the state symbol for organic compounds? are they usually all just (l)? for instance, what are the state symbols for the species involved in the production of ethyl ethanoate? all (l)? is there a way to 'work out' the symbols?

thanks

[Aurelian]

1. when asked to give the semistructural formula for ethene, do we write CH2CH2 or CH2=CH2? is the double bond symbol '=' necessary? if so, must we write H2C=CH2, since technically it is a C=C double bond?
2. will we be expected to know the state symbol for organic compounds? are they usually all just (l)? for instance, what are the state symbols for the species involved in the production of ethyl ethanoate?

thanks

1) I suspect that CH₂CH₂ would be accepted, but I would write in the double bond to be safe. Examiners won't care whether you write CH₂=CH₂ or H₂C=CH₂, though personally I think the latter is good chemical practice.

2) Yes you will need to know the states of all organic reactions in the course, and no they are not usually all just (l). There isn't really a way to "figure out" states for these reactions, but some common-sense will come in handy (you're never going to be dealing with solid ethanol, for instance).

[brightsky]

okay but what if you had something more complicated, like some ester or an alkene with multiple C=C double bonds? H2C=CH2 is a bit unwieldy...

with regard to state symbols, for combustion reactions, is it safe to assume that the H2O produced will always be in the gaseous state? or will it sometimes be in the liquid state as well?

[Aurelian]

okay but what if you had something more complicated, like some ester or an alkene with multiple C=C double bonds? H2C=CH2 is a bit unwieldy...

with regard to state symbols, for combustion reactions, is it safe to assume that the H2O produced will always be in the gaseous state? or will it sometimes be in the liquid state as well?

1. For an ester you would be safe to omit the double bonds. For ethyl butanoate, for example, the following would be satisfactory: CH₃CH₂CH₂COOCH₂CH₃. I'm fairly sure that you could also explicitly draw the carbonyl oxygen with a double bond pointing vertically above the semi-structural formula (if you know what I mean). From memory, VCAA really just considers "semi-structural formula" to be anywhere in between full structural formula and molecular formula. They're fairly lenient so long as you respect those two parameters (probably double check this with Ms. F though).

2. VCAA accept both H₂O (l) and H₂O (g) as the products for combustion reactions (according to past exam reports).

[brightsky]

I see, I see. Cheers, Aurelian!

Another concern: I have a sneaky suspicion that there is something wrong with SA Q6 d ii) of Lisachem 2013 Trial Exam. (The question's a bit involved so hopefully people have access to the exam.) In the solutions, they write n(Na2S2O3) = 0.00625 mol and n(HCl) = 0.010 mol. Then they say that Na2S2O3 is in excess and HCl is the limiting reagent. Unless there is a malicious demon who is employing all of his/her energies towards deceiving me as to the magnitude of positive real numbers, 0.00625 < 0.010...How can HCl be the limiting reagent?

Thanks!

[SocialRhubarb]

I think they react in a two to one ratio.

[brightsky]

Ahhh of course. Thanks SocialRhubarb!

[brightsky]

What would be the mass of a polymer made from 5 mol of glucose monomers?

okay so the solutions say that when you condense 5 glucose molecules, you form 4 ether links and lose 4 water molecules. therefore, when you condense 5 mol of glucose molecules, you form 4 mol ether links and lose 4 mol water molecules. therefore, the mass is (5*180) - (4*18) = 828. but my question is, how is it that when you condense 5 mol glucose molecules, you lose 4 mol water? surely the number of water molecules you lose is 5*N_A - 1?

[lzxnl]

Yeah, there was an argument about that question in my chemistry classroom.
The question means to test you on how much water is lost when five molecules of glucose react, but literally, five moles of glucose means 5*6.02*10^23 molecules of glucose. The question sucks.

[brightsky]

so what would you put as an answer? surely if that were the case, then the real answer would be closer to 810?

[lzxnl]

I would go for interpreting five moles as five molecules, although I'm really not sure.

[brightsky]

8.2 g of C6H10 was treated with 60 g of Br2 dissolved in a suitable solvent. On completion of the reaction, 16.2 g of HBr had been evolved and 12 g of Br2 remained in excess. Calculate the number of bromine atoms in each molecule of C6H10 which are involved in addition.

now it says bromine atoms (Br) not bromine molecules (Br2). so you calculate n(Br) total by multiplying by 2 the n(Br2) you originally had, subtract n(Br) involved in substitution (which is the same as n(HBr)) and then subtract n(Br) excess which is basically 2 times n(Br2) remaining. this calculates to n(Br) addition = 0.40 mol. then you divide this number by 0.10 mol (the amount of C6H10 present originally for reaction) to get 4 bromine ATOMS involved in substitution, which makes sense since C6H10 has 2 C=C double bonds. but STAV says that the answer is 2. where did I go wrong?

for those who have done this exam, it's 2013 STAV short answer Q14 b.