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October 26, 2021, 05:04:40 am

Author Topic: VCE Chemistry Question Thread  (Read 1708849 times)  Share 

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Harrycc3000

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Re: VCE Chemistry Question Thread
« Reply #9195 on: August 22, 2021, 02:37:58 pm »
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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, Specialist Mathematics, Psychology, Chemistry, English Language

Harrycc3000

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Re: VCE Chemistry Question Thread
« Reply #9196 on: August 22, 2021, 02:39:33 pm »
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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]
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Jimmonash1991

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Re: VCE Chemistry Question Thread
« Reply #9197 on: August 22, 2021, 03:36:08 pm »
+7
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

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Re: VCE Chemistry Question Thread
« Reply #9198 on: August 22, 2021, 05:18:04 pm »
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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, Specialist Mathematics, Psychology, Chemistry, English Language

Billuminati

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Re: VCE Chemistry Question Thread
« Reply #9199 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
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saransh

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Re: VCE Chemistry Question Thread
« Reply #9200 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
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peachteaaa

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Re: VCE Chemistry Question Thread
« Reply #9201 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

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Re: VCE Chemistry Question Thread
« Reply #9202 on: September 21, 2021, 11:34:38 pm »
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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.

wingdings2791

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Re: VCE Chemistry Question Thread
« Reply #9203 on: September 22, 2021, 11:00:16 am »
+3
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 »
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miyukiaura

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Re: VCE Chemistry Question Thread
« Reply #9204 on: September 22, 2021, 02:19:01 pm »
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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.
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wingdings2791

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Re: VCE Chemistry Question Thread
« Reply #9205 on: September 23, 2021, 01:34:05 pm »
+2
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 :)
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wingdings2791

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Re: VCE Chemistry Question Thread
« Reply #9206 on: September 24, 2021, 10:57:25 pm »
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Hi everyone,
I'm doing my U4 AOS3 (prac) SAC on fuels, as my school has not assessed most of U3 AOS1. The prac focuses on producing a biodiesel from vegetable oil, then determining and comparing the density and energy content on each fuel.

Because of lockdown, we weren't able to complete the experiment ourselves, so the teachers provided the following sample data (to analyse as practice):

Spoiler
For the energy content determination (combusting fuels with a spirit burner to heat water, recording \(\Delta T\) of water):
- \(v(H_2O)=50\ mL\)
- \(v(fuel)=10\ mL\)
- \(\Delta m(fuel)=3.66g\) (for both fuels)
- \(\Delta T=+28.5^\circ C, +31.0^\circ C\) (oil and biodiesel respectively)

Using this data, I determined the energy content of the oil and biodiesel each as follows:



These values seem very inaccurate, given that fats/oils typically have a heat of combustion \(\sim 37kJg^{-1}\). Am I forgetting something and doing a step wrong, or is this data just unrealistic? I've checked over the units, given data, and calculations a hundred times but still can't figure out what's wrong. Under the results column, the rubric does specify that 'any outliers should be identified (if appropriate)', but I'm hesitant to declare this a huge systematic error caused by a terribly calibrated thermometer or similar.

Any help would be greatly appreciated.
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Mackenzie Aps

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Re: VCE Chemistry Question Thread
« Reply #9207 on: September 28, 2021, 09:06:01 am »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions

Mackenzie Aps

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Re: VCE Chemistry Question Thread
« Reply #9208 on: September 28, 2021, 09:06:42 am »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions for 2021

lm21074

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Re: VCE Chemistry Question Thread
« Reply #9209 on: September 28, 2021, 12:51:11 pm »
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Does anyone have the NEAPS Chemistry 3/4 Trial exam and solutions for 2021
Hi Mackenzie,

It is against the forum rules to distribute copyrighted materials, including the NEAP trial exams and solutions.

Does your school have access to these resources?