hi eliot, thank you for everything you do. as you see im an a2 who just learned how to work out moles, a day before my first chemistry exam, youre a real g and way better than my teacher chris
I've learnt SO MUCH!! using your videos, and gosh, I don't think I would have survived Chemistry without your videos. I wish you was my teacher. You're such a gifted teacher! :) I wanted to ask, have you covered the empirical and molecular formulae and Percentage yields in Amounts of substance? I have a mock this week and i'm really stuck on it :(
***** Hi, Diana. Firstly, thanks for the kind words! Secondly, no I haven't covered those. I'll try and go through them properly quick now though... Empirical formula is just the simplest ratio of atoms in a compound. E.g. C2H4 is not an empirical formula, but CH2 is. The empirical formula does not have to actually be able to exist! In terms of how to calculate it, all you need to do is work out the moles of the elements you are given. You do this by calculating moles as mass / Mr as per usual. Note that you may be given the masses of the elements in the compound, OR you might be given the percentages of each element. Either way, just multiply the mass or % by the Mr and you've got your moles! Then it's a case of finding the simplest ratio. E.g. 3 C to 6 O would be 1 C to 2 O etc. % yield is basically just an efficiency calculation. You put the amount you actually got, in g, over the theoretical amount you could get (from a calculation) and multiply by 100! E.g. I got 6g of a substance but I found through calculation that the maximum yield of that substance could have been 10g. My calculation is (6/10) x 100 = 60%. That helped at all?!
Yes, it's helped me a lot! Thank you very much! Thank you very much for the time you took to reply also. :) I now understand Empirical formula and the rest of it! I had a question earlier. You know how when you calculate the number of moles from the Ideal gas equation and add up the numbers in front of each compound in the products of the reaction and divide by the number (I can't really explain it), is it the same if I was asked to work out, say, the volume? Or is it just for moles? Do I still add up those numbers and divide? Thanks.
***** Hi Diana. Sorry for my late reply, your comment slipped past me! I'm struggling to understand which numbers you are referring to... Can you try and be a bit more specific, please?
Omg whenever I watch your videos i actually like chemistry !!!!!!!Im like depending on your videos to pass the exam!!! also can you do a video on the mcq's related to the other gas laws like Boyles law and Charles law because they are very tricky!
I've been doing the exam questions on the video before you started to work on them. I got them all right apart from the one at the end. Because I forgot to convert from m^3 to dm^3 🙊. But thanks for pointing out, I will be extra careful if something like this appears in the exam.
I don't get why you divide by 5 in the boron trichloride question (10:50), I know there are 5 moles on the right, but isn't there 7 moles of stuff on the left? Why do you not divide by 7? So, in these type of questions do you only ever look at the moles on the right hand side?
skwli Using the ideal gas equation, which accounts for all the gases that are produced in this particular question (the question is talking about both), the values for the moles is also taking into account both gases. In total, there are 5 moles of gas. Looking at the ratio from the equation, you should be able to see that the ratio of boron oxide to the gases is 1:5. So, to get the moles of boron oxide, we divide by 5! Does that make any sense?
I'm quite confused on one of the questions you did, the second one. PV=nRT aids in discovering the total moles of the gases produced.. And am I right in saying the the total moles of Reactants will equal total moles of Products? You divided it by five because there were five moles on the right and one mole on the left, However I would have divided it by 7, because I would have assumed that the reactant's total moles would have equaled 0.004 ( or whatever the sum came to). And the ratio there is 1:7, I know I'm wrong, I'm just wondering why :)! Thank you!
Jam Head It helps if you can link to the point in the video! No, you're over simplifying things slightly. You may find that the ratio of compounds shows that there are more moles on one side than the other, but it's what each compound contains that makes the difference. For example, the reaction of carbon and oxygen, C + O2 --> CO2, shows 2 moles becoming one. However, all of the atoms have remained, they are now just in one compound rather than being separate. In the example in the video, it is the products, NOT the reactants that we are looking at - the give-away is when it says that the gases PRODUCED occupy the volume of 5000 cm3 etc. However, the fact that both products are gases means that we can't separate the number and work things out for just one. Instead, we work it out for all and then divide (this works because a mole of ANY gas occupies the same volume - given constant temperature and pressure). Has that helped at all?
E Rintoul Right, I think I understand. So because we know the volume of the gases produced, and not the reactants, we divide it by the total moles of the GASES produced. Also, re-watching the video, I've realised that you mentioned that the gases all occupy the same volume - So I should be thinking of the gases as one ( when it comes to doing the ratio). Thank you, that's helped tremendously!
Jam Head Thinking of the gases as one big cloud of whatever id probably the easiest way to think about it. Then when you've calculated bits, you can divide by the ratios to single out the individuals. That OK?
In the second question y did u do 0.2019/5 and take the value of one mole to the other side when the other side =7 moles and the right side =5 moles. Don't they have to be same?
10cm3 gaseous hydrocarbon reacts completely with 40cm3 oxygen to produce 30cm3 carbon dioxide. How do you find the number of carbon atoms in the formula of the hydrocarbon??
+Chiddy01 You need to think about where the carbon dioxide comes from... All carbons in the carbon dioxide come from the initial hydrocarbon. Once you know the moles of the carbon dioxide you can work out the ratio and get the numbers.
Dakota fireshard Yep, I never got round to covering the atom economy and other bits. I felt as though they were easier and so I left them out and did what I thought was some of the harder stuff! I will eventually get it done, though. And you certainly DO need to know it!
Could you help me how to figure this answer out? 'If the percentage yield of ammonia in the Haber Process is 15% what mass of ammonia is produced from 5 tonnes of hydrogen?'
hi love ur vids but my as chem exam is in Friday and I was wondering if you could make a video on titration and percentage purity before that if you can please
The question told you the information to work out 'the two gases produces' which gave you the answer of 0.2019 for the total number of moles on the products side of the equation. And even though the ration of moles of ALL the reactants to all the products is 7:5, the only reactant the question cares about is boron oxide ('calculate the mass of boron oxide produced') which is only 1 mole. You worked out that 0.2019=5parts So 0.2019/5 will give you one part. Make sure to make your final answer into grams (by multiplying by the Mr)
@@farah-vb4tihey how you been? got mocks in November 23 :( if we don't do the actual exams they'll use those results) - ngl i was stuck on it again, same part too 😂. i got stuck with the 1 mole = 0.2019/5
Went to from an E to a B hella quick with these videos.
Your videos and a ton more revision in general helped me go from a U to a C so far. Thanks so much for putting these out.
SleepingWithBunnies i have a u now any tips
@@saidoosman4381 This is not what Eliot Rintoul would have wanted
@@AK-yc3tv hey man he's trying why else would he be here?
Atom economy//percentage yield??
5:48 holy Jesus that smack of the lips actually near deafened me
You make me love chemistry again and enjoy doing calculations. Thank You :)
hi eliot, thank you for everything you do. as you see im an a2 who just learned how to work out moles, a day before my first chemistry exam, youre a real g and way better than my teacher chris
I've learnt SO MUCH!! using your videos, and gosh, I don't think I would have survived Chemistry without your videos. I wish you was my teacher. You're such a gifted teacher! :) I wanted to ask, have you covered the empirical and molecular formulae and Percentage yields in Amounts of substance? I have a mock this week and i'm really stuck on it :(
***** Hi, Diana. Firstly, thanks for the kind words! Secondly, no I haven't covered those.
I'll try and go through them properly quick now though...
Empirical formula is just the simplest ratio of atoms in a compound. E.g. C2H4 is not an empirical formula, but CH2 is. The empirical formula does not have to actually be able to exist!
In terms of how to calculate it, all you need to do is work out the moles of the elements you are given. You do this by calculating moles as mass / Mr as per usual. Note that you may be given the masses of the elements in the compound, OR you might be given the percentages of each element. Either way, just multiply the mass or % by the Mr and you've got your moles! Then it's a case of finding the simplest ratio. E.g. 3 C to 6 O would be 1 C to 2 O etc.
% yield is basically just an efficiency calculation. You put the amount you actually got, in g, over the theoretical amount you could get (from a calculation) and multiply by 100! E.g. I got 6g of a substance but I found through calculation that the maximum yield of that substance could have been 10g. My calculation is (6/10) x 100 = 60%.
That helped at all?!
Yes, it's helped me a lot! Thank you very much! Thank you very much for the time you took to reply also. :) I now understand Empirical formula and the rest of it! I had a question earlier. You know how when you calculate the number of moles from the Ideal gas equation and add up the numbers in front of each compound in the products of the reaction and divide by the number (I can't really explain it), is it the same if I was asked to work out, say, the volume? Or is it just for moles? Do I still add up those numbers and divide? Thanks.
E Rintoul Don't you mean Moles = Mass/Mr
henchhitmens You are correct! Typos are the bane of my life...
***** Hi Diana. Sorry for my late reply, your comment slipped past me!
I'm struggling to understand which numbers you are referring to... Can you try and be a bit more specific, please?
You're an actual legend!!!I missed a bunch of lessons this year, and this is the best make up I could wish for, Thanks.
TheDocs95 Thanks!
bro thanks so much. I already decided that you will have full custody of my first born child, keep it up... or else. xx
remember
@@el5847 yeah I do lol
Omg whenever I watch your videos i actually like chemistry !!!!!!!Im like depending on your videos to pass the exam!!! also can you do a video on the mcq's related to the other gas laws like Boyles law and Charles law because they are very tricky!
patiently waiting for the rest of the amount of substance series...
Literally
still waiting.....
The way you explain the content is fabulous and I hope you realise what you are doing here
Many Thanks, helped a lot especially as I'm struggling to find revision material due to the new specification!
+Shreya Gupta Yeah, it's a bit of a pain this whole changing thing. Hopefully these videos will help though and I shall be making more!
The work you do is much appreciated. Thank you.
Thank you so much. Finally someone who explains decently
I've been doing the exam questions on the video before you started to work on them. I got them all right apart from the one at the end. Because I forgot to convert from m^3 to dm^3 🙊. But thanks for pointing out, I will be extra careful if something like this appears in the exam.
Jen xX It's pretty unlikely that there will be a question where there is a horrible conversion - but it can happen! And well done!
Do you have videos for the full A level couse or for AS only.
I don't get why you divide by 5 in the boron trichloride question (10:50), I know there are 5 moles on the right, but isn't there 7 moles of stuff on the left? Why do you not divide by 7?
So, in these type of questions do you only ever look at the moles on the right hand side?
skwli Using the ideal gas equation, which accounts for all the gases that are produced in this particular question (the question is talking about both), the values for the moles is also taking into account both gases. In total, there are 5 moles of gas.
Looking at the ratio from the equation, you should be able to see that the ratio of boron oxide to the gases is 1:5. So, to get the moles of boron oxide, we divide by 5!
Does that make any sense?
E Rintoul Yes thank you!
Thank you so much!! The exam questions that you go through are REALLY helpful!
aamazzzinnnggggggg!!!!!!!!!!!!11 more please!!
Colonel Gains There will be more - watch this space!
Can you do a video on titrations? Please
+EXO K&M I certainly can! The concentration stuff is one that is on the list to be done soon! Watch this space...
Still waiting for your titration video! it would be really helpful if you did so :(
are you gonna do it
Thanks for all these videos that you have made! It would be awesome if you could also make a titrations video.
Really really helpful! Thanks! :)
You are a life saver thank you for your videos
Do you have any videos that would help with the EMPA for A2 (titration calcs)?
+Olivia Batterbee Not at this point, I'm afraid.
8:02
Well said good sir.
Well said indeed
I'm quite confused on one of the questions you did, the second one. PV=nRT aids in discovering the total moles of the gases produced.. And am I right in saying the the total moles of Reactants will equal total moles of Products? You divided it by five because there were five moles on the right and one mole on the left, However I would have divided it by 7, because I would have assumed that the reactant's total moles would have equaled 0.004 ( or whatever the sum came to). And the ratio there is 1:7, I know I'm wrong, I'm just wondering why :)! Thank you!
Jam Head It helps if you can link to the point in the video!
No, you're over simplifying things slightly. You may find that the ratio of compounds shows that there are more moles on one side than the other, but it's what each compound contains that makes the difference.
For example, the reaction of carbon and oxygen, C + O2 --> CO2, shows 2 moles becoming one. However, all of the atoms have remained, they are now just in one compound rather than being separate.
In the example in the video, it is the products, NOT the reactants that we are looking at - the give-away is when it says that the gases PRODUCED occupy the volume of 5000 cm3 etc.
However, the fact that both products are gases means that we can't separate the number and work things out for just one. Instead, we work it out for all and then divide (this works because a mole of ANY gas occupies the same volume - given constant temperature and pressure).
Has that helped at all?
E Rintoul Right, I think I understand. So because we know the volume of the gases produced, and not the reactants, we divide it by the total moles of the GASES produced. Also, re-watching the video, I've realised that you mentioned that the gases all occupy the same volume - So I should be thinking of the gases as one ( when it comes to doing the ratio). Thank you, that's helped tremendously!
Jam Head Thinking of the gases as one big cloud of whatever id probably the easiest way to think about it. Then when you've calculated bits, you can divide by the ratios to single out the individuals. That OK?
I do ocr and honestly your videos are the best
This video was so helpful, appreciate it.
In the second question y did u do 0.2019/5 and take the value of one mole to the other side when the other side =7 moles and the right side =5 moles. Don't they have to be same?
Exactly what I'm confused about...
that's exactly what's confusing me too
The question asked for the boron oxide specifically so you only relate it to that.
@@DJ-tc6me ohhh thanx
10cm3 gaseous hydrocarbon reacts completely with 40cm3 oxygen to produce 30cm3 carbon dioxide. How do you find the number of carbon atoms in the formula of the hydrocarbon??
+Chiddy01 You need to think about where the carbon dioxide comes from... All carbons in the carbon dioxide come from the initial hydrocarbon. Once you know the moles of the carbon dioxide you can work out the ratio and get the numbers.
LIFE SAVER.
Haha thanks!
Amazing video!!!! How dyu convert from dm3 to m3?
Errmm one question. How would you change the *c ( degrees Celsius) to k
+naveed ehsan Add 273 e.g. 25 C would be 298 K.
+E Rintoul thnx
how do you know thers 5?
+ghetto child Errr. Do you want to be a little less vague...? Haha.
Will these videos go in par with the new A-level syllabus?
ye
Will you be adding atom economy/ percentage yield/ titrations and back titrations?
Exam tomorrow and I legit just found u now
For the last question, when you wrote 0.111242 m cubed. How did u get m cubed
V is given as m cubed in the ideal gas equation.
Great video😄😄😄😃😄
+naveed ehsan Thanks very much :)
Hi there, would you be able to make a video covering those really tough titration calculations please : / would be a massive help!
for the first question , wouldn't you use the NO in reaction three by finding out how much NO2 is made, just wondering
haven't you missed a large part of the amount of substance section? or do we not need to really know that stuff? (atom economy/ percentage yield etc)
Dakota fireshard Yep, I never got round to covering the atom economy and other bits. I felt as though they were easier and so I left them out and did what I thought was some of the harder stuff! I will eventually get it done, though. And you certainly DO need to know it!
ok thank you very much your videos you have done certainly are amazing :D
12:50 I got 2.79 using full values
awesome video
Thank you, this was very much appreciated :)
Have you done a video on atom economy yet? and percentage yield?
Could you help me how to figure this answer out? 'If the percentage yield of ammonia in the Haber Process is 15% what mass of ammonia is produced from 5 tonnes of hydrogen?'
Do you have anything on Kc??
hello I didn't get how you got the 273 for the temperature bit? would you mind explaining it a bit? Thank you!
Which lart
Part*
in 6:45 why is it devided by ten ? and if it is bc ( 6+4=10) then why did you multiply 4 again/ :) ?
because out of the total moles of product only four tenth of it is NO
Please do make a video on Percentage Purity and Yield
is there a video on percentage yield ?????????
hi love ur vids but my as chem exam is in Friday and I was wondering if you could make a video on titration and percentage purity before that if you can please
*10:06** where my 2019 gang at* 🤭
cringe but ay yo!
was just going to comment that 👆
Sir can you finish the amount of substance series😭😭
where is the part 3 of amount of substance?
plz make it
Where's the empirical formula video?
hello, how come the molar mass of bromine you used was 10.8 ,the periodic table says 79.904
Boron and Bromine aren’t the same thing.
Thanks for the help!
Can you make a video on the ideal gas chapter in terms of what its about and what's behind it? It would be really helpful if you could :D
can you do ionic equations please
How do i convert dm3 to m3?
During the actual exam, can we use highlighters to highlight the important parts of questions, or is that not allowed?
Yup u can do that
where is empirical and molecular formula as part of this topic?
Avogardo's constant???
i am doing AS chemistry unit 1
is this included in our sylabus ??
ye
does anyone know how the moles of B2O3 is 0.2019/5? its there 7 moles on the left-hand side, so how does it equate to 5 moles on the other side
The question told you the information to work out 'the two gases produces' which gave you the answer of 0.2019 for the total number of moles on the products side of the equation. And even though the ration of moles of ALL the reactants to all the products is 7:5, the only reactant the question cares about is boron oxide ('calculate the mass of boron oxide produced') which is only 1 mole.
You worked out that 0.2019=5parts
So 0.2019/5 will give you one part.
Make sure to make your final answer into grams (by multiplying by the Mr)
@@farah-vb4ti thank you🙏 I did kind of figure it out later, just forgot to revome the comment
@@farah-vb4ti lmao im here revisiting this topic and i got stuck on the exact same question again
@@Zen_Zen_Zense lmaooo. U understand it now right.? When r ur exams?
@@farah-vb4tihey how you been?
got mocks in November 23 :( if we don't do the actual exams they'll use those results) - ngl i was stuck on it again, same part too 😂. i got stuck with the 1 mole = 0.2019/5
For the first exam question, don’t you have to worry about the mole ratio on the left hand side, why?
Thanks
273 kelvin
For the last question, i got 11.6 dm cubed because the calculation came to 0.011622366 m cubes and then multiplied by 1000 gave 11.622366?
Layla A nah I get 11.1 dm^3
Layla A yeah you multiplied by 0.42 instead of 0.402
Oh I see now thanks for letting me know :)
I'm love COCAïNE
U
I have a crush in you
Iiuu
Great video😄😄😄😃😄
+naveed ehsan No problem :)