VCE Chemistry Question Thread

[soNasty]

is the titre the volume of the thing inside the burette?

also with the mols this question says calculate the amount of mol present in the 25ml sample. So in the beginnging of the question they has a 25ml sample diluted to 250ml and 20 ml aliquot

ive got the mol of the aliquot but why do i need to multiply by 500/20 to find mol in 25ml sample. wouldnt it be 250/25 since the 25ml sample is diluted. i.e mol in 20ml x 250/25?

again this brings me to my point, the moles do change? but i have been explained before that moles do not change, if i do multiply by 250/20 or 25 it does actually change??

whenever you have something like this, you multiply by (total diluted volume) / (volume of aliquot taken)

[jessica666]

Hey guys, could anybody help me with these 3 questions? Thanks heaps

[sin0001]

Hey guys, could anybody help me with these 3 questions? Thanks heaps

1) Find the amount (mol) of Barium Sulfate precipitated, and hence, n(S): n(BaSO4) = 0.493/233.39 g/mol = 0.00211 mol. 

-Hence n(S) = n(BaSO4) = 0.00211 mol

-Finding the mass of Sulfur precipitated: m(S) = 0.00211 mol x 32.1 g/mol = 0.0678 grams -> Meaning this is the mass of Sulfur extracted from coal

-The sulfur composition of coal can now be found: m(S)/m(coal) = 0.0678/2.00 x 100% = 3.39% composition

2) Reaction that takes place is: NaOH + CH3COOH --> NaCH3COO + H2O

-Finding the mol of Sodium Hydroxide reacted and acetic acid present: n(acetic acid or CH3COOH) = n(NaOH) = 0.02735 L x 0.0942 M = 0.00258 mol

-Since this is only the amount of acetic acid present in the aliquot taken, we can find n(acetic acid) in the original vinegar sample by multiplying this by 250/20: n(acetic acid) = 0.00258 mol x 250/20 = 0.0322 mol in the original sample

-The mass of acetic acid: m(acetic acid) = 0.0322 mol x 60.05 g/mol = 1.93 grams

-d(vinegar) = 1 just means that the vinegar sample weighed 40 grams, therefore the composition of acetic acid in the original sample = 1.93 g / 40 g x 100% = 4.83% composition

3) Another titration...*sigh*
-Always write the balanced equation first: 2HCl + Na2CO3 --> 2NaCl + H20 + CO2

-Finding the amount of reacted Na2CO3 also gives the amount of HCl acid in the 25 mL aliquots taken: n(HCl) = 2 x n(Na2CO3) = 2 x (0.117 M x 0.01783 L) = 0.00417 mol of HCl present in each aliquot (on average, remember!)

-n(HCl) present in the 5 mL sample of cleaner = 0.00417 mol x 250/25 = 0.0417 mol of HCl; m(HCl) = 1.52 g

(a) If there is 1.52 g of HCl acid present in 5 mL of the sample, there will be 304 grams of HCl present in every litre of the cleaner (304 g per L)

(b) Molar concentration = n(HCl)/0.005 L = 8.34 M concentration

Hope this is all correct, haven't touched chem in 2 years

[paper-back]

In intepreting Infa-red spectra graphs how would you distinguish between the troughs that are important and the ones that aren't?
I'm talking about those small troughs that occur outside of the fingerprint region. I sometimes mistake them for an actual bond

Thanks

[doher109]

In intepreting Infa-red spectra graphs how would you distinguish between the troughs that are important and the ones that aren't?
I'm talking about those small troughs that occur outside of the fingerprint region. I sometimes mistake them for an actual bond

Thanks

Hi,

This at the start of Year 12 seems really difficult, and a lot of practice questions from external companies will be downright confusing.

With the exception of a few almost all external (NON VCAA) exams are not indicative of the graphs you'll get on the real exam. This is because VCAA wants to leave very little to chance. They do not want to have to deliberate whether they will award a mark for a mis-interpreted question because of a really 'full' graph.

So in short. The graphs in the exam are typically simple and all will be important. There will not be heaps of conflicting troughs.

Typically look for

• Alkene
• Carboxy
• Alcohol

99% of graphs will only be testing these 4.

Attached is a question from the 2014 exam, the IR graph as you can see is very simple. With only 2 clear troughs present.

[paper-back]

Thank you Doher, but in the case that one does show up would you then proceed identify it as not important by looking at its trough size or?

[doher109]

The only two that can sometimes appear in the same area are the alcohol and carboxy.

From memory one is broad and one is narrow (that's how you differentiate). Can anyone remember which is which? 

[Zues]

Why is it necessary to measure the peak areas produced
by a number of standards?

[doher109]

Why is it necessary to measure the peak areas produced
by a number of standards?

This allows a calibration curve to be determined.

You can only plot experimental data between points from standard solutions (this ensures accuracy)

Typically 4 or 5 is used. So in simple terms it ensures accuracy of the calibration points and allows test data to be plotted on an accurate and true curve.

[jessica666]

1) Find the amount (mol) of Barium Sulfate precipitated, and hence, n(S): n(BaSO4) = 0.493/233.39 g/mol = 0.00211 mol. 

-Hence n(S) = n(BaSO4) = 0.00211 mol

-Finding the mass of Sulfur precipitated: m(S) = 0.00211 mol x 32.1 g/mol = 0.0678 grams -> Meaning this is the mass of Sulfur extracted from coal

-The sulfur composition of coal can now be found: m(S)/m(coal) = 0.0678/2.00 x 100% = 3.39% composition

2) Reaction that takes place is: NaOH + CH3COOH --> NaCH3COO + H2O

-Finding the mol of Sodium Hydroxide reacted and acetic acid present: n(acetic acid or CH3COOH) = n(NaOH) = 0.02735 L x 0.0942 M = 0.00258 mol

-Since this is only the amount of acetic acid present in the aliquot taken, we can find n(acetic acid) in the original vinegar sample by multiplying this by 250/20: n(acetic acid) = 0.00258 mol x 250/20 = 0.0322 mol in the original sample

-The mass of acetic acid: m(acetic acid) = 0.0322 mol x 60.05 g/mol = 1.93 grams

-d(vinegar) = 1 just means that the vinegar sample weighed 40 grams, therefore the composition of acetic acid in the original sample = 1.93 g / 40 g x 100% = 4.83% composition

3) Another titration...*sigh*
-Always write the balanced equation first: 2HCl + Na2CO3 --> 2NaCl + H20 + CO2

-Finding the amount of reacted Na2CO3 also gives the amount of HCl acid in the 25 mL aliquots taken: n(HCl) = 2 x n(Na2CO3) = 2 x (0.117 M x 0.01783 L) = 0.00417 mol of HCl present in each aliquot (on average, remember!)

-n(HCl) present in the 5 mL sample of cleaner = 0.00417 mol x 250/25 = 0.0417 mol of HCl; m(HCl) = 1.52 g

(a) If there is 1.52 g of HCl acid present in 5 mL of the sample, there will be 304 grams of HCl present in every litre of the cleaner (304 g per L)

(b) Molar concentration = n(HCl)/0.005 L = 8.34 M concentration

Hope this is all correct, haven't touched chem in 2 years

Thank you so much!!

[Zues]

i know the basic answers of these questions but can someone give me a more informational answer


a Why must the level of the solvent be lower than the origin
where spots of the mixture are originally placed?
b Why are Rf values always less than one?

i know a is because then the components will mix with the solvent and b is because the solvent front carries the components so components cant go past this but whats a "better response"

also how do i do b, in attached?

[lArcdeTriomphe]

a. If the level of the solvent is higher than the origin, the substance/components would just dissolve in the solvent and form a solution (which would defeat the whole purpose of chromatography).
b. Rf = distance travelled by component from origin/distance travelled by solvent front from origin
The component can never move more than the solvent front (because the solvent front effectively provides the path for the component to move). OR, you could think of it like this. If the component travelled further than the solvent, how can it actually move? There isn't any solvent there.

Since distance travelled by component < distance travelled by solvent front from origin, therefore the fraction is always less than one (think of it like ratios).

(A more in-depth solution would probably consider the fact that hte component moves only because it forms 'bonds' to the mobile phase. Without the mobile phase, the component's only intermolecular forces are with the stationary phase, so it can ONLY be adsorbed to the chromatogram (and, hence, not move). That is why it makes no sense for the component to travel further than the mobile phase - if it moves further, there is no mobile phase, meaning it can only form bonds to the stationary phase; therefore it must stay still). <-- I do realise that I've made it quite convoluted (and possibly circular) haha Hopefully you get the gist of it)

[Zues]

Thanks

btw can someone help explain this

•   The mass of the atoms attached to each bond also affects the type of radiation energy absorbed. The lower the mass of the atoms in the bond, the higher the absorption. For example a C-H bond will absorb higher frequencies of infared than a C-O bond. Each bond in the molecule is the part that is vibrating, so bond types affect this.

why is the lower the mass the higher the absorption? lower mass means less to absorb hence lower absorption ?

[doher109]

 

Thanks

btw can someone help explain this

•   The mass of the atoms attached to each bond also affects the type of radiation energy absorbed. The lower the mass of the atoms in the bond, the higher the absorption. For example a C-H bond will absorb higher frequencies of infared than a C-O bond. Each bond in the molecule is the part that is vibrating, so bond types affect this.

why is the lower the mass the higher the absorption? lower mass means less to absorb hence lower absorption ?

Whilst I can't explain this i'm sure someone like pi or thushan can.

But to be honest you will not need to know this for VCE Chem, if you interest you'd like to know that is fantastic! 

[psyxwar]

Thanks

btw can someone help explain this

•   The mass of the atoms attached to each bond also affects the type of radiation energy absorbed. The lower the mass of the atoms in the bond, the higher the absorption. For example a C-H bond will absorb higher frequencies of infared than a C-O bond. Each bond in the molecule is the part that is vibrating, so bond types affect this.

why is the lower the mass the higher the absorption? lower mass means less to absorb hence lower absorption ?

In the context of IR Spectroscopy, we can use a spring analogy for the bond between two atoms. A higher frequency of radiation means the spring vibrates a lot/quickly, while a lower frequency of radiation means that it does not vibrate as much.

If you have say, a C-H bond, the lower mass of the hydrogen atom means the spring is able to vibrate much more because its so light. Think about it -- if we have the much heavier O attached to our carbon, then its going to vibrate much more slowly (ie. lower frequency of radiation) simply because it's harder for it to vibrate as much.

Furthermore, just like with springs, the stiffness matters. A stiffer spring can contract much faster than one that isn't as stiff. Hence stronger bonds will absorb more energy than weaker bonds, which is why for example C=O absorbs at a higher frequency than C-O.