The Student Room Group

OCR Chemistry F322 June 2016

Scroll to see replies

Did anyone of u get CH3CH2CH2CH2Br as the the isomer. Because I calculated the empirical formula which gave me C3H7Br. Then to find the molecular formula I did Mr which was 150 divided by the Mr of empirical formula which gave me like 1.22... multiplied this with the empirical formula. Which gave me C3H9BR. Is this wrong?
Original post by Apple Core
Did anyone of u get CH3CH2CH2CH2Br as the the isomer. Because I calculated the empirical formula which gave me C3H7Br. Then to find the molecular formula I did Mr which was 150 divided by the Mr of empirical formula which gave me like 1.22... multiplied this with the empirical formula. Which gave me C3H9BR. Is this wrong?


Hey! Mr wasn't 150. They said the maximum value the mr could be was 150 but didn't specify its value. The mr of the empirical formula was somewhere between 110-130. As a result, there are no other ratios of atoms in the molecule that molecular formula can be because they would all be bigger than the max of 150. So, the molecular formula was the same as the empirical formula. The atoms in C3H9Br are in a different ratio than in the empirical formula
Original post by ali_t9001
Hey! Mr wasn't 150. They said the maximum value the mr could be was 150 but didn't specify its value. The mr of the empirical formula was somewhere between 110-130. As a result, there are no other ratios of atoms in the molecule that molecular formula can be because they would all be bigger than the max of 150. So, the molecular formula was the same as the empirical formula. The atoms in C3H9Br are in a different ratio than in the empirical formula



Yo! What did you get for number of repeat units?
Did anyone of u get CH3CH2CH2CH2Br as the the isomer. Because I calculated the empirical formula which gave me C3H7Br. Then to find the molecular formula I did Mr which was 150 divided by the Mr of empirical formula which gave me like 1.22... multiplied this with the empirical formula. Which gave me C3H9BR. Is this wrong?
Can anyone remember the phrasing of the percentage yield question? Have no recollection of it at all - worried I missed a page!
Reply 225
Original post by KJB97
Can anyone remember the phrasing of the percentage yield question? Have no recollection of it at all - worried I missed a page!


It was to do with how many grams of Mg would you need for a reaction, if 5g of (can't remember) needs to be produced and % yield is 45%?
Original post by Randall13
Yo! What did you get for number of repeat units?


Of the polymer? 50 000 / 70 I think. Can't remember very clearly. But, if I remember correctly, the individual monomers were C5H10 -- meaning the number of monomers in mr= 50 000 would be 714.285. I rounded that down to 714 as the actual number of monomers would be a whole number
Original post by ali_t9001
Of the polymer? 50 000 / 70 I think. Can't remember very clearly. But, if I remember correctly, the individual monomers were C5H10 -- meaning the number of monomers in mr= 50 000 would be 714.285. I rounded that down to 714 as the actual number of monomers would be a whole number


How did you get C5H10 in my last post the repeating unit is C6H12 so the Mr would be 84.
Reply 228
Original post by nimbusninja
I got 1,1-dimethylpropanoatesimply because the O-H group was in the middle, so there would have been a methyl on the top and the bottom of the middle carbon


Congrats, you win the nobel prize! 1,1-dimethylpropanoate isn't an IUPAC name for an ester noice.

It should be 1-methylethyl propanoate or methylethyl propanoate

Propanoic Acid + Propan-2-ol

Google is your friend :P

Also called isopropyl propanoate (this is the most correct naming)

As iso naming isn't taught at A Level we don't need to name it
(edited 7 years ago)
Original post by Randall13
How did you get C5H10 in my last post the repeating unit is C6H12 so the Mr would be 84.


Looking at your diagram, I think you've put your double bond in the wrong place, as the different monomers in the polymer always attach to eachother via covalent bonds from the carbons that originally share the double bond, if that makes sense. Plus I think you may have included an extra CH2 on the right hand side of your structure ( or I didn't put one in in mine). That would account for the difference of 14.
It was supposed to be 50,000/70
Hi everyone, I also had propanoic acid and propan-2-ol to make an ester however i didn't name the ester as the question didnt ask us to name it.
It would be good to see an unnoficial mark scheme. What did everyone get for the first question?
Original post by Randall13
How did you get C5H10 in my last post the repeating unit is C6H12 so the Mr would be 84.


Also can you remember the M/z fragment ion values?
Original post by ali_t9001
Looking at your diagram, I think you've put your double bond in the wrong place, as the different monomers in the polymer always attach to eachother via covalent bonds from the carbons that originally share the double bond, if that makes sense. Plus I think you may have included an extra CH2 on the right hand side of your structure ( or I didn't put one in in mine). That would account for the difference of 14.


Exactly
Original post by slugmonkey97
how many marks was the whole of 7 worth?

It was 13 marks
Original post by Randall13
85 for an A. Easier than last year so if an AS Paper would be 81 for an A but it was only A2 people so reckon 85 is fair.


It was not only a2 people lol, some schools are still doing last years course so there was some AS students. Also it wouldn't matter if it was just A2, the grade boundaries wouldn't change
thankyou !
Original post by mercuryman
image.jpg Guys does this confirm that nitrogen dioxide can also be a source of NO oxides? Like it is technically right? Right? I Identified thunderstorms and no2 hopefully I don't lose yet another mark :frown:

No2 itself is an nitrogen oxide
Don't suppose you can remember how many marks that question was worth?

Quick Reply