VCE Chemistry Question Thread

[Bri MT]

Hello,

Is this organic compound named as 'fluoro-prop-1-ene', or '1-methyl fluoroethene'.
When naming, do you take into consideration the fact that the longest carbon chain is 3, or that there is a CH3 attached to one of the carbons?

1. Your first step should be to identify the longest possible carbon chain. In this case, it's 3 carbons long
2. Then identify the different parts of the molecule. In this case, there's a halogen and a double bond
3. We start numbering the carbons from the direction that gives us the lowest numbers. In this case we start numbering the carbons from the double bond
4. We name the molecule. In this case, 2-fluoropropene  as ArtyDreams indicated

[Chocolatemilkshake]

Hi, just a few questions if anyone feels like answering 

1. So Liquified petroleum gas (LPG) can be derived from both wet natural gas deposits as well as crude oil?

2. Does coal seam gas need to be refined?

3. Is coal seam gas classified as a type of natural gas?

Thanks everyone 

[Erutepa]

Hi, just a few questions if anyone feels like answering 

1. So Liquified petroleum gas (LPG) can be derived from both wet natural gas deposits as well as crude oil?

2. Does coal seam gas need to be refined?

3. Is coal seam gas classified as a type of natural gas?

Thanks everyone 

1) This is correct!

2) coal seem gas is a form of raw natural gas which will be processed to remove contaminants and hydrocarbons other than methane inorder to produce natural gas with a very high methane concentration. The other hydrocarbons that were removed can then be recovered and refined to produce such products as LPG.

I would like to note that the processing of this raw natural gas is something beyond VCE chem level and I have never seen anything assessing such content. I belive knowing the process of fracking is sufficient.

3) Yes, Coal seam gas is a type of natural gas as natural gas has other sources (fracking being one of them). If you want to read breifly through some of the other sources of natural gas, wikipedia here seems to do a good job of it

[Amith2002]

Hi, i'm not a tech expert and I couldn't find a send a new message so clicked the next best thing and clicked reply.\

Anyways, I was wondering whether you can bring two scientific calculators to the exam?

[sweetiepi]

Hi, i'm not a tech expert and I couldn't find a send a new message so clicked the next best thing and clicked reply.\

Anyways, I was wondering whether you can bring two scientific calculators to the exam?

Hey! It's okay, I can see your post, so I can respond!

I assume you cannot bring two scientific calculators into an exam, however, if you're concerned, I'd suggest replacing the batteries before an exam!


Edit: VCAA states that in Enviro, Physics, and Chemistry, only one can be brought in. See here: https://www.vcaa.vic.edu.au/assessment/vce-assessment/materials/Pages/index.aspx

Hope this helps!

[Lakshi]

Oh aww, that's a shame, but thank you

[jnlfs2010]

When 120 ml of a gaseous hydrocarbon was burnt in 400mL Oxygen, which was more than sufficient, 280ml of gas was present at the end of the reaction after the products were cooled to SLC. The gas mixture was bubbled through concentrated LiOH, which absorbed all the CO2 present, leaving 40ml of gas. What was the molecular formula of the hydrocarbon, if all gas volumes had been measured at SLC?

Hello, a question anyone help me out?

[Erutepa]

When 120 ml of a gaseous hydrocarbon was burnt in 400mL Oxygen, which was more than sufficient, 280ml of gas was present at the end of the reaction after the products were cooled to SLC. The gas mixture was bubbled through concentrated LiOH, which absorbed all the CO2 present, leaving 40ml of gas. What was the molecular formula of the hydrocarbon, if all gas volumes had been measured at SLC?

Hello, a question anyone help me out?

Hey Jnlfs!
I am more than happy to answer your question, but before I do its important for you to have a go at it first and try to explain your current aproach to the question. Don't worry about being wrong as thats all part of the learning process.
Having a go at the question first and showing us your thinking helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question.

[jnlfs2010]

Hey Jnlfs!
I am more than happy to answer your question, but before I do its important for you to have a go at it first and try to explain your current aproach to the question. Don't worry about being wrong as thats all part of the learning process.
Having a go at the question first and showing us your thinking helps people answering your question figure out what specifically you don't quite get so that we can better help you towards understanding the question.

I understand all volumes are at SLC. Thus volume of CO2 in reaction is 280-40=240ml. Therefore ratio of 2*n(CxHy) = N(CO2)
We can find the length of the hydrcarbon which is 2 carbons long. The thing im confused about is how to find the ratio for hydrogen

[sweetcheeks]

I understand all volumes are at SLC. Thus volume of CO2 in reaction is 280-40=240ml. Therefore ratio of 2*n(CxHy) = N(CO2)
We can find the length of the hydrcarbon which is 2 carbons long. The thing im confused about is how to find the ratio for hydrogen

You know the amount of CO2 produced in the reaction. This tells you how much of the oxygen gas that reacted with the carbon content of the hydrocarbon. You are also told how much oxygen is leftover from the reaction. From there, you should be able to work out the volume of oxygen that reacted with hydrogen to form water and similar to how you worked out the carbon, you can determine the hydrogen content.

[Erutepa]

I understand all volumes are at SLC. Thus volume of CO2 in reaction is 280-40=240ml. Therefore ratio of 2*n(CxHy) = N(CO2)
We can find the length of the hydrcarbon which is 2 carbons long. The thing im confused about is how to find the ratio for hydrogen

You got the first part correct in calculating the volume of carbon dioxide gas as 240ml (280-40ml).
Thus we have 120ml of hydrocarbon : 240ml of CO2.
Since all gasses are at SLC we can then say the molar ration of hydrocarbon to CO2 is 1:2 (this can be explained by applying the universal gas equation PV=nRT, where n=PV/RT with V/RT being the same for both gasses (due to identical pressure and temperature conditions) such that something of twice the volume will be twice the molar amount)

We can also calculate the volume of oxygen that reacted with the hydrocarbon by subtracting the remaining 40ml of gas (which is excess oxygen) from the 400ml of oxygen initially present. This gives us a volume of 360ml of oxygen that reacted with the unknown hydrocarbon.
Since this oxygen is also at SLC, we can say the ration of oxygen:hydrocarbon:CO2 is 3:1:2.

We can now apply this information to the general formula for the complete combustion of a hydrocarbon:

1CxHy + 3O2 --> 2CO2 + ?H2O

First we can balance the carbons:
C2Hy + 3O2 --> 2CO2 + H2O

Now we can balance the oxygens:
C2Hy + 3CO2 --> 2CO2 + 2H2O

And lastly, we can balance the hydrogens:
C2H4 + 3CO2 --> 2CO2 + 2H2O

Thus we arrive at the formula for the hydrocarbon!
It seems you got confused with the ratio of hydrocarbon:CO2 and flipped it the wrong way around - remember that there was twice the amount of CO2 produced as there was hydrocarbon initially.
I think the main reason you got stumped after that was perhaps that you didn't realise that the remaining 40ml of gas were oxygen. Since oxygen was in excess, it would be expected that there would be oxygen remaining (hence the 40ml of gas that were remaining). This 40ml of remaining gas would not have been composed of any other gas as the complete combustion of a hydrocarbon only produces CO2 as a product and water (as another product of the combustion reaction) would have existed in liquid form at SLC.

Hopefully this clarifies things!

[jnlfs2010]

You got the first part correct in calculating the volume of carbon dioxide gas as 240ml (280-40ml).
Thus we have 120ml of hydrocarbon : 240ml of CO2.
Since all gasses are at SLC we can then say the molar ration of hydrocarbon to CO2 is 1:2 (this can be explained by applying the universal gas equation PV=nRT, where n=PV/RT with V/RT being the same for both gasses (due to identical pressure and temperature conditions) such that something of twice the volume will be twice the molar amount)

We can also calculate the volume of oxygen that reacted with the hydrocarbon by subtracting the remaining 40ml of gas (which is excess oxygen) from the 400ml of oxygen initially present. This gives us a volume of 360ml of oxygen that reacted with the unknown hydrocarbon.
Since this oxygen is also at SLC, we can say the ration of oxygen:hydrocarbon:CO2 is 3:1:2.

We can now apply this information to the general formula for the complete combustion of a hydrocarbon:

1CxHy + 3O2 --> 2CO2 + ?H2O

First we can balance the carbons:
C2Hy + 3O2 --> 2CO2 + H2O

Now we can balance the oxygens:
C2Hy + 3CO2 --> 2CO2 + 2H2O

And lastly, we can balance the hydrogens:
C2H4 + 3CO2 --> 2CO2 + 2H2O

Thus we arrive at the formula for the hydrocarbon!
It seems you got confused with the ratio of hydrocarbon:CO2 and flipped it the wrong way around - remember that there was twice the amount of CO2 produced as there was hydrocarbon initially.
I think the main reason you got stumped after that was perhaps that you didn't realise that the remaining 40ml of gas were oxygen. Since oxygen was in excess, it would be expected that there would be oxygen remaining (hence the 40ml of gas that were remaining). This 40ml of remaining gas would not have been composed of any other gas as the complete combustion of a hydrocarbon only produces CO2 as a product and water (as another product of the combustion reaction) would have existed in liquid form at SLC.

Hopefully this clarifies things!

omg thanks i couldn't do it because i thought the 40ml gas had water vapour as well. So if its cooled to SLC the vapour in the combustion returns to liquid right?

[Erutepa]

omg thanks i couldn't do it because i thought the 40ml gas had water vapour as well. So if its cooled to SLC the vapour in the combustion returns to liquid right?

That's right!
SLC is 25 degrees Celsius. Rember that water's melting pont is at 0 degrees Celsius and it's boiling point is at 100 degrees Celsius, meaning it will be a liquid at 25 degrees Celsius (SLC)

[jnlfs2010]

Hey Erutepa I think I have another question.

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only C, H, and O. In one experiment, 0.4953 g of the acid was burnt in excess oxygen. The gases produced were passed through CaCl2 and NaOH. It was found that the mass of the CaCl2 increased by 0.1613 g and the mass of the NaOH increased by 1.0495 g. What is the empirical formula of terephthalic acid?

So thoughts on this one->
>0.4953g of the acid will contain masses of C, H, O (excess oxygen, thus complete combustion)
>I don't understand how we can deduce grams/mols of combustion products with the information given by CaCl2 and NaOH. It says mass of them increased when CO2/H2O reacts with them, how does that work?

[Erutepa]

Hey Erutepa I think I have another question.

Terephthalic acid is an important chemical used in the manufacture of polyesters and plasticisers. It contains only C, H, and O. In one experiment, 0.4953 g of the acid was burnt in excess oxygen. The gases produced were passed through CaCl2 and NaOH. It was found that the mass of the CaCl2 increased by 0.1613 g and the mass of the NaOH increased by 1.0495 g. What is the empirical formula of terephthalic acid?

So thoughts on this one->
>0.4953g of the acid will contain masses of C, H, O (excess oxygen, thus complete combustion)
>I don't understand how we can deduce grams/mols of combustion products with the information given by CaCl2 and NaOH. It says mass of them increased when CO2/H2O reacts with them, how does that work?

Hi jnlfs2010 again!
Good work on explaining your thought process so far!

Its helpfull to start by writing out the equation to begin with:
CxHyOz + ?O2 --> ?CO2 + ?H2O

This next bit I did find a littly tricky and had to do a quick bit of googling- It might just be me but I do think this is a tad bit beyond a VCE level (would love others inputs on it though)
The resulting gas mixture will be consisting of carbon dioxide gas, water vapour, and excess oxygen. When this gas mixtuture is passed though CaCl2, this anhydrous calcium chloride (meaning a calcium chloride compound without associated water molecules) readily absorbs the water vapour. As such, the change in madd of the CaCl2 represents the mass of water produced in the combustion of Terephthalic acid. Passing the gas mixture though NaOH would then cause the CO2 in the gas to react with the NaOH to result in the formation of a sodium bicarbonate (NaHCO3) and as such the change in NaOH mass would represent the mass of the CO2 produced from the combustiuon of the Terephthalic acid.
Now that you know the mass of the products, you can figure out the mole amount of each product and do the math from there to get the empiracal formula of the acid - I will let you try to do this on your own though.
let me know if you can work it out (if not I can help you further)

As I said, I think the question expects a bit too much interms of wanting you to know how each of the products will react with the CaCl2 and the NaOH, but I would love to hear what others think in terms of how hard it is.