VCE Chemistry Question Thread

[rg_123]

As there is a chloro substituent, you would go with 3-chloroprop-1-ene, otherwise the term would be "chloropropene", which appears ambigious (as the chloro group could be anywhere on the molecule).

(Feel free to correct me here )

Thanks for the quick reply insanipi! Some people were just arguing that the location of the double bond did not need to be there because the molecule only has three carbons and I was a little worried whether by putting the location of the double bond I would lose marks. But that makes sense, thanks again!

[j.wang]

Helloo

80.0 g of propan-1-ol was oxidised to propanoic acid using an acidified solution of K2Cr2O7. The propanoic acid obtained at the end of the reaction had a mass of 55.0 g. Calculate the percentage yield of this oxidation reaction.

I calculated the mol of  propan-1-ol (using n=m/M > 80/60=1.333) which equals the mol of propanoic acid by stoich (1.333). The mass of propanoic acid is 1.333*74=98.6666, so I got (55/98.666)*100= 55.7% for the percentage yield!

But the ans is 55.9% :/ does anyone know where I went wrong? thanks!

[sweetiepi]

Helloo

80.0 g of propan-1-ol was oxidised to propanoic acid using an acidified solution of K2Cr2O7. The propanoic acid obtained at the end of the reaction had a mass of 55.0 g. Calculate the percentage yield of this oxidation reaction.

I calculated the mol of  propan-1-ol (using n=m/M > 80/60=1.333) which equals the mol of propanoic acid by stoich (1.333). The mass of propanoic acid is 1.333*74=98.6666, so I got (55/98.666)*100= 55.7% for the percentage yield!

But the ans is 55.9% :/ does anyone know where I went wrong? thanks!

Hey there!

To get 55.9%, I found that I had to round off the number of moles from 4/3 (1.3333333) to 1.33. I feel that this is a rounding error, as carrying over the exact number gives me your answer of 55.7%, which in my opinion is more correct.

[j.wang]

Hey there!

To get 55.9%, I found that I had to round off the number of moles from 4/3 (1.3333333) to 1.33. I feel that this is a rounding error, as carrying over the exact number gives me your answer of 55.7%, which in my opinion is more correct.

Wow :O So incredible that you even figured out how the textbook got that answer! Thanks so much

[j.wang]

 Hi again Why is 3-Bromobutane and 3-chloropentane immiscible? They're both haloalkanes (which are slightly polar), and since like dissolves like, shouldn't they be miscible?

Thanks

Edit: A question about atom economy! In the attached pic, I worked out the atom economy of the preparation of ethlyene oxide (the product) by 44*2/ (28+14+142+74.1), then multiplied by 100, which equals 34.10%

But the answer is 32%! Where have I gone wrong :O Thanks again

[sweetiepi]

Hi again Why is 3-Bromobutane and 3-chloropentane immiscible? They're both haloalkanes (which are slightly polar), and since like dissolves like, shouldn't they be miscible?

Thanks

Hey again!

I'm not 100% on this, however, if you mean why aren't those two molecules aren't miscible in water, it's due to the lack of ability to form hydrogen bonds. This is due to the energy required to break the bonds in water are higher than the energy released when new bonds between the water and the haloalkane are formed.

As both these molecules have symmetry (as in the halogen group is on carbon 3), the molecule is considered overall non-polar; as the dipole between the carbon and the haloalkane is small, therefore the water molecules don't "mix" with the haloalkane.

(I could have some concepts mixed up here, anyone else who'd like to chip in, please do so! ^_^)

[MisterNeo]

Hi again Why is 3-Bromobutane and 3-chloropentane immiscible? They're both haloalkanes (which are slightly polar), and since like dissolves like, shouldn't they be miscible?

The polarity of halo alkanes decreases down the halogen group, and longer hydrocarbon chains increase the non-polar nature as more dispersion forces are present.

[tinagranger]

Hey, just wondering when describing trends of organic molecules, when you talk about BPs/viscosity/etc. increasing because of bonds, do you say there is an increase in the STRENGTH of dispersion forces, or an increase in the AMOUNT of dispersion forces?

[Shadowxo]

Hey, just wondering when describing trends of organic molecules, when you talk about BPs/viscosity/etc. increasing because of bonds, do you say there is an increase in the STRENGTH of dispersion forces, or an increase in the AMOUNT of dispersion forces?

You'd say an increase in the strength of dispersion forces. The dispersion forces get stronger or weaker, or are weak / strong

[tinagranger]

Ok thanks Do we need to know about colour changes in oxidation reactions? If so, could someone summarise what we need to know?

[VanillaRice]

Ok thanks Do we need to know about colour changes in oxidation reactions? If so, could someone summarise what we need to know?

If you're referring to the redox titrations, the most you would probably need to know is that some redox titrations don't require indicators, since one of the reactants will change colour on its own when it becomes oxidised/reduced. You won't need to remember the different colours for different substances in reduced/oxidised forms.

[gnaf]

Hi

What have I done wrong in 4a) in the photo?

The answer is 0.658M but I got 2.63M (mol of NaOH= 0.026 and mol of C2H2O4= 0.0526)!

And how do I got about working out 4b)? Is % m/m referring to the mass of C2H2O4 divided by the the mass of NaOH then multiplied by 100? Or is it something else?

[VanillaRice]

Hi

What have I done wrong in 4a) in the photo?

The answer is 0.658M but I got 2.63M (mol of NaOH= 0.026 and mol of C2H2O4= 0.0526)!

And how do I got about working out 4b)? Is % m/m referring to the mass of C2H2O4 divided by the the mass of NaOH then multiplied by 100? Or is it something else?

Let's first write the equation (remembering that oxalic acid is diprotic)
\(\ce{C2H2O4 + 2NaOH -> Na2C2O4 + 2H2O}\)
You calculated n(NaOH) = 0.026 mol
n(oxalic acid) = n(NaOH) x 1/2 = 0.026 x 1/2 = 0.01315 mol
c(oxalic acid) = n/V = 0.01315/0.02 = 0.6577M = 0.658M as per the answer
Most likely, you might've mixed up your molar ratio. Since oxalic acid is diprotic, every 1 mol of oxalic acid will react with 2 mol of NaOH. I find it helps to write out the equation in situations like this 

%m/m would be the mass of oxalic acid, divided by the total mass of the solution, multiplied by 100.

Hope this helps 

[gnaf]

Sure does helps, thanks VanillaRice (saving me in chem AND spesh)

Some questions from the same exercise:

To find the % m/m, is m(solution)= m(NaOH) + m(C2H2O4)? I got ((0.01315*90)/(0.01315*90 + 0.026*90)*100=53% but the ans is 5.92%

For oxalic acid + water, I know the products have to be salt + water, but how do you find the chemical formula of the salt?

Is knowing which acids are monoprotic e.g. ethanoic acid assumed knowledge?

I know that the average titre is the volume of solution delivered by the burette. Is this the standard solution?

Is it assumed knowledge that ethanamine is a weak base?

In the first photo, I got n(MnO4-)=0.125*0.01414=0.0012675, so n(CH3OH)=5/4 * n(MnO4-)=0.002. So C(CH3OH)=0.002/0.01= 0.2209M, but the answer is 0.201M

In the second photo, n(Cr2O72-)= 0.1*0.01798= 0.001798
So n(CH3CH2OH)= (3/2)* n(Cr2O72-)= 0.002697
C(CH3CH2OH in 20ml aliquot)= 0.002697/ 0.02= 0.13485
C(CH3CH2OH in 500ml)= (500/20)*0.13485=3.37125
Using C1V1=C2V2, 3.37125*0.500=C2 * 0.01
So C2= 168M (which doesn't even look right) :/ The correct answer is 6.74M

What have I done wrong? :O Thanks!

[Shadowxo]

Sure does helps, thanks VanillaRice (saving me in chem AND spesh)

Some questions from the same exercise:

To find the % m/m, is m(solution)= m(NaOH) + m(C2H2O4)? I got ((0.01315*90)/(0.01315*90 + 0.026*90)*100=53% but the ans is 5.92%

For oxalic acid + water, I know the products have to be salt + water, but how do you find the chemical formula of the salt?

Is knowing which acids are monoprotic e.g. ethanoic acid assumed knowledge?

I know that the average titre is the volume of solution delivered by the burette. Is this the standard solution?

Is it assumed knowledge that ethanamine is a weak base?

In the first photo, I got n(MnO4-)=0.125*0.01414=0.0012675, so n(CH3OH)=5/4 * n(MnO4-)=0.002. So C(CH3OH)=0.002/0.01= 0.2209M, but the answer is 0.201M

In the second photo, n(Cr2O72-)= 0.1*0.01798= 0.001798
So n(CH3CH2OH)= (3/2)* n(Cr2O72-)= 0.002697
C(CH3CH2OH in 20ml aliquot)= 0.002697/ 0.02= 0.13485
C(CH3CH2OH in 500ml)= (500/20)*0.13485=3.37125
Using C1V1=C2V2, 3.37125*0.500=C2 * 0.01
So C2= 168M (which doesn't even look right) :/ The correct answer is 6.74M

What have I done wrong? :O Thanks!

So for the first photo:
I think your answer is right. I got 0.221 as well.

Second photo:
First of all, for a question like this I'd leave the concentration until the end
n(CH₃CH₂OH) in aliquot= 0.002697 as you worked out
n(CH₃CH₂OH) in 500mL = 0.002697*500/20 = 0.067425
n(CH₃CH₂OH) in sample = n(CH₃CH₂OH) in 500mL=0.067425
c=n/V = 0.067425/0.01=6.7425

Now, with your error in that question, it's supposed to be
c(CH₃CH₂OH) in aliquot = c(CH₃CH₂OH) in 500mL as the aliquot is taken from the diluted 500mL, so it has the same concentration (but a lower number and volume)

I can't answer all of your questions but I believe you're expected to know the common acids and bases (eg NaOH, H2SO4, CH3COOH, HCl). Ethanoic acid falls under the "common acids" category
The solution delivered by the burette is usually the standard solution but not always, sometimes it's the unknown solution but they'll tell you.
I don't believe ethanamine is one of the ones they expect you to know (you can double check the study design if needed)

For %m/m, it's referring to the concentration in terms of weight, ie mass of oxalic acid divided by mass of solution. The NaOH isn't used to calculate anything

Edit: added answer for first question