Login

Welcome, Guest. Please login or register.

March 28, 2024, 08:51:56 pm

Author Topic: VCE Chemistry Question Thread  (Read 2313007 times)  Share 

0 Members and 8 Guests are viewing this topic.

Billuminati

  • Science Games: Gold
  • Forum Obsessive
  • ***
  • Posts: 300
  • Respect: +127
Re: VCE Chemistry Question Thread
« Reply #9180 on: August 20, 2021, 04:31:24 pm »
+1
Hi,

for mass spec if they ask the fragmentation for the parent ion, do I still have to write + at the back of the ion I give?

You absolutely have to include the + charge ie for a CH3 fragment, you should write [CH3]+ in order for it to be marked correct by VCAA
VCE 2016-2018

2017: Biology [38], Further Maths [44]

2018: Methods [37], French [38], Chem [40], English [44]

UMAT: 56/43/80, 57th percentile (LLLLOOOOOOOOOLLLLLLLL)

ATAR: 98.1

2019-2021: Bachelor of Biomedical Science at Monash (Scholars), minoring in Chemistry

GAMSAT September 2021: 65/67/86, 76 overall (98th percentile)

2022: Chilling

2023+: Transfer to teaching degree

amyzzwq

  • Trailblazer
  • *
  • Posts: 34
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9181 on: August 20, 2021, 06:26:19 pm »
0
You absolutely have to include the + charge ie for a CH3 fragment, you should write [CH3]+ in order for it to be marked correct by VCAA

Not sure if I was clear in my previous question. For eg, if they give you a mass spec of the fragmentation of CH3COOH and they ask you the fragment ion of 60m/z and that would be CH3COOH itself, do I still have to write CH3COOH+ or I can just leave it as CH3COOH?

Billuminati

  • Science Games: Gold
  • Forum Obsessive
  • ***
  • Posts: 300
  • Respect: +127
Re: VCE Chemistry Question Thread
« Reply #9182 on: August 20, 2021, 09:07:33 pm »
+1
Not sure if I was clear in my previous question. For eg, if they give you a mass spec of the fragmentation of CH3COOH and they ask you the fragment ion of 60m/z and that would be CH3COOH itself, do I still have to write CH3COOH+ or I can just leave it as CH3COOH?

You still need to write it in square brackets and the + charge even when it's the full molecule (this is your parent ion)
VCE 2016-2018

2017: Biology [38], Further Maths [44]

2018: Methods [37], French [38], Chem [40], English [44]

UMAT: 56/43/80, 57th percentile (LLLLOOOOOOOOOLLLLLLLL)

ATAR: 98.1

2019-2021: Bachelor of Biomedical Science at Monash (Scholars), minoring in Chemistry

GAMSAT September 2021: 65/67/86, 76 overall (98th percentile)

2022: Chilling

2023+: Transfer to teaching degree

Jimmonash1991

  • Adventurer
  • *
  • Posts: 19
  • Respect: +2
Re: VCE Chemistry Question Thread
« Reply #9183 on: August 22, 2021, 11:38:13 am »
0
G'day,

I have just started at a tutoring agency, and have been given some chemistry students. What ideally should chemistry students be up to at this point? I am trying to gauge what chapters I should focus on immediately, so I have notes to provide my students. The agency isn't great letting me know what the students are up to!

Cheers,

Jim

Harrycc3000

  • Forum Regular
  • **
  • Posts: 53
  • Respect: +8
Re: VCE Chemistry Question Thread
« Reply #9184 on: August 22, 2021, 02:37:58 pm »
0
Hi guys,
me and my friends were looking at the molecule ethylene glycol and the nmr spectrum had two peaks but what we weren't sure of was why the two peaks were both singlets. Wouldn't the adjacent middle carbon atoms be triplets considering that their neighbouring carbon atoms have 2 protons directly attached? I'll attach the structure of ethylene glycol and its nmr spectrum below. Cheers.
VCE 2020: Biology [50]
VCE 2021: Mathematical Methods [44], Specialist Mathematics [43], Psychology [45], Chemistry [45], English Language [49]
ATAR: 99.90
UCAT: 3240 (99th)
2022-2024: UoM Bachelor of Science Majoring in Data Science (Planning on going into DDS)

Harrycc3000

  • Forum Regular
  • **
  • Posts: 53
  • Respect: +8
Re: VCE Chemistry Question Thread
« Reply #9185 on: August 22, 2021, 02:39:33 pm »
0
G'day,

I have just started at a tutoring agency, and have been given some chemistry students. What ideally should chemistry students be up to at this point? I am trying to gauge what chapters I should focus on immediately, so I have notes to provide my students. The agency isn't great letting me know what the students are up to!

Cheers,

Jim
We're up to around spectroscopy starting onto food chemistry
VCE 2020: Biology [50]
VCE 2021: Mathematical Methods [44], Specialist Mathematics [43], Psychology [45], Chemistry [45], English Language [49]
ATAR: 99.90
UCAT: 3240 (99th)
2022-2024: UoM Bachelor of Science Majoring in Data Science (Planning on going into DDS)

Jimmonash1991

  • Adventurer
  • *
  • Posts: 19
  • Respect: +2
Re: VCE Chemistry Question Thread
« Reply #9186 on: August 22, 2021, 03:36:08 pm »
+8
Hi guys,
me and my friends were looking at the molecule ethylene glycol and the nmr spectrum had two peaks but what we weren't sure of was why the two peaks were both singlets. Wouldn't the adjacent middle carbon atoms be triplets considering that their neighbouring carbon atoms have 2 protons directly attached? I'll attach the structure of ethylene glycol and its nmr spectrum below. Cheers.


G'day Harry,

Since you have scratched my back, I am going to scratch yours.

Look at the Ethylene Glycol molecule through the centre between the methylene (R-CH2-R) groups. You can fold it in half, which means both sides of this imaginary line are equivalent. Whenever you look at the molecular formula provided, and the number of H-NMR signals, if the latter is less than the number of carbons, then it is a rule of thumb you are probably dealing with a molecule that has a degree of symmetry.

Then we know we only have to look at one side of the molecule. The hydroxyl (R-OH) will always give a singlet in the H-NMR spectrum. Thus, the protons on the R-CH2-R isn't actually near any neighbouring proton environments, so these two equivalent R-CH2-R environments give a singlet.

Now to the chemical shift. We have now realised we have two signals, and the R-CH2-R should be taller than for the hydroxyl group, because the two equivalent proton environments have four protons, whereas the hydroxyl groups only have two. This gives an integration of 2:1. These methylene protons also more deshielded, due to the electronegative oxygen to which the same carbon is attached. This "deshielding" means it withdraws electron density.

Given the normal chemical shift for an R-CH2-R is about 1.3 ppm, and being attached to the same carbon that is attached to an oxygen, will cause the chemical signal to be shifted to about 3.7 ppm. The additional 2.4 ppm added to chemical shift applies to any proton environment that neighbors an electronegative oxygen atom.

The hydroxyl proton signal is usually around 2-5 ppm, however it depends on the solution and reagents present, particularly if there is hydrogen bonding occuring, or whether the pH is neutral or not.

Hope this helps,

Jim
« Last Edit: August 22, 2021, 03:45:19 pm by Jimmonash1991 »

Harrycc3000

  • Forum Regular
  • **
  • Posts: 53
  • Respect: +8
Re: VCE Chemistry Question Thread
« Reply #9187 on: August 22, 2021, 05:18:04 pm »
0
G'day Harry,

Since you have scratched my back, I am going to scratch yours.

Look at the Ethylene Glycol molecule through the centre between the methylene (R-CH2-R) groups. You can fold it in half, which means both sides of this imaginary line are equivalent. Whenever you look at the molecular formula provided, and the number of H-NMR signals, if the latter is less than the number of carbons, then it is a rule of thumb you are probably dealing with a molecule that has a degree of symmetry.

Then we know we only have to look at one side of the molecule. The hydroxyl (R-OH) will always give a singlet in the H-NMR spectrum. Thus, the protons on the R-CH2-R isn't actually near any neighbouring proton environments, so these two equivalent R-CH2-R environments give a singlet.

Now to the chemical shift. We have now realised we have two signals, and the R-CH2-R should be taller than for the hydroxyl group, because the two equivalent proton environments have four protons, whereas the hydroxyl groups only have two. This gives an integration of 2:1. These methylene protons also more deshielded, due to the electronegative oxygen to which the same carbon is attached. This "deshielding" means it withdraws electron density.

Given the normal chemical shift for an R-CH2-R is about 1.3 ppm, and being attached to the same carbon that is attached to an oxygen, will cause the chemical signal to be shifted to about 3.7 ppm. The additional 2.4 ppm added to chemical shift applies to any proton environment that neighbors an electronegative oxygen atom.

The hydroxyl proton signal is usually around 2-5 ppm, however it depends on the solution and reagents present, particularly if there is hydrogen bonding occuring, or whether the pH is neutral or not.

Hope this helps,

Jim
Hi Jim,
Thanks for the very detailed answer! Actually helped a lot. I just had one question about the symmetry part of your answer and was just wondering why if the molecule was symmetrical that you just disregard the other half, or why you don't account for the hydrogens on the ch2 group next to it.
Thank you!
VCE 2020: Biology [50]
VCE 2021: Mathematical Methods [44], Specialist Mathematics [43], Psychology [45], Chemistry [45], English Language [49]
ATAR: 99.90
UCAT: 3240 (99th)
2022-2024: UoM Bachelor of Science Majoring in Data Science (Planning on going into DDS)

Billuminati

  • Science Games: Gold
  • Forum Obsessive
  • ***
  • Posts: 300
  • Respect: +127
Re: VCE Chemistry Question Thread
« Reply #9188 on: August 22, 2021, 05:27:28 pm »
+6
Hi Jim,
Thanks for the very detailed answer! Actually helped a lot. I just had one question about the symmetry part of your answer and was just wondering why if the molecule was symmetrical that you just disregard the other half, or why you don't account for the hydrogens on the ch2 group next to it.
Thank you!

You ignore the other half in a symmetrical molecule as the signals in those environments will integrate as a single signal by overlapping with the half you already accounted for. The CH2 can't couple with the other CH2 as protons in identical environments can't couple to each other
VCE 2016-2018

2017: Biology [38], Further Maths [44]

2018: Methods [37], French [38], Chem [40], English [44]

UMAT: 56/43/80, 57th percentile (LLLLOOOOOOOOOLLLLLLLL)

ATAR: 98.1

2019-2021: Bachelor of Biomedical Science at Monash (Scholars), minoring in Chemistry

GAMSAT September 2021: 65/67/86, 76 overall (98th percentile)

2022: Chilling

2023+: Transfer to teaching degree

saransh

  • Fresh Poster
  • *
  • Posts: 3
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9189 on: August 25, 2021, 02:13:46 pm »
+2
Hey everyone, just had a quick one about some titration theory.
"A student conducts an experiment to determine the ethanoic acid concentration in a commercial
brand of vinegar. An outline of her procedure and her measurements are provided below.
230.0 mL deionised water added to a 250 mL volumetric flask
20.0 mL of vinegar added to the flask
20.00 mL aliquots added to flasks and titrated against 0.150 M NaOH."

The solutions said that an error in this experimental design was that "20 mL added to 230 mL of water will not necessarily give 250 mL. Some liquids are miscible in each other. All concentration calculations will be subsequently affected," and suggested that "It should be made up to the mark and not have 20 mL added to 230 mL."

Would miscibility not result in an error regardless if we're filling the volumetric flask up to the mark or adding 20 to 230? Cheers
[2020] Accounting- 45
[2021] Chemistry Methods Specialist English Psych :(

peachteaaa

  • Fresh Poster
  • *
  • Posts: 3
  • Respect: 0
Re: VCE Chemistry Question Thread
« Reply #9190 on: September 21, 2021, 05:49:26 pm »
0
Is identification of monomers from polymers (other than those in food chemistry) out of the study design?

lm21074

  • MOTM: JAN 19
  • Victorian Moderator
  • Forum Leader
  • *****
  • Posts: 589
  • Respect: +594
Re: VCE Chemistry Question Thread
« Reply #9191 on: September 21, 2021, 11:34:38 pm »
+3
Is identification of monomers from polymers (other than those in food chemistry) out of the study design?
I believe so! We aren't required to name polymers, which would involve knowing what the monomers are.
2021: VCE
2022: Science / Arts @ Monash

wingdings2791

  • Trendsetter
  • **
  • Posts: 117
  • I'm going back to the start
  • Respect: +108
Re: VCE Chemistry Question Thread
« Reply #9192 on: September 22, 2021, 11:00:16 am »
+4
Hey everyone, just had a quick one about some titration theory.
"A student conducts an experiment to determine the ethanoic acid concentration in a commercial
brand of vinegar. An outline of her procedure and her measurements are provided below.
230.0 mL deionised water added to a 250 mL volumetric flask
20.0 mL of vinegar added to the flask
20.00 mL aliquots added to flasks and titrated against 0.150 M NaOH."

The solutions said that an error in this experimental design was that "20 mL added to 230 mL of water will not necessarily give 250 mL. Some liquids are miscible in each other. All concentration calculations will be subsequently affected," and suggested that "It should be made up to the mark and not have 20 mL added to 230 mL."

Would miscibility not result in an error regardless if we're filling the volumetric flask up to the mark or adding 20 to 230? Cheers

Hi saransh,
I think the main error described here is that the vinegar to the volumetric flask before the deionised water.

The analyte (vinegar) should always be added first to ensure that the correct concentration is achieved. When deionised water is added directly to the vinegar already in the flask, up to the \(250 mL\) mark, you know that the volume totals as close to \(250 mL\) as possible, and that the \(n(vinegar)\) is as close to \(20 mL\) as possible.

If the deionised water is added to the volumetric flask first, then the separately measured \(20 mL\) of vinegar is added to the deionised water, you risk decreasing accuracy. As the uncertainty of the volumetric flask and the pipette used to measure the vinegar might result in inaccurate measurements, separately measuring the deionised water and vinegar and then adding them together combines the potential for error from two pieces of equipment. That is: when vinegar is added first, although inaccuracy can still come from the uncertainty of the \(n(vinegar)\) measurement, the total \(v\) wouldn't be affected as much.
If the deionised water is added first, both the vinegar and water would be more likely to be measured inaccurately, compounding the error. I'll list the possible errors from measurement in each case here:

Water after vinegar
High vinegar: less water added to achieve \(v=250 mL\), inaccurately high \([CH_3COOH]\)
Low vinegar: more water added to acheive \(v=250 mL\), inaccurately low \([CH_3COOH]\)

Vinegar after water
Low water/low vinegar: \(v<250 mL\), inaccurately low \([CH_3COOH]\) calculated as \(v=250 mL\) will be used
Low water/high vinegar: \(v(solution)\approx\ 250 mL\), aliquots have higher \([CH_3COOH]\), high \([CH_3COOH]\) calculated
High water/low vinegar: \(v(solution)\approx\ 250 mL\), aliquots have lower \([CH_3COOH]\), low \([CH_3COOH]\) calculated
High water/high vinegar: \(v>250 mL\), high \([CH_3COOH]\) calculated as \(v=250 mL\) will be used
High water/accurate vinegar: low \([CH_3COOH]\) from \(v>250 mL\) and aliquot \([CH_3COOH]<\) actual
Low water/accurate vinegar: high \([CH_3COOH]\) from \(v<250 mL\) and aliquot \([CH_3COOH]>\) actual
Accurate water/high vinegar: high \([CH_3COOH]\) from \(v>250 mL\) and aliquot \([CH_3COOH]>\) actual
Accurate water/low vinegar: low \([CH_3COOH]\) from \(v<250mL\) and aliquot \([CH_3COOH]<\) actual

As you can see, measuring the deionised water and vinegar separately, then adding the water first creates many more potential errors than adding the vinegar first and topping up the volume with water accordingly. This makes vinegar first, water second a more accurate approach.

Other errors
I don't see an error that would result from the miscibility of liquids. If anything, I would've thought that the analyte and titrant being immiscible would pose a problem, as this suggests that they would not react with each other and sit in the conical flask as a heterogenous solution.

Some other errors you could discuss, in case you need more ideas:
- No mention of equipment rinsing or swirling the solution to dissolve particles
- Indicator is not added to the analyte prior to titration
- The phrasing used is 'aliquots added to flasks and titrated against \(0.150M NaOH\)', which falsely implies that this is a back titration
- No mention of the titration process beyond 'titrate vinegar': should specify that three concordant titres must be achieved/results must be recorded (precision)

Anyways, that's what I think is going on in this question; it does seem a little strange. Hope I could help, let me know if there are any errors (see what I did there) :)
« Last Edit: September 22, 2021, 02:29:42 pm by wingdings2791 »
ATAR: 99.75
UCAT: 95th

2022-2025: B. Radiography and Medical Imaging (Honours) @ Monash

miyukiaura

  • Trendsetter
  • **
  • Posts: 108
  • per aspera ad astra
  • Respect: +3
Re: VCE Chemistry Question Thread
« Reply #9193 on: September 22, 2021, 02:19:01 pm »
0
Hey guys,

I just had a question about determining the calibration factor by electrical calibration.
For a temperature-time graph (attached), how do you know if the calorimeter is well-insulated or poorly-insulated, and whether to extrapolate back to when the current was turned on, or to simply read off the graph to determine the change in temperature? I know that for poorly-insulated calorimeters, the graph should decrease and be less linear, however, if it isn't as obvious, how would you know which method to use?

Thanks so much.
Offering 50 raw English tutoring, PM for details

wingdings2791

  • Trendsetter
  • **
  • Posts: 117
  • I'm going back to the start
  • Respect: +108
Re: VCE Chemistry Question Thread
« Reply #9194 on: September 23, 2021, 01:34:05 pm »
+3
Hey guys,

I just had a question about determining the calibration factor by electrical calibration.
For a temperature-time graph (attached), how do you know if the calorimeter is well-insulated or poorly-insulated, and whether to extrapolate back to when the current was turned on, or to simply read off the graph to determine the change in temperature? I know that for poorly-insulated calorimeters, the graph should decrease and be less linear, however, if it isn't as obvious, how would you know which method to use?

Thanks so much.

Hello miyukiaura,
Determining the quality of insulation for a calorimeter largely depends on what happens to \(T\) after the current is removed, as well as the ratio of \(ΔT\)(experimental) to \(ΔT\)(theoretical). For telling whether a calorimeter is well-insulated or poorly-insulated, I don't think you could within the scope of VCE, as it doesn't appear anywhere on the study design and there are no indications I can find for what is considered high/low (like how \(K>M^4\) is a large equilibrium constant). Any determination of a calorimeter's insulation quality would probably have to be comparative.

Of course the better insulated a calorimeter system is, the less thermal energy will be lost to the environment, so just to reiterate:

Well-insulated:
- Less decline in \(T\) after current is turned off (some is inevitable since no calorimeter is 100% insulated)
- Higher initial \(ΔT\) (at instant current is removed), since less heat is lost during the reaction
- Lower CF (specific heat capacity of calorimeter is lower, less energy required to achieve same \(ΔT\))

Poorly insulated:
- More decline in \(T\) after current is removed (continues to lose thermal energy to environment)
- Lower initial \(ΔT\), since more heat is lost to the environment during reaction
- Higher CF (more energy required to achieve same \(ΔT\) as thermal waste is greater)

In general, I think that extrapolation is a safer approach (unless you're dealing with an ideal calorimeter). Some heat loss is always inevitable, so going with a line of best fit will probably produce the most reliable CF; if the calorimeter is close to ideal, the extrapolation will closely match anyway.

Hope this helps :)
ATAR: 99.75
UCAT: 95th

2022-2025: B. Radiography and Medical Imaging (Honours) @ Monash