Chemistry qn
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Originally posted by d3sT1nY:
Oops press wrongly.
uhhh i think your calculation for the number of moles is wrong. for Al should be 27 instead of 13. same for Cr2O3 i think.
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as in the Molar mass (:
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As long as you get the idea of how to do it. I was using the periodic table from wikipedia.
Maybe I see wrongly. I didn't really think much into the numbers.
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2. Ca(OH)2 reacts with the CO2 from the combustion to produce CaCO3. So I think the 10.00g is CaCO3. Find the balanced equation then work backwards to find the number of mole of CO2 produced. Then with the number of mole of water produced, you should be able to find the empirical formula using the standard combustion balanced equation.
i'm halfway through the question. stuck after equating the no. of moles of CaCO3 to CO2 in combustion. after that anyone know how?
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A hydrocarbon was burnt completely and the products were bubbled into excess of Ca(OH)2 solution. the resulting mixture was filtered and the residue dried and weighed. The weight of the residue was found to be 10.00g. In a seperate experiment, the same amount of hydrocarbon was burnt and the mass of water produced was 2.25g. Determin the empirical formula of the hydrocarbon.
Ans:
No. of mol of CaCO3 = 10/(40+12+48)= 0.100 mol
CO2 + Ca(OH)2 ---> CaCO3 + H2O
No of mol of CO2 = 0.100
No of mol of water = 2.25/18= 0.125
Therefore no of mol of carbon = 0.100 and no of mol hydrogen in water = 0.250
taking whole number ratio, no of mol of carbon = 1 no of mol of hydrogen = 0.25/0.1 = 2.5
Empirical formula is C2H5
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It feels weird. The C2H5 doesn't conform to any of the hydrocarbon ratio I know.
It's not alkane nor alkene.
O.o
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There are only 2 (unstable, transient, reactive, intermediate) species with such a ratio of C to H; they are :
ethyl carbocation, C2H5+
ethyl carbanion, C2H5-
The only structural difference between them, is the presence or absence, of one lone pair on the valence shell of C.
For carbon (group IV),
3 bond pairs, 0 lone pair = +ve formal charge.
3 bond pairs, 1 lone pair = -ve formal charge.
Ethyl carbocation is a primary carbocation, and hence is very unstable, and cannot be easily formed even with an excellent leaving group (eg. halogen) in nucleophilic substitutions. Hence SN2 rather than SN1 will dominate. Free radical substitution reactions (eg. mixing an alkane with a halogen under UV radiation) will also favour the formation of tertiary and secondary carbocation intermediates, rather than primary carbocations.
A Grignard reagent (utilizing an electropositive metal) may be used to generate a carbanion intermediate. Another possible mechanism to generate the carbanion, would be seen in the past year 'A' level exam question here :
>>> When trichloroethanal is heated with aqueous sodium hydroxide, two products are formed in a single reaction pathway. One of the products is trichloromethane. Draw the mechanism for this pathway and hence identify the other product. <<<
Solution :
1) Hydroxide ion nucleophile attacks delta +ve carbonyl carbon.
2) Electron density shifts up; pi-bond pair becomes a lone pair on oxygen.
3) When electron density shifts back down (lone pair becomes pi-bond pair, reforming the carbonyl group), the trichloromethyl group leaves as an anion.
4) The trichloromethyl anion is capable of being a leaving group (ie. it is stable enough to exist on its own if only briefly), only because of the 3 highly electronegative and hence strongly electron-withdrawing Cl groups.
5) However, carbanions are still more unstable compared to alkoxide or alkonoate ions, hence the next step is
6) proton transfer (ie. acid-base reaction) from methanoic acid to the carbanion (ie. lone pair on carbanion becomes bond pair with proton (H+); bond pair between proton (H+) and oxygen becomes lone pair on oxygen), forming
7) trichloromethane and sodium methanoate (negative charge of HCOO- counterbalanced by Na+; note that these ions are aqueous meaning dissolved in water, having a hydration shell, ion-dipole interactions with polar water molecules).
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er so chim! don't understand leh.. haha
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Originally posted by chocolates-xed:
er so chim! don't understand leh.. haha
The above paragraph is 'A' level. So all he's trying to say is that it's possible to have C2H5 as an answer.
That's all.
But those are good knowledge. Awesome.
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Actually its is possible the alkane was C2H5. Its an empirical formula which is a formula of the atoms in the molecule in the simplest ratio. The alkane could have been C4H10 which is just butane.
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Yes, good point, Chemfreak022. Well done.
Heartening to see you're someone who appreciates Chemistry for it's own sake, and not just for the sake of exams and paper chase qualifications, d3sT1nY! Hopefully more students will share your healthy attitude (I'm aware your took your 'A' levels a long time ago).
chocolates-xed, you can look forward to studying 'A' levels. Trust me when I say, that's where the fun (really, no sacarsm intended) really begins!
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No the fun begins in university. haha!
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Originally posted by Chemfreak022:
No the fun begins in university. haha!
True. But since for my 'A' level tuition students, I start teaching them University level stuff, so to be precise, the fun begins in my tuition classes! haha!