VCE Chemistry Question Thread

[paper-back]

How would you prepare an ester from acetic acid and butan-1-ol? give an equation for the reaction

Was that the exact question for the first one?

Reaction type:
Esterification/Condensation reaction
CH3COOH+OHCH2CH2CH2CH3 (+H2SO4 catalyst) --> CH3COOCH2CH2CH2CH3 (Butyl ethanoate) + H2O

[cosine]

Permethrin, contains 0.480g of carbon. Work out the mass of Chlorine

Stuck on this one

[IndefatigableLover]

Permethrin, contains 0.480g of carbon. Work out the mass of Chlorine

Stuck on this one

n(C)=0.480/12=0.04

To work out the moles of Cl, we look at mole ratios where it'd be unknown over known (that is n(Cl)/n(C)) which we can tell by the number of moles of carbon atoms and chlorine atoms. Therefore it'll be 2/21 which is equivalent to n(Cl)/n(C).

However we already know n(C) which we can substitute in for and find that n(Cl) = 0.00381 mol
From here we work out the Mass of Chlorine where we multiply it with 35.5 (Molar mass of Chlorine) to get a mass of 0.135g

[Redoxify]

Permethrin, contains 0.480g of carbon. Work out the mass of Chlorine

Stuck on this one

work out the mole of carbon,
when you have worked out mole of carbon
do mole ratios for chlorine,
once you have found out the mole of chlorine, use
mass=mole x molar mass to work out mass

edit: beaten by IndefatigableLover haha

[cosine]

n(C)=0.480/12=0.04

To work out the moles of Cl, we look at mole ratios where it'd be unknown over known (that is n(Cl)/n(C)) which we can tell by the number of moles of carbon atoms and chlorine atoms. Therefore it'll be 2/21 which is equivalent to n(Cl)/n(C).

However we already know n(C) which we can substitute in for and find that n(Cl) = 0.00381 mol
From here we work out the Mass of Chlorine where we multiply it with 35.5 (Molar mass of Chlorine) to get a mass of 0.135g

Where are you pulling those formulas from? haha

Can you do it logically, by observing the data o.O?

Thanks for the help

[warya]

Cosine, they're just using n=m/Mr

And for mole ratios its just want/got or unknown/known
Maybe look through your year 11 notes or watch Chemisode unit 1/3 videos on youtube, they'll refresh your memory

[IndefatigableLover]

Where are you pulling those formulas from? haha

Can you do it logically, by observing the data o.O?

Thanks for the help

Umm they're formulas you would have dealt with before if you've done gravimetric/volumetric analysis (n=m/M, mole ratios etc.)

And well you are using the data for the answer but working it out without a calculator won't be worth the time and effort.

[cosine]

Sorry guys, I meant the n(Cl)/n(C) is this a formula to use or is it just acceptable in our scenario cos i've never dealt with it?

[cosine]

During an acid-base titration, why do we use an indicator and how does it affect the end point?

Also how do we know what indicator to use with a specific reaction?

Thanks

Anyone?

[warya]

The equivalence point is when the acid and base are present in the titration flask in the same mole ratios as the chemical equation indicates.

The end point is the point at which the colour changes i.e. where the acid and base are at the same mole ratios and the solution in flask is now neutral. so the indicator doesn't affect the end point, it just tells us when it occurs so we know that we have reached neutrality (?) and that we should stop

If you have a stronger base and a weak acid, your are more likely to have a higher pH value as end point, similarly, if you have a strong acid and weak base you'll have a low pH value at the end point (~point where solution is neutral). Strong acid+strong base will yield pH of 7.

So as you can see, you have to use an indicator which has a colour change at a specific range of pH values which you know the end point of your reaction will occur, no use using an indicator which changes colour at pH 8 when you are reacting a strong acid with a weak base and more likely to have a colour change at 5...

Heinemann explains it really well

Sorry if I made mistakes or something lol this was like 2 months ago for my cohort

[cosine]

The equivalence point is when the acid and base are present in the titration flask in the same mole ratios as the chemical equation indicates.

The end point is the point at which the colour changes i.e. where the acid and base are at the same mole ratios and the solution in flask is now neutral. so the indicator doesn't affect the end point, it just tells us when it occurs so we know that we have reached neutrality (?) and that we should stop

If you have a stronger base and a weak acid, your are more likely to have a higher pH value as end point, similarly, if you have a strong acid and weak base you'll have a low pH value at the end point (~point where solution is neutral). Strong acid+strong base will yield pH of 7.

So as you can see, you have to use an indicator which has a colour change at a specific range of pH values which you know the end point of your reaction will occur, no use using an indicator which changes colour at pH 8 when you are reacting a strong acid with a weak base and more likely to have a colour change at 5...

Heinemann explains it really well

Sorry if I made mistakes or something lol this was like 2 months ago for my cohort

Thanks heaps!!

[alt-x]

Are we required to know the process of how a mass spectrometer works? Thanks!

[warya]

Not so much the specifics of the instrumentation but you should know the processes i.e. Ionisation-sample gets bombarded with electrons and consequently, atoms become charged cations, usually +1, Acceleration- ions are accelerated using an electric field, Deflection- ions are deflected in other words bent according to m/z (heavier ions= less deflection, think basket ball vs ping pong ball) and Detection where the ions are collected and and read by way of an electric current.

Download the analysis notes in the notes section in the forum, they should help

[chocolate.cake.1]

Hello

Why is it that when the length of a molecule's hydrocarbon tail increases in length, its solubility in water decreases?

Thanks

[Kel9901]

Hello

Why is it that when the length of a molecule's hydrocarbon tail increases in length, its solubility in water decreases?

Thanks

Assuming you're talking about a molecule with a polar functional group like alkanols/chloroalkanes/etc, it's because as the hydrocarbon length increases, the molecule becomes 'less polar' and 'more non-polar' (as length increases, the non-polar hydrocarbon tail is longer/more significant and the polar functional group shorter/less significant). Water is a polar liquid, and polar mixes with polar (and non-polar doesn't mix well with polar), so a less polar compound generally means a less water-soluble compound.