Chemistry question from BMAT

Hello! This question is from the BMAT 2014 and I was wondering how to work with the ratios. It seems simple but it confuses me :sadhug:
What I did was finding the number of moles of the compound in 6g. That would be 0.1 mol.
Then I inferred that 1 mol would be 60gr, so the Mr of the compound is 60. Do I just have to guess the molecular formula with the ratios or there’s any calculation I could do?
Thank you so much in advance! :sunflower:


Hope my answer’s correct!


hello! :sunflower: why did you multiply by 0,5 ? :pleading_face:

I considered each part as a gram.
Like 6 part is 6 grams…
So 6 part is 0.5 mole in carbon atom.

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thank you so much :two_hearts::ok_woman:

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You’re welcome :smiling_face_with_three_hearts::heart::pray:t2:


hi everyone!

i’m really sorry i don’t understand the worked solution, could someone try explaining it differently? How did you get 0,5 from 6 parts?

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Hi Juliette!
This is what Dorsa did:
firstly she calculated the moles of C,H and O from the given parts in the question.(considering it given mass)
C= 6/12=0.5
O= 8/16=0.5.
After that we calculated/found the lowest whole no. ratio of the different components(by dividing each mole value with the lowest no. which here is 0.5)
from here we have the empirical formula which is CH2O.
from here we will find the molar mass of the gas.
6g of the gas has a volume of 2.4dm^3.
considering molar volume(24dm^3) we have 60g of the gas.
empirical formula is 30g.
using Molecular mass=n x empirical formula mass.
Thus molecular formula=2(CH2O)=C2H4O2.
Hope it helps:)

hey! now that I re read the exercise I dont get where it says that the compound weights 60grams :pleading_face:

Thank you so much Ujjwal!!!
it’s much more clear now!

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yes! but how did you know that we have 1 mole of the compound? I mean, the question mentions that in the case we have 6grams, we have 0.1 moles.
in other words, when should I infer that the exercise asks for 1 mole of the compound?

yes🙆‍♀️ I understood that 1 mole occupies 24dm3. I was confused because I thought this was a general rule for gases at room temperature. So the problem is telling us, in that sentence, that not only that rules apply here but that we also have 1 mole of the compound? (in this exercise) What I mean is that the problem itself does not states specifically that we have 1 mol of the compound, so we just infer it from the general rule? Thank you soo much x your patience

Ya you’re right that’s a general rule that 1 mol of any gas occupies 24dm^3 at room temperature and pressure.
And here we have 0.1 moles you can infer it from the fact that we have 2.4dm^3 of gas present and molar volume is 24dm^3.
Thus, n=2.4/24=0.1moles.
Now, you can use this mole to calculate the molar mass of the given gas.
Mr = 6/0.1=60g.
I think my above explanation was bit sketchy this should make things clear for you :slight_smile:

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OMG i cant believe I missed that. Im sincerely grateful to you, thank you sooo much for your kidness and patience

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